University of Juba

School of Applied and Industrial Sciences

Department of Chemistry and Industrial Chemistry

Course title:

Solution and Phase Equilibrium

By/ Dr. Omer Eisa

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa

Course outlines

 Chemical potential
 Gibbs free energy
 Simple mixtures
 Binary mixtures
 Activity coefficient
 Fugacity
 Solution phase equilibrium

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa
 Introduction to Simple Mixture
Solution:
is referred to as a homogeneous mixture composed of solute and solvent

Solutions categories:
1- Dilute solution
2- Concentrated solution
3- Unsaturated solution

Notes:
 Solution can be a pure substance, but mixtures can not.
 Substance can be an element or component.

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa

Introduction to Simple Mixture

Mixtures: Are a homogeneous solutions that containing a compound

substances.

Thermodynamic descriptions of mixtures:

 The concept of chemical potential as an example of partial molar quantity.

 Descripting the physical properties of mixtures using the chemical
potential.
 Study the application of Raoult’s law and Henry’s law.

William Henry François- M. Raoult

1774-1836 Solution thermodynamics and Phase 1830 – 1901
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Equilibrium, By/ Dr.Omer Eisa
 Introduction to Simple Mixture
The Properties of Solutions:
Application of chemical potential to study the effects of solute on the
thermodynamic properties of solution including:

1. Lowering the vapor pressure of solvent.

2. Elevation of boiling point.
3. Depression of its freezing point.
4. Originating the osmotic pressure.

Notes:
 Models can possibly constructed for real solution (regular solution).
 Regular solution properties deriving from those of ideal solutions..

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa

Introduction to Simple Mixture

Phase diagrams for binary liquid systems:

1. Summarize the equilibrium properties of mixtures in phase diagram.

2. Describing the system of liquids with gradually increasing of complexity
(composition).
3. Phase diagram summarize the empirical observation on the conditions
under which the liquid and vapor of system are stable.

Phase diagrams for ternary systems:

1. Types of phase diagrams describing the system with three components.

2. The diagrams represented by the triangular phase diagrams

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa
 Introduction to Simple Mixture
Activity and Activity coefficient:

Activity: is the concept of chemical potential to real solution involves

introducing an effective concentration.

 Activity also can be interoperated in terms of intermolecular

interactions.
 Activity can present the deviation of solution behavior from ideal
(dilute) extended to more concentrated solutions.

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa

Partial Molar Quantities

Partial Molar Volume: it is the quantity which describes the variation of an
extensive property of a solution or mixture with changes in the molar
composition of mixture at constant T and P.

1 mole H2O 1 mole H2O

H2 O Ethanol
1000 cm3 1000 cm3

The new volume is: The new volume is:

(1000+18)cm3 = 1018 cm3 (1000+18)cm3 = 1014 cm3

The quantity 14 cm3 called the partial molar volume

Solution thermodynamics and Phase
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Equilibrium, By/ Dr.Omer Eisa
 Partial Molar Quantities

Extensive properties vary with the amount of the substance and include.
1. Mass
2. Weight
3. Volume
4. U, S, H,

Intensive properties, in contrast, do not depend on the amount of the

substance; they include;
1. Colour
2. Melting point
3. Boiling point
4. Electrical conductivity
5. Physical state at a given temperature.

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa

Partial Molar Quantities

Mathematically the partial molar volume (Vρ) of the substance (J) at some
general composition defined as:

, ,

n: the amount of substance

n’: the amount of other substances presented are constant.
P: pressure
T: Temperature
Here, mentioned (P,T, and n’) are constants.

: partial molar quantity.

: : partial molar volume.
V: total volume of solution.

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa
 Partial Molar Quantities

The partial molar volumes of water and ethanol at 25°C. Note the different
scales (water on the left, ethanol on the right).

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa

Partial Molar Quantities

The partial molar volume of a substance is the slope of the variation of the total
volume of the sample plotted against the amount of that substance. In general,
partial molar quantities vary with the composition, as shown by the different slopes
at a and b. Note that the partial molar volume at b is negative: the overall volume of
the sample decreases as A is added.

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa
 Partial Molar Quantities
For binary mixtures:
The composition changed by addition of A (dnA) and B (dnB) .

The total volume of mixture changed by:

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa

Partial Molar Gibbs Energy

Definition:
The partial molar Gibbs free energy for substance in a mixture is
equal to the chemical potential of that substance. From the following
definition of chemical potential,

, ,

n’: donated that amount of all other components of the mixture are
held constant.

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa
 Partial Molar Gibbs Energy
For a pure substance (J)
G = nJGJ,m

In this case, the chemical potential is:

μJ = GJ,m

Here the chemical potential is simply equal to the molar Gibbs energy
of the substance:

For binary mixture A+B, the total Gibbs free energy presented as;

G= nAμA +nB μB

μA and μB are the chemical potentials for the composition on the mixture

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa

Partial Molar Gibbs Energy

In thermodynamics, :the change in Gibbs Energy depends on the
change of the P&T,

dG = Vdp - SdT

Then the a general change in G becomes

dG = Vdp - SdT + µAdnA + µBdnB + …

Fundamental equation of chemical thermodynamics

At constant pressure and temperature,

dG = µAdnA + µBdnB + …

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa
 Partial Molar Gibbs Energy
At constant P&T, the change in Gibbs Energy equal to the change of
system work (w),

dG = dw add,max

Then,
dw = µAdnA + µBdnB + …

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa

Chemical potential and internal energy

Given,
G = U + pV – TS
and therefore
U = -pV + TS + G
The general form of an infinite small change in U for a system of
variable composition is:

dU = -pdV - Vdp + SdT + TdS + dG

= -pdV - Vdp + SdT + TdS+ (Vdp - SdT + µAdnA + µBdnB + …)
= -pdV + TdS + µAdnA + µBdnB +…
At constant volume and entropy

dU = µAdnA + µBdnB +…

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa
 Chemical potential and internal energy
Hence that,

At constant entropy, volume and composition

Chemical potential can also presented as a function of enthalpy (H)

and mass (A).

At constant entropy, pressure and composition

At constant temperature, volume and composition

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa

The Gibbs-Duhem Equation

The total Gibbs energy of a binary mixture is given by:

G = nAµA + nBµB

 Here, the chemical potential depends on the on the composition

(n).
 When (n) changed, the Gibbs energy also changed.

So, for the multi component mixture,

dG = µAdnA + µBdnB + nAdµA + nBdµB

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa
 The Gibbs-Duhem Equation
At constant T & P, Gibbs energy presented as;

G = nAµA + nBµB

 G is state function
 The two expressions of (dG) must be equal.

At constant T & P,
nAµA + nBµB= 0

This equation is a special case of the Gibbs–Duhem equation:

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa

The Gibbs-Duhem Equation

Example-2
If the composition of a mixture is such that nA = 2nB, and a small
change in composition results in µA changing by ∆µA = +1 J mol-1, find
the change in µB ?

Solution:

Given,
nA /nB = 2
Then,
∆ µB = -2(+1 J/mole)
= -2 J/mole

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa
 The Thermodynamics of Mixing
Mixture system definitions:

 In chemistry, a mixture is a martial made up of two or more

different chemical substances which are not chemically bonded.

 A mixture is the physical combination of two or more substances

in which the identities are retained and mixed in the form of
solution or suspension and colloids.

# Mixtures can be liquids or gases #

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa

The Thermodynamics of Mixing

Gibbs Energy of mixtures:

 The dependence of Gibbs energy of mixture on its composition is

given previously by the following relation;

G= nAμA +nB μB

The simple example of such kind of systems is;

[two gases in container]

 When the mixture is spontaneous, so it must correspond to a

decrease in Gibbs Energy.
 Molecules of the two gases will arranged to get lower Gibbs
energy.

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa
 The Thermodynamics of Mixing
Gibbs Energy of Mixing of Perfect Gases:

 We assume we have tow gases A and B, at conditions of T&P.

 Before mixing, the chemical potential of both gases will have
(pure) values.
 By applying the definition µ = Gm ; to the equation
Gm(p) = G⦵m + RT ln p/p⦵

It will give:
μ= μ⦵ + RT ln p/p⦵
# the variation of chemical potential with pressure for perfect gas #
μ⦵ :the standard chemical potential
p⦵ : the standard pressure (reference pressure = 1 bar)
Solution thermodynamics and Phase
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Equilibrium, By/ Dr.Omer Eisa

The Thermodynamics of Mixing

At standard pressure p⦵ =1 bar,

p/p⦵ = p/1 = p

The relation become:

μ= μ⦵ + RT ln p

Perfect gas or ideal gas whose specific heat capacities cp and cv are
constant.

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa
 The Thermodynamics of Mixing
 Refreshing the basic concentration calculations, molar and molal
fractions:

Let us take:
A: as a solvent
B: as solute
The molar concentration (CB)

CB = nB/V [mole/L]

nB is the amount of solute in mole

V is the solution volume

 The standard concentration

c⦵ = 1mole/L

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa

The Thermodynamics of Mixing

 The molality concentration (bB):

bB = solute b in mole/solvent in kg [mole/kg]

bB = nB/mA

 The standard concentration

b⦵ = 1mole/kg

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa
 The Thermodynamics of Mixing
 The relation between molality & mole fraction;
 Consider one solution with amount (n) mole of solute (B)

 Mole fraction of solute is (xB)

Amount of solute B;
nB = xB × n
 Mole fraction of solvent A is (xA)
xA = 1 – xB
 Amount of solvent;
nA= xA × n
=(1-xB)n

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa

The Thermodynamics of Mixing

 The mass (mA) of the solvent with molar mass (MA);
mA(s) = nA (mol) × MA (g/mol)
= [(1-xB) × n] × MA

 The molality, bB, of a solute:

then;

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa
 The Thermodynamics of Mixing

2-Molality & molarity linked with density:

If;
Total volume is (V)
Solution mass is (m)
The mass density of solution is:
ρ = m/V
The amount of solute is;
nB = cB V (through molar concentration)
The mass of solute:
mB = nB MB (MB is the molar mass)
mB = cB V MB

Solution thermodynamics and Phase

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The Thermodynamics of Mixing

Also the mass of solvent :

mA = m – mB Which (m) is the total mass
= ρV – cBVMB
= V (ρ – cBMB)
Therefor, the molality of solute is therefore ;

the molar concentration in terms of the molality, is

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa
 The Thermodynamics of Mixing
3-The relation between molar concentration and mole fraction

with xA = 1 − xB.
For a dilute solution in the sense that
xB MB xAMA

If, moreover, xB 1,
so xA ≈1,
Then;

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa

The Thermodynamics of Mixing

Back to Gibbs Energy of Mixing of Perfect Gases:

The last equation was;

μ= μ⦵ + RT ln p Here, p⦵ = 1 bar

The Gibbs free energy of separated gases given by;

Gi = nAµA + nBµB

= nA(µA⦵ + RT ln p) + nB(µB⦵ + RT ln p)

After mixing of A&B;

pA + pB = p

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa
 The Thermodynamics of Mixing
The total Gibbs Energy changed to (Gf);

Gf = nA(µA⦵ + RT ln pA) + nB(µB⦵ + RT ln pB)

The total changed Gibbs Energy (ΔmixG);

ΔmixG = Gf - Gi

Substituting the value of Gi and Gf;

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa

The Thermodynamics of Mixing

By replacing the values of nA and nB by xAn and xBn, and xA , xB
replaced by pA/p, pB/p respectively .we can get;
Gibbs energy
of mixing
[perfect gas]

Note that,
 lnxA and lnxB will give negative (-) values.
 ΔmixG will be negative (-) too.
 So that refers to perfect mixing in all properties.

 The Gibbs Energy of mixing is

negative for all composition.
 So, the perfect mixing of gases
spontaneously in all properties.

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa
 The Thermodynamics of Mixing
Example-3
A container is divided into two equal compartments (Fig. below).
One contains 3.0 mole H2(g) at 25°C; the other contains 1.0 mole
N2(g) at 25°C. Calculate the Gibbs energy of mixing when the
partition is removed. Assume that the gases are perfect.

Solution:
Given that the pressure of nitrogen is p, the
pressure of hydrogen is 3p. Therefore, the
initial Gibbs energy is:

Gi = (3.0 mole)[µ⦵(H2) + RT ln 3p]

+ (1.0 mole)[µ⦵(N2) + RT ln p]

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa

The Thermodynamics of Mixing

When the partition is removed and each gas occupies twice the
original volume, the final total pressure is 2p. The partial pressure of
nitrogen falls to 1/2 p and that of hydrogen falls to 3/2 p. Therefore.

The Gibbs energy changes to:

Gf = (3.0 mole)[µ⦵(H2) + RT ln 3/2 p] + (1.0 mole)[µ⦵(N2) + RT ln 1/2 p]

The Gibbs energy of mixing is the difference of these two quantities:

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa
 The Thermodynamics of Mixing
Home work-1
Suppose that 2.0 mole H2 at 2.0 atm and 25°C and 4.0 mole N 2 at 3.0
atm and 25°C are mixed by removing the partition between them.
Calculate ∆mixG.

Solution:
p total =pH2+pN2
= 2 atm+ 3 atm =5 atm
∆mixG = - RT Ʃ [ni × ln p/pi
= - 8.31 KJ/mole. K× 298 K [ (2mole × ln 5/2) +(4 mole × ln 5/3)]
= -9.598 KJ/mole

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa

Other Thermodynamics of Mixing Functions

The entropy of mixing, (∆mixS):

Given in thermodynamics;
(∂G/∂T)p = −S
for a mixture of perfect gases initially at constant pressure, the
entropy of mixing, ∆mixS, is:
Entropy of mixing
[perfect gases, constant
T and p]

Notes,
 Under the give conditions, the (∆mixS ) always have a positive
(+)value because the value of (ln xA) is negative (-).
 ΔmixS > 0.
 The increases of (∆mixS ) value mentioned to the dispresing of one
gas into the other gas, which use to increase the randomness (S).
Solution thermodynamics and Phase
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Equilibrium, By/ Dr.Omer Eisa
 Other Thermodynamics of Mixing Functions
Example-4
Calculate the entropy of mixing for tow perfect gas molecules have
an equal amount of a total number of 2 moles are mixed at the same
pressure. set (yA = yB = ½).

∆mixS = (-)nR [½ ln ½ + ½ ln ½ ] =(+) nR ln(2)

with n = 2 moles

∆mixS = (2 mole) × (8.31 J/mole.K ) ln(2) = +11.5 J/K

Note:
An increase in entropy is expected when one gas disperses into the
other and the disorder (untidiness) increases.

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa

Other Thermodynamics of Mixing Functions

The enthalpy of mixing, (∆mixH):

For a perfect gases at constant Temperature and pressure, the

enthalpy of mixing, (∆mixH = 0) :

∆H = cp . ∆T ∆T = 0
From; ∆G = ∆H - T ∆S
∆G = - T ∆S

Reasons,
 No chemical interaction between the perfect gas molecules during
the mixing process.
 No temperature change for the surroundings.

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa
 The Chemical Potentials of Liquids
a- Ideal solutions

Assumption:
For liquid solutions/ mixtures; at equilibrium, the chemical
potential of a substance present as a vapor must be equal to its
chemical potential in the liquid.

Definition of Ideal solution:

In chemistry, an ideal solution or ideal mixture is a solution that
shows thermodynamic properties similar to those at a mixture of
ideal gases.

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa

The Chemical Potentials of Liquids

The ideal solution thermodynamic properties are:

1. ∆mixH= 0
2. ∆mixG = able to measure.
3. ∆mixS = able to measure.
4. The vapor pressure of the solute and solvent are obey Raoult’s
and Henry’s laws respectively.
5. The activity coefficient (deviation from ideality) is equal to one
for each component.

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa
 The Chemical Potentials of Liquids
 Pure substances are denoted by a subscript (*)
 Standard values subscript ( Ɵ)

The step 1:
The chemical potential of pure substance (A):

(μ*A) and also as (μ*A (l) ) it its in liquid phase

Vapor pressure of pure liquids (p*A)

The chemical potential of (A) in vapor phase is:

μ A(g) = μƟA(g) + RT ln pA
p*A =pA/pƟA
μ*A (l) = μƟA (g) + RT ln p*A

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa

The Chemical Potentials of Liquids

The chemical potential of substance (A) in liquid phase is:
μA (l) = μƟA (g) + RT ln pA

That’s occurs only when: vapor and solvent are still are at equilibrium..

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa
 The Chemical Potentials of Liquids
The step 2:
The chemical potential of gas at standard condition (μƟA(g) ) can be
eliminated to combine it with the chemical potential of single component.

μA (l) = μ*A (l) - RT ln p*A + RT ln pA

= μ*A (l) + RT ln pA/p*A
The step 3:
Using Rauolt’s law for ideal solution;

pA = xA p*A pA : is the pressure of solution

p*A : is the pressure pure solvent
xA = pA /p*A xA : mole fraction of solvent

μA (l) = μ*A (l) + RT ln xA

Chemical potential for ideal solution

Solution thermodynamics and Phase
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Equilibrium, By/ Dr.Omer Eisa

The Chemical Potentials of Liquids

b- Ideal- dilute solution:

 In ideal solutions, the solute and solvent obeys Rauolt’s law.

 Henry had experimentally discover that,
For real solutions at low concentrations,
 The vapor pressure of the solute is proportional to its mole fraction.
 The constant of proportionality is not the vapor pressure of the pure
substance.

pB = xB KB

Henry’s law [ideal–dilute solution]

pB : vapor pressure of solute
xB : mole fraction of solute
KB : an experimental constant [dimensions of pressure]
Solution thermodynamics and Phase
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Equilibrium, By/ Dr.Omer Eisa
 Checklist of equations

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa

The Properties of Solutions

1- liquid Mixtures
For component (J) in ideal mixture/solution, the chemical potential
(μJ) represented as:

μJ = μ*J + RT ln xJ

Chemical potential for ideal solution

μ*J : chemical potential for pure (J).

xJ : mole fraction I the mixture.

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa
 The Properties of Solutions
2- Ideal Solutions
Gibbs Energy of mixing:
Give in the first section before liquid mixed: (*, for pure liquid)

Gi = nAµ*A + nBµ * B

The total Gibbs Energy :

Gf = nA(µ*A + RT ln xA) + nB(µ*B + RT ln xB)

The Gibbs energy of mixing:

∆mixG = nRT(xA ln xA + xB ln xB) where n = n A + nB

Gibbs energy of mixing [ideal solution]

Solution thermodynamics and Phase

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The Properties of Solutions

The ideal entropy of mixing:

∆mixS = −nR(xA ln xA + xB ln xB)

Entropy of mixing [ideal solution]

Then from the relation:

ΔmixG = ΔmixH – T ΔmixS
For ideal mixture, the change in the ideal volume is zero, because
the enthalpy of mixing is zero (ΔmixH = 0).

Solution thermodynamics and Phase

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 The Properties of Solutions
Explanation:
V = (∂G/∂p)T
→ ΔmixG is depend on pressure (the derivative with respect to pressure =0)
ΔmixV = (∂ΔmixG/∂Δmixp)T = 0
Notes;
 The equations of (∆mixG ) and (∆mixS ) are the same as the mixed
two gases, because the enthalpy of mixing ∆mixH =0 and there is
no change in the entropy of the surroundings
 In a perfect gas there are no interactions between the molecules.
 In ideal solutions there are interactions, but the average energy of
A–B interactions in the mixture is the same as the average energy
of A–A and B–B interactions in the pure liquids.
Solution thermodynamics and Phase
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Equilibrium, By/ Dr.Omer Eisa

The Properties of Solutions

∆mixG = nRT (xA ln xA + xB ln xB) ∆mixS = (−)nR (xA ln xA + xB ln xB)

The Gibbs energy of mixing of two The entropy of mixing of two liquids
liquids that form an ideal solution that form an ideal solution

Solution thermodynamics and Phase

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 The Properties of Solutions
Example-5
Consider a mixture of benzene and methylbenzene, which form an
approximately ideal solution, and suppose 1.0 mole C6H6(l) is mixed
with 2.0 mole C6H5CH3(l). For the mixture, the mole fraction of
benzene is x = 0.33 and for methyl-benzene is x = 0.67. Calculated the
Gibbs energy and entropy of mixing at 25°C, when RT = 2.48 kJ/mole.
Solution:
∆mixG = nRT (xA ln xA + xB ln xB)
∆ mixG/n = (2.48 kJ mole-1) × (0.33 ln 0.33 + 0.67 ln 0.67)
= -1.6 kJ mole-1 (for n=1 mole)
for (n=3) → 3× (-1.6 kJ mole-1 ) = - 4.8 kJ mole-1

∆mixS = (−)nR (xA ln xA + xB ln xB)

∆ mixS/n = (-)(8.3145 J K-1 mole-1) × (0.33 ln 0.33 + 0.67 ln 0.67)
= +5.3 J K-1 mole-1 (for n=1 mole)
for (n=3) → 3× (5.3 J K-1) = 15.9 J K-1 mole-1
Solution thermodynamics and Phase
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Equilibrium, By/ Dr.Omer Eisa

Excess Functions and Regular Solutions

Excess functions, (XE ):
Excess function defines as difference between the observed
thermodynamic function and the ideal solution function.

XE=∆mixX - ∆mixXideal

Excess function [definition]

Excess entropy:
SE = ∆mixS - ∆ mixSideal

Excess enthaply:
HE = ∆mixH – 0 because ∆ mixHideal =0

Excess volume:
VE = ∆mixV – 0 because ∆ mixVideal =0
Solution thermodynamics and Phase
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 Excess Functions and Regular Solutions

Notes;
 The deviation of the enthalpy from zero indicates the range to
which the solutions are non-ideal. In this concentration the useful
model is a regular-solution,
when, HE ≠ 0 , and SE = 0
 Excess enthalpy depending on composition,
HE = n ξ R T xA xB
ξ : is dimensionless parameter that measure of the energy of A-B
interaction relative to A-A and B-B interactions.

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Excess Functions and Regular Solutions

Then from the relation:

ΔmixG = ΔmixH – T ΔmixS
ΔmixG = n ξ R T xA xB – T[-n R(xA ln xA + xB ln xB)]
= n R T(xA ln xA + xB ln xB + ξ xA xB )

Solution thermodynamics and Phase

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 Excess Functions and Regular Solutions
Example-5

Identify the value of (ξ) for equimolar mixture of benzene and

cyclohexane at 25oC, and estimate the Gibbs energy of mixing. If the
total number of mole is (1 mole) and excess enthalpy (HE) is close to
701 J/mol.

Solution:

Use the excess enthalpy to find the ξ value,

Given:
HE = n ξ R T xA xB
701 J mol-1 = 1 mol × ξ × 8.314 5 J K-1mol -1× 298 K × ½ × ½
701 J mol−1
ξ
1 mol × 8.314 5 J K−1mol −1× 298 K × ½ × ½

Solution thermodynamics and Phase

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Excess Functions and Regular Solutions

∆mixG/(1) = ½ RT ln ½ + ½ RT ln ½ + 701 J mol-1

∆mixG/(1) = (-) (8.3145 J K-1 mol-1× 298 K × ln2) + 701 J mol-1
= 2418 J/mol
∆mixG/(1) =2.418 KJ/mol

Solution thermodynamics and Phase

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Equilibrium, By/ Dr.Omer Eisa