Fugacity: It is derived from Latin, expressed as fleetness or escaping ...

Fugacity: It is derived from Latin, expressed as fleetness or escaping tendency. It is used 

to study extensively phase and chemical reaction equilibrium. 

We know that 

dG VdP SdT -(1) 

For isothermal condition 

dG VdP 

For ideal gases 

RT 

V 

P 

RT 

dG . dP 

P 

dG RTd ln P 

To find Gibbs free energy for an real gases. True pressure is related by effective 

pressure. Which we call fugacity(f). 

dG RTd ln f -------(2) 

Applicable for all gases (ideal or real) 

On differentiation 

G RT ln f 

Is an constant depends on temperature and nature of gas. 

fugacity has same units as pressure for an ideal gas. 

For ideal Gases 

dG VdP SdT 

For Isothermal conditions 

dG VdP 

from equation (2) 

RTd ln f VdP 

V 

d ln f dP 

RT 

d ln f 

dp 

p 

d ln f d ln p 

f p Fugacity = Pressure for Ideal gases

Fugacity coefficient ( ) : Fugacity coefficient ids defined as the ratio of fugacity of a 

component to its pressure. 

f 

 

P 

Is the measure of non ideal behavior of the gas. 

Standard State: Pure gases, solids and liquids at temperature of 298k at 1 atmosphere are 

said to exist at standard condition. The property at this condition are known as standard 

state property and is denoted by subscript’o’. 

o 

G = Standard Gibbs free energy 

o 

f = Standard fugacity 

Estimation of fugacity for gases 

I method 

dG VdP SdT 

For Isothermal conditions 

dG VdP 

from equation (2) 

RTd ln f VdP 

V 

d ln f dP 

RT 

Integrating the above equation with the limits 0 to f and pressure 1 to P 

 

d ln f 

 

P 

 

1 

V 

RT 

dP 

Lower limit is taken as P =1 atm 

At 1 atm assuming the gases expected to behave ideally 

P 

1 

ln f VdP 

RT 1 

if PVT relations are known , we can find fugacity at any temperature and pressure

II method 

Using compressibility factor 

dG VdP 

RTd ln f VdP --------------(2) 

V in terms of compressibility terms it is given as V 

Substitute V in equation 2 

ZRT 

RTd ln f dP 

p 

d ln f Zd ln p 

subtracting both sides by dlnP 

d ln f d ln P Zd ln P d ln P 

f 

d ln ( z 1) 

d ln P 

P 

d ln ( Z 1) 

d ln P 

Integrating the equation from 1 to and 0 to P 

 

 

1 

d ln 

 

P 

P 

 

0 

( Z 1) 

d ln P 

dP 

ln ( Z 1) 

P 

0 

ZRT 

P 

Using generalized charts: using reduced properties a similar chart as compressibility chart 

is predicted for fugacity. 

f Z 1 

ln 

dPr 

P P 

r

Fugacity is plotted against various reduced pressure at various reduced temperature 

Using Residual Volume(): The residual volume is the difference between actual 

volume (V) and the volume occupied by one mole of gas under same temperature and 

RT RT 

pressure 

V , V 

P 

P 

RT 

dG 

dP 

RTd ln f 

P 

dP 

RT dP RTd ln f 

RT P 

 

dP d ln 

RT 

f 

P 

 

 

 

f 

ln dP 

P RT 

Problem 

For isopropanol vapor at 200 o C the following equation is available 

Z=1- 9.86 x 10 -3 P-11.45 x 10 -5 P 2 

Where P is in bars. Estimate the fugacity at 50 bars and 200 o C 

PV 

3 

5 

2 

Z 1 9. 

86 10 

P 11. 

4110 

P 

RT 

V 

 

ln f 

ln f 

ZRT 

p 

 

 

1 

RT 

1 

RT 

 

 

P 

 

1 

P 

 

1 

RT 

( 1 

P 

VdP 

 

9. 

86 10 

3 

RT 

( 1 

9. 

86 10 

P 

3 

3 

9. 

86 10 

dP 

P 11. 

4110 

5 

P 11. 

4110 

50 50 

50 

dp 

5 

ln f 

 

11. 

4110 

PdP 

1 

p 

1 

1 

5 

P 

2 

) 

2 

P ) dP

50 

 

ln f ln 

9. 

86 10 

1 

f=26.744 bar 

f 26. 

744 

0. 

5348 

P 50 

50 1 

Fugacity for liquids and solids 

General expression for fugacity is 

P 

3 

11. 

4110 

5 

2 2 

( 50 1 

) 

2 

1 

ln f VdP 

RT 1 

For solids and liquids at constant temperature the specific volume does not change 

appreciably with pressure, therefore the above equation is integrated by taking volume 

constant. Integrating the above equation from condition 1 to 2 

f2 

ln f 

f1 

f 

V 

RT 

V 

P2 

 

P1 

dP 

P 

2 

ln 2 P1 

f1 

RT 

Problem: 

Liquid chlorine at 25 o C has a vapour pressure of 0.77Mpa, fugacity 0.7Mpa and Molar 

volume 5.1x 10 -2 m 3 /kg mole. What is the fugacity at 10 Mpa and 25 o C 

P 

1 

P 

2 

6 

0. 

77 10 

Pa 

1010 

f 

V 

6 

Pa 

P 

2 

ln 2 P1 

f1 

RT 

f=0.846Mpa 

f 

1 

6 

0. 7 10 

Pa T 298K 

R 8314 

J 

Kgmole

Activity(a): 

It is defined as the fugacity of the existing condition to the standard state fugacity 

f 

a o 

f 

Effect of pressure on activity 

The change in Gibbs free energy for a process accompanying change of state from 

standard state at given condition at constant temperature can be predicted as 

G RT ln 

f 

o 

G RT ln 

o f 

G 

G G RT ln RT ln a 

o 

f 

 

 

 

at constant temperature dG VdP 

G 

dG V 

G 

P 

O o 

P 

dP 

o 

o 

G V ( P P ) RT ln a V ( P P ) 

f 

o 

V 

o 

ln a ( P P ) This equation predicts the effect of pressure on activity 

RT 

Effect of Temperature on Activity 

G 

G G 

o 

RT ln a 

o 

G G 

R ln a 

T T 

Differentiating the above equation with T at constant P 

G 

 

 

d ln a T 

R 

 

 

dT 

P T 

 

 

P 

o G 

 

 

 

 

T 

 

 

 

 

T 

 

 

 

 

d ln a 

R 

 

dT 

H 

2 

RT 

o 

H 

2 

RT 

P 

P

d ln a 

R 

 

dT 

P 

o 

H H 

This equation predicts the effect of temperature on activity. 

2 

RT

Properties of solutions 

The relationships for pure component are not applicable to solutions. Which needs 

modification because of the change in thermodynamic properties of solution. The 

pressure temperature and amount of various constituents determines an extensive state. 

The pressure, temperature and composition determine intensive state of a system. 

Partial Molar properties: 

The properties of a solution are not additive properties, it means volume of solution is not 

the sum of pure components volume. When a substance becomes a part of a solution it 

looses its identity but it still contributes to the property of the solution. 

The term partial molar property is used to designate the component property when it is 

admixture with one or more component solution. 

A mole of component i is a particular solution at specified temperature and pressure has 

got a set of properties associated with it likeVP , Sietc 

. These properties are partially 

responsible for the properties of solution and it is known as partial molar property 

It is defined as 

nM 

 

M i 

ni 

 

T , P, 

n 

j 

j i 

M i = Partial molar property of component i. 

M = Any thermodynamic property of the solution 

n = Total number moles in a solution 

ni=Number of moles of component I in the solution 

This equation defines how the solution property is distributed among the components. 

Thus the partial molar properties can be treated exactly as if they represented the molar 

property of component in the solution. 

The above expression is applicable only for an extensive property using 

We can write 

 

x 

n 

 

n 

i i M n nM i 

xi 

M 

i i M 

xi=Mole fraction of component i in the solution.

Measuring of partial molar properties 

To understand the meaning of physical molar properties consider a open beaker 

containing huge volume of water in one mole of water is added to it, the volume increase 

is 18x 10 -6 m 3 If the same amount of water is added to pure ethanol the volume increased 

is approximately 14 x 10 -6 m 3 this is the partial molar volume of H2O in pure ethanol. 

The difference in volume can explain the volume applied by water molecules depending 

on water molecules surrounding to them. When water is added to large amount of 

ethanol, ethanol molecules hence volume occupied surround all the water molecules will 

be different in ethanol. 

If same quantity of water is added to an equimolar mixture of H2O and ethanol, the 

volume change will be different. Therefore Partial molar property change with 

composition. The intermolecular forces also changes with change in thermodynamic 

property. 

Let V w = Partial molar volume of the water in ethanol water solution 

V w = Molar volume of pure water at same temperature and pressure 

t 

V =Total volume of solution when water added to ethanol water mixture and allowed 

for sufficient time so that the temperature remains constant 

 

V 

t 

V 

w 

V 

 

n 

n 

t 

w 

w 

V 

w 

In a process a finite quantity of water is added which causes finite change in composition. 

V w = property of solution for all infinitely small amount of water. 

Vw limv0 

V 

n 

t 

w 

V 

 

n 

t 

w 

 

 

 

Temperature pressure an number of moles of ethanol remains constant during addition of 

water. 

t V 

 

Vw 

nE- no of moles of ethanol 

nw 

 

T , P, 

nE 

The partial molar volume of component i 

V 

i 

V 

 

n 

i 

t 

 

 

 

T , 

P, 

n i 

j

Partial molar properties and properties of the solution 

Consider any thermodynamic extensive property (V i, Gi etc) for homogenous system can 

be determined by knowing the temperature, pressure and various amount of constituents. 

Let total property of the solution 

M nM 

t 

n n1 

n2 

n3 

 

1,2,3 represents no of constituents 

f T, 

P, 

n , n , n n 

Thermodynamic property is a 

1 

2 

For small change in the pressure and temperature and amount of various constituents can 

be written as 

t 

t 

t 

t 

t M 

M 

M 

 

M 

 

dM dP dT dn1 

dni 

 

P 

P 

P 

 

P 

T , n 

p, 

n 

T , p, 

n n 

 

2 , 3 

P, 

T , n j i 

At constant temperature and pressure dP and dT are equal to zero. 

The above equation reduces to 

i n t 

t M 

dM dni 

i ni 

P T n j i 

 

 

 

1 

. , 

dM t in terms of partial molar property 

n 

t 

dM 

i1 

M 

i 

dn 

i 

M i is an extensive property depends on composition and relative amount of constituents. 

All constituent properties at constant temperature and pressure are added to give the 

property of the solution. 

dM M 1dn1 

M 2dn2 

M 3dn3 

 

t 

dM M x M x M x 

dn 

t 

 

M 

M t 

 

 

3 

1 1 2 2 3 3 

M 1x1 

M 2 x2 

M 3x 

3 n 

M 1n1 

M 2n 

M 2 3n 

3 

t 

niM 

i 

Problem 

A 30% mole by methanol –water solution is to be prepared. How many m 3 of pure 

methanol (molar volume =40.7x10 -3 m 3 /mol) and pure water (molar volume = 

18.068x10 -6 m 3 /mol) are to be mixed to prepare 2m 3 of desired solution. The partial molar 

volume of methanol and water in 30% solution are 38.36x10 -6 m 3 /mol and 17.765x10 -6 

m 3 /mol respectively. 

j

Methanol =0.3 mole fraction 

Water=0.7 mole fraction 

V t =0.3 x38.36x10 -6 +0.7x17.765x10 -6 

=24.025x10 -6 m 3 /mol 

For 2 m 3 soolution 

2 

3 

83. 

246 10 

mol 

6 

24. 

02510 

Number of moles of methanol in 2m 3 solution 

=83.246x10 3 x0.3= 24.97x10 3 mol 

Number of moles of water in 2m 3 solution 

=83.246x10 3 x07= 58.272x10 3 mol 

Volume of pure methanol to be taken 

= 24.97x10 3 x 40.7x10 -3 =1.0717 m 3 

Volume of pure water to be taken 

= 58.272x10 3 x 18.068x10 -6 =1.0529 m 3

Estimation of Partial molar properties for a binary mixture : 

Two methods for estimation 

Analytical Method and Graphical Method(Tangent Intercept method) 

Analytical Method: The general relation between partial molar property and molar 

property of the solution is given by 

M 

M i M xk 

x 

K T , P, 

nk 

 

 

 

 

 

For binary mixture 

i 1, k 2 

k i x1 

M 

M 1 M x2 

x 

2 T , P 

 

 

 

 

2 x1 

1 

 

At constant Temperature and pressure 

M 

 

M 

 

 

 

1 M ( 1 

x1) 

------(a) 

x1 

 

M 

 

M 

 

 

 

2 M x 

1 

-----------(b) 

x1 

 

x , x2 1 x1 

, x2 

x1 

The partial molar property (e xtensive property) for a binary mixture can be estimated 

from the property of solution using above equations (a) and (b). 

Tangent Intercept method: 

This is the graphical method to estimate partial molar properties. If the partial molar 

property (M) is plotted against the composition we get the curve as shown in the figure.

Suppose partial molar properties of components required at any composition, and then 

draw a tangent to this point to the curve. The intercept of the tangent with two axis x1=1 

and x1=0 are I1 and I2. 

Slope of the tangent 

M I x 

2 

2 

I M x 

1 

1 

dM 

dx 

1 

dM 

dx 

1 

dM M I 

 

dx x 

Comparing I2 with equation (b) 2 2 M I 

dM I1 

M 

and also (right hand side) 

dx1 

1 x1 

I M 1 

x 

1 

I 

1 

dM 

dx 

1 

dM 

1 

1 M 1 

x1 

Comparing with equation(a) 1 1 

dx1 

M I 

The intercept of the tangent gives the partial molar properties. 

1 

2

Limiting cases: For infinite dilution of component when at x1=0 a tangent is drawn at 

x1=0 will give the partial molar property of component 1 at infinite dilution M ) and 

tangent is drawn at x2=0 or x1=1 will give infinite dilution M ) of component 2 

M 2 

 

Problem 

The Gibbs free energy of a binary solution is given by 

G x 150x 

x x ( 10x 

x ) 

100 1 

2 1 2 1 2 

cal 

mol 

(a) Finnd the partial molar free energies of the components at x2=0.8 and also at infinite 

dilution. 

(b) Find the pure component properties 

Sol: G x 150x 

x x ( 10x 

x ) 

100 1 

2 1 2 1 2 

Substitute 1 

3 

x2 1 x and simplifying 

G 9x1 

8x1 

49x1 

150 

G 

1 

1 

G 

 

G ( 1 

x 

 

 

 

1) 

x1 

 

G 

2 

27x1 

16x1 

49 

x 

G 

G 

G 

1 

2 

2 

3 

2 

2 

18x1 35x1 

16x1 

101 

G 

 

G x 

 

 

1 

x1 

 

3 

1 

18x 

8x 

2 

1 

150 

cal 

mol 

To find the partial molar properties of components 1 and 2 

x2=0.8, x1=1-0.8 = 0.2 

( 2 

 

( 1

G 

1 

3 

2 

18x1 35x1 

16x1 

101 

G 102. 

944 

G 

1 

2 

3 

1 

cal 

mol 

18x 

8x 

G 149. 

824 

2 

2 

1 

cal 

mol 

At infinite dilution 

G 

1 

 

G atx 

1 

 

G1 101 

1 

cal 

mol 

0 

150 

G G atx 1 or x2=0 

2 

 

2 

 

G2 160 

1 

cal 

mol 

To find the pure component property 

G 

1 

G atx 

1 

1 

cal 

G1 100 

mol 

G 

2 

G atx 

2 

1 

cal 

G1 150 

mol 

1 

0 

Chemical Potential:

It is widely used as a thermodynamic property. It is used as a index in chemical 

equilibrium, same as pressure and temperature. The chemical potential i of component i 

in a solution is same as its partial molar free energy in the solution i G 

The chemical potential of component i 

i 

G 

t 

G 

dG 

i 

t G 

 

 

ni 

 

T , P, 

n 

 

, , n n T P 

1 2 , f 

t 

t G 

 

 

P 

 

T , n 

j 

t G 

 

dP 

T 

 

P , n 

dT 

i n 

i 

1 

t G 

 

ni 

t 

t 

t G 

G 

 

dG dP dT i 

dni 

P 

T 

 

T , n 

P, 

n 

For closed system there will be no exchange of constituents (n is constant) 

dG 

t 

t t 

V dP S dT 

at constant temperature 

G 

 

 

V 

P 

 

 

T 

at constant pressure 

G 

 

 

T 

 

 

P 

t 

S 

t 

 

t t t 

dG V dP S dT i 

dni 

At constant temperature and pressure 

t T , 

P 

dG 

idni 

 

 

 

T , n 

j 

dn 

i

For binary solution G x11 

x2 

2 

Effect of temperature and pressure on chemical potential 

Effect of temperature: 

We know that 

t G 

 

i Gi 

-------------------(1) 

ni 

 

T , P, 

n 

j 

differentiating equation (1) with respect to T at constant P 

2 

 

i G 

 

T 

 

Tdn 

P, 

n 

dG VdP SdT ---------(3) 

i 

---------------------(2) 


G 

 

 

T 

 

 

P 

S 

differentiating again w r t ni 

2 

t 

G S 

 

 

 

Tni 

ni 

 

P, 

n 

j 

S 

Si is partial molar entropy of component I 

i 

 

 

 

T 

 

 

T 

 

 

 

 

 

P, 

n 

T Si 

i 

2 

 

T 

G H TS 

i 

 

T 

 

T 

 

 

2 

T 

In terms of partial molar properties 

i 

i

G H TS 

 

i 

i 

H 

i 

H 

i 

i 

 

 

i 

 

T 

 

 

TS 

i 

P, 

n 

i 

i 

TS 

H 

 

T 

i 

i 

2 

This equation represents the effect of temperature on chemical potential. 

Effect of Pressure: 

We know that 

t G 

 

i Gi 

-------------------(4) 

ni 

 

T , P, 

n 

j 


2 

 

i G 

 

P 

 

Pdn 

T , n 

dG VdP SdT ---------(3) 

i 

---------------------(5) 

differentiating equation (3) with respect to P at constant T 

G 

 

 

P 

 

 

T 

V 


2 

G 

Pn 

i 

V 

 

 

ni 

 

T , n 

j 

V , 

i

i 

 

P 

 

 

T , n 

V 

i 

This equation represents the effect of pressure on chemical potential 

Fugacity in solutions 

For pure fluids fugacity is explained as 

dG RTd ln 

f 

lim P0 

1 

P 

f 

The fugacity of the component i in the solution is defined as analogously by 

d RTd ln f 

i 

lim 

P0 

f 

P 

i 

i 

i 

1 

i is Chemical potential 

f i is partial molar fugacity 

For ideal gases Pi yi 

PT 

PT – Total pressure. 

Fugacity in Gaseous solutions 

We know that 

t G 

 

i Gi 

------(1) 

ni 

 

T , P, 

n 

j 


2 

 

i G 

 

P 

-----(2) 

Pdn 

T , n 

i

dG VdP SdT -----(3) 

differentiating equation (3) with respect to P at constant T 

G 

 

 

P 

 

 

T 

V 


2 

G 

Pn 

i 

 

i 

 

P 

 

 

 

i 

 

V 

 

 

ni 

 

T , n 

V i 

V 

P 

i 

T , n 

j 

V , 

i 

RTd f i VidP 

ln 

Subtracting both sides by i P 

Vi 

d ln fi 

dP 

RT 

d ln 

i 

d ln f i d ln P i dP d ln 

 

f 

 

Pi 

 

V 

d ln P d ln P d ln y 

i 

V 

RT 

i i 

d ln dP d ln 

RT 

Composition is constant dln yi=0 

The above equation can be written as 

d ln Pi 

d ln P 

P 

dP 

P 

i 

i 

P 

i

Modifying equation (a) 

RT 

V 

dP 

1 

d ln i 

i 

RT 

i - Represents fugacity of the component i in the solution. 

Ideal solutions 

An ideal mixture is one, which there is no change in volume due to mixing. In other 

words for an ideal gaseous mixture partial molar volume of each component will be equal 

to its pure component volume at same temperature and pressure. 

i i V V and Vi xi 

xiVi 

P 

V --- Raoults law 

Ideal solutins are formed when similar components or adjacent groups of group are 

mixed i i V V 

Eg: Benzene-Toluene 

Methanol- Ethanol 

Hexane- Heptane 

Solutin undergo change in volume due to mixing are known as non ideal solutions 

i i V V 

Eg: Methanol-Water 

Ethanol-water. 

Ideal solutions formed when the intermolecular force between like molecules and unlike 

molecules are of the same magnitude. 

Non-ideal solutions are formed when intermolecular forces between like molecules and 

unlike molecules of different magnitude. 

For ideal solutions 

t 

V niVi 

-------------------(1) 

Vi is the molar volume of pure component I 

t V 

 

Vi T 

, P , n j Vi 

----------------- (2) 

ni 

 

 

The residual volume for the pure component is 

RT 

Vi 

p 

Therefore we know from reduced properties 

P 

f i 1 RT 

ln Vi 

dP ----(3) 

P RT P 

0 

For component i in terms of partial molar properties

P 

f i 1 RT 

ln V 

i dP ----(4) 

P RT i P 

0 

Subtracting equation (3) from (4) 

V i V 

P 

f i P 1 

ln i dP ----(5) 

f P RT 

i 

i 

we know that Pi yi 

P 

equation (5) reduces to 

P 

f i 1 

ln V i Vi 

dP 

f y RT 

i 

i 

comparing equation (2) 

f i 

1 

f y 

i 

i 

0 

0 

f i f i yi 

--Lewis Randal rule 

Lewis Randal rule is applicable for evaluating fugacity of components in gas mixture. 

Lewis Randal rule is valid for 

1. At low pressure when gas behaves ideally. 

2. When Physical properties are nearly same. 

3. At any pressure if component present in exess. 

Henrys Law 

This law is applicable for small concentration ranges. For ideal solution Henrys law is 

given as 

f x k 

i 

P x k 

i 

i 

i 

i 

i 

ki- Henrys Constant, f i -Partial molar fugacity, 

Pi -Partial pressure of component i. 

Non Ideal solutions 

For ideal solutions f i xik 

i ------------(a)

For non ideal solutions f i i xik 

i ----------(b) 

Where i is an activity coefficient of component i. 

Comparing equation (a) and (b) 

f i 

i Non ideal solution / Ideal solution 

f 

i 

For both ideal and Non ideal solutions the fugacity of solution is given by 

equation 

f i 

ln f xi 

ln 

x 

 

ln 

ln 

xi i 

i 

Problem: 

A terinary gas mixture contains 20mole% A 35mole% B and 45mole% C at 60 atm and 

75 o C. The fugacity coefficients of A,B and C in this mixture are 0.7,, 0.6 and 0.9. 

Calculate the fugacity of the mixture. 

Solution: 

ln x ln 

x ln 

x ln 

A 

A 

ln 0. 

2ln( 

0. 

7) 

0. 

35ln( 

0. 

6) 

 

ln 0. 

2975 

0. 

7426 

f 

, f =44.558atm 

P 

Gibbs Duhem Equation 

B 

B 

c 

c 

0. 

45ln( 

0. 

9) 

Consider a multi component solution having ni moles of component I the property of 

solution be M in terms of partial molar properties 

t 

nM nM 

-----------------------(1). 

M 

i 

Where n is the total no of moles of solution 

Differentiating eq (1) we get 

d( nm) 

nid 

M i 

We know that 

M idni 

----------------------------(2) 

 

, , n n P T f nM 

 

1 2 ,

( 

nM ) 

( 

nM ) ( 

nM ) 

( 

nM ) 

( 

nM ) 

 

dT 

 

dP dn1 

dn 

T 

P, 

n , n 

P 

T , n , n 

n1 

 

n2 

 

n( 

M ) 

( 

nM ) 

T 

 

1 

2 

P, 

n , n 

1 

2 

n( 

M ) 

dT 

P 

 

From the definition of partial molar properties 

1 

2 

T , n , n 

1 

2 

T , P1 

, n2. 

n3 

( 

nM ) 

dP 

ni 

 

T , P, 

n 

n( 

M ) n( 

M ) 

( 

nM ) 

 

dT 

 

dP M idni 

----------(3) 

T 

P 

 

P, 

n , n 

Subtracting equation (2) from equation (3) 

n( 

M ) 

 

T 

 

 

n( 

M ) 

dT 

 

P 

 

 

dP 

P, 

n , n 

1 

2 

1 

2 

T , n , n 

1 

2 

T , n , n 

1 

 

2 

n d M 

Equatin (4) is the fundamental form of Gibbs Duhem equation 

i 

i 

j 

dn 

0 ------------------(4) 

i 

T , P, 

n1 

, n3 

 

Special Case 

At constant temperature and pressure dT and dP are equal to zero. The equation becomes 

i i 0 M d x 

 

For binary solution at constant temperature and pressure the equation becomes 

x d M x d M 0 

1 

1 

2 

2 

x d M 1 

x ) d M 0 ------------(5) 

1 

1 

( 1 2 

dividing equation(5) by dx1 

x 

1 

d M 

dx 

1 

1 

 

( 1 

x 

1 

) 

d M 

dx 

1 

2 

0 

The above equation is Gibbs Duhem equation for binary solution at constant temperature 

and pressure in terms of Partial molar properties. 

Any Data or equation on Partial molar properties must satisfy Gibbs Duhem 

equation. 

2

Problem: 

Find weather the equation given below is thermodynamically consistent 

G x 150x 

x x ( 10x 

x ) 

G 

G 

G 

G 

1 

2 

1 

2 

dG 

dx 

100 1 

2 1 2 1 2 

G 

G ( 1 

x1) 

x1 

G 

G x1 

x 

3 

1 

2 

18x1 35x1 

16x1 

101 

3 

1 

18x 

8x 

1 

1 

2 

2 

1 

150 

54x1 70x1 

16 

dG 

2 

2 

54x1 16x1 

dx1 

G D equation 

x 

1 

d M 

dx 

1 

1 

2 

 

( 1 

x 

1 

) 

d M 

dx 

1 

2 

0 

x1 ( 54 x1 

70 x1 

16 ) ( 1 x1 

)( 54 x1 

16 x1 

) 0 

It satisfies the GD equation, the above equation is consistent. 

2

Phase Equilibrium 

Criteria for phase equilibrium: If a system says to be in thermodynamic equilibrium, 

Temperature, pressure must be constant and there should not be any mass transfer. 

The different criteria for phase equilibrium are 

At Constant U and V: An isolated system do not exchange mass or heat or work with 

surroundings. Therefore dQ=0, dW=0 hence dU=0. A perfectly insulated vessel at 

constant volume dU=0 and dV=0. 

dS 

U , V 

0 

At Constant T and V: Helmoltz free energy is given by the expression 

A U TS 

U A TS 

on differentiating 

dU dA TdS SdT 

we know that 

dU TdS PdV 

TdS PdV dA TdS SdT 

dA PdV 

SdT 

Under the restriction of constant temperature and volume the equation simplifies to 

dAT 

, V 

0 

At Constant P and T 

The equation defines Gibbs free energy 

G H TS 

G U PV TS 

dG dU PdV VdP TdS SdT 

dU dG PdV VdP 

TdS SdT 

dU dG dG PdV TdS 

Under the restriction of constant Pressure and Temperature the equation simplifies to 

dGT 

, P 

0 

This means the Gibbs free energy decreases or remains un altered depending on the 

reversibility and the irreversibility of the process. It implies that for a system at 

equilibrium at given temperature and pressure the free energy must be minimum.

Phase Equilibrium in single component system 

Consider a thermodynamic equilibrium system consisting of two or more phases of a 

single substance. Though the individual phases can exchange mass with each other. 

Consider equilibrium between vapor and liquid phases for a single substance at constant 

temperature and pressure. Appling the criteria of equilibrium 

dG 0 

a 

dG 

b 

dG 0 

a 

dG and 

b 

dG are chane in free energies of the phases a and b respectively. 

We know that 

i i dn G SdT VdP dG 

For phase a 

a 

dG 

a a a a a a 

V dP S dT G dn 

For phase b 

b 

dG 

b b b b b b 

V dP S dT G dn 

At constant temperature and pressure 

a 

dG 

a a 

G dn , 

b 

dG 

b b 

G dn 

For a whole system to be at equilibrium 

a 

dn 

b 

dn 0 

or 

dn 

a 

dn 

a b a 

G G dn 0 

b 

a b 

G G 

When two phases are in equilibrium at constant temperature and pressure, Gibbs free 

energies must be same in each phase for equilibrium. 

a 

b 

RT ln f C RT ln f C 

a 

f 

f 

b 

Clapeyron Clausius Equation 

Clapeyron Clausius Equation are developed two phases when they are in equilibrium 

(a) Solid- liquid 

(b) Liquid-Vapor 

(c) Solid Vapor

dP 

dT 

dP 

dT 

Consider any two phases are in equilibrium with each other at given temperature and 

pressure. It is possible to transfer some amount of substance from one phase to other 

in a thermodynamically reversible manner(infinitely slow).The equal amount of 

substance will have same free energies at equilibrium . 

Consider GA is Gibbs free energy in phase A and GB is Gibbs free energy in phase B 

at equilibrium. 

GA=GB ------(1) 

G = GA -GB =0 ------(2) 

At new temperature and pressure the free energy / mole of substance in phase A is 

G A dG A for Phase B GB dGB 

From basic equation 

dG VdP SdT -----(3) 

dG A V AdP 

S AdT 

dGB VBdP 

S BdT 

VAdP S AdT 

VBdP 

S BdT 

V V 

dTS 

S 

dP 

B 

A 

B 

A 

dP S 

 

dT V 

V represents the change in volume when one mole of substance pases from 

phase A to Phase B 

Q 

S 

 

T 

Q 

 

TV 

Q 

----(4) 

T 

V V 

 

B 

A 

This is a basic equation of Clapeyorn Clasius equation. 

B—Vapor state A- Liquid state 

Q=molar heat of vaporization = V H 

VB is molar volume in vapor state, VA is molar volume in liquid state,

dP 

dT 

T 

 

Q 

V V 

g 

l 

Vg>>>Vl (Gas volume is very high when compared to liquid volume) 

dP H 

 

dT TV 

V 

g 

H V (Latent heat of vaporization), 

dP P 

 

dT TRT 

dP dT 

2 

P R T 

V g 

RT 

P 

Integrating the above equation from T1to T2 and pressure from P1 to P2. 

P 

2 

dP 

 

P R 

P 

1 

P 

ln 

P 

2 

1 

1 

 

R T1 

 

1 

T 

2 

 

 

 

This equation is used to calculate the vapor pressure at any desired temperature. 

Problem: 

The vapor pressure of water at 100 o C is 760mmHg. What will be the vapor pressure 

at 95 o C. The latent heat of vaporization of water at this temperature range is 

41.27KJ/mole. 

P 760mmHg 

P 

1 

2 

T 

 

T 

2 

1 

 

dT 

T 

T 

T 

1 

2 

100 273 

95 273 

373K 

368K

P 

ln 

P 

2 

1 

P2 

 

ln 

 

760 

 

1 

 

R T1 

 

1 

T 

2 

 

 

 

41. 

27 10 

8. 

314 

P2 634. 

3mmHg 

3 

1 

 

373 

 

1 

368 

 

Phase Equilibrium in Multi component system 

An heterogeneous system contains two or more phases and each phase contains two or 

more components in different proportions. Therefore it is necessary to develop phase 

equilibrium for multi component system in terms of chemical potential. 

The partial molar free energy or chemical potential is given as 

G 

 

i 

Gi 

 

ni 

T , P, 

n j 

For a system to be in equilibrium with respect to mass transfer the driving force for 

mass transfer( Chemical potential) must have uniform values for each component in 

all phases. 

Consider a heterogeneous system consists of phase , , and 

containing various components 1,2,3-------C , that constitutes the system. 

k 

i 

The symbol represents chemical potential of component “i”in phase k. 

At constant temperature and pressure, for the criterion for equilibrium is 

dG=0 ----(1) 

Free energy for multi component system given by the expression 

SdT VdP dG ------(2) 

 

i i dn 

At constant Temperature and pressure 

dG ----(3) 

i i dn 

Comparing equation 1 and 3 

0 

 

i i dn 

For multi component system 

C 

 

 

i1 k 

 

k k 

i i dn 

0 

Expanding the above equation

1 

 

2 

 

c 

dn 

dn 

dn 

 

1 

 

2 

 

c 

 

 

 

1 

 

 

2 

 

c 

dn 

dn 

 

1 

dn 

 

2 

 

c 

 

 

1 

 

 

 

2 

 

c 

dn 

 

1 

dn 

dn 

 

 

2 

 

c 

 

 

 

 

1 

 

 

dn 

 

2 

 

c 

 

1 

dn 

dn 

 

 

2 

 

c 

 

0 

-----(4) 

Since the whole system is closed it should satisfy the mass conservation equation 

dn 

dn 

 

 

dn 

 

1 

 

2 

 

C 

dn 

dn 

 

1 

dn 

 

2 

 

C 

 

 

 

dn 

 

1 

dn 

 

2 

dn 

 

 

C 

 

0 

----(5) 

The variation in number of moles dni is independent of each other. However the sum of 

change in mole in all the phases must be zero. For the criterion of equilibrium is that the 

chemical potential of each component must be equal in all phases. 

 

 

 

 

 

 

1 

 

2 

 

C 

1 

 

 

2 

 

C 

 

1 

 

 

2 

 

C 

 

At constant temperature and pressure the general criterion for thermodynamic 

equilibrium in closed system for heterogeneous multi component system 

At constant T and P 

 

i 

 

i 

i 

for i=1,2,3--------C 

Since G RT ln f 

i i 

i 

The above equation is also satisfied in terms of fugacity 

 

i 

 

 

f i f i for i=1,2,3--------C 

f

Phase Diagram for Binary solution 

Constant pressure equilibrium 

Consider a Binary system made up of component A and B. Where it is assumed to be 

more volatile than B where vapour pressure ‘A’ is more than ‘B’. When the pressure is 

fixed at the liquid composition can be changed the properities such as temperature and 

vapour compositions get quickly determined VLE at constant pressure is represented on 

T-xy diagram. 

Boiling point diagrams 

When temperature is plotted against liquid(x) and vapour(y) phase composition. The 

upper curve gives and lower curve gives. The lower curve is called as bubble point 

curve and upper curve is called as Dew point curve. The mixture below bubble point is 

sub cooled liquid and above the Dew point is super heated vapour. The region between 

bubble point and Dew point is called mixture of liquid and vapour phase. 

Consider a liquid mixture whose composition and temperature is represented by 

point ‘A’. When the mixture is heated slowly temperature rises and reach to point ‘B’, 

where the liquid starts boiling, temperature at that point is called boiling point of whether 

heating mixture reaches to point ‘G’ when all liquids converts to vapour the temperature 

at that point is called as Dew point. Further heating results in super heated vapour. The 

number of tic lies connects between vapour and liquid phase. For a solution the term 

boiling point has no meaning because temperature varies from boiling point to Dew point 

at constant pressure.

Effect of pressure on VLE 

The boiling point diagram is drawn from composition x=0 to x=1, boiling point of pure 

substances increases with increase in pressure. The variation of boiling point diagrams 

with pressure is as shown in diagram. The high pressure diagrams are above low 

pressure.

Equilibrium diagram 

Vapor composition is drawn against liquid comp at constant pressure. Vapour is always 

rich in more volatile component the curve lies above the diagonal line. 

Constant temperature Equilibrium 

VLE diagram is drawn against composition at constant temperature. The upper curve is 

and lower curve is drawn at vapour comp(y). Consider a liquid at known pressure and 

composition at point ‘A’ as the pressure is decreases and reaches to point ‘B’ where it 

starts boiling further decreases in pressure reaches to point ‘C’ when all the liquid 

converts to vapour, further decreases in pressure leads to formation of super heated 

vapour. In between B to C both liquid and vapour exists together

Non ideal solution 

An ideal solution obeys Raoults law and p-x line will be straight. Non ideal solution do 

not obey Raoults law. The total pressure for non ideal solutions may be greater or lower 

than that for ideal solution. When total pressure is greater than pressure given by Raoults 

law, the system shows positive deviation from Raoults law. 

Eg: Ethanol-toluene 

When the total pressure at equilibrium is less than the pressure given by Roults law the 

system shows negative deviation from Raoults law. 

Eg: Tetrahydrofuron-ccl4 

Azeotropes 

Azeotropes are constant boiling mixtures. When the deviation from Raoults law is very 

large p-x and p-y curve meets at this point y1=x1 and y2=x2 

A mixture of this composition is known as Azeotrope. Azeotrope is a vapour liquid 

equilibrium mixture having the same composition in both the phases. 

Types of Azeotropes 

Azeotropes are classified into two types 

1. Maximum curve pressure, minimum boiling azeotropes 

2. Minimum pressure maximum boiling azeotrope 

The azeotrope formed when negative deviation is very large will exhibit minimum 

pressure or maximum boiling point, this is known as minimum pressure Azeotrope. 

Eg : Ethanol –water, benzene-ethanol 

The azeotrope formed when –ue deviation is very large will exhibit minimum pressure or 

maximum boiling point, this is known as minimum pressure azeotrope. 

Phase diagram for both types of Azeotropes 

Minimum Temperature azeotropes

Maximum Tem – Min pressure Azeotropes 

Calculation of VLE for ideal solution 

Ideal solution: Ideal solution is one which obeys Raoults law. Raoults law states that the 

partial pressure is equal to product of vapor pressure and mole fraction in liquid phase. 

PA PVAx 

A 

PA- Partial pressure 

PVA- Vapor pressure

xA- Mole fraction of component A 

For binary solution (For component A and B) 

We know that PT PA 

PB 

-------(1) 

PA PVAx 

A , PB PVB 

xB 

Substitute PA and PB in equation 1 

P P x P x 

T VA A VB B 

1 x A xB 

, 1 

x A xB 

PT PVAx 

A PVB 

( 1 

x A ) 

P P P x P --------(2) 

x 

T 

A 

 

VA VB A VB 

P 

P 

T 

VA 

P 

VB 

P 

VB 

Assuming the vapor phase is also ideal 

PA 

PVAx 

A 

y A 

PT 

PT 

Substituting PT from equation (2) 

y 

A 

 

P 

VA 

x 

PVA PVB 

x A PVB 

A 

Dividing numerator and denominator by PVB 

y 

y 

A 

A 

 

P 

 

 

P 

 

VA 

VB 

P 

P 

VA 

VB 

x 

A 

 

1 

 

x 

A 1 

 

x 

A 

1x 

1 

A 

The above equation relates x and y 

is known as relative volatility of component A with respect to component B

Problem: The binary system acetone and acetone nitrile form an ideal solution. Using the 

following data prepare 

P-x-y diagram at 50 o C 

T-x-y diagram at 400mm Hg 

x-y diagram at 400 mm Hg 

T PV1 PV2 

38.45 400 159.4 

42 458.3 184.6 

46 532 216.8 

50 615 253.5 

54 707.9 295.2 

58 811.8 342.3 

62.3 937.4 400 

Solution: PT PV 1 PV 

2 x1 PV 

2 

, 

x1 PT y1 

0.0 253.5 0.0 

0.2 325.8 0.377 

0.4 398.1 0.6179 

0.6 470.4 0.784 

0.8 542.7 0.906 

1.0 615 1 

T-x-y diagram at 400 mm Hg , PT=400mmHg 

T PV1 PV2 x1 y1 

38.45 400 159.4 1 1 

42 458.3 184.6 0.7869 0.9015 

46 532 216.8 0.5812 0.7729 

50 615 253.5 0.405 0.6226 

54 707.9 295.2 0.2539 0.449 

58 811.8 342.3 0.122 0.2475 

62.3 937.4 400 0 0 

Calculation of VLE for non Ideal solution 

For non ideal solution 

Partial pressure is given by Pi i xi 

PVi 

i - Activity coefficient 

xi- Mole fraction 

PVi – vapor pressure 

y 

P 

x 

V1 

1 

A , PV1= 615, PV2 = 253.5 

PT 

x 

P 

P 

T V 2 

1 , 

PV 

1 PV 

2 

y 

A 

 

P 

P 

T 

x 

V1 

1

For binary mixture 

P x P 

1 1 1 V1 

P 

x 

P 

2 2 2 V 2 

P T 

T 

P P 

1 

P x P 

2 

x 

P 

1 1 V1 

2 2 V 2 

P1 

y1 

 

PT 

1x1P 

V1 

 

1x1P 

V1 

2 x2 

PV 

2 

1 

 

2 x2 

PV 

2 

1 

1x1P 

V1 

The above equation relates x and y for non ideal solution 

Activity coefficients are functions of liquid composition x, many equations are available 

to estimate them. The important equations are 

Vanlaar equation 

Wilson equation 

Margules Equation 

Vanlaar Equation: 

Estimation of activity coefficient 

2 

Ax2 

ln 

1 

2 

A 

 

x1 

x2 

 

B 

ln 

2 

2 

1 

Bx 

 

B 

 

x1 

x2 

 

A 

2 

Where A and b are known as vanlaar constants, if the constants are known the 

activity coefficients can be estimated. 

Estimation of Vanlaar constants 

1 st Method: If the activity coefficients are known at any one composition, then vanlaar 

constants A and B can be estimated by rearranging the equation 

A 

 

x 

 

ln 

 

2 2 

ln 1 1 

 

x1 

ln 

1 

2

B 

 

x ln 

 

1 1 

ln 2 1 

 

x2 

ln 

2 

2 

2 nd Method 

For a systems forming azeotrope, if the temperature and pressure s are known at 

azeotropic composition then activity coefficients can be calculated as shown below. 

x P 

P 

1 1 1 V1 

P y x 

T 

P 

1 1 1 V1 

At azeotrpic composition x1=y1 

PT 

P 

1 , 

PT 

P 

2 

V1 

V 2 

Van laar constants A and B can calculated using the above equations. 

Problem: The azeotope of ethanol and benzene has composition of 44.8mol% C2H5OH at 

68 o C and 760mmHg.At 68 o C the vapor pressure of benzene and ethanol are 517 and 506 

mmHg. 

Calculate 

Vanlaar constants 

Prepare the graph of activity coefficients Vs composition 

Assuming the ratio of vapor pressure remains constant, prepare equilibrium 

diagram at 760mmHg. 

Solution: 

At azeotropic composition x1=y1=0.448, x2=y2=0.552, PV1=506mmHg, 

PV2 =517mmHg 

A 

P 

P 

T 

V1 

760 

506 

 

P 

 

T 

1 , 2 

PV 

2 

760 

1 1. 

501, 

2 1. 

47 

517 

x 

 

ln 

 

2 2 

ln 1 1 

 

x1 

ln 

1 

( 0. 

552) 

ln( 1. 

47) 

 

A 

ln( 1. 

501) 

1 

 

( 0. 

448) 

ln( 1. 

501) 

 

 

 

2 

2

A=0.829 

B 

 

x ln 

 

1 1 

ln 2 1 

 

x2 

ln 

2 

( 0. 

448) 

ln( 1. 

501) 

 

B ln( 1. 

47) 

1 

 

( 0. 

552) 

ln( 1. 

47) 

 

 

 

B=0.576 

2 

2 

Make use of the equations given below for plotting graph of activity coefficients Vs 

composition. 

ln 

y 

1 

2 

2 

Ax2 

Bx1 

1 

, ln 

2 

2 

2 

T 

A 

 

x 

B 

1 

 

x2 

 

 

P1 

1x1P 

V1 

 

P x P x P 

x1 

1 1 V1 

x2 

2 

2 

V 2 

B 

 

x1 

x2 

 

A 

1 

 

2 x2 

P 

1 

x P 

1 2 

V 2 

1 1 V1 

0 1 6.745 1 0 

0.2 0.8 2.4 1.095 0.384 

0.4 0.6 1.644 1.374 0.4384 

0.6 0.4 1.21 1.721 0.5077 

0.8 0.2 1.0112 2.618 0.609 

1.0 0 1 3.723 1 

Margules equation 

Eetimation of activity coefficients 

ln x 

2 

A 2 B A x 

1 

2 

2 

2 

1 

 

B 2A 

B 

 

ln x x 

1 

2 

The constant A in the above equation is terminal value of ln1 at x1=0 and constant B is 

the terminal value of ln2 at x2=0 

y1

When A=B 

2 

ln Ax , 

1 

2 

ln Ax 

2 

2 

1 

The above equations are known as Margules suffix equation. 

Wilson Equation: 

Wilson proposed the following equation for activity coefficients in binary solution 

12 

21 

ln 

1 lnx1 

12 

x2 

x2 

 

x1 

12 

x2 

21x1 

x2 

 

ln 

2 

ln 

 

 

12 

21 

x2 21x1 

x1 

 

x1 

12 

x2 

21x1 

x2 

 

Wilson equation have two adjustable parameters 12 

component molar volumes. 

 

 

12 

21 

V2 11 2 

12 

 

V 

1 

 

V 

12 

 

exp 

RT 

12 

 

exp 

RT 

V 

 

V 

1 

V 

 

V 

a 

exp 

RT 

V1 22 1 

21 

2 

2 

 

 

 

a 

exp 

RT 

V1 and V2 – molar volumes of pure liquids 

- Energies of interaction between molecule 

 

 

 

 

 

and 21 

. These are related to pure 

Wilson equation suffers main disadvantages which is not suitable for maxima or minima 

on ln versus x curves. 

Consistency of VLE data 

Gibbs duhem equation in terms of thermal consistency 

x 

1 

d ln 

dx 

1 

1 

 

( 1 

d ln 

x1) 

dx 

1 

2 

0 

Plot of logarithmic activity coefficients Vs x1 of component in binary solution

According to Gibbs Duhem equation both slopes must have oppsite sign then only the 

data is consistent, otherwise it is inconsistent. 

For the data to be consistent it has satisfy the following condition. 

1. If one of ln curves has maximum at certain concentration and the other curve 

2. 

should be minimum at same composition. 

If there is no maximum or minimum point both must have + ue or –ue on entire 

range. 

Co-existence equation 

The general form of Gibbs duhem equation at constant temperature and pressure 

0 d x ---(1) 

 

i i M 

In terms of partial molar free energies 

0 d x ------(2) 

 

i i G 

dividing equation (2) by dx1

x 

i 

dG 

dx 

1 

i 

0 -----(3) 

Gi RT ln f i 

at constant temperature 

dGi RT ln f i 

----(4) 

dx1 

dx1 

substituting equation (4) in equation (3) 

RT ln f 

xi 

dx 

d ln f 

xi 

dx 

1 

1 

i 

i 

0 

0------(5) 

For binary system 

d ln f1 

d ln f 2 

x1 x2 

0 ----(6) 

dx dx 

1 

x d f x d ln f 0 -----(7) 

1 

ln 1 2 2 

1 

Equation (5) (6) and (7) are Gibbs Duhem equation in terms of fugacites and thus 

applicable for both liquid and vapor phase 

For liquids 

f x f 

i 

i 

i 

i 

i 

ln f ln 

ln x ln f 

i 

differentiating with respect to x1 

1 

1 

i 

d ln f i d ln 

i d ln xi 

d ln f i 

-------(8) 

dx dx dx dx 

1 

i 

fi is a pure component ffugacity it does not vary with x1 

Substituting equation (8) in equation (5) 

d ln 

i d ln x 

xi 

xi 

dx dx 

1 

1 

i 

=0 

0 

1 

=0

d ln 

xi 

dx 

1 


d ln 

1 d ln 

x1 

x2 

dx dx 

1 

i 

0 ----------------(9) 

1 

2 

0 ---------(10) 

x d x d ln 

0 -----------(11) 

1 

ln 1 2 2 

Equation (9) (10) and (11) are Gibbs Duhem equation in terms of activity coefficients 

For ideal vapor 

f i Pi 

From equation 5 

d ln Pi 

xi 

0 ------------(13) 

dx 


1 

d ln P1 

d ln P2 

x1 x2 

0 ----------(14) 

dx dx 

1 

1 

x d P x d ln P 0 ---------------(15) 

1 

ln 1 2 2 

Equation (13) (14) and (15) are Gibbs Duhem equation in terms Partial pressure. 

For ideal vapor 

P1 y1PT 

, P2 y2 

PT 

x1d 

ln( y1PT 

) x2d 

ln( y2 

PT 

) 0 

x1d 

ln y1 

x1d 

ln y2 

( x1 

x2 

) d ln PT 

0 

x x 1 

1 

x d 

1 

2 

ln y1 

2 

x1d 

ln y d ln PT 

0 

dP 

 

x1d 

ln y1 

x2d 

ln y 

P 

2

dP 

P 

x 

1 

dy 

y 

1 

1 

x 

2 

dy 

y 

2 

2 

y 2 y1 

0 , dy2 dy1 

on simplification 

dP 

x1 

( 1 

x1) 

 

dy1 

 

P y1 

( 1 

y1) 

 

Further simplification 

dP y1 

x1 

 

P 

 

dy1 

y( 

1 

y1) 

 

This is known as coexistence equation. It relates P,x,y for binary VLE system. This 

equation is used to rest consistency of VLE data. 

Consistency test for VLE data 

x1d 

ln 1 x2d 

ln 

2 0 

x1d 

ln 

d ln 

2 

x 

d ln 

d ln 

2 

2 

2 

1 

x1d 

ln 

 

( 1 

x ) 

1 

1 

x1d 

ln 

 

( 1 

x ) 

Redlich Kister Test 

1 

1 

The excess free energy of mixing for a solution is given as 

e 

G G G 

ideal 

 

G 

Nonideal 

e 

RT xi 

ln i 


G RT x ln x ln 

e 

 

 

 

1 

1 

2 

2

differentiating with respect to x1 

dG e 

dx 

1 

 

RT x 

 

1 

d ln 

dx 

1 

1 

ln 

From Gibbs Duhem Equation 

d ln 

1 d ln 

2 

x1 

x2 

0 

dx dx 

dG e 

dx 

dx 

1 

dG e 

dx 

1 

dG e 

x1 

1 

 

x1 

0 

1 

dG 

1 

 

RT ln 

1 ln 

 

dx dx 

2 

RT 

1 

1 

2 

ln ln 

 

1 

 

RT ln 

 

e 

RT 

 

1 

2 

x1 

1 

x1 

0 

 

 

 

 

ln 

 

1 

2 

2 

dx 

dx 

 

dx 

 

1 

2 

1 

1 

x 

 

 

 

2 

d ln 

dx 

1 

2 

ln 

e 

The two limits indicates pure component for which G 0 , where LHS is zero for both 

limits. 

We can write 

x1 

1 

1 

0 ln 

dx 

x 0 

2 

1 

1 

This can be checked graphically. Net area should be equal to zero for 

consistency(Area=0). 

2 

dx 

dx 

2 

1

Problem 

Verify whether the following data is consistent 

X1 1 2 

0 0.576 1.00 

0.2 0.655 0.985 

0.4 0.748 0.930 

0.6 0.856 0.814 

0.8 0.950 0.626 

1.0 1.00 0.379 

Solution: 

We know that the Redlich kister test is 

x1 

1 

1 

0 ln 

dx1 

x 0 

2 

1 

1 

2 

1 

ln 

2 

0.576 -0.552 

0.665 -0.408 

0.804 -0.218 

1.052 0.051 

1.518 0.417 

2.639 0.97 

Plot 

ln 

1 

Vs x1 

2 

Area under the curve is zero. Data is consistent.

CHEMICAL REACTION EQUILIBRIUM 

Energy change accompanies all chemical reactions. Because of this energy 

change the temperature of the products may increase or decrease depending on the 

exothermic or endothermic nature of the reaction. The energy change may be 

expressed in terms of heat of reaction, heat of combustion and heat of formation. 

Heat of reaction is the change in enthalpy of a reaction under pressure of 1.0 

atmosphere, starting & ending with all materials at a constant temperature T 

Heat of combustion is the heat of reaction of a combustion reaction. 

Heat of formation is the heat of reaction of a formation reaction. A formation 

reaction is one which results in the formation of one mole of a compound from the 

elements. 

Eq. H2 + ½ O2 H2O 

C +O2 CO2 

C + ½ H2 + 1/2 Cl2 CHCl3 

The standard heat of reaction, standard heat of formation, and standard heat of 

formation are respectively the heat of reaction, heat of combustion and heat of formation 

under 1 atmosphere starting and ending with all materials at constant temperature of 

25 0 C. 

In chemical industries, processes are carried out under isothermal conditions 

and this requires the addition or removal of heat from the reactor. Heat of reaction 

values will give the amount of heat to be removed or added. Knowing of this heat 

value helps to design the heat exchange equipment. 

Standard heat of reaction 

Calculation of ∆H 298 form heats of formation data: 

The standard heat of reaction accompanying any chemical change is equal to 

the algebraic sum of the standard heats of formation of products minus the 

algebraic sum of the standard heat of formation of reactants. 

∆Hr = ∑∆H f 298 products - ∑∆Hf 298 reactants 

Heat of formation of any clement is zero

Heat effects of chemical Reaction: 

For the reaction, aA + bB ↔ cC + dD 

Let heat capacity be Cp = α + βT + γT 2 

then we have CpA = αA + βAT + γAT 2 

CpB = αB + βBT + γBT 2 

CpC = αC + βCT + γCT 2 

CpD = αD + βDT + γDT 2 

And ΔH 0 298 be standard heat of reaction at 298 K and standard heat of reaction at any 

other temperature can be found by Kirchoff’s rule 

o 

d H 

dT 

C where 

= [c(αC + βCT + γCT 2 ) + d( αD + βDT + γDT 2 ) ] - [a (αA + βAT + γAT 2 ) + b (αB + βBT + 

γBT 2 )] 

 

C C C 

p p p 

products reac tan ts 

Cp 

cCpC dC pD 

aCpA bC pB then substituting the values of 

= [(c αC +d αD ) – (a αA + b αB ) ]+ [ (c βC + d βDT) – (a βA + b βB)]T + [ c γC + dγD ) – 

(a γA+ b γB )] T 2 


ΔCp = Δα + Δβ T + ΔγT 2 

Substituting in Kirchoff’s rule 

 

 

Pr oducts Re ac tan ts 

cC dC aC bC 

 

C D A B 

 

 


 

 


p 

C pC , C pD , and C pA, C pB

H T T T 

0 

H 

T 

0 2 

 

 

H298 

0 2 

d 

H T T 

dT 

298 

 

 

2 3 

OR in general 

o 

2 3 

H T 

T T I 

P 

 

0 2 

H C dT T T dT 

 

= 

2 3 

2 3 

T T T I 

A chemical reaction proceeds in the direction of decreasing free energy.The sum of the 

free energies of the reactants should be more than the sum of energies of products. For a 

reaction to take place 

 

G G G or G 0 

Re action Re action 

products reaction 

Therefore ΔG should be less than zero for a reaction to occur. When equilibrium is 

reached free energies of the reactants equal free energies of products,. Therefore the 

criterion for reaction equilibrium is ΔG = 0 

ΔGReaction Calculation: Consider the reaction aA + bB ↔ cC + dD 

Let GA, GB, GC , and GD be the partial molar free energies of A,B,C, and D in the 

reaction mixture. 

 

 

2 3 

o 

2 3 

H T 

T T I 

f f 

G G RT ln but a the activity of A 

0 A A 

A A 0 

fA 0 

fA 

A 

G G RT lna 

A 

0 

A A 

G G RT lna 

B 

0 

B B 

G G RT lna 

C 

0 

C C 

G G RT lna 

D 

0 

D D

G G G 

Re action 

 

At equilibrium ΔG = 0 and 

 

products Re ac tan ts 

cGC dGD aGA bGB 

 

0 GC RT 

0 

aC d GD RT aD 

 

0 

a GA 0 

RT ln aA b GB RT lnaB 

 

cGC dGD aGA bGB 

 

RT lnaC lnaD lnaA lnaB 

 

 

= c ln ln 

 

 

 

 

0 

GRe action G RT 

ln C 

c d 

a a 

D 

 

a b 

aA a 

B 

c d 

a a 

C D 

K K is equilibrium constant of the reaction 

a b 

aA a 

B 

a a 

G RT G RT K 

c d 

0 

0 

 

C ln a 

aA D 

b 

aB 

 

 

0 

ln 

 

at equilibrium

EFFECT OF TEMPERATURE ON EQUILIBRIUM CONSTANT K 

Gibb’s - Helmoholtz equation at constant pressure is given 

o G 

 

d 

T H 

o 

but G RT 

ln K 

2 

dT T 

RT 

ln K 

d 

T 

 

H d R ln K 

H 


 

2 2 

dT T dT T 

d ln K 

H 

Van't Hoff equation. 

2 

dT RT 

This may be integrated to find the effect of temperature on K 

K2 

T2 

 

When H 

is contant d lnK 

K1 

T1 

K2 ln 

K1 

H 

1 1 

 

R T2 T1 

 

if K is known at T ,then 

K can be calculated at other T 

1 1 

ln K 

H 

dT 2 

RT 

2 2 

When H varies with T 

d 

dT 

0 

H 

2 

RT 

 

T 

I 

d ln K 

dT 2 

RT 2R 3R 

 

RT 

0 

 

 

ln T 

ln K = 

R 2R 6R 

RT 

G RT ln K 

2 

T T I 

M 

Where M is a consatnt of integration 

2 

0 ln T T T 

I 

G RT M 

 

R 2R 6R 

RT 

T T 

G T I MRT 

2 6 

2 3 

 

0 

ln T

For the reasction C2H4 + ½ O2 ↔ C2H4O, develop equations for ΔH 0 , K, and ΔG 0 , find 

ΔG 0 at 550 K 

Cp (Cal / mol K) Data: C2H4 = 3.68 + 0.0224 T 

Standard heat of reaction ΔH 0 298 

C2H4O = - 12 190 and C2H4 = - 12 500, Cal / mol 

Re action 298 

In this problem C 

p 

 

H H H 


T 

O2 

And ΔG 0 298 = -19070 Cal / mol 

1 

= 1.59 3.68 6.39 5.285 


2 

= 6.39 + 0.0021 T 

C2H4O = 1.59 + .00332 T 

= 1(-12190) – 1( -12500) = 310 Cal / mol 

1 

= 0.0332 0.0224 0.0021 0.00975 


2 

Then the variation of standard heat of reaction with temperature is

o 2 

3 0.00975 2 

H T T T I 5.285 T T I 

2 3 2 

0 0.00975 2 

Given H298 =310 = 5.285 T T I then I = 1452 

2 

o 

0.00975 2 0 

H 5.285 T T 1452 This is variation of H298 

with temparature 

2 

2 I 5.285 

0.00975 1452 

lnK = ln T T T M = lnT 

T M 

R 2R 6R RT 2 2(2) 2T 

0 

0 G298 19070 

but G298 RT ln K298 

or lnK298 32 

RT 2(298) 

5.285 

0.00975 1452 

32 = ln298 298 M therefore M = 48.763 

2 2(2) 2(298) 

5.285 

0.00975 1452 

lnK = lnT T 48.763 

2 4 2T 

0 

G RT lnK 2T 

lnK 

5.285 0.00975 1452 

2T lnK 2T lnT T 48.763 

2 4 2T 

 

 

0 3 

2 

G 5.285T lnT 4.875(10)() T 1452 97.526() T 

G 5.285(550)ln550 4.875(10)(550) 1452 97 

.526(550) 

0 

550 

3 

2 

G 35320.62 

Cal / mol 

0 

550 

50 mol% each of SO2 and O2 is fed to a converter to form SO3. 

Show the variation of i) Standard heat of reaction with T ii) Equilibrium constant 

with T. The ΔHf and ΔGf at 298 K are 

Component 

ΔHf 

k Cal / k 

mol 

ΔGf 

k Cal / k 

mol 

SO2 -70960 -71680 

O2 0 0 

SO3 -94450 -88590

Feed contains 50 mole% SO2 and O2.This means for every one mole SO2 there will be 

one mole O2 in the reactant side and ½ mole O2 in the product side, according to SO2 + 

½ O2 SO3 

Reactants: 1 mol SO2 + 1 mol O2 Products: 1 mol SO3 + ½ mol O2 (excess) 

1 

= 6.0377 + 6.148 7.116 6.148 - 4.152 

2 

Pr oducts reac tan ts 

-3 

= 23.53(10) + 

Pr oducts reac tan ts 

0 2 

3 

H T T T I 

2 3 

-3 

12.474x10 2 

= -4.152T+ 

T I 

2 

1 

3 

-3 

3.102(10) - 9.512 (10) + 3 

2 

 

-3 

.102 (10) 

=12.474 

x10 

-3 

0 0 

But H at 298 is given by H 

0 

H 

0 

H 94450 70960 23490 

k Cal / k mol 

298 

products Re ac tan ts 

-3 2 

-23490=-4.152 (298 ) + 6.237 x 10(298) 

 

I = -22818.19 

0 

3 

2 

H 4.152 T 6.237x10 T 22818.19 

Next for K we have 

 

 

 

2 I 

lnK= lnT 

T T M 

R 2R 6R 

RT 

3 

4.152 

12.474x10 22818.19 

lnK= lnT 

T M 

2 2xR 2T 

I 

CP data: 

Component a b(10 3 ) 

SO2 7.116 9.512 

O2 6.148 3.102 

SO3 6.0377 23.537

0 

0 G298 

We have G 

298= -RTln K298 

or lnK 298= 

2(298) 

G = G - G 

= - 88590 - (- 71680) = - 16910 k Cal / k mol 

0 

298 

 

Products Reactants 

16910 

lnK 298= 

28.372 

2(298) 

3 

4.152 

12474x10 22818.19 

28.372 ln298 

298 M 

2 2x2 

2x298 M 0.9842 

11409.09 

3 

lnK 2.076lnT 3.1185x10 T 0.9842 

Derive the general equation for the standard free energy formation for ½ N2 + 3/2 

H2 NH3 

The absolute entropies in ideal gas state at 298 K and 1.0 atm are 

NH3 = 46.01 g Cal / g mol K 

N2 = 45.77 

H2 = 31.21 

ΔH 0 298 (g Cal / g mol) = - 11040 

Heat capacity Data is given by Cp= a + bT + cT 2 the values are 

Component a bx10 2 

T 

cx10 5 

NH3 6.505 0.613 0.236 

N2 6.903 - 0.038 0.193 

H2 6.952 - 0.46 0.096 

Compute the free energy change at 1500K. Is reaction feasible?

0 o 0 

We have G and 

0 

298 

H S S S S 

298 

 


G 11040 298( 23.69) 3980.38 

g Cal / g mol K 

1 3 

=46.01(45.77) + (31.21) = 23.69 g Cal / g mol K 

2 2 

 

 

a a a 


1 6.505(6.903) 

2 3 

6. 

2 

 

952 7.3745 

 

b b b 


x 

1 

2 x 

3 

 

2 

x 

 

 

 

c c c 


0 b 2 c 

3 

H298 aT T T I 

2 3 

1 0.236x10(0.193 10 

2 x 

3 

)(0.096 10) x 

2 

 

0.045 10 

 

x 

11040( 7.3745)298 

3 7.01X10 298( 

2 

6 

2 0.045 

X10 

 

3 

3 

298) I OR I = -9153.26 

a b c 

2 I 

lnK ln T T T M and R =2 gCal / g mol K 

R 2R 6R 

RT 

0 

-3980.38 

G298 RT 

ln K298 

or ln K 298 = = 6.6784 

-(2x298) 

2 2 

2 3 

0.613 10( 0.038 10) ( 0.046 10 ) 7.01x10 5 5 5 6 

3 6 

7.3745 7.01x10 0.045x10 

2 9153.26 

6.6784 ln298 298(298) M 

2 2x2 6x2 2x298 M 11.7987 

The iation of G G 

0 

var with T is given by, = - RT lnK 

3 6 

0 -7.3745 7.01x10 0.045x10 

2 9153.26 

G = -2T lnT T T 11.7987 

 

2 2x2 6x2 2T 

 

3 6 

0 -7.3745 7.01x10 0.045x10 

2 9153.26 

G1500 = -2x1500 ln1500 1500 1500 11.7987 

 

2 2x2 6x2 

2x1500 

G G G 

0 

1500 

 


g Cal 

28488.81 g mol 

The reaction is not feasible ΔG is highly +ve: For feasibility it should equal to zero

The hydration of Ethylene to alcohol is given by C2H2 + H2O C2H5OH 

Temperature 0 C 145 6.8 x 10 -2 

Equilibrium Constant 320 1.9 x 10 -3 

The heat capacity data for the component can be 

represented by Cp = a + bT where T is in kelvin 

and Cp is in cal / g mol 0 C 

Develop general expression for the equilibrium constant and standard Gibb’s free 

energy change as function of temperature. 

 

a a a = 6.99(2.83 7.3) 3.14 


 

b b b 39.741x10(28.6 10 x 2.46 10) x 8.68 10 x 


a b c 

I 

lnK = ln 

R 2R 6R 

RT 

2 

T T T M 

3 3 3 3 

3 

3.14 

8.68x10 I 

lnK = lnT 

T M 

2 2x2 2T 

3 

-2 3.14 

8.68x10 I 

At 418 K, 

ln 6.8x10 = ln 418 418 M 

2 2x2 2x418 

Solving, 

I = -9655.807 & M = -5.667 

3 

-3 3.14 

8.68x10 I 

593 K, ln 1.9x10 ln593 593 

At M 

2 2x2 2x593 

3 

3.14 

8.68x10 -9655.807 

lnK = lnT 

T 

-5.667 

2 2x2 2T 

3 

3.14 8.68x10 -9655.807 

But G = -2T lnT T -5.667 

2 2x2 2T 

 

-3 2 

G = 3.14T ln T - 4.34x10 T 9655.807 11.334T 

Component a bx10 3 

C2H4 2.83 28.601 

H2O 7.3 2.46 

C2H5OH 

6.99 39.741

THERMODYNAMIC FEASIBILITY OF REACTIONS 

The equilibrium constant K is a measure of the concentration of the products 

formed at equilibrium. It is related by the equation ΔG 0 = - RT ln K. The more the value 

of K, more will be the equilibrium conversion of products. When the value of ΔG < 0, 

means the value of K should be very large. Hence ΔG is the measure of feasibility of a 

chemical reaction. 

i) If ΔG 0 < 0, there can be appreciable conversion of reactants in to products. 

The more the –ve value, more will be the feasibility of the reaction 

ii) If ΔG 0 is +ve but less than 10 000 k Cal / mol, the reaction is not feasible, 

at atmosphere , may be feasible at any other pressure 

iii) If ΔG 0 is greater than 10 000 kCal / mol the reaction is not at all feasible 

under any condition. 

Equilibrium Calculations ( Homogeneous gas Phase reactions): 

aA (g) + bB (g) ↔ cC (g) + dD (g) 

c 

c y c c p 

 

We have 

f 

Activity a E f 

0 

f 

K 

c d 

a a 

C D 

 

a b 

a A a 

B 

for gases, 

0 

as = p = 1 atmosphere: a = f 

f = y but a f x p = y p 

c c c c c c 

c c c 

K = 

a y p 

D D D D 

a y p 

A A A A 

a y p 

B B B B 

 

 

 

 

 

 

 

 

 

 

 

 

A A A B B B A B A B A B 

 

 

D y DD p 

 

c D 

x 

y c yD x 

c D 

x 

p p 

y p y p y y p p 

c + d - (a + b) n 

()()()(P) y =()()()(P) where = activit y 

y co efficient for gas phase 

 

 

 

 

d c d c d c d c d 

a b a b a b a b a b 

K k k k k k k 

y = mole fraction 

= Fugacity Co efficient for gas phase 

P = Total pressure

Effect of variables on equilibrium : 

Effect of temperature: 

G RT lnK 

since ΔG depends only on temperature as the 

0 

298 298 

pressure is fixed at 1.0 atmosphere, K value varies with temperature. It is not affected by 

Pressure, Concentration, etc,. Variation of K with T is given by 

ln K 

d H 

2 

dT RT 

Van't Hoff equation. 

For endothermic reactions ΔH 0 is + ve as T increases K also increases. This 

means that the equilibrium conversion is more at higher pressure. 

For exothermic reactions ΔH 0 is –ve. Therefore K increases with T. Hence the 

equilibrium conversion decreases as T increases. Eg. SO2 + 1/2 O2 ↔ SO3 

Effect of pressure: Consider a reaction of ideal gases, then K = KyP Δn or Ky = K / P Δn 

When Δn > 0. An increase in pressure decreases Ky. Hence equilibrium product 

yield is less at high pressures 

When Δn< 0. An increase in pressure increases Ky and equilibrium product 

yield 

pressure is required 

C2H4 + H2O C2H5OH Δn = 1 – (1+1) = -1, Δn < 0 

N2 + 3H2 2NH3 Δn = 2 – 4 = – 2 Δn < 0 Here high 

When Δn = 0 Here pressure has no effect on reaction. For ideal gases the effect 

of pressure depends on the variation of γ and Φ with pressure 

Effect of Inert: Presence of inert has the opposite effect of increase of pressure. 

Therefore when Δn > 0, addition of inert increases Ky and equilibrium 

conversion 

And Δn < 0, addition of inert decreases equilibrium conversion. 

Effect of excess reactants: Presence of excess reactants increases equilibrium 

conversion of the limiting reactant 

Presence of products in feed: decreases the equilibrium conversion of reactants. 

Eg. CH3COOH + C2H5OH CH3COOC2H5 + H2O 

If water is added by 1.0 mole to the feed, equilibrium conversion of CH3COOH 

reduces from 30% to 15%

HCN is produced by the reaction N2 (g)+ C2H2 (g) 2HCN (g). The reactants are 

taken in stoichiometric ratio at 1.0 atmosphere and 300 0 C. At this temperature ΔG 0 = - 

30100 k J / k mol. Calculate equilibrium composition of product stream and 

maximum conversion of C2H2 

30100 

8.314(573) 

0 3 

Δn=2– (1+1) = 0 we have G RT 

ln K or lnK = or K = 1.8029x10 

Component Moles 

in feed 

Moles 

reacted 

Moles 

present 

Mole 

fraction 

N2 1 X 1 – X (1 – X) / 2 

C2H2 1 X 1 – X (1 – X) / 2 

HCN -- 2X 2X 

K K K K P K K P P 

n n 

0 

y Assume 1 and 1 

y X 

K K 

Solving for X 

2 2 

3 

1.8029(10) y 

HCN 

1 1 

yN y 

2 C2H2 1 X 1 

X 

, X = 0.0207 

Equilibrium Composition: 

Equilibrium Conversi on of C H 

 

2 

 

2 

 

 

1 X 1 0.0207 

N2 100 100 48.965% 

2 

 

2 

 

 

1 X 1 0.0207 

C2H2 100 100 48.965% 

2 

 

2 

 

 

HCN ()1 X 00 2.007% 

2 2 

Equilibrium conversion is max imum for reversible reaction 

Moles of C H reacted X 0.0207 

2 2 

Max. Conversion of C2H2 100 100 100 2.07% 

Moles of C2H2 in feed 1 1 

(2X) / X = 

Workout the problem, when the pressure is changed to 203 bar. Fugacity co efficients 

of N2,,C2H2 and HCN are 1.1, 0.928, and 0.54 Assume KΦ= 1 

X

y 

 

K K K K P 

 

c c 

y 

n 

C C 

0 

= (1) 

 

203 where a,b, and c are stoichiometric co efficients 

a b a b 

A 

B yA 

y 

B 

0.54 X 

 

(1.1)(0928) 1-X 1-X 

 

2 

 

2 

 

 

2 2 

-3 

and K is known already = 1.8029x10 = .(1) 

solving 

X = 0.0382 

then equilibrium Composion of C H is ( Max reversible reaction) 

2 2 

2 2 

moles C2H2reacted 0.0382 

= x 100 = 3.82% 

moles of C H in feed 1 

Calculate the K at 673K & 1.0 bar for N2 (g) + 3 H2 (g) 2NH3 (g). 

Assume heat of reaction remains constant. Take standard heat of formation and 

standard free energy of formation of NH3 at 298 K to be - 46110 J / mol and – 16450 J 

/ mol respectively. 

G 2( 16450) 32900 J 

0 

298 

0 

298 

 

2 (-46110) = -92 220 J 

0 

G298 RT 

ln K298 

H 

G RT 

ln K 

0 

298 1 

- 32900 = - 8.314 (298) ln K 

K 564861.1 

0 

K2 H 

1 1 

we have ln 

K1 

R T2 T1 

 

K2 92220 1 1 

ln 

584861.1 8.314 

 

673 298 

 

 

K 5.75 10 

2 

4 

X 

Acetic acid is esterified in the liquid phase with ethanol at 373 K and 1.0 bar according 

to 

CH3COOH (L) + C2H5OH (L) CH3COOC2H5(L) + H2O (L) 

The feed consists of 1.0 mol each of acetic acid & ethanol, estimate the mole fraction of 

ethyl acetate in the reacting mixture at equilibrium. The standard heat of formation 

and standard free energy of formation at 298 K are given below. 

Assume that the heat of reaction is independent of T and liquid mixture behaves as 

ideal solution. 

1 

1

Δf 0 f 

(J) 

ΔG 0 f 

(J) 

ΔG 0 298 = – 318218 – 237130 + 389900 + 174780 = 9332 J 

We have, ΔG 0 298 = – RT ln K 

ΔH 0 298 = – 463250 – 285830 + 277690 + 484500 = 13110 J 

9332 = – 8.314 (298) ln K1 or K1 = 0.02313 

and for K2 at 373K we have 

Component 

CH3COOH 

(L) 

Moles 

in feed 

C2H5OH 

(L) 

Moles 

reacted 

Moles 

present 

Mole 

fraction 

CH3COOH 1 X 1 – X (1 – X) / 2 

C2H5OH 1 X 1 – X (1 – X) / 2 

CH3COOC2H5 -- X X X / 2 

H2O -- X X X / 2 

2 

CH3COOC2H5(L) H2O (L) 

- 484500 - 277690 - 463250 - 285830 

- 389900 -174780 - 318218 - 237130 

K1 ln 

K2 H 

1 1 

 

R T1 T2 

 

0.02313 12110 1 1 

ln OR K 2 

K 8.314 298 373 

0.067 

K2 0 

K y P 

x 

 

2 

 

 

0.067= 

2 

1 x 

 

2 

 

 

2 

x 

2 

(1) x 

solving, x = 0.252 

x 0.252 

mole fraction of CH3COOC2H 5 0.126 

2 2 

2 

2

The standard free energy change for the reaction C4H8 (g) C4H6 (g) + H2 (g) 

is given by ΔGT 0 =1.03665 X 10 5 – 20.9759T ln T 12.9372 T, where range of ΔGT 0 in J 

/ mol and T in K 

a) Over what range of T, is the reaction promising from thermodynamic view 

point? 

b) For a reaction of pure butene at 800K, calculate equilibrium conversion at 1.0 

bar & 5.0 bar 

c) Repeat part (b) for the feed with the 50 mol% butene and the rest inerts 

a. For the reaction at ΔG = 0, T ≈ 812.4 K, above this temperature ΔG < 

0, reaction is promising below 550 K, ΔG > 0, reaction is unfavorable, 

b. At 800K, ΔG 0 = 1842.16 J / mol 

ΔG 0 = - RT ln K, or 1842.16 = - 8.314 ( 800 ) ln K or 

K = 1.3191 

K = ky P Δn = ky P ( 2- 1 ) = ky P 

Therefore at 1.0 bar K = kyP 1.3191 = ky ( 1 ) 

ky = 1.31910At 5 bar, 1.3191 = ky ( 5 ) 

ky = 0.26382 

Moles in 

feed 

C4H8 1 1 - X 

Moles in product stream yi 

(1 – X ) / (1 + 

C4H6 X X / (1 + X) 

H2 X X / (1 + X) 

1 + X 

X)

x x 

Y Y 

x 

k or 1.3191 = or x = 0.7541 

' ' 2 

C4H6 H2 1 x 1 x 

Y 

 

' 2 

Y 1 x 

C4H8 

1 x 

 

1 x 

Conversion of betene = 75.41% 

2 

x 

At 5.0 bar, k y = 0.26382 = , x = 0.4569, Conversion of betene 

= 45.69 % 

2 

1 x 

C. 

Moles in 

feed 

Moles in product stream 

C4H8 1 1 - X (1 – X ) / (2 + X) 

C4H6 X X / (2 + X) 

H2 X X / (2 + X) 

Inerts 1.0 1 1 / (2 + X) 

2 + x 

2 x 

2 

x 

yi 

2 

x 2 x 

Y 2 

 

at 1.0 bar, k K 1.3191 

1 x 

x = 0.8194 

K 

2 

x 2 x 

Y 2 

 

at 5.0 bar k 0.26382 

5 1 x 

x = 0.5501

Fugacity: It is derived from Latin, expressed as fleetness or escaping ...

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