Chapter 11:
Other Types of
Phase Equilibria in
Fluid Mixtures
(selected topics)
Partitioning a solute among
two coexisting liquid phases
• Two partially miscible or completely
immiscible liquids
• How the solute distributes between the
two phases
• Purification
– LL extraction, Partition chromatography
• Drug distribution
– Lipids, body fluids
• Pollutant distribution
– air.,water,soil
Partition of a solute
Distribution coefficient
Concentration of solute in phase I
K
Concentration of solute in phase II
The LLE equilibrium condition is:
x x 
I
i
Then:
K x ,i
I
i
II
i


II
i
 T , P, x
x

 I
I
x
 i T , P, x
I
i
II
i
II
i
II


Case I: solute does not affect
solubility of the solvents
• Little amount of solute or totally
immiscible liquids
• I-a) Ni moles of solute completely
dissolved and distributed between
immiscible solvents
Case I-b: some undissolved solute (solid
or gas) in equilibrium with two
immiscible solvents
Case Ic: partially miscible
liquids
Case II: solute affects the LLE
(partially miscible solvents)
Liquid-liquid equilibrium (LLE)
Extraction problems involve at least three components: the
solute and two solvents. It is usual to represent their phase
behavior in triangular diagrams.
Liquid-liquid equilibrium (LLE)
Extraction problems involve at least three components: the
solute and two solvents. It is usual to represent their phase
behavior in triangular diagrams.
Plait point
Tie line
Binodal curve
Liquid-liquid equilibrium (LLE)
Reading the scale in a triangular diagram
Example 4
It is desired to remove some acetone from a mixture that
contains 60 wt% acetone and 40 wt% water by extraction with
methyl isobutyl ketone (MIK). If 3 kg of MIK are contacted with 1
kg of this acetone+water solution, what will be the amounts and
compositions of the phases in equilibrium?
Solution
Example 4
60 wt% acetone and 40 wt% water (1 kg)
Pure MIK (3 kg)
Example 4
Next…
Find the point that represents the global composition of the
system.
Based on the information given, the total amounts of acetone,
water, and MIK are equal to 0.6 kg, 0.4 kg, and 3 kg.
The corresponding weight fractions are: 0.15, 0.10, and 0.75.
Locate this point in the diagram.
Example 4
Global composition
Example 4
Approximated tie line
Example 4
Approximated tie line
MIK-rich phase
80.5% MIK
15.5% Acetone
4.0% Water
Water-rich phase
2.0% MIK
8.0% Acetone
90.0% Water
Example 4
Calculation of the phase amounts (LI and LII)
Global mass balance
LI  LII  4  LII  4  LI
One component mass balance (water for example)
0.04 LI  0.90 LII  0.04 LI  0.90  4  LI  
0.4kg  L  3.721kg
I
Liquid-liquid equilibrium (LLE)
Liquid-liquid equilibrium (LLE)
Osmotic equilibrium
Consider two cells at the same temperature, separated by a
membrane permeable to some of the species present, but
impermeable to others.
For simplicity, assume a binary solute+solvent system and that
the membrane is permeable to the solvent only. Cell I contains
the pure solvent and cell II contains the mixture.
At equilibrium, the following equation is valid:
f solvent T , P
I
 f
II
solvent
Osmotic equilibrium
f solvent T , P
f solvent T , P
I
x
I
II
solvent
 f

II
solvent
II
solvent
f solvent T , P
II
 PI V L

I
sat
sat
f solvent T , P   solvent Psolvent exp   solvent dP 
RT
sat
 Psolvent

 PII V L

sat
sat
f solvent T , P II   solvent
Psolvent
exp   solvent dP 
RT
sat
 Psolvent


Osmotic equilibrium
L


V solvent
II
II
1  xsolvent  solvent exp  
dP 
 P I RT

P II
Assuming the liquid is incompressible:
V
II
II
1  xsolvent  solvent exp 

L
solvent
P
 P 

RT

II
I
Osmotic equilibrium
Applying logarithm:
0  ln x
II
solvent
P  P
II
I
 ln 
   
II
solvent

RT
V
L
solvent
V
L
solvent
P
II
P
I

RT
II
II
ln
x

ln

 solvent
solvent 
 : osmotic pressure
Example 5
Compute the osmotic pressure at 298.15 K between an ideal
aqueous solution 98 mol% water and pure water.
For an ideal solution, the activity coefficient is equal to 1 and
the molar volume of water is approximate equal to 18x10-6
m3/mol.
Solution
Example 5
Compute the osmotic pressure at 298.15 K between an ideal
aqueous solution 98 mol% water and pure water.
For an ideal solution, the activity coefficient is equal to 1 and
the molar volume of water is approximate equal to 18x10-6
m3/mol.
Solution
J
8.314
 298.15 K
mol.K
 
 ln 0.98
3
6 m
18 10
mol
Osmotic equilibrium
For ideal solutions:
 
RT
V
L
solvent
II
solvent
ln x
For a dilute ideal solution, by using a Taylor series expansion of
the logarithm of the solvent mole fraction, the following
approximated expression can be derived:

RT
V
L
solvent
1  x
II
solvent
V
RT
L
solvent
II
solute
x
Osmotic equilibrium

RT
L
V solvent
II
xsolute

RT
L
V solvent
II
nsolute
 II

II
nsolvent  nsolute
II
nsolute
II
Vsolution
RT
 II
L
II
V solvent nsolvent  nsolute
II
Vsolution
Assuming:
II
II
Vsolution
 Vsolvent
and
II
II
II
nsolvent
 nsolute
 nsolvent
II
nsolute
II
II
II
Vsolution
nsolute
Csolute
RT
 L
 II
 RT  II
 RT
Vsolution
msolute
V solvent nsolvent
II
Vsolvent
Osmotic equilibrium
For simplicity, let us drop superscript II, then:
Csolute
  RT
msolute
where msolute is the solute’s molar mass.
A practical application of this equation is to use it to find the molar mass
of polymers and proteins.
Osmotic equilibrium
Schematics of an osmometer
Example 6
Polyvinyl chloride (PVC) is soluble in cyclohexanone. At 25oC, if a
solution of PVC batch with 2 g/L of solvent is placed in an
osmometer, the height h in the osmometer is 0.85 cm. Knowing
that the density of pure cyclohexanone is 0.98 g/cm3, estimate
the molar mass of this PVC batch.
Example 6
Solution
At the membrane, in the
mixture side:
P II  Patm  solution g  h  H 
H
At the membrane, in the
pure solvent side:
P I  P atm   solvent gH
Assuming the density of the solution
and of the solvent are equal:
P II  P I     gh
Example 6
Solution
H
g
1kg 106 cm3
m
1m
   gh  0.98 3 

 9.81 2  0.85cm 
 81.72 Pa
3
cm 1000 g
1m
s
100cm
Example 6
Solution
Csolute
  RT
msolute
H
msolute
msolute  RT
Csolute

J
g 1000 L 1kg
8.314
 298.15K  2 

3
kg
mol.K
L
m
1000 g

 60.67
81.72 Pa
mol
Recommendation
Read the sections of chapter 11 covered in these notes and review the
corresponding examples.