Phase Equilibrium
A phaseis a part of a system, uniform in chemical composition and physical properties, that is separated from other homogeneous parts
6.1 Phase Diagrams of One-Component Systems
The P V− −T surface for a one-component system projected onto the P V− plane or the P−T plane
The phase diagram for water
The phase diagram for water at high pressures The ice I melts at –22℃ when the pressure is raised to 2000 bar.
F = C – p + 2
For a one-component system, F = 3 – p: more than three phases can not be in equilibrium
Phase diagrams
The phase diagram for carbon dioxide The equilibrium between solid and gas
at 1 bar and –78℃. The phase diagram for carbon
The equilibrium between graphite and diamond at room temperature only at pressures above 10,000 bar.
Chemical potential of phases
Dependence of the chemical potentials of solid, liquid and gas phases on temperature at constant pressure.
The dashed lines are for a lower pressure.
The plots should be slightly concave downwaed, since the entropy increases with increasing temperature..
,
g l s
P
g l s
T
S S S S
T
V V V V
P µ µ
∂ = − > >
∂
∂ =
∂
Types of phase transitions
First order phase transition
continuous chemical potential (free energy) with a discontinuity in the first derivatives of the chemical potential. Involves a latent heat and coexistence of phases.If is discontinuous, then and
If is discontinuous, then
, 0
P
T
S S S H H H
T
V V V
P
α β
α β α β
α β
µ
µ
→
∂ = − ≠ ≠ ∆ ≠
∂
∂ ∂ = ≠
Second order phase transition
continuous chemical potential (free energy) and its first derivative but with a discontinuity in the second derivatives of the chemical potential. Continuous transitions such as the ferromagnetic transition, the superfluid transition, and Bose-Einsteincondensation.
6.2 The Clapeyron Equation
For a one-component system with two phases, α and β
At equilibrium,
Along the coexistence curve,
d d
d d d d
d d d d
d d
G V P S T
V P S T V P S T
P S S S H
T V V V T V
α β
α β
α α β β
β α
β α
µ µ
µ µ
µ
=
=
= = −
− = −
− ∆ ∆
= = =
− ∆ ∆
Coexistence curve for a one-component system. Along the line µα= µβ. If
temperature is changed by dT, the pressure has to be changed by dP as indicated to maintain equilibrium.
The Clapeyron equation
Example 6.1
What is the change in boiling point of water at 100℃ per Pa change in atmospheric pressure?vap 2 1
3 3 1 3 3 1
2 2
vap 1
3 3 1
At 100 C and 1.01325 bar
and
(H O) 40.69 kJ mol
(H O, l) 0.019 10 m mol (H O, g) 30.199 10 m mol
d 40,690 J mol
3,613 Pa K d (
g l) (373.15 K)(30.180 10 m mol )
H
V V
P H
T T V V
−
− − − −
− −
− −
∆ =
= × = ×
= ∆ = =
− ×
1
4 1
d 2.768 10 K Pa d
T P
− −
= ×
The Clapeyron equation
Example 6.3
Using the thermodynamic data, determine the vapor pressure of H2O(l) st 298.15K.1
f 298 2
1
f 298 2
1 vap 298
(H O, l) 237.129 kJ mol (H O, g) 228.572 kJ mol
ln 228.572 ( 237.129) 8.857 kJ mol 0.03169 bar
G G
G RT P P P
−
−
−
∆ = −
∆ = −
∆ = − = − − − =
=
H
2O(l) = H
2O(g)
The Clapeyron equation
Example 6.4
Calculate the equilibrium pressure for the conversion of graphite to diamond at 25℃.The densities of graphite and diamond may be taken to be 2.25 and 3.51 g cm–3, respectively, independent of pressure.
C(graphite) = C(diamond)
2 1
1 f 298
1
f 298 f 298
6 3 1 6 3 1
2
2 1 2
1
(diamond) 2, 900 J mol
(diamond) (graphite) 2, 900 0 2, 900 J mol
1 1
12 10 m mol 1.91 10 m mol
3.15 2.25
, d
Pd (
T P
G
G G G
V
G V G G G V P V P P
P
−
−
− − − −
∆ =
∆ = ∆ − ∆ = − =
∆ = − × = − ×
∂∆ = ∆ ∆ = ∆ − ∆ = ∆ = ∆ −
∂
∫ ∫
1
1
2 1 5
2 1 6 3 1
9 4
2
) 0 2, 900 J mol
10 Pa 1.91 10 m mol
1.52 10 Pa 1.52 10 bar
G G
P P
V P
−
− −
∆ − ∆ −
= + = +
∆ − ×
= × = ×
6.3 The Clausius – Clapeyron Equation
g l
vap vap
2 g
vap 2
vap
2
vap
Assuming the ideal gas and
d d
d d ln d
d ln d
ln
V RT V P
H P H
P
T TV RT
P P H
P P RT T
P H T
P R T
P H
RT C P
=
∆ ∆
= =
= = ∆
= ∆
= − ∆ +
∫ ∫
Vapor pressure of water
2 2
1 1
/ vap
/ 2
2 vap
1 2 1
d ln 1 d
1 1
ln
P P T
P P T
P H
R T
P T
P H
P R T T
= ∆
∆
= − −
∫
∫
vap vap
2 vap
d ln d ln
d
Pd(1/ )
PP P
H RT R
T T
∆ = = −
The enthalpy of
vaporization approaches zero as the temperature approaches the critical temperature.
2 vap
could be used.
H A BT CT
∆ = + +
6.4 Vapor-Liquid Equilibrium of Binary Liquid Mixtures
When a binary liquid mixture is in equilibrium, Assume that the vapor is an ideal gas, The chemical potential of a
(g) (l)
(g) (g) ln
i i
i
i i
RT P P
µ µ
µ µ
=
=
+
component in a liquid mixture,
µ
i(l) = µ
i(l) + RT ln a
i*
* *
for the pure liquid phase,
(g) ln (l) ln
1 (g) ln (l)
ln ln
i
i i i
i
i i i
i i
i i
i i
RT P RT a
P
a RT P
P
P P
RT RT a a
P P
µ µ
µ µ
+ = +
= + =
= =
The activity of a component of a solution is equal to the ratio of its partial pressure above the solution to the vapor pressure of the pure liquid.
*
*
Raoult's law
(l) (l) ln
i i i
i
i i
i
i i i
P x P a P x
P
RT x
µ µ
=
= =
=
+
For a pair of liquids A and B, A–A, A–B, and B–B interactions are all the same.
The definition of an ideal solution
Vapor-liquid equilibrium of binary liquid mixtures
*
* * * * *
1 2 1 1 2 2 2 1 2 1
* *
1 1 1 1 1
1 * * * * *
1 2 1 1 2 2 2 1 2 1
* 1 2
1 * * *
1 2 1 1
* * 1 2
* * *
1 2 1 1
Raoult's l
For an ideal liquid mixture aw
,
( )
( )
( )
( )
i i i
P x P
P P P x P x P P P P x
P x P x P
y P P x P x P P P P x x y P
P P P y P P P
P P P y
=
= + = + = + −
= = =
+ + + −
= + −
= + −
the bubble point line
the dew point line
Benzen(1)–toluene(2) system at 60℃.
the tie line The lever rule If the mole fraction of toluene is x,
the ratio of the number of moles of liquid to the amount of vapor is equal to (x – v) / (l – x).
Ideal solution
*
*
Raoult's law
(l) (l) ln
i i i
i
i i
i
i i i
P x P a P x
P
RT x
µ µ
=
= =
=
+
The definition of an ideal solutionln ln
i i i i i i
i i i i
i i
i i
G x x RT x x
S x S R x x H x H
V x V
µ µ
= = +
= −
=
=
∑ ∑ ∑
∑ ∑
∑
∑
2
( / )
[ ( / ) / ]
( / )
P
P
T
S G T
H T G T T
V G P
= − ∂ ∂
= − ∂ ∂
= ∂ ∂
mix
mix
mix
mix
ln ln 0
0
i i
i i
G RT x x
S R x x
H V
∆ =
∆ = −
∆ =
∆ =
∑
∑
An ideal solutions involves ideal mixing.
There is no volume change or heat evolution when liquids are mixed to form ideal solutions at constant temperature and pressure.
Ideal binary liquid mixtures
t/℃ 80.1 88 90 94 98 100 104 110.6
P1*/bar - 0.508 0.543 0.616 0.698 0.742 0.836 1.013
P2*/bar 1.013 1.285 1.361 1.526 1.705 1.800 2.004 - Vapor pressures of Tolume(1) and Benzene(2)
Benzene(2)–toluene(1) boiling point diagram.
The liquid boils at the temperature given by the lower curve.
At 100℃
* 2
1 * *
1 2
* 1 1 1
1.013 bar 1.800 bar
0.744 0.742 bar 1.800 bar
(0.744)(0.742 bar)
0.545 1.013 bar
P P x P P y x P
P
− −
= = =
− −
= = =
Two-component liquid and vapor system
P–T–composition diagram for a two-component liquid and vapor system The plot consists of two surfaces, bubble point surface and dew point surface.
Fractional distillation
Bubble-cap fractionating column
When a binary solution is partially vaporized, the component that has the higher vapor pressure is concentrated in the vapor phase. This vapor may be condensed, and the vapor obtained by partially vaporizing this condensate is still further enriched in more volatile component. The process of successive
vaporization and condensation is carried in a fractionating column.
6.5 Nonideal Mixtures
Liquid mixture with a negative azeotrope: chloroform(1) – acetone(2) at 35.17℃
Boiling point curve at constant
pressure for a maximum boiling point
Liquid mixture with a positive azeotrope: carbon disulfide(1) – acetone(2) at 35.17℃
Boiling point curve at constant pressure for a minimum boiling point Azeotrope the vapor has the
same composition as the liquid
A–B interactions are stronger than A–A and B–B interactions due to a weak hydrogen bond
A–A and B–B interactions are stronger than A–B interactions
Cl
Cl Cl
H O
C H3 C H3
Henry’s Law
Raoult’s law is approached for a component as its mole fraction approaches unity. As the mole fraction approaches zero, its partial pressure is given by the Henry’s law.
i i i i
P = y P = K x
In sufficiently dilute solutions the environment of the minor component is constant, and its partial pressure is proportional to its mole fraction.
Vapor pressure curve for one component of a binary liquid mixture at constant temperature
Evaluation of Henry’s law constant K2
Dilute real solutions
If the vapor pressure of a solute follows Henry’s law, in dilute solutions its chemical potential follows an equation like that of an ideal solution.
*
*
(l) (g) (g) ln (g) ln ln (l) ln
(l) (g) ln
i i i
i i i i i i i
i
i i
K x K
RT RT RT x RT x
P P
RT K P
µ µ µ µ µ
µ µ
= = + = + + = +
= +
The chemical potential of solute i in its standard state in the liquid. The solute at xi= 1 has the same properties as in very dilute solution, where each molecule is surrounded only by molecules of solvent.
Henry’s law and Raoult’s law
If Henry’s law holds for the solute, Raoults’ law holds for the solvent.
2 2
2 2 2
2
2 2
2
1 1 2 2
1 2 2 1
2
1 2 2 1 1
1 1 1
1 1
1 1 1
1 1 1
ln ln
d d
The Gibbs-Duhem equation d d 0 1, d d
d d d d d ln
ln constant If 1, then
ln
i
RT a RT K x
P RT x
x
x x
x x x x
x RT RT
x x RT x
x x x
RT x x
RT x
µ µ µ
µ
µ µ
µ µ
µ
µ µ µ µ
= + = +
=
+ =
+ = = −
= − = − = =
= +
= =
= +
6.6 Activity Coefficients
The activity coefficient γi for nonideal solutions,
a
i= γ
ix
iAs x
i→ 1, a
i→ 1 and γ
i→ 1
Based on Raoult’s law
* * * for ideal gases
(l) (l) ln
i i i i
i i i
i i i i
i i i i i
RT x
P P y P
a x
P x P x P
µ µ γ
γ γ
= +
= = = =
Partial pressure of ether(1) – acetone(2) solution at 30℃
x2 P1/kPa x1P1*/kPa γ1 P2/kPa x2P2*/kPa γ2 K2x2/kPa γ’2
0 86.1 86.1 1.00 0.00 0.00 … 0.00 (1.000)
0.2 71.3 68.9 1.04 12.0 7.50 1.60 15.7 0.77
0.4 58.7 51.7 1.14 19.7 15.1 1.31 31.4 0.63
0.5 52.1 43.1 1.21 22.4 18.9 1.19 39.2 0.57
0.6 44.3 34.4 1.28 25.3 22.7 1.12 47.0 0.54
0.8 26.9 17.3 1.56 31.3 30.1 1.04 62.7 0.50
1.0 0.00 0.00 … 37.7 37.7 1.00 78.4 (0.48)
Table 6.5 Activity coefficients for ether(1) – acetone(2) solution at 30℃
Activity coefficient based on Henry’s law
* '
*
' '
(l) (l) ln
(l) (g) (g) ln
(l) (g) ln
i i
i i i i
i
i i i
i
i i
i i
i i i
i i i i
RT x RT P
P RT K
P
P y P P K x
x K x K
µ µ γ
µ µ µ
µ µ
γ γ
= +
= = +
= +
= = =
Evaluation of Henry’s law constant K2for
acetone in ether(1) – acetone(2) solution at 30℃
6.7 Colligative Properties
Colligative = bind together (L) properties that all depend on the number of particles but not the kind:
freezing point depression, boiling point elevation, osmotic pressure and vapor pressure lowering The ideal solution of a solute B and a solvent A, out of which pure crystalline A freezes.
A A A A A
fus A
A A
A
fus A
fus ,A fus A fus
two phases are equilibrium at
the Gibbs energy of fusion at
Assume and that and
(s, ) (l, , ) (l, ) ln
( ) (s, ) (l, )
ln
( )
0
P
T
T
T T x T RT x
G T
T T
x RT RT
G T
C H
µ µ µ
µ µ
= = +
∆
= − = −
∆
∆ = ∆ ∆
fus,A
A
fus A
fus A fus A fus A fus A
fus,A
fus,A
fus A fus A fus A f
A 2 B
fus,A fus,A
are independent of temperature.
( )
1 1
ln ln(1 )
BS
G T H T S H T H
T
T T
H H H T
x x x
R T T R T T RT
∆ = ∆ − ∆ = ∆ − ∆
−
∆ ∆ ∆ ∆
= − − = − = − = − = −
Freezing point depression
2 fus,A
f B
fus A
B
B B A A B
A B A
2
fus,A A
f B f B
fus A
2
fus,A A f
fus A
freezing
To use molal concentration,
The point constant
( 1 )
1 /
T RT x
H
m x m m M m m
M m M
RT M
T m K m
H
RT M
K H
∆ = ∆
= = = >>
+
∆ = =
∆
= ∆
Example 6.11
What is the value of the freezing constant for water? The enthalpy of fusion at 273.15K is 6.00 kJ mol–1.1 1 2 3 1
f 1 1
(8.3145 J K mol )(273.15 K) (18.02 10 kg mol ) K 1.86
6000 J mol mol kg
K
− − − −
− −
=
×=
The Osmotic pressure
l l 1 l 1
1 1
1 l l
*
l l 1
* 1
Assuming an ideal solution at constant temperature ,
Assuming that 1 is constant,
( , ) ( , , ) ( , ) ln
d d ( , ) ( , ) d
( , ) ( , )
P P
T
V
P T P T x P T RT x
V P P T P T V P
P T P T V
V RT
µ µ µ
µ µ µ
µ µ
+Π
= + Π = + Π +
= + Π = +
+ Π = + Π
Π = −
∫
1 2 2
2 2 * 2
2 1 2 2
1
2
2 1 1 2
2
ln ln(1 )
, ,
x RT x RT x
n n V m
x V n V n RT
n n n
m M
n M
RT V
= − − =
= ≅ = = Π =
+
Π =
6.8 Two-Component Systems of Solid and Liquid Phases
Components are completely miscible in the liquid state and completely immiscible in the solid state
Tm(Bi) = 273℃, Tm(Cd) = 323℃
Eutectic temperature 140 ℃ at 40% Cd The eutectic is not a phase, but a mixture of two solid phase with a fine grain structure.
Above JKL, one liquid phase (the solution region). F = 2 – p + 1 = 2
Along JK, Bi freezes out; Under JK and above K, solid Bi and a solution. F = 1
Along LK, Cd freezes out; Under LK and above K, solid Cd and a solution. F = 1 The relative amount of the two phases is determined by the lever rule on the tie line.
K. the eutectic point. F = 0, an invariant point.
Below K solid Bi and solid Cd, F = 1 (T)
fus A A
fus,A
1 1
exp H
x R T T
∆
= − −
A reacting binary system
The components of a binary system react to form a solid compound that exists with liquid.
congruently melting compound
Maximum at 50%, 33%, 25%, and so on,
corresponding to 1:1, 1:2, 1:3, and so on binary compounds.
J at 650℃, 0.6 mol Mg and 0.4 mol Zn solution
K at 470℃. MgZn2 separates from the solution.
The freezing point is lowered as the solution becomes richer in Mg till 347℃.
D at 347℃ 74 mol% Mg and 26 mol% Zn, the whole solution freezes and solid MgZn2and solid Mg come out together.
A solid solution
The freezing point curve for the liquid solution
The melting point curve for the solid solution
At 1400℃, 70 mol% Pt solid solution and 28 mol% Pt liquid solution coexist on the tie line.
6.9 Effect of Surface Tension on the Vapor Pressure
A curved surface of a liquid, or a curved interface between phases, exerts a pressure so that the pressure is higher in the phase on the concave side of the interface.
Spherical droplet of a pure liquid in contact with its vapor in a closed container at temperature T
Consider the liquid and vapor phases separately at constant volume and constant temperature
l l l l l s
v v v v v
3 2 2
l l s s
s l
l v v l v l l v
v l l l
l v
(d ) d d d
(d ) d d
4 , d =4 r d ; 4 , d 8 d 3
d 2 d
, d d , , and d d
(d ) ( )d 2 d 0 at equilibrium
2
T
T
T
A P V n A
A P V n
V r V r A r A r r
A V
r
A A A V V n n
A P P V V
r P P
r
µ γ
µ
π π π π
µ µ γ
γ
= − + +
= − +
= = =
=
= + = − = = −
= − + =
− =
l v2
for bubble
.
c f P P
r
− = − γ
Effect of Surface Tension on the Vapor Pressure
s l l l
' '
l l
l , l l l l l l l
v , v v
' l
l v , l l v v
' l
v l l
Assuming that the vapor is an ideal gas, v
2 2
d d d
2 2
(d ) d d ( )d d
(d ) d
, (d ) ( 2 )d d 0 at equilibrium 2
T P
T P
T P
A V V n
r r
G n V n V n n
r r
G n
G G G G V n n
r r V
R
γ γ
µ µ µ
µ
µ γ µ
µ µ γ µ
µ
= =
= + = + =
=
= + = + + =
= + =
+
'l l
*
* '
v l
* l
ln 2
lim (for flat surface), ln the Ke lvin equation ln 2
r
T P V
P r
P P RT P
P
RT P V
P r µ γ
µ µ
γ
→∞
= +
= + =
=
Consider the liquid and vapor phases separately at constant pressure and constant temperature