Thermodynamics

1
Thermodynamics

• 33J thermodynamics
• Concerned with transformation of energy.

2
Systems and surroundings

• UNIVERSE
• Open system can exchange matter and energy with
  surroundings
• Closed system exchange energy only with
  surroundings.
• Isolated system No exchange of anything with
  surroundings.

System open, Closed and isolated
Surroundings every where else in universe
3
First law of thermodynamics

• ?U q w - 1st law where U
  internal energy, q quantity of energy
  transferred and
• 
  w work done on system.
• For an isolated system
• When q 0, w 0 ? ?U is constant.
• W hen w ? 0 or q ? 0 ? Exothermic
  reaction
• w ? 0 or q ? 0 ?
  Endothermic reaction
• State Functions such as H, U, S, V and p
  are path independent i.e. does not dependent on
• the path taken but on the initial and final
  states of the system.
• Let us look at some of the properties of these
  state functions.

4
Properties of state function

• Internal energy, U
• ?U q w
• For an infinitesimal change in U d U d q
  d wexpansion d we -extra work
• At constant V, both d w s 0, and hence
• d u d q V
• For a measurable change,
• ?U q V

5
Measurement of ?U

• Since ?U q at constant volume, can use an
  instrument in which volume cannot change such as
  
• a Bomb Calorimeter to determine ?U ( see lab.
  manual for details). Need first to determine the
• Calorimeter constant, C ( heat capacity) C
  heat supplied / increase in temperature J
  K-1. .
• To measure U for an unknown substance, we measure
  the temperature rise when the substance
• is combusted in the same calorimeter Hence
• ?U heat output (q V) C x increase in
  temperature J K-1 x K J.
• In practice we measure C by combusting Benzene
  (heat output of 3227 kJ mol-1) and measure the
• increase in temperature, ?T during the
  combustion.

6
Heat capacity

• Hence
• Heat capacity, C q / ?T
• Specific heat capacity C / mass of sample
  J K-1 g-1
• Molar heat capacity C / no. of mol
  J K-1 mol-1
• Can define two heat capacities CV ( for a
  system that cannot expand) and Cp ( for a
  system that
• is free to expand) The molar heat capacities, C
  V,m and Cp,m are related
• Cp,m C V,m R.
• Since ?U q V and q V C ?T ? ?U
  C V ?T at constant volume.

7
Enthalpy, H

• Defined as the heat supplied at constant
  pressure i.e.
• d H d qp and ? H qp .
• Can also show that H U p V where U
  internal energy p pressure and V volume.
• Since p V n RT where n no. of mols,
• H U n RT and ? H ? U
  ? n RT
• ?n n (products) - n (reactants)
• As T? H ? I.e. H Cp ? T
• Hence Cp (? H / ? T)p and CV (
  ? U / ? T)V .

8
Standard Enthalpy , H?

• H? defined as standard state I.e. 1 mol at 1bar
  pressure ( 100, 000 Pa for gases) at any
  temperature ( usually 298 K). For solutes
  activity 1 .
• Used to carry out thermochemical calculations
• ?rxn H S nP ? H? (products) - S nR ? H?
  (reactants)
• Suppose need ?subH? for the transformation
  Solid ? Gas. Process will be
• Solid ? Melt ? Gas , hence ?subH?
  ?meltH? ?vapH?
• ?? H (indirect route) ? H (direct route)
  Hesss Law

9
Free Energy, G

• G defined as maximum work from system .
• For reversible work (Equilibrium) d u d qrev
  d wrev and
• d S d qrev / T
• ? d U T d S d wrev
  Integrate to give
• i.e. A U - T S where A Helmholtz
  Free energy.
• At constant temperature d A d U - T d S
• d wrev d A.
• At Equilibrium d A 0 i.e. no work at
  constant T and V.

10
G at constant pressure

• More useful for G to be defined at constant p not
  volume.
• G H - T S U p V - T S
• ? d G d U p d V V d p - T d S - S d
  T
• At const, T p V d p 0 and S d T
  0
• d G d U p d V - T d S
• If T p change then ? d G d U p d V
  V d p - T d S - S d T and since
• d G d U p d V - T d S
• d G V d p - S d T important
  thermo eq.

11
Free energy

• i.e .G is a function of p T I. e. G (p,
  T )
• ? d G ( ?G / ? p )T d p ( ? G / ? T )p
  d T
• V ( ?G / ? p )T and - S ( ? G
  / ? T )p ( see Maxwells relations)
• For an open system exchange of matter and energy
  with surroundings can occur
• i.e. G ( p, T, n1, n2,)
• d G (? G / ? T ) p,n1,n2 d T ( ? G / ? p )
  T, n1,n2, d p (? G / ? n1 ) p,T,n2 dn1
• (? G / ? n2 ) p,T, n1 dn2,
• where (? G / ? n1 ) p,T,n2 and (? G / ? n2 )
  p,T, n1 are the partial molar free energies or
  chemical
• potential, µ.

12
Chemical potential, µ

• The chemical potential is defined as chemical
  effects of components.
• Hence for an open system d G V d p - S d
  T µ1 dn1 µ2 dn2 µ3 dn3
• ? d G V d p - S d T Si µI d
  ni . i.e.
• ( ? G / ? p )T V and (
  ? G / ? T ) p - S .
• G vs p at constant temperature
• d G V d p - S d T
• At const. Temp. d T 0 and hence,
• d G V d p

13
Properties of G

• Since for a perfect gas p V n R T ?
• V n R T / p
• ? d G n R T d p /p

G - G ? n R T l n ( p / p ? )
i.e. G G ? n R T l n ( p / p ? )
P ? 1 atm.
14
Variation of G with Temperature

• Since ( ? G / ? T ) p - S ( G - H
  ) / T
• Or ( ? ? G / ? T ) p ( ? G - ? H ) / T
  Rearrange to give
• ( ? / ? T ( ? G / T) ) p - ? H / T 2 or
• ( ? ( ? G / T) / ? T ) p - ? H / T 2
  Gibbs-Helmholtz equation
• Also ? r G? R T l n Q where Q
  products / reactants reaction Quotient and
• ? r G? is the standard free energy change for
  reaction.
• Hence ? r G ? r G? R T l n Q
• At equilibrium, ? r G 0 and 0 ? r
  G? R T l n K where K is the equilibrium
  constant.

15
G with T Contd.

• At a temperature, T1
• l n K1 - ? r G? / R T1 - ? r H ? / R
  T1 ? r S ? / R
• At another temperature , T2
• l n K2 - ? r G? / R T2 - ? r H ? / R
  T2 ? r S ? / R and
• l n K2 - l n K1 ? r H ? / R ( I / T1 - 1 /
  T2 )
• For a reaction ? r G? S n p ? G? p - S n R
  ? G? R
• NB. At eqlb. ? G 0 when ? G - ve (
  spontaneous rxn. )

16
PHASE EQUILIBRA

• What happens when G ? L ? S
• NB. The phase with the lowest G under any
  condition is the most stable.
• For solids H is large ( because of strong
  bonds) and S is low ( molecules have little
  freedom)
• ? Stable phase at low temperature.
• For Gases H 0 and S ve. ?
  stable phase at high temperatures.
• Intermediate for Liquids. Hence the driving
  force is S i.e. ( ? G / ? T ) p - S
• Where (? G / ? p) T is the gradient of the graph
  of G versus T.
• Can see this better on a graph of G versus T.

17
Graph of G versus T
G
s
( ? G 0)
Phase boundaries
l
g
T
18
Phase Equilibria

• Since ? G ? H - T ? S and at phase
  boundaries we get equilibrium between two phases
  i.e.
• ? G 0
• ? ? H T ? S ( at phase boundary )
• Hence ? fusion S ? fusion H / T fusion and
  ? vap S ? vap H / T vap
• G versus T not convenient to represent phase
  boundaries because difficult to measure G.
• Better to use pressure versus Temperature i.e. p
  versus T graph
• E. g. look at p vs. T graph for water.

19
P vs. T for a one phase system e.g. water
p
l
s
g
Triple point
T
20
p vs. T contd.

• At the phase boundaries we get equilibrium
  between phases
• e.g. s / l boundary.
• i.e. G (s) G (l) at constant T and p
  Hence,
• d G (s) d G (l)
• Since d G V d p - S d T ,
• For a phase change a ? ß , at
  phase boundary d G 0
• ? V d p S d T and
• ( V ß - V a ) d p ( S ß - S a ) d T or
• ? V d p ? S d T

21
Clapeyron equation

• Since ? S ? H / T
• d p / d T ? H / T ? V
  Clapeyron Eq.
• The eq. applies for any phase transition e.g. s /
  l boundary
• d p / d T ? fus H / T ? fus V where ? fus
  V is molar volume change on melting.
• ?
• d p ? fus H / ? fus V . d T / T .
• p2 p1 ? fus H / ? fus V l n ( T2 / T1)
• l n (T2 / T1 ) ( (T2 - T1)) /
  T1)

22
Clausius- Clapeyron Equation
p

• For l ? g (vapourization) and
• s ? g ( sublimation)
• Suppose l ? g
• d p / d T ? vap H / T ? vap V
• ? vap V V (g) - V (l)

s
l
T
23
Clausius-Clapeyron Equation

• Since V (g) ?? V (l) ? ? vap V V
  (g) R T / p
• ? d p / d T ? vap H / R T 2 . p or
• d p / p . 1 / T ? vap H / R T 2
  ( Rem. d p / p l n p )
• ? d l n p / d T ? vap H / R T 2
  Clausius Clapeyron EQ
• ? d l n p ? vap H / R . ? d T / T 2
  
• l n p - ? vap H / R T
  constant
• y m x c

24
Clausius-Clapeyron

• Explains why s / l boundary is a straight line.

l n p
Slope - ? vap H / R
1 / T
25
Application of Clausius and Clapeyron EQ

• For a finite change
• And l n ( p2 / p1) - ?vap H / R ( 1 / T2 -
  1 / T1 )
• If assume that ? vap H is independent of
  temperature then
• p2 p1 e x where x ? vap H / R ( 1
  / T2 - 1 / T 1 )
• i.e. Exponential relationship between p and T

l
s
g
26
Applications

• For the phase change a ? ß
• At phase boundary (d G a, m ) p, T ( d G
  ß, m ) p, T
• ? µ a µ ß
• ( ? µ ß / ? p ) T - ( ? µa / ? p ) T
  V ß, m - V a, m ? V Trans and
• ( ? µ ß / ? T ) p - ( ? µa / ? T ) p
  - S ß, m S a, m - ? S Trans
  - ? trans H / T Trans

H
V
discontinuity
T t
T t
T
T
27
Applications contd.

• NB First derivatives are discontinues at phase
  boundaries where the gradient of the slopes
• change i.e. at the transition temperatures, T t.
• These are all called first order transitions.
  However transitions can occur where the first
• derivative Is continuous but the second
  derivative is discontinuous e.g. slope before and
  after
• transition dont change. e g.

H
V
C p
T
T
T
28

• These are all 2 nd order transitions e.g.
  change in molecular symmetry I.e. a phase
  transition from
• say tetragonal phase ? cubic phase (
  involves no change in V, S, H etc).. Note
  discontinuity for
• C p versus T and hence only way to detect 2
  nd
• transitions.
• 
  PHASE RULE
• What are the conditions for equilibrium between
  phases?
• For an open system d G V d p - S
  d T Si µ I d n I
• ? ( d G ) T, p Si µ I d n I
• For a ? ß

29
Phase Rule

• At Equilibrium
• i.e. chemical potential same in both phases.
• Also T a T ß and
• p a p ß
• ?

30
Phase Rule

• Equilibrium between phases is governed by the
  phase rule
• P F C 2 ,
• where P no. of phases F no. of degree of
  freedom (e.g. T , p , composition i.e.no. of
  variables) C no. of components.
• ? F C - P 2
• For water, C 1 ? F 3 - P
• If P 1, F 2 i.e. both p and T can be
  varied independently within the phase.
• If two phases are in equilibrium, F 1 i.e.
  can only alter p or T i.e. on the line in phase
  diagram.

31
Phase Rule

• If all 3 phases are in equilibrium, F 0 (
  critical point is established at a definite T
  and p)

F 1
Tie line
b
T
a
F 2
p
l
c
d
a
c
b
e
d
s
g
e
e
critical point
(F 0)
time
T
At b, g l at d, l s
32
Liquid Mixtures

• What happens when two or more liquids are mixed?
• Mole fraction
• x I n I / S n
  where x I mole fraction of I component in
  mixture n no. of mols S n n1 n2,
  n3 n 4,..
• For two component mixtures x 1 ( w 1/ M 1) /
  ( w 1 / M 1 w 2 / M 2 ) and
• x
  2 ( w 2/ M 2) / ( w 1 / M 1 w 2 / M 2 )
  and
• 
  x 1 x 2 1
  
• Molar concentration no. of mols. solute per
  dm 3 (1000 cm3) of solvent
• Molal concentration no. of mols. solute per
  kg of solvent

33
Mixtures

• i
• We know ( d G ),T,p S µ I d n i
• ?
• For a two component mixture ( binary mixture) of
  n j and n k keep n j constant and vary n k
• ( d G ) T, p, n j µ
  k d n k and
• µ k ( ? G / ? n k ),
  T, p, n j
• For a pure substance we integrate to give
  G µ n
• What then is an ideal mixture? Must obey
  Raoults Law ( R - L ) at all concentrations i.
  e.
• p i x i p i ,
  where p i partial vapour pressure of
  component I in mixture
• 
  x Ii mole fraction of component I
  in mixture
• 
  p vapour pressure of pure
  substance

34
Raoults Law

• e.g.

benzene
p
p ( benzene)
Total pressure
p (toluene)
1
x
0
toluene
35
R - L

• Consider a liquid A in a mixture in equilibrium
  with its vapour at a partial pressure, p A

µ A ( g )
B g
A g
µ A ( l )
A l

B l
mixture
At equilibrium µ A (l) µ A ( g )
36
R - L

• We know G m G? m R T l n ( p / p ? )
  ( Rem. G m µ )
• At equilibrium µ A (l) µ A? R T l n (
  p A / p ? )
• R - L p A x
  A p A
• µ A (l) µ A? R
  T l n ( x A pA / p ? )
• µ A (l) µ A?
  R T l n ( p A / p ? ) R T l n x A
• When x A 1,, µ A? R T l n ( p A /
  p ? ) constant µA
• since for the pure liquid, µ A (l) µA
  i.e. µ A is the chemical potential of pure
  liquid
• ? µ A
  (l) µ A R T l n x A
• For non-ideal solutions

37
Non-ideal solutions

• a a concentration, c ? a ? c
  where ? activity coefficient
• For pure liquid, ideal or very dilute
  solutions, a 1. Can have two types of
  non-ideal phase diagrams

ve deviation
Acetone / chloroform
CS2 / Acetone
overall
- ve deviation
VP
overall
R-L
R-L
1
0
1
x CS2
0
x CHCl3
38
Non-ideal behaviour

• Ideal means a I x I and ? I 1 ( ?
  a / x )
• - v e means that x I ? a I i.e. ? I a I /
  x I lt 1 ? less tendency to escape into
  vapour phase.
• v e means that a I ? x I and ? I ? 1 ?
  greater tendency to escape into vapour phase.
• NB that at low mole fractions ( concentrations )
  vapour pressure of solute a mole fraction of
  solvent ? Henrys Law

Ideal dilute solution
K B
p B
R-L
1
0
x B
Henrys Law p B K b x B
39
Partial Molar Volume

• Useful in determining amount of oxygen (solute )
  in river water (solvent ).
• e.g. x O2 p O2 / K B where p
  O2 partial pressure of oxygen and K B from
  tables
• Partial molar volume
• Large volume of pure water 1 mol of water
  ? volume increase by 18 cm3 ( i.e.molar volume of
  water is 18 cm3 mol-1)
• Large volume of ethanol I mol of water ?
  volume increase by 14 cm3 ( i.e. 14 cm3 mol-1
  is the partial molar volume of water in ethanol )
• The partial molar volume, Vi of a substance i is
  defined Vi ( ? V / ? n I ) T, p, n J
• For a two component system i.e add d n A and d
  n B volume changes to

40
Molar Volume

• d V ( ? V / ? n A ) T, p, nB d n A
  ( ? V / ? n B ) T, p, nA d nB
• ? d V VA d nA VB d nB
  Integrate to give
• V VA nA VB nB
  ,, where V total volume.
• Partial molar quantities can be measured in
  several ways Do a V versus composition graph
  e.g.

V
V A
V B
Partial molar volume varies with composition a
is v e b is v e
b
a
Composition, n A
41
Partial molar volume

• Use curve fitting program to get V as a function
  of n A ( Use a computer )
• e.g. V A B n A C ( n2A - 1 ),
  where A, B and C are constants (EQ is shape
  dependent
• Hence at a VA ( ? V / ? n A ) T, p, nB
  B 2 C nA and at b
• VB ( V - nA VA ) /
  nB A - ( n2A 1) C / nB
• Another way to determine molar volume is
• We know that V n A VA n B V B
  Differentiate to give
• d V n A d VA n B d VB V A d n A
  V B d n B and as before
• d V V A d n A V B d n B

42
Gibbs Duhem Equation

• Equalize equations to give n A d VA n B d V
  B 0 Gibbs Duhem equation
• i.e S i n
  i d V i 0
• Hence d VB -
  ( n A / n B ) d V A and can find V B by
  integration
• V B
  V B - ( n A / n B ) ? d V A ,.. where
  V B is the integration constant.
• Integrate between m ( molality) 0 and molality
  of interest once V versus composition EQ. is
  known
• Can also show that S i n i d µ i 0
  Gibbs- Duhem
• i.e. for a two component system n A d µ A n
  B d µB 0 and
• d µ B - (n
  A / n B ) d µ A i.e. When µ A ? µ B ?

43
Thermodynamics of Mixing

• What about ? mix H , ? mix S and ? mix G on
  mixing ?
• ? mix G G f -
  G i . where, i initial
  and f final
• For an ideal mixture G i n A µ A n B
  µB
• G f n A
  ( µA R T l n x A ) n B ( µB R T l n
  x B )
• ? mix G n R T ( x
  A l n x A x B l n x B ).where, n A n B
  n
• I.e. ? mix G n R T S i n
  i l n x i
• Since ( ? G / ? T ) p,
  n - S
• ? mix S -
  ( ? ? G / ? T ) p, n A, n B - n R ( x A
  l n x A x B l n x B )
• i.e. ? mix S -
  R S i n l n x i

44
Thermodynamics of mixing

• ? mix H 0 (ideal solution mixture) and ? mix
  V 0 ( at const. T p ).
• Because x 1, l n x is - ve and therefore ?
  mix G is - ve ( spontaneous driving force for
  mixing)
• Also ? mix S is ve ( also a driving force for
  mixing) Best seen on a graph

? mix S (non-ideal)
? mix S / nR is ve ( ideal)
? mix H / n R T (ideal) 0
? mix H (non-ideal)
1
? mix G / n R T is - ve (ideal)
0
45
Excess function

• Can define an excess function, X E for non-ideal
  mixtures
• S E ? mix S - ? mix S ( ideal )
• H E ? mix H - ? mix ( ideal ) etc.
• e.g. Benzene/ Cyclohexane
  tetrachloromethane / cyclopentane

endothermic
V E
H E
1
1
0
0
? mix V 0 (ideal)
? mix H 0 (ideal)
x(C6H6)
x (C2Cl4)
46
Liquid Vapour Phase Diagrams

• Suppose one component of the mixture is volatile?
  What would be compositions of the liquid and
• vapour phases? Do they differ? Assume a two
  component mixture, A and B and that A and B are
• miscible.
• p p A p B..where p
  total vapour pressure
• 
  p A and p B
  partial vapour presssures.
• If ideal, R L obeyed and
• p x A (l) p A x B (l) pB
• Dalton law states (applies to gases) x i (g)
  p i / p i .e. x A p A / p x B
  p B / p

47
Phase diagrams

• Hence x A (g) x A (l) . p A / ( x A (l)
  . pA x B (l) . pB ) and
• x B (g) x B (l) . p B / ( x A
  (l) . p A x B (l) . p B ) . Best seen
  on a V p versus x graph

Liquid line ( R L )
V p
Tie line
T
Gas line ( Dalton )
1
0
x A
x A (l)
x A (g)
48
Phase diagram

• NB The vapour is richer in x A than the liquid.
  Basis for distillation. However, it is more
• convenient to measure at constant pressure and
  plot Temperature versus composition, x i.

Distillation
NB gas phase is high temp. phase
p
B pt.
T
v
T
l v
vapour line
l
liquid line
x A(l)
x A
x A(g)
x A
49
Distillation

• Can improve separation by Fractional distillation
  on columns.
• For non ideal mixtures get an Azeotropic
  mixture at a particular mole fraction

Ethanol / Benzene - ve
deviation
Acetone / chloroform ve
deviation
Azeotropic point
T
T
v
v
l v
l v
l v
l v
l
Azeotropic point
l
1
0
x C6H6
x CHCl3
0
1
50
Steam Distillation

• NB We cannot distil beyond azeotropic point.
  Hence could not separate acetone from a mixture
• with chloroform. To do this would have to use
  Steam Distillation to separate immiscible
• liquids.
• e.g. Mixture boils when p A p B
  1 atm.
• i.e. at a temperature below the boiling point of
  each component.
• Blow steam through the mixture high temperature
  vapour immiscible with water and is condensed
  ? get separation.
• Useful technique when substances are unstable at
  high temperatures.

51
Other Non ideal Phase Diagrams ( Partially
Miscible Liquids )

• 
  
  
  

Nicotine / Water
Butanol / Water
One phase
( P 1)
One phase
( P 1)
T
T
A
l ß
Two phases
l a
( P 2 )
x A
x A
52
Partially miscible liquids

• What would be the composition of the phase at A?
  
• Use n a / n ß l ß / l a
  Lever Rule ..where n composition of phase.
• Eutectics Consider solid / solid mixture (
  Naphthalene / Benzene )

F
T
solution
solid N
Solid B solution
solution
E
Eutectic composition, ( lowest M pt. )
solid N solid B
1
0
x B
53
Eutectic mixture

• Since F C - P 2
• At eutectic point F 2 - 3 2 1
  ( invariant point) ( P 3 1l 2 s
  phases in equilibrium )
• Solubility
• When a solid left in contact with a solvent it
  dissolves until solution is saturated. Can
  estimate solubility of solute ( solid ).

A solvent B solute
B (dissolved in A)
?B (l)
?B (s)
B (s)
54
Solubility

• At equilibrium ?B (s) ?B (l)
• For an ideal solution R L is obeyed and
• ?B (s ) ?B (l)
  ? B (l) R T l n x B ,,.. where x B is
  the mole fraction of B in solution.
• If process is s ? l ( melt )
• ? ? B (l) - ? (s) - R T l n x B
  ? melt G m
• ? ? melt G m / T - R l n x B
  ( Rem. Gibbs Helholtz ( ? ( ? G / T ) / ? T
  ) p - ? H / T 2
• ? ( ? l n x B / ? T ) p - ? melt H m / R
  T 2

55
Solubility

• For a finite change
• l n x B - ? melt H m / R ( 1 / T - 1 /
  T )
• NB. T melting point of pure solid ie. when
  x B 1, T T
• i.e. l n x B - ? melt H m / R ( 1 / T -
  1 / T )
• x B e - ? melt H m / R ( 1 / T -
  1 /T )
• Over a narrow temperature range
• l n x B - ? melt H m / R . 1 / T
  constant
• y m
  x c

56
Solubility
Slope - ? melt H m / R
l n x B
1 / T
57
Colligative Properties

• The effect of non-volatile matter in solution on
• 1. Boiling point elevation
• 2. Freezing point depression
• 3. Osmotic pressure
• NB. Colligative properties depends on the amount
  of solute and not its nature or chemical
  composition. e.g. Boiling point elevation.

pure solvent
Vp
solution
1 atm.
T B.pt. Of pure solvent
d T T - T (elevation of B.pt.)
T
T
Temperature
58
Colligative Properties

• Elevation of boiling point Let solvent A
  B involatile solute

? A (g)
A (g )

• A (l)

A (l) B
At Equilibrium ? A (l) ? A (g) ?A
(l) R T l n x A
Since l ? g , ( vapourization )
59
Colligative Properties

• l n x A ( ?A (g) - ? A (l) ) / R T
  ? vap G m / R T
• Applying Gibbs Helmotz ( ? ( ? G / T ) / ? T
  ) p - ? H / T 2
• ( ? l n x A / ? T ) p - ? vap H
  m / R T 2
• when x A 1 , T T ( ie, pure solvent)
  and
• - l n x A - ? vap H m / R ( 1 /T
  - 1 / T )
• Interested in solvent effect and since x A
  x B 1, x A 1 - x B
• ? l n ( 1 - x B ) ? vap H m / R ( 1 /T -
  1 / T )
• NB. l n ( 1 - x B ) - x B

60
Colligative Property

• ? x B ? vap H m / R ( 1 /T - 1 /
  T )
• Because T T , ( 1 / T - 1 / T )
  ( T - T ) / T T d T / T 2
• x B ? vap H m / R ( d T / T 2 )
  so that
• d T ( T 2 R / ? vap H m ) . x B
• ie. d T a x B and d T K x B
  where K R T 2 / ? vap H m or
• ? T K x B
• Since x B a molality, b then
• ? T K B b where K B
  ebullioscopic constant of the solvent.
• Similar equation for freezing point depression ?
  T K / x B

61
Colligative Properties

• Osmotic pressure

p
p p
h
p osmotic pressure
Is pressure applied to stop flow of solvent
Pure solvent,A
Solution

?A ( p) x A
? A ( p p ) x B
Semipermeable membrane
Permeable to solvent not solute
62
Osmotic Pressure

• x B lowers the chemical potential on the
  solution side. Hence
• ? A gt ? A
• ? pure solvent diffuses into the solution side
  ( RHS of the chamber).
• Ar Equilibrium ? A ( p ) ? A ( p p
  )
• Assume x B in low concentration so as to apply
  R - L
• ? A ?A R T l n x A
  and
• ? A - ? A - R T l n
  x A
• On the RHS of the chamber chemical potential
  increases with pressure.

63
Osmotic Pressure

• Since d G V d p S d T
• ( d G m ) T V m d p and
  hence ( ? ? / ? p ) T V m

Now
64
Colligative Property

• Since x B n B / ( n A n B ) and x
  B n B / n A
• p n B R T / n A V m
  n B R T / V where V total volume of
  solvent.
• N B n B / V B concentration of
  B.
• ? p B R T
• The equation applies only to dilute solutions
  where ideal behaviour is observed and finds
• application in osmometry e.g. to determine molar
  masses of macromolucules.
• Activites
• For a real or ideal solution ? A ? A
  R T l n ( p A / p A ) ,
• where p A Vapour pressure of A in mixture and
  p A vapour pressure of pure A.

65
Activities

• For an ideal solution, Raoults Law is obeyed and
  
• ? A ? A R T l n x A ,
  , where x A mole
  fraction of A in the solution.
• For a non- ideal solution ? A ? A
  R T l n a A , ,
• ? a A ( p A / p A ) and
  since all solvents obey R L
• x A ( p A / p A ) .
• By definition a A ? A x A ,, where ? A
  activity coefficient of A
• ? A ? 1 as x A ? 1
• ? A ? A R T l n ? A x A
  , , and ? A ? A R T l n x A , R
  T l n ? A,

66
Solute Activity

• (1) Ideal dilute solution
• If solute, B obeys Henrys Law p B K B x B
• ? ? B ? B R T l n ( p B / p B )
  ? B R T l n ( K B / p B ) R T l n
  x B, , , ,
• In order to compare equation with that for
  solvent, we combine all the constants
• ? B ? B R T l n ( K B / p B ) so
  that
• ? B ? B R T l n x B
• (2) Real Solutes
• Deviates from ideality.

67
Activity of solute and relation with molalities

• a B p B / K B and a B ? B x B
• NB all the deviations from ideality captured in ?
  B.
• As x B ? 0 ? B ? 1 and a B
  x B i.e as zero concentration is approached
  deviation
• from ideality disappears.
• (3) With molalities ( mol kg-1 )
• Usual to replace x B with molality, b.
• Since x B n B / ( n A n B ) and
  n A gt n B i.e. x B n B / n A and
• x B a n B i.e. x B
  k n B / b ? , where k is the
  proportionality constant and b ? 1.

68
Molalities

• ? ? B ? B R T l n k R T l n
  b and
• ? B ?? B R T l n b , where
  ?? B ? B R T l n k
• NB For a solvent R L activity a p / p
  and ? a / x ? ? 1 as
  x ? 1
• For a solute Henrys Law activity a
  p / K and ? a / x ? ? 1 as x
  ? 0.

69