Slide 1

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 2

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 3

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 4

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 5

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 6

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 7

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 8

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 9

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 10

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 11

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 12

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 13

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 14

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 15

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 16

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 17

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 18

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 19

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 20

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 21

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 22

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 23

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 24

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 25

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 26

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 27

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 28

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 29

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 30

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 31

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 32

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 33

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 34

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 35

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 36

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 37

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 38

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 39

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 40

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 41

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 42

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 43

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 44

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 45

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 46

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 47

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 48

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 49

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 50

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 51

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 52

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 53

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 54

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 55

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 56

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 57

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 58

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59


Slide 59

Department of Chemistry

Electrochemistry & Solutions
1. Solutions and Mixtures

Year 1 – Module 3
8 Lectures

Dr Adam Lee
afl2@york.ac.uk

1

Aims
To:
• Understand physical chemistry of
solutions and their thermodynamic
properties
 predict/control physical behaviour
 improve chemical reactions

• Link electrochemical properties to
chemical thermodynamics
 rationalise reactivity.

2

Synopsis
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

Phase rule
Clapeyron & Clausius-Clapeyron Equations
Chemical potential
Phase diagrams
Raoults law (Henry’s law)
Lever rule
Distillation and Azeotropes
Osmosis
Structure of liquids
Interactions in ionic solutions
Ion-ion interactions
Debye-Huckel theory
Electrodes
Electrochemical cells
Electrode potentials
Nernst Equation
Electrode types

Recommended Reading

R.G. Compton and G.H.W. Sanders, Electrode Potentials
Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry,
OUP, 3rd Edition, Chapters 5, 6 & 9.
3

P. W. Atkins, Physical Chemistry,
OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.

Phase Diagrams
Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

4

Significance of Areas, Lines and
Points
Areas: single phase exists
(specify p and T
 state of matter)

P


S

P
L
G

T
T

Lines: 2 phases coexist in ium
(specify p or T
 need to know
what 2 phases present)

P


S



L


G
T

G

G

L

S

L
S

5

S-L line  change in m.pt. with pressure
S-G line 
L-G line 

"

in sub.pt. with pressure

Vacuum
"
in b.pt. with pressure
Pump

(isoteniscope explores how pressure affects
boiling of liquid)

Triple Point: all 3 phases coexist in ium
(unique value of p and T)

Critical Point: conditions (T and p) above
which liquid state matter ceases to exist

6

Slope of Solid/Liquid Line
Ice-water system:
Slopes Backward: (rare)
 i.e.

dT/dP = - ve

m.pt.  pressure
- ice-skating

 also note ice floats on water

(ice< water)

CO2 system:
Slopes Forward: (usual)
 i.e.

dT/dP = + ve

m.pt.  pressure
L
P

 solid CO2 sinks in liquid CO2

S
G
T

7

These phenomena can be understood
by simple (!) thermodynamics

BUT FIRST
The Phase Rule
 obtained from experiment
 derived/proven by TD
 allows rigorous discussion of phase
behavior

8

The Phase Rule
It states: (only at ium)

F = c - p + 2
No. of degrees
No. of phases
of freedom
No. of components
Components
- smallest no. of substances needed to
describe system
e.g.
H2 O
= 1 (ice, water and steam all H2O)
EtOH + H2O = 2
H2O + NaCl = 2 (despite NaCl  Na++Cl-)
9

Phases
- no. of uniform, homogeneous regions
within the phase diagram
e.g.

ice = 1
water + ice = 2
water + ice + steam = 3

Degrees of Freedom
- smallest no. of independent State Variables
needed to describe ium system
(e.g. p, V, T)
A state function (or function of state) depends
only on the current state of the system, and
not on the route by which it was reached.

10

Consider Water Diagram
L

P

S



Y



Z G

X
T

 At X

 At Y

P = 1 (solid)
F=C-P+2
= 2

must specify 2 variables to
locate position on diagram

P = 2 (L + G)
F=1

position determined by
only 1 variable

 At Z
P = 3 (S + L + G)
F=0

can only co-exist for
fixed p and T
11

Thermodynamics of Two-Phase
Equilibria of a Single Component:
The Clapeyron Equation
L

P

S

concerned with
equilibria along lines

G

phase 

e.g. ice/water
at m.pt.

T

phase 

Thermodynamic eqns. needed

Josiah Willard Gibbs

1. G = H - TS
2. G = 0

ium (S=H/T)
Gibbs Freeat
Energy

3. dG = Vdp - SdT (fundamental eqn. for G)

American mathematical physicist developed
theory of chemical thermodynamics. First US
engineering PhD…later Professor at Yale.

12

1839 -1903

Derivation of Clapeyron Eqn.:


convert
1 molPaul
from
state
 to state 
Benoit
Emile
Clapeyron
at ium (i.e. @ transition)




G.F.E = G


G

G = G - G = 0
G = G
 dG = dG

1799 - 1864
Parisian engineer and mathematician.
Derived differential equation for determining
heat of melting of a solid
Since,

dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT

13



dp
dT

=

(S   S  )
(V   V  )

S

=

V

BUT
G = H - TS = 0
so



at ium

S = H / T

dp
dT

=

H
TV

Clapeyron
Equation

(slope of a line on the phase diagram)

14

Application of Clapeyron Eqn. to
Solid/Liquid Phase
 Process: fusion (melting)
T = melting point, Tm
H = enthalpy

(fH)

V = volume

(fV)

dp
dT m

= slope of line =

 fH
T m  fV

 Inverting:
dT m
dp

= change in Tm = T m  f V
 fH
with p

15

 What is fV?

V = 1/density = 1/ = 1/gcm-3 = cm3g-1
fV = 1/l - 1/s

 Sign of slope:
Water

dp/dT = - ve =

and


fH = positive
fV = - ve

s

<

l

 fH
T m  fV

hence ice floats
16

Modified eqn. for Liquid/Gas and
Solid/Gas Lines:
Clausius-Clapeyron Eqn.
Rudolf Julius Emmanuel Clausius

Maths

dx/x = dlnx

n +1
x
 x dx =
n +1
n

dp/p = dlnp

dx

 2 +1
x
= 1
 2 =
x
x
2+1


dT

1822 -1888T 2

dp =
dT

H
TV

= 1
T

Clapeyron
Equation

17

Derivation
e.g. liquid/gas system
V = Vg - Vl
Vg >> Vl
 V  Vg

For an Ideal Gas,
pVg = RT (per mol)
Vg = RT/p = V

substituting into Clapeyron eqn
dp
dT
dp
p

=
=

H p
T RT
 H dT
RT

2

18

Hence,
d ln p

 H dT

dlnp =

R T

dT

2

=

H
RT

2

Clausius-Clapeyron
Equation

Integrating,
lnp = y

H  1 
 
R T 

=

m

x

+ constant
+

c

Isoteniscope expt.
lnp

slope = -H/R

X
X
X
X
X

1/T
NB
- normal boiling point is T at 1 atmosphere
- T is always in K not C
19

Example Problems
The densities of a substance in solid & liquid states at the
melting point (23 C) are 0.875 & 0.901 gcm3 respectively.
Heat of fusion is 276 Jg-1. At what temperature will the
substance melt at an applied pressure of 100 atmos.?

Use,

dp
dT m

 fH

=

T m  fV

fV = 1/l - 1/s
dT m
dp

= 1/0.901 - 1/0.875
= -0.033 cm3g-1
V

T

= (23+273) x (-0.033)
H

276

(cm3g-1)
(Jg-1)

convert cm-3m-3..... x 10-6
dT m
dp

99

= -296 x 0.033x10-6 KPa-1
276
= -0.35 K
20

The vapour pressure of liquid napthelene is 15.5 mmHg at
95 C and 505.7 mmHg at 200.5 C. Calculate the heat of
vapourisation and the normal boiling point of napthalene.

Use,

 p1 


p
 2

ln 

  vH  1
1 



R
T
T
 1
2 

=

505 . 7 

 15 . 5 

ln 

=

  vH  1
1 

8 . 314  473 . 5 338 

H = 47,900 Jmol = 47.9 kJmol-1

Normal b.pt.
760 

 15 . 5 

ln 

(i.e. Tb for p = 1atm = 760 mmHg)

=  47 , 900  1  1 


8 . 314
T
338

Tb = 490 K



b



(217C)

21

Topics Covered
 1-component systems
- phase diagrams
Ttriple, Tmelt ... physical props

- the phase-rule
 Phase equilibria

(1-component)

- solid-liquid (Clapeyron eqn.)
H, V ... physical props/
behaviour

- solid-gas/liquid-gas (Claus-Clap)
H, S ... physical props

22

23

24

Consider ideal gas at constant temperature:

dG = Vdp – SdT = Vdp
Since pV = nRT,
G 



2

Vdp 

1



2

1

nrTdp

 nRT ln

p

G 2  G + nRT lnp

p2
p1



2

If initial state 1 = STP (1 atm)

In general for a mixture AB:

GA = GoA + nART ln pA, GB = GoB + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
25

Phase Diagrams in Binary
Systems
Composition:
mol fraction of A = nA / nA+nB
for liquids - xA, xB
xA =

xA + xB = 1

w A / MA
w A / M A + w B / MB

for vapour phase - yA, yB
nA  pA, nB  pB

yA + yB = 1

(pA,pB - partial pressures)

p = pA + pB

yA = pA / (pA + pB) = pA / p
26

yA

yA

yB

yB

Ginitial = nAA + nBB

nA

nB

nA+ nB

= nA[ + RTlnp] + nB[B + RTlnp]
A

B

Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]

G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
Initial
Final
ya

ya

yb

yb

Gmixing

ya
yb
27

yA

yA

yA

yA

yB

yB

yB

yB

Gmixing

Smixing

ya
yb

0
1

1
0

ya
yb
28

Chemical Potential (in English!)

Molecules acquire
more spare energy
Gibbs
Free
Energy

Greater “chemical potential”



G  ln(pressure)
Pressure
Low Pressure

High Pressure

Constant Temperature

Effect of environment
on this free energy

Gmolar = G molar + RT lnp
Energy free for molecules
to “do stuff”at STP

29

Why do we use Chemical Potential?
Gibbs Free Energy (G) is total energy in entire system
available to “do stuff”
- includes all molecules, of all substances, in all phases

G = nAA + nBB
For single component
e.g. pure H2O

No real need to use 

G = nH2OH2O
Free energy only comes
from H2O

For mixtures
e.g. H2O/C2H5OH

Free energy from 2 sources

G = nH2OH2O+nEtOHEtOH
 tells us how much from
H2O versus C2H5OH

30

Why do we different molecules have
different Chemical Potentials?

Involatile

Volatile

Ethanol can soak up much more energy in extra vibrational
modes and chemical bonds
- will respond differently to pressure/temperature increases

Chemical Potential :
1. A measure of "escaping tendency" of components in a
solution
Gas-phase molecule
Free
Energy
(G)
&

Liquid-phase
Solid-state
Pressure

31
2. A measure of the reactivity of a component in a solution

Raoult's law
Ideal solns. obey Raoult's Law:

Partial pressures pA and pB vary linearly with
the mol fractions xA and xB in a liquid.

i.e.
eqns. of
straight lines
passing thru
origin

pA = poA xA
pB =

poB

Raoult's law

xB

y = m . x

+

c (=0)

Volatility

p = pA + pB =

poA xA + poB xB

Total pressure above boiling liquid

32

xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
 this means mixing ALWAYS gives a negative lnxA
0

Mixing always
lowers 

lnxA
Pure B

Pure A
Dilution
0

xA

1

33

Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.

(a) Vapour pressure vs. liquid composition
(the liquid line)

p

= p A + pB =

poA xA + poB xB

= poA xA + poB (1 - xA)
= poA xA + poB - poB xA
= poB + (poA - poB) xA
34

If applied pressure > p
vapour  liquid

P
A + B

BUT
p depends on liquid composition (x)

p oA 

Liquid

p

Pure A
Volatile



1

x oA

0

0

x oB

1

Pure B
Involatile
35

however
(b) Vapour pressure vs. vapour composition
p is not a linear function of y
p oA 

p

 po B
Vapour

Pure A
Volatile

1

yoA

0

0

yoB

1

Pure B
Involatile

36

The Vapour-Pressure Diagram

poA 

lliiqq
uuiidd
lliinn
ee

vap

p

liquid
only

L+
V
our

lin

e

 p

o

B

liquid+
vapour

vapour
only

A

mol fraction

B

37

Deviations from Ideality

Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)

Strength of interaction  heat of mixing
i.e. mH for A-B system

3 cases
Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes)

B
A

B
A

mixH = 0 Ideal mix

38

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
39

Summary: Raoult’s Law for Solvents
Proportionality constant

pA = xA . pAΘ
Total pressure

Volatile
high vapour pressure

p oA 

Liquid

Involatile
low vapour pressure

pBΘ

p
Partial
pressure
of B



1

x oA

0

0

x oB

1

Partial
pressure of
A

pB = xB . pBΘ
A

B
40

High p0ө(A)

Low p0ө(A)

High order: low S

Less order: higher S

A

A

A

A

Strong desire to  S

Less need to  S

Boiling of A
favoured

A happier in liquid

p0ө = vapour pressure
= tendancy of system to increase S

41

Proportionality constant

42

Amount in solution

Dissolution is
EXOTHERMIC

For dissolution of oxygen in water, O2(g)
O2(aq), enthalpy
change under standard conditions is -11.7 kJ/mole.

43

pO2
pH2O

Solvent: H2O

Solute: O2
H2O

O2

Henry's law accurate for gases dissolving in liquids when
concentrations
and partial in
pressures
Consider O2 dissolution
water: are low.
AsImportant
conc. and
partialChemistry
pressuresforincrease,
in Green
selective deviations
oxidation from
Henry's law become noticeable
Cinnamaldehyde

Cinnamic Acid
Cinnamyl
Alcohol of gases
Similar
to behavior

- deviate from the ideal gas law at high P and low T.
Aspects of Allylic Alcohol Oxidation
Adam
F. Lee Henry's
et al, Green
2000,often
6, 279called 44
Solutions
obeying
lawChemistry
are therefore

ideal dilute solutions.

45

The Tie-Line
Horizontal line thru a point on V-P diagram
defining p and composition of the system

p oA 

A

L


p

yA

 poB

xA

V
A

mol fraction
(liq, vap)

B

join compositions of co-existing
phases at equilibrium

46

The Lever Rule
 tie-lines  composition of phases
But
 how much total material is in each phase
Define no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
poA


v a

l

p
A

mol fraction

Lever Rule:

n

l

nv




B

length

av

length

al

poB

47

Volatile

Involatile

B

A

Lever Rule

Tie-line

B

B

B

GAS

A

A
B
B
A
A A
B
B
A
A
A
B

A-B Composition

GAS
LIQUID
LIQUID

Liquid-Gas Distribution

48

Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(for normal b.p. , p = 1atm.)

2. b.p. is  1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P

Thus
Fixed T
poB

Fixed p
vapour

liquid

l-lin

-l
dew

e

L+V

v -li

p

poA

ne

L+V

TB,B

bu

e -l
bbl

vapour

0

xA

TB,A

in e
in e

T

liquid

1

0

xA

1
49

Example Problem
The following temperature/composition data were obtained
for a mixture of octane (O) and toluene (T) at 760 Torr,
where x is the mol fraction in the liquid and y the mol
fraction in the vapour at equilibrium
x (T o lu e n e ) y(T o lu e n e ) T e m p e ra tu re / C
0 .9 0 8
0 .9 2 3
1 1 0 .9
0 .7 9 5
0 .8 3 6
112
0 .6 1 5
0 .6 9 8
114
0 .5 2 7
0 .6 2 4
1 1 5 .8
0 .4 0 8
0 .5 2 7
1 1 7 .3
0 .3
0 .4 1
119
0 .2 0 3
0 .2 9 7
120
0 .0 9 7
0 .1 6 4
123

Liquid

Boiling/
Condensation
Temperature

Vapour

The boiling points are 110.6 C for toluene and 125.6 C for
octane. Plot the temperature/composition diagram of the
mixture. What is the composition of vapour in equilibrium
with the liquid of composition:
1. x(T) = 0.25
2. x(O) = 0.25
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
P.W. Atkins, Elements of Phys.Chem. page 141

50

Principle of Distillation



T

boiling

y1

cool

y2
x2

0

x1

mol fraction

1

Take liquid x1 & heat to bubble line
 vapour composition y
 Cool vapour
 liquid composition x2
REPEAT!!

51

Extreme Deviations from Ideality

Result in:
maxima/minima in B.Pt + V.P. diagrams

AZEOTROPES
a - without
zeo - boil
tropos - change
means composition does not change
during boiling
52

Case 2: -ve deviation
A more attracted by B
(e.g. CHCl3 + acetone)
b.pt. > ideal

B
A
A

mixH = < 0

Case 3: +ve deviation
A less attracted by B
(e.g. EtOH + water)

b.pt. < ideal

B

B
A

B
A

mixH = > 0
53

+-ve

-ve

Vapour-Pressure diagrams
Liquid

Liquid

P

L+V
L+V

L+V
L+V

Vapour

Vapour

A composition B

A composition B

Boiling-Point diagrams
Vapour

T

azeotropic
composition

Vapour
L+V
L+V

L+V
L+V

Liquid

A composition B

Liquid

A composition B
54

Distillation of Azeotropes
V

T

Distillate always ends up
at azeotropic composition

L+V

Residue is pure A or B

A

A
L

A

P

B B

composition

B

-EtOH cannot be dried by
distillation
P = 96% EtOH, 4% H2O

V

T

Distillate always ends up
as pure A or B

L+V

Residue is mixture with
azeotropic composition

B B
A A
L

A
Residue

composition

B
Distlllate
55

Topics Covered (lectures 2-4)
 Chemical Potential
- A(l) = A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
 Mol fractions
- A = nA / nA+nB
 Raoult’s Law
- pA = poA xA

pB = poB xB

- ideal solutions
- +ve/-ve deviations
 Vapour-pressure diagrams
- Tie-lines
- Lever Rule
56

Solutions Equations
Phase Rule
F = c - p + 2
degrees of freedom

Clapeyron Equation
dp
dT



H
TV

c = components
p = no. of phases

H = enthalpy of phase change
V = volume change associated
with phase change

Clausius-Clapeyron Equation
H
d ln p
p   H1
or

ln   
2
p 
R  T

dT
RT
1

2

v

1



1 
T 2 

Mol fractions
xA = nA / nA+nB
yA = p A / p A + p B

ni = mols of i
pi = partial pressure of i

Raoults Law
pA = poA xA and

pB = poB xB

Lever Rule (for tie-line joining phases via point a)
nl
length av n =no. moles in liquid phase 57
l

nv
length al nv =no. moles in liquid phase

A(solution) < A(solvent)

A

Contains solute (e.g. NaCl, glucose)

WHY?!!!
Solvent A
A(l) = A(l) + RTlnxA

58

59