Liquid – Liquid, Liquid – Solid,
Gas – Solid Equilibrium
Chapter 11
Applications
 Distillation is the major application of
vapor – liquid equilibrium
 Larger variety of applications for
other equilibria




Extraction
Decantation
Vapor Phase Deposition
Metallurgy
Liquid–Liquid Equilibrium
 The most common combination is
water and organic compounds
 For a binary combination of liquids
 Totally miscible
 Partly miscibile
 Practically Immiscible
 Depends on molecular interactions
Solubility Data
Temperature
0C
0
Solubility of
Benzene in
Water
0.04 mole%
Solubility of
Water in
Benzene
0.133 mole%
20
0.04 mole%
0.252 mole%
50
0.0474 mole% 0.664 mole%
70
0.0615 mole% 1.19 mole%
Solubility usually increases with temperature
Practically Insoluble Liquid Pairs
 Water is practically insoluble in
hydrocarbons that do not contain
oxygen or nitrogen
 Those hydrocarbons are practically
insoluble in water
 Adding oxygen increases the solubility
 Adding nitrogen further increases the
solubility
Miscible Pairs
 If we lower the number of carbons in
an organic compound, we increase
the solubility
 The low molecular weight alcohols are
completely miscible
 Glycols are completely miscible
Other Miscible pairs
 Similar chemically
 All straight chain liquid hydrocarbons are
miscible with each other
 Chlorinated hydrocarbons
 Most molten metals
 Many molten salts
 They do not necessarily have an activity
coefficient of one
 Most form Type II solutions
 May have an azeotrope
Ternary LLE
 Mixtures of three liquids
 Might be completely miscible
 Might form three liquid phases, each
primarily one of the species with small
amounts of the other species dissolved
 Might form two liquid phases
 Water
 Ethanol
 Benzene
Completely miscible
Completely miscible
Immiscible pair
Three Component Phase Diagrams
So far we have only looked at 2
component phase diagrams (like those
in chapter 8)
Three component phase diagrams can
be very complicated – they require a
three dimensional phase diagram
A 2-D slice can tell us a lot though
Z
X
20%
40%
60%
Wt % Y
80%
Y
Pure
Ethanol
One Phase
Region
Two Phase
Region
Pure
Water
Pure
Benzene
Not to Scale
Pure
Ethanol
Tie lines
connect
equilibrium
concentrations
from each
phase
Plait
Point
The plait point
represents the
concentration
where the
compositions
of the two
phases
becomes the
same
Tie
Lines
Pure
Water
Pure
Benzene
Not to Scale
If the overall
composition is:
40% benzene
40% water
20 % ethanol,
How many
phases are
present?
Pure
Water
Pure
Ethanol
What is the
composition of
each of the
phases?
How much of
each phase?
Use the lever
law
Pure
Benzene
Remember that solubility is
temperature dependent
 You need a triangular phase diagram
like the one on the previous slide for
each temperature (Figure 11.6)
 For binary systems, a solubility
diagram may be useful
Solubility is temperature dependant
Solubility temperature diagram
Single
phase
region
Temperature
Solubilty
diagrams
that
represent
several
different
behavior
patterns are
shown in
Figures
11.3, 11.4
and 11.5
Two
phase
region
Mole fraction, xa
Elementary Theory of LLE
 There is a lot of literature describing liquidliquid equilibria
 However, theoretical models give fair to
good estimates of liquid-liquid behavior –
and it’s a lot easier to calculate than to
search through the literature.
 Usually implemented on computers
 Simplified calculations can be done by hand
Back to basics
fi
fi
 fi
( Liquid1)
( L1)
 fi

( L 2)
P x
( L1) ( L1) 0
i
i
i
x
( Liquid 2 )

( L 2) ( L 2) 0
i
i
i
P
The pure component vapor pressure is the same in
each phase
This is true for any number of
species and any number of liquid
phases

( L1) ( L1)
i
i
x
( L1)
i
x

x

( L 2) ( L 2)
i
i

( L 2) ( L 2)
i
i
( L1)
i
x

Now consider only a binary system
in two liquid phases
( L1)
i
x


( L 2) ( L 2)
i
i
( L1)
i
x
If the species under
consideration are
practically insoluble
our math gets easier

If phase two is almost pure i, then it’s mole fraction is
almost one and it’s activity coefficient is almost one in
phase two
( L1)
i
x

1

( L1)
i
Section
8.5
This is a great way to find activity
coefficients for practically insoluable species
– but doesn’t work for species with
intermediate solubility – our approach must
go back to some sort of an activity
coefficient estimating equation– Like the
Van Laar equation or any of the equations
described in Chapter 9. We’ll also need to
return to Gibbs Free Energy calculations,
like those in chapter 6.
Gibbs free energy of two
imaginary chemicals
Remember
this?
Gibbs Free Energy, g,
kJ/mole
4
3
xa=.833
in phase 2
xa=.166
in phase 1
2
a
b
1
0
-1
0
0.2
0.4
0.6
0.8
Mole fraction of component a
1
Example 11.5
 Consider a binary system, with compounds
a and b
 The pure species Gibbs free energies are:
 ga0=2 kJ/mol
 gb0=1 kJ/mol
 Their liquid phase activity coefficients can
be represented by the symmetric or two
suffix Margules equation (Chapter 9) –
which is simpler than the van Laar equation
Example 11.5 cont
 Prepare a Gibbs free energy plot for
this mixture at 298 K, for values of
the constants in the activity
coefficient correlation of:




a
a
a
a
=
=
=
=
0
1
2
3
Equation for Gibbs free Energy
g mixture  g ideal  g
E
Developed in Chapter 9
g mixture   xi g   xi ( RT ln xi )   xi ( RT ln  i )
0
i
from the symmetric equation….
ln  i  a(1  xi )
2
So… if we know a and x, we can find g of the mixture!!
The simplest case is for a=0
ln  i  a (1  xi )  0
2
which means that the activity
coefficient is 1 .. and we have
an ideal solution!!
0
g mixture   xi g   xi ( RT ln xi )   xi ( RT ln  i )
0
i
Ideal solutions do not separate into two liquid phases
Gibbs free energy of the
mixture, kJ/mole
Gibbs Free Energy of a Binary Mixture
2.5
2
1.5
1
a=0
0.5
0
-0.5
0
0.2
0.4
0.6
0.8
mole fraction of component a
1
What happens when a=1?
ln  i  a (1  xi )  (1  xi )
2
 when xa is 1, γa is 1
 when xa is 0, γa is 2.718
2
Gibbs free energy of the
mixture, kJ/mole
Gibbs Free Energy of a Binary Mixture
2.5
2
1.5
a=1
1
0.5
0
-0.5
0
0.2
0.4
0.6
0.8
mole fraction of component a
1
What happens when a=2?
ln  i  a (1  xi )  2(1  xi )
2
 when xa is 1, γa is 1
 when xa is 0, γa is 7.389
2
Gibbs free energy of the
mixture, kJ/mole
Gibbs Free Energy of a Binary Mixture
2.5
2
a=2
1.5
1
0.5
0
-0.5
0
0.2
0.4
0.6
0.8
mole fraction of component a
1
What happens when a=3?
ln  i  a(1  xi )  3(1  xi )
2
 when xa is 1, γa is 1
 when xa is 0, γa is 20.09
2
Gibbs free energy of the
mixture, kJ/mole
Gibbs Free Energy of a Binary Mixture
2.5
a=3
2
1.5
1
0.5
0
-0.5
0
0.2
0.4
0.6
0.8
mole fraction of component a
1
Gibbs free energy of the
mixture, kJ/mole
Gibbs Free Energy of a Binary Mixture
2.1
1.9
1.7
1.5
1.3
1.1
0.9
0.7
0.5
Two phase region
0
0.2
0.4
0.6
0.8
mole fraction of component a
1
Remember that equality of the
partial molal Gibbs free energy, is
what determines the equilbrium
 However, the values of pure
component Gibbs Free energies does
not influence whether or not there are
two phases
g mixture   xi g   xi ( RT ln xi )   xi ( RT ln  i )
0
i
This is the quantity usually plotted in figures
g mixture   xi g   xi ( RT ln xi )   xi ( RT ln  i )
0
i
You can use this procedure with whatever
activity coefficient equation matches the
expermental data best
 Example 11.6 asks us to:
 Find the equilibrium compositions for the
n-butanol water system at 92 C
 Use the van Laar equation to find the
activity coefficients
g mixture - sum of the pure
component terms, kJ/mole
A Binary Mixture of n-butanol and water
0
-0.2
-0.4
-0.6
Two phase
region
-0.8
-1
0
0.2
0.4
0.6
mole fraction of water
0.8
1
This approach only gives fair
results
 In the n-butanol - water example we
predicted two phases with
compositions of
 0.47 water and .97 water
 Experimental results show
 0.67 water and 0.98 water
 This is common – computer programs
use more complicated activity
coefficient correlations
Effect of pressure on LLE
 Pressure has very little effect on
liquid – liquid equilibria
Effect of temperature
g mixture   xi g   xi ( RT ln xi )   xi ( RT ln  i )
0
i
 Temperature has effect on the Gibbs
Free energy of the mixture, and so
we would expect it to have an effect
on the solubility of liquids
Solubility is a function of
temperature
Distribution coefficients
 Consider a multi component system,
like the benzene – ethanol – water
system
water rich phase
Ethanol
benzene rich phase
Ethanol
x
K
x
Recall…
fi
fi
 fi
( Liquid1)
( L1)
 fi

( L 2)
P x
( L1) ( L1) 0
i
i
i
x
( Liquid 2 )

( L 2) ( L 2) 0
i
i
i
P
Thus…
( L1)
i
( L 2)
i
x
x



( L 2)
i
( L1)
i
K
Distribution Coefficients
Liquid Solid Equilibria
 Solids often dissolve in liquids, but
liquids rarely dissolve in solids
 For example, consider salt (NaCl) and
water
 Solid NaCl dissolves in water
 Water does not dissolve into the solid
salt significantly
 Solubility products are usually used to
quantify solubilities
Effect of temperature
 Solids are usually more soluble as
the temperature goes up – but not
always
 Pressure rarely has a significant effect
 Most solids
increase in
solubility with
temperature
 Gypsum becomes
less soluble
Solid – Liquid Phase Diagrams
 The salt – water phase diagram is
fairly simple
 It contains a eutectic point, and an
intermediate compound
 NaCl·2H2O
Iron-Iron Carbide Phase Diagram
1 atm
1600 C
d, ferrite
Molten Metal -- Liquid
1400 C
γ + liquid
Eutectic
,
austenite
1200
C
cementite
+ liquid
1000 C
γ + cementite
γ+
ferrite
Eutectoid
800 C
600 C
a, ferrite
ferrite + cementite
400 C
Fe
1% C
2% C
3% C
4% C
Cementite
(Fe3C
5% C
6% C
6.70% C
Gas – solid equilibria (GSE)
 Pressure does not change the
properties of the solid very much, but
it does change the properties of the
gas
 Temperature affects the solid some –
but it affects the gas significantly
 First let’s look at low pressures
Low pressure GSE
 Very similar to VLE
 sublimation instead of vaporization
 ends at the triple point
 We need tables of data, like the
steam tables
 The Antoine equation does not apply
Examples
 If you hang your laundry out to “dry”
when it is below freezing outside,
eventually the lce sublimes
 If you place a glass mug in the
freezer it develops a layer of frost –
vapor deposition
 Vapor deposition is used industrially
in the semiconductor business
GSE at High Pressures
 The gas is affected, not the solid
 At pressures above the critical pressure,
the resulting “dense fluid” or “supercritical
fluid” display significantly different
properties
 They often serve as good solvents for solids
 Supercritical fluids are used in the production of
coffee to remove impurities that affect taste