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

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<strong>Fugacity</strong>: <strong>It</strong> <strong>is</strong> <strong>derived</strong> <strong>from</strong> <strong>Latin</strong>, <strong>expressed</strong> <strong>as</strong> <strong>fleetness</strong> <strong>or</strong> <strong>escaping</strong> tendency. <strong>It</strong> <strong>is</strong> used<br />

to study extensively ph<strong>as</strong>e and chemical reaction equilibrium.<br />

We know that<br />

dG VdP SdT -(1)<br />

F<strong>or</strong> <strong>is</strong>othermal condition<br />

dG VdP<br />

F<strong>or</strong> ideal g<strong>as</strong>es<br />

RT<br />

V <br />

P<br />

RT<br />

dG . dP<br />

P<br />

dG RTd ln P<br />

To find Gibbs free energy f<strong>or</strong> an real g<strong>as</strong>es. True pressure <strong>is</strong> related by effective<br />

pressure. Which we call fugacity(f).<br />

dG RTd ln f -------(2)<br />

Applicable f<strong>or</strong> all g<strong>as</strong>es (ideal <strong>or</strong> real)<br />

On differentiation<br />

G RT ln f <br />

Is an constant depends on temperature and nature of g<strong>as</strong>.<br />

fugacity h<strong>as</strong> same units <strong>as</strong> pressure f<strong>or</strong> an ideal g<strong>as</strong>.<br />

F<strong>or</strong> ideal G<strong>as</strong>es<br />

dG VdP SdT<br />

F<strong>or</strong> Isothermal conditions<br />

dG VdP<br />

<strong>from</strong> equation (2)<br />

RTd ln f VdP<br />

V<br />

d ln f dP<br />

RT<br />

d ln f <br />

dp<br />

p<br />

d ln f d ln p<br />

f p <strong>Fugacity</strong> = Pressure f<strong>or</strong> Ideal g<strong>as</strong>es


<strong>Fugacity</strong> coefficient ( ) : <strong>Fugacity</strong> coefficient ids defined <strong>as</strong> the ratio of fugacity of a<br />

component to its pressure.<br />

f<br />

<br />

P<br />

Is the me<strong>as</strong>ure of non ideal behavi<strong>or</strong> of the g<strong>as</strong>.<br />

Standard State: Pure g<strong>as</strong>es, solids and liquids at temperature of 298k at 1 atmosphere are<br />

said to ex<strong>is</strong>t at standard condition. The property at th<strong>is</strong> condition are known <strong>as</strong> standard<br />

state property and <strong>is</strong> denoted by subscript’o’.<br />

o<br />

G = Standard Gibbs free energy<br />

o<br />

f = Standard fugacity<br />

Estimation of fugacity f<strong>or</strong> g<strong>as</strong>es<br />

I method<br />

dG VdP SdT<br />

F<strong>or</strong> Isothermal conditions<br />

dG VdP<br />

<strong>from</strong> equation (2)<br />

RTd ln f VdP<br />

V<br />

d ln f dP<br />

RT<br />

Integrating the above equation with the limits 0 to f and pressure 1 to P<br />

<br />

d ln f<br />

<br />

P<br />

<br />

1<br />

V<br />

RT<br />

dP<br />

Lower limit <strong>is</strong> taken <strong>as</strong> P =1 atm<br />

At 1 atm <strong>as</strong>suming the g<strong>as</strong>es expected to behave ideally<br />

P<br />

1<br />

ln f VdP<br />

RT 1<br />

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


II method<br />

Using compressibility fact<strong>or</strong><br />

dG VdP<br />

RTd ln f VdP --------------(2)<br />

V in terms of compressibility terms it <strong>is</strong> given <strong>as</strong> V <br />

Substitute V in equation 2<br />

ZRT<br />

RTd ln f dP<br />

p<br />

d ln f Zd ln p<br />

subtracting both sides by dlnP<br />

d ln f d ln P Zd ln P d ln P<br />

f<br />

d ln ( z 1)<br />

d ln P<br />

P<br />

d ln ( Z 1)<br />

d ln P<br />

Integrating the equation <strong>from</strong> 1 to and 0 to P<br />

<br />

<br />

1<br />

d ln<br />

<br />

P<br />

P<br />

<br />

0<br />

( Z 1)<br />

d ln P<br />

dP<br />

ln ( Z 1)<br />

P<br />

0<br />

ZRT<br />

P<br />

Using generalized charts: using reduced properties a similar chart <strong>as</strong> compressibility chart<br />

<strong>is</strong> predicted f<strong>or</strong> fugacity.<br />

f Z 1<br />

ln<br />

dPr<br />

P P<br />

r


<strong>Fugacity</strong> <strong>is</strong> plotted against various reduced pressure at various reduced temperature<br />

Using Residual Volume(): The residual volume <strong>is</strong> the difference between actual<br />

volume (V) and the volume occupied by one mole of g<strong>as</strong> under same temperature and<br />

RT RT<br />

pressure<br />

V , V <br />

P<br />

P<br />

RT <br />

dG <br />

dP<br />

RTd ln f<br />

P <br />

dP <br />

RT dP RTd ln f<br />

RT P <br />

<br />

dP d ln<br />

RT <br />

f<br />

P<br />

<br />

<br />

<br />

f <br />

ln dP<br />

P RT<br />

Problem<br />

F<strong>or</strong> <strong>is</strong>opropanol vap<strong>or</strong> at 200 o C the following equation <strong>is</strong> available<br />

Z=1- 9.86 x 10 -3 P-11.45 x 10 -5 P 2<br />

Where P <strong>is</strong> in bars. Estimate the fugacity at 50 bars and 200 o C<br />

PV<br />

3<br />

5<br />

2<br />

Z 1 9.<br />

86 10<br />

P 11.<br />

4110<br />

P<br />

RT<br />

V<br />

<br />

ln f<br />

ln f<br />

ZRT<br />

p<br />

<br />

<br />

1<br />

RT<br />

1<br />

RT<br />

<br />

<br />

P<br />

<br />

1<br />

P<br />

<br />

1<br />

RT<br />

( 1<br />

P<br />

VdP<br />

<br />

9.<br />

86 10<br />

3<br />

RT<br />

( 1<br />

9.<br />

86 10<br />

P<br />

3<br />

3<br />

9.<br />

86 10<br />

dP <br />

P 11.<br />

4110<br />

5<br />

P 11.<br />

4110<br />

50 50<br />

50<br />

dp<br />

5<br />

ln f<br />

<br />

11.<br />

4110<br />

PdP<br />

1<br />

p<br />

1<br />

1<br />

5<br />

P<br />

2<br />

)<br />

2<br />

P ) dP


50 <br />

<br />

ln f ln<br />

9.<br />

86 10<br />

1 <br />

f=26.744 bar<br />

f 26.<br />

744<br />

0.<br />

5348<br />

P 50<br />

50 1<br />

<strong>Fugacity</strong> f<strong>or</strong> liquids and solids<br />

General expression f<strong>or</strong> fugacity <strong>is</strong><br />

P<br />

3<br />

11.<br />

4110<br />

5<br />

2 2<br />

( 50 1<br />

)<br />

2<br />

1<br />

ln f VdP<br />

RT 1<br />

F<strong>or</strong> solids and liquids at constant temperature the specific volume does not change<br />

appreciably with pressure, theref<strong>or</strong>e the above equation <strong>is</strong> integrated by taking volume<br />

constant. Integrating the above equation <strong>from</strong> condition 1 to 2<br />

f2<br />

ln f <br />

f1<br />

f<br />

V<br />

RT<br />

V<br />

P2<br />

<br />

P1<br />

dP<br />

P <br />

2<br />

ln 2 P1<br />

f1<br />

RT<br />

Problem:<br />

Liquid chl<strong>or</strong>ine at 25 o C h<strong>as</strong> a vapour pressure of 0.77Mpa, fugacity 0.7Mpa and Molar<br />

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

P<br />

1<br />

P<br />

2<br />

6<br />

0.<br />

77 10<br />

Pa<br />

1010<br />

f<br />

V<br />

6<br />

Pa<br />

P <br />

2<br />

ln 2 P1<br />

f1<br />

RT<br />

f=0.846Mpa<br />

f<br />

1<br />

6<br />

0. 7 10<br />

Pa T 298K<br />

R 8314<br />

J<br />

Kgmole


Activity(a):<br />

<strong>It</strong> <strong>is</strong> defined <strong>as</strong> the fugacity of the ex<strong>is</strong>ting condition to the standard state fugacity<br />

f<br />

a o<br />

f<br />

Effect of pressure on activity<br />

The change in Gibbs free energy f<strong>or</strong> a process accompanying change of state <strong>from</strong><br />

standard state at given condition at constant temperature can be predicted <strong>as</strong><br />

G RT ln<br />

f<br />

o<br />

G RT ln<br />

o f <br />

G<br />

G G RT ln RT ln a<br />

o<br />

f<br />

<br />

<br />

<br />

at constant temperature dG VdP<br />

G<br />

dG V <br />

G<br />

P<br />

O o<br />

P<br />

dP<br />

o<br />

o<br />

G V ( P P ) RT ln a V ( P P )<br />

f<br />

o<br />

V<br />

o<br />

ln a ( P P ) Th<strong>is</strong> equation predicts the effect of pressure on activity<br />

RT<br />

Effect of Temperature on Activity<br />

G<br />

G G<br />

o<br />

RT ln a<br />

o<br />

G G<br />

R ln a <br />

T T<br />

Differentiating the above equation with T at constant P<br />

G <br />

<br />

<br />

d ln a T<br />

R<br />

<br />

<br />

dT <br />

P T<br />

<br />

<br />

P<br />

o G <br />

<br />

<br />

<br />

<br />

T <br />

<br />

<br />

<br />

<br />

T<br />

<br />

<br />

<br />

<br />

d ln a <br />

R<br />

<br />

dT <br />

H<br />

2<br />

RT<br />

o<br />

H<br />

2<br />

RT<br />

P<br />

P


d ln a <br />

R<br />

<br />

dT <br />

P<br />

o<br />

H H<br />

Th<strong>is</strong> equation predicts the effect of temperature on activity.<br />

2<br />

RT


Properties of solutions<br />

The relationships f<strong>or</strong> pure component are not applicable to solutions. Which needs<br />

modification because of the change in thermodynamic properties of solution. The<br />

pressure temperature and amount of various constituents determines an extensive state.<br />

The pressure, temperature and composition determine intensive state of a system.<br />

Partial Molar properties:<br />

The properties of a solution are not additive properties, it means volume of solution <strong>is</strong> not<br />

the sum of pure components volume. When a substance becomes a part of a solution it<br />

looses its identity but it still contributes to the property of the solution.<br />

The term partial molar property <strong>is</strong> used to designate the component property when it <strong>is</strong><br />

admixture with one <strong>or</strong> m<strong>or</strong>e component solution.<br />

A mole of component i <strong>is</strong> a particular solution at specified temperature and pressure h<strong>as</strong><br />

got a set of properties <strong>as</strong>sociated with it likeVP , Sietc<br />

. These properties are partially<br />

responsible f<strong>or</strong> the properties of solution and it <strong>is</strong> known <strong>as</strong> partial molar property<br />

<strong>It</strong> <strong>is</strong> defined <strong>as</strong><br />

nM<br />

<br />

M i <br />

ni<br />

<br />

T , P,<br />

n<br />

j<br />

j i<br />

M i = Partial molar property of component i.<br />

M = Any thermodynamic property of the solution<br />

n = Total number moles in a solution<br />

ni=Number of moles of component I in the solution<br />

Th<strong>is</strong> equation defines how the solution property <strong>is</strong> d<strong>is</strong>tributed among the components.<br />

Thus the partial molar properties can be treated exactly <strong>as</strong> if they represented the molar<br />

property of component in the solution.<br />

The above expression <strong>is</strong> applicable only f<strong>or</strong> an extensive property using<br />

We can write<br />

<br />

x<br />

n<br />

<br />

n<br />

i i M n nM i<br />

xi<br />

M<br />

i i M<br />

xi=Mole fraction of component i in the solution.


Me<strong>as</strong>uring of partial molar properties<br />

To understand the meaning of physical molar properties consider a open beaker<br />

containing huge volume of water in one mole of water <strong>is</strong> added to it, the volume incre<strong>as</strong>e<br />

<strong>is</strong> 18x 10 -6 m 3 If the same amount of water <strong>is</strong> added to pure ethanol the volume incre<strong>as</strong>ed<br />

<strong>is</strong> approximately 14 x 10 -6 m 3 th<strong>is</strong> <strong>is</strong> the partial molar volume of H2O in pure ethanol.<br />

The difference in volume can explain the volume applied by water molecules depending<br />

on water molecules surrounding to them. When water <strong>is</strong> added to large amount of<br />

ethanol, ethanol molecules hence volume occupied surround all the water molecules will<br />

be different in ethanol.<br />

If same quantity of water <strong>is</strong> added to an equimolar mixture of H2O and ethanol, the<br />

volume change will be different. Theref<strong>or</strong>e Partial molar property change with<br />

composition. The intermolecular f<strong>or</strong>ces also changes with change in thermodynamic<br />

property.<br />

Let V w = Partial molar volume of the water in ethanol water solution<br />

V w = Molar volume of pure water at same temperature and pressure<br />

t<br />

V =Total volume of solution when water added to ethanol water mixture and allowed<br />

f<strong>or</strong> sufficient time so that the temperature remains constant<br />

<br />

V<br />

t<br />

V <br />

w<br />

V<br />

<br />

n<br />

n<br />

t<br />

w<br />

w<br />

V<br />

w<br />

In a process a finite quantity of water <strong>is</strong> added which causes finite change in composition.<br />

V w = property of solution f<strong>or</strong> all infinitely small amount of water.<br />

Vw limv0<br />

V<br />

n<br />

t<br />

w<br />

V<br />

<br />

n<br />

t<br />

w<br />

<br />

<br />

<br />

Temperature pressure an number of moles of ethanol remains constant during addition of<br />

water.<br />

t V<br />

<br />

Vw<br />

nE- no of moles of ethanol<br />

nw<br />

<br />

T , P,<br />

nE<br />

The partial molar volume of component i<br />

V<br />

i<br />

V<br />

<br />

n<br />

i<br />

t<br />

<br />

<br />

<br />

T ,<br />

P,<br />

n i<br />

j


Partial molar properties and properties of the solution<br />

Consider any thermodynamic extensive property (V i, Gi etc) f<strong>or</strong> homogenous system can<br />

be determined by knowing the temperature, pressure and various amount of constituents.<br />

Let total property of the solution<br />

M nM<br />

t <br />

n n1<br />

n2<br />

n3<br />

<br />

1,2,3 represents no of constituents<br />

f T,<br />

P,<br />

n , n , n n<br />

Thermodynamic property <strong>is</strong> a <br />

1<br />

2<br />

F<strong>or</strong> small change in the pressure and temperature and amount of various constituents can<br />

be written <strong>as</strong><br />

t<br />

t<br />

t<br />

t<br />

t M<br />

M<br />

M<br />

<br />

M<br />

<br />

dM dP dT dn1<br />

dni<br />

<br />

P<br />

P<br />

P<br />

<br />

P<br />

T , n<br />

p,<br />

n<br />

T , p,<br />

n n <br />

<br />

2 , 3<br />

P,<br />

T , n j i<br />

At constant temperature and pressure dP and dT are equal to zero.<br />

The above equation reduces to<br />

i n t<br />

t M<br />

dM dni<br />

i ni<br />

P T n j i<br />

<br />

<br />

<br />

1<br />

. ,<br />

dM t in terms of partial molar property<br />

n<br />

t<br />

dM <br />

i1<br />

M<br />

i<br />

dn<br />

i<br />

M i <strong>is</strong> an extensive property depends on composition and relative amount of constituents.<br />

All constituent properties at constant temperature and pressure are added to give the<br />

property of the solution.<br />

dM M 1dn1<br />

M 2dn2<br />

M 3dn3<br />

<br />

t<br />

dM M x M x M x<br />

dn<br />

t<br />

<br />

M<br />

M t<br />

<br />

<br />

3<br />

1 1 2 2 3 3<br />

M 1x1<br />

M 2 x2<br />

M 3x<br />

3 n<br />

M 1n1<br />

M 2n<br />

M 2 3n<br />

3<br />

t<br />

niM<br />

i<br />

Problem<br />

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

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

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

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

m 3 /mol respectively.<br />

j


Methanol =0.3 mole fraction<br />

Water=0.7 mole fraction<br />

V t =0.3 x38.36x10 -6 +0.7x17.765x10 -6<br />

=24.025x10 -6 m 3 /mol<br />

F<strong>or</strong> 2 m 3 soolution<br />

2<br />

3<br />

83.<br />

246 10<br />

mol<br />

6<br />

24.<br />

02510<br />

Number of moles of methanol in 2m 3 solution<br />

=83.246x10 3 x0.3= 24.97x10 3 mol<br />

Number of moles of water in 2m 3 solution<br />

=83.246x10 3 x07= 58.272x10 3 mol<br />

Volume of pure methanol to be taken<br />

= 24.97x10 3 x 40.7x10 -3 =1.0717 m 3<br />

Volume of pure water to be taken<br />

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


Estimation of Partial molar properties f<strong>or</strong> a binary mixture :<br />

Two methods f<strong>or</strong> estimation<br />

Analytical Method and Graphical Method(Tangent Intercept method)<br />

Analytical Method: The general relation between partial molar property and molar<br />

property of the solution <strong>is</strong> given by<br />

M<br />

M i M xk<br />

x <br />

K T , P,<br />

nk<br />

<br />

<br />

<br />

<br />

<br />

F<strong>or</strong> binary mixture<br />

i 1, k 2<br />

k i x1<br />

M<br />

M 1 M x2<br />

x <br />

2 T , P<br />

<br />

<br />

<br />

<br />

2 x1<br />

1<br />

<br />

At constant Temperature and pressure<br />

M<br />

<br />

M <br />

<br />

<br />

<br />

1 M ( 1<br />

x1)<br />

------(a)<br />

x1<br />

<br />

M<br />

<br />

M <br />

<br />

<br />

<br />

2 M x<br />

1<br />

-----------(b)<br />

x1<br />

<br />

x , x2 1 x1<br />

, x2<br />

x1<br />

The partial molar property (e xtensive property) f<strong>or</strong> a binary mixture can be estimated<br />

<strong>from</strong> the property of solution using above equations (a) and (b).<br />

Tangent Intercept method:<br />

Th<strong>is</strong> <strong>is</strong> the graphical method to estimate partial molar properties. If the partial molar<br />

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


Suppose partial molar properties of components required at any composition, and then<br />

draw a tangent to th<strong>is</strong> point to the curve. The intercept of the tangent with two ax<strong>is</strong> x1=1<br />

and x1=0 are I1 and I2.<br />

Slope of the tangent<br />

M I x<br />

2<br />

2<br />

I M x<br />

1<br />

1<br />

dM<br />

dx<br />

1<br />

dM<br />

dx<br />

1<br />

dM M I<br />

<br />

dx x<br />

Comparing I2 with equation (b) 2 2 M I <br />

dM I1<br />

M<br />

and also (right hand side) <br />

dx1<br />

1 x1<br />

I M 1<br />

x<br />

1<br />

I <br />

1<br />

dM<br />

dx<br />

1<br />

dM<br />

1<br />

1 M 1<br />

x1<br />

Comparing with equation(a) 1 1<br />

dx1<br />

M I <br />

The intercept of the tangent gives the partial molar properties.<br />

1<br />

2


Limiting c<strong>as</strong>es: F<strong>or</strong> infinite dilution of component when at x1=0 a tangent <strong>is</strong> drawn at<br />

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

tangent <strong>is</strong> drawn at x2=0 <strong>or</strong> x1=1 will give infinite dilution M ) of component 2<br />

M 2<br />

<br />

Problem<br />

The Gibbs free energy of a binary solution <strong>is</strong> given by<br />

G x 150x<br />

x x ( 10x<br />

x )<br />

100 1<br />

2 1 2 1 2<br />

cal<br />

mol<br />

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

dilution.<br />

(b) Find the pure component properties<br />

Sol: G x 150x<br />

x x ( 10x<br />

x )<br />

100 1<br />

2 1 2 1 2<br />

Substitute 1<br />

3<br />

x2 1 x and simplifying<br />

G 9x1<br />

8x1<br />

49x1<br />

150<br />

G<br />

1<br />

1<br />

G<br />

<br />

G ( 1<br />

x <br />

<br />

<br />

<br />

1)<br />

x1<br />

<br />

G<br />

2<br />

27x1<br />

16x1<br />

49<br />

x<br />

G<br />

G<br />

G<br />

1<br />

2<br />

2<br />

3<br />

2<br />

2<br />

18x1 35x1<br />

16x1<br />

101<br />

G<br />

<br />

G x <br />

<br />

<br />

1<br />

x1<br />

<br />

3<br />

1<br />

18x<br />

8x<br />

2<br />

1<br />

150<br />

cal<br />

mol<br />

To find the partial molar properties of components 1 and 2<br />

x2=0.8, x1=1-0.8 = 0.2<br />

( 2<br />

<br />

( 1


G<br />

1<br />

3<br />

2<br />

18x1 35x1<br />

16x1<br />

101<br />

G 102.<br />

944<br />

G<br />

1 <br />

2<br />

3<br />

1<br />

cal<br />

mol<br />

18x<br />

8x<br />

G 149.<br />

824<br />

2 <br />

2<br />

1<br />

cal<br />

mol<br />

At infinite dilution<br />

G<br />

1<br />

<br />

G atx<br />

1<br />

<br />

G1 101<br />

1<br />

cal<br />

mol<br />

0<br />

150<br />

G G atx 1 <strong>or</strong> x2=0<br />

2<br />

<br />

2<br />

<br />

G2 160<br />

1<br />

cal<br />

mol<br />

To find the pure component property<br />

G<br />

1<br />

G atx<br />

1<br />

1<br />

cal<br />

G1 100<br />

mol<br />

G<br />

2<br />

G atx<br />

2<br />

1<br />

cal<br />

G1 150<br />

mol<br />

1<br />

0<br />

Chemical Potential:


<strong>It</strong> <strong>is</strong> widely used <strong>as</strong> a thermodynamic property. <strong>It</strong> <strong>is</strong> used <strong>as</strong> a index in chemical<br />

equilibrium, same <strong>as</strong> pressure and temperature. The chemical potential i of component i<br />

in a solution <strong>is</strong> same <strong>as</strong> its partial molar free energy in the solution i G<br />

The chemical potential of component i<br />

i<br />

G<br />

t<br />

G<br />

dG<br />

i<br />

t G<br />

<br />

<br />

ni<br />

<br />

T , P,<br />

n<br />

<br />

, , n n T P<br />

1 2 , f<br />

t<br />

t G<br />

<br />

<br />

P<br />

<br />

T , n<br />

j<br />

t G<br />

<br />

dP <br />

T<br />

<br />

P , n<br />

dT <br />

i n<br />

i<br />

1<br />

t G<br />

<br />

ni<br />

t<br />

t<br />

t G<br />

G<br />

<br />

dG dP dT i<br />

dni<br />

P<br />

T<br />

<br />

T , n<br />

P,<br />

n<br />

F<strong>or</strong> closed system there will be no exchange of constituents (n <strong>is</strong> constant)<br />

dG<br />

t<br />

t t<br />

V dP S dT<br />

at constant temperature<br />

G<br />

<br />

<br />

V<br />

P<br />

<br />

<br />

T<br />

at constant pressure<br />

G<br />

<br />

<br />

T<br />

<br />

<br />

P<br />

t<br />

S<br />

t<br />

<br />

t t t<br />

dG V dP S dT i<br />

dni<br />

At constant temperature and pressure<br />

t T ,<br />

P <br />

dG <br />

idni<br />

<br />

<br />

<br />

T , n<br />

j<br />

dn<br />

i


F<strong>or</strong> binary solution G x11<br />

x2<br />

2<br />

Effect of temperature and pressure on chemical potential<br />

Effect of temperature:<br />

We know that<br />

t G<br />

<br />

i Gi<br />

-------------------(1)<br />

ni<br />

<br />

T , P,<br />

n<br />

j<br />

differentiating equation (1) with respect to T at constant P<br />

2<br />

<br />

i G<br />

<br />

T<br />

<br />

Tdn<br />

P,<br />

n<br />

dG VdP SdT ---------(3)<br />

i<br />

---------------------(2)<br />

differentiating equation (3) with respect to T at constant P<br />

G<br />

<br />

<br />

T<br />

<br />

<br />

P<br />

S<br />

differentiating again w r t ni<br />

2<br />

t<br />

G S<br />

<br />

<br />

<br />

Tni<br />

ni<br />

<br />

P,<br />

n<br />

j<br />

S<br />

Si <strong>is</strong> partial molar entropy of component I<br />

i<br />

<br />

<br />

<br />

T<br />

<br />

<br />

T<br />

<br />

<br />

<br />

<br />

<br />

P,<br />

n<br />

T Si<br />

i<br />

2<br />

<br />

T<br />

G H TS<br />

i<br />

<br />

T<br />

<br />

T<br />

<br />

<br />

2<br />

T<br />

In terms of partial molar properties<br />

i<br />

i


G H TS<br />

<br />

i<br />

i<br />

H<br />

i<br />

H<br />

i<br />

i<br />

<br />

<br />

i <br />

<br />

T<br />

<br />

<br />

TS<br />

i<br />

P,<br />

n<br />

i<br />

i<br />

TS<br />

H<br />

<br />

T<br />

i<br />

i<br />

2<br />

Th<strong>is</strong> equation represents the effect of temperature on chemical potential.<br />

Effect of Pressure:<br />

We know that<br />

t G<br />

<br />

i Gi<br />

-------------------(4)<br />

ni<br />

<br />

T , P,<br />

n<br />

j<br />

differentiating equation (4) with respect to T at constant P<br />

2<br />

<br />

i G<br />

<br />

P<br />

<br />

Pdn<br />

T , n<br />

dG VdP SdT ---------(3)<br />

i<br />

---------------------(5)<br />

differentiating equation (3) with respect to P at constant T<br />

G<br />

<br />

<br />

P<br />

<br />

<br />

T<br />

V<br />

differentiating again w r t ni<br />

2<br />

G<br />

Pn<br />

i<br />

V<br />

<br />

<br />

ni<br />

<br />

T , n<br />

j<br />

V ,<br />

i


i <br />

<br />

P<br />

<br />

<br />

T , n<br />

V<br />

i<br />

Th<strong>is</strong> equation represents the effect of pressure on chemical potential<br />

<strong>Fugacity</strong> in solutions<br />

F<strong>or</strong> pure fluids fugacity <strong>is</strong> explained <strong>as</strong><br />

dG RTd ln<br />

f<br />

lim P0<br />

1<br />

P<br />

f<br />

The fugacity of the component i in the solution <strong>is</strong> defined <strong>as</strong> analogously by<br />

d RTd ln f<br />

i<br />

lim<br />

P0<br />

f<br />

P<br />

i<br />

i<br />

i<br />

1<br />

i <strong>is</strong> Chemical potential<br />

f i <strong>is</strong> partial molar fugacity<br />

F<strong>or</strong> ideal g<strong>as</strong>es Pi yi<br />

PT<br />

PT – Total pressure.<br />

<strong>Fugacity</strong> in G<strong>as</strong>eous solutions<br />

We know that<br />

t G<br />

<br />

i Gi<br />

------(1)<br />

ni<br />

<br />

T , P,<br />

n<br />

j<br />

differentiating equation (1) with respect to T at constant P<br />

2<br />

<br />

i G<br />

<br />

P<br />

-----(2)<br />

Pdn<br />

T , n<br />

i


dG VdP SdT -----(3)<br />

differentiating equation (3) with respect to P at constant T<br />

G<br />

<br />

<br />

P<br />

<br />

<br />

T<br />

V<br />

differentiating again w r t ni<br />

2<br />

G<br />

Pn<br />

i<br />

<br />

i <br />

<br />

P<br />

<br />

<br />

<br />

i<br />

<br />

V<br />

<br />

<br />

ni<br />

<br />

T , n<br />

V i<br />

V<br />

P<br />

i<br />

T , n<br />

j<br />

V ,<br />

i<br />

RTd f i VidP<br />

ln<br />

Subtracting both sides by i P<br />

Vi<br />

d ln fi<br />

dP<br />

RT<br />

d ln<br />

i<br />

d ln f i d ln P i dP d ln<br />

<br />

f <br />

<br />

Pi<br />

<br />

V<br />

d ln P d ln P d ln y<br />

i<br />

V<br />

RT<br />

i i<br />

d ln dP d ln<br />

RT<br />

Composition <strong>is</strong> constant dln yi=0<br />

The above equation can be written <strong>as</strong><br />

d ln Pi<br />

d ln P <br />

P<br />

dP<br />

P<br />

i<br />

i<br />

P<br />

i


Modifying equation (a)<br />

RT <br />

V<br />

dP<br />

1<br />

d ln i<br />

i<br />

RT <br />

i - Represents fugacity of the component i in the solution.<br />

Ideal solutions<br />

An ideal mixture <strong>is</strong> one, which there <strong>is</strong> no change in volume due to mixing. In other<br />

w<strong>or</strong>ds f<strong>or</strong> an ideal g<strong>as</strong>eous mixture partial molar volume of each component will be equal<br />

to its pure component volume at same temperature and pressure.<br />

i i V V and Vi xi<br />

xiVi<br />

P<br />

V --- Raoults law<br />

Ideal solutins are f<strong>or</strong>med when similar components <strong>or</strong> adjacent groups of group are<br />

mixed i i V V <br />

Eg: Benzene-Toluene<br />

Methanol- Ethanol<br />

Hexane- Heptane<br />

Solutin undergo change in volume due to mixing are known <strong>as</strong> non ideal solutions<br />

i i V V <br />

Eg: Methanol-Water<br />

Ethanol-water.<br />

Ideal solutions f<strong>or</strong>med when the intermolecular f<strong>or</strong>ce between like molecules and unlike<br />

molecules are of the same magnitude.<br />

Non-ideal solutions are f<strong>or</strong>med when intermolecular f<strong>or</strong>ces between like molecules and<br />

unlike molecules of different magnitude.<br />

F<strong>or</strong> ideal solutions<br />

t<br />

V niVi<br />

-------------------(1)<br />

Vi <strong>is</strong> the molar volume of pure component I<br />

t V<br />

<br />

Vi T<br />

, P , n j Vi<br />

----------------- (2)<br />

ni<br />

<br />

<br />

The residual volume f<strong>or</strong> the pure component <strong>is</strong><br />

RT<br />

Vi <br />

p<br />

Theref<strong>or</strong>e we know <strong>from</strong> reduced properties<br />

P<br />

f i 1 RT <br />

ln Vi<br />

dP ----(3)<br />

P RT P<br />

0 <br />

F<strong>or</strong> component i in terms of partial molar properties


P<br />

f i 1 RT <br />

ln V<br />

i dP ----(4)<br />

P RT i P<br />

0 <br />

Subtracting equation (3) <strong>from</strong> (4)<br />

V i V <br />

P<br />

f i P 1<br />

ln i dP ----(5)<br />

f P RT<br />

i<br />

i<br />

we know that Pi yi<br />

P <br />

equation (5) reduces to<br />

P<br />

f i 1<br />

ln V i Vi<br />

dP<br />

f y RT<br />

i<br />

i<br />

comparing equation (2)<br />

f i<br />

1<br />

f y<br />

i<br />

i<br />

0<br />

0<br />

f i f i yi<br />

--Lew<strong>is</strong> Randal rule<br />

Lew<strong>is</strong> Randal rule <strong>is</strong> applicable f<strong>or</strong> evaluating fugacity of components in g<strong>as</strong> mixture.<br />

Lew<strong>is</strong> Randal rule <strong>is</strong> valid f<strong>or</strong><br />

1. At low pressure when g<strong>as</strong> behaves ideally.<br />

2. When Physical properties are nearly same.<br />

3. At any pressure if component present in exess.<br />

Henrys Law<br />

Th<strong>is</strong> law <strong>is</strong> applicable f<strong>or</strong> small concentration ranges. F<strong>or</strong> ideal solution Henrys law <strong>is</strong><br />

given <strong>as</strong><br />

f x k<br />

i<br />

P x k<br />

i<br />

i<br />

i<br />

i<br />

i<br />

ki- Henrys Constant, f i -Partial molar fugacity,<br />

Pi -Partial pressure of component i.<br />

Non Ideal solutions<br />

F<strong>or</strong> ideal solutions f i xik<br />

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


F<strong>or</strong> non ideal solutions f i i xik<br />

i ----------(b)<br />

Where i <strong>is</strong> an activity coefficient of component i.<br />

Comparing equation (a) and (b)<br />

f i<br />

i Non ideal solution / Ideal solution<br />

f<br />

i<br />

F<strong>or</strong> both ideal and Non ideal solutions the fugacity of solution <strong>is</strong> given by<br />

equation<br />

f i<br />

ln f xi<br />

ln<br />

x<br />

<br />

ln<br />

ln<br />

xi i<br />

i<br />

Problem:<br />

A terinary g<strong>as</strong> mixture contains 20mole% A 35mole% B and 45mole% C at 60 atm and<br />

75 o C. The fugacity coefficients of A,B and C in th<strong>is</strong> mixture are 0.7,, 0.6 and 0.9.<br />

Calculate the fugacity of the mixture.<br />

Solution:<br />

ln x ln<br />

x ln<br />

x ln<br />

A<br />

A<br />

ln 0.<br />

2ln(<br />

0.<br />

7)<br />

0.<br />

35ln(<br />

0.<br />

6)<br />

<br />

ln 0.<br />

2975<br />

0.<br />

7426<br />

f<br />

, f =44.558atm<br />

P<br />

Gibbs Duhem Equation<br />

B<br />

B<br />

c<br />

c<br />

0.<br />

45ln(<br />

0.<br />

9)<br />

Consider a multi component solution having ni moles of component I the property of<br />

solution be M in terms of partial molar properties<br />

t<br />

nM nM<br />

-----------------------(1).<br />

M <br />

i<br />

Where n <strong>is</strong> the total no of moles of solution<br />

Differentiating eq (1) we get<br />

d( nm)<br />

nid<br />

M i<br />

We know that<br />

M idni<br />

----------------------------(2)<br />

<br />

, , n n P T f nM<br />

<br />

1 2 ,


(<br />

nM ) <br />

(<br />

nM ) (<br />

nM ) <br />

(<br />

nM ) <br />

(<br />

nM ) <br />

<br />

dT <br />

<br />

dP dn1<br />

dn<br />

T<br />

P,<br />

n , n <br />

P<br />

T , n , n <br />

n1<br />

<br />

n2<br />

<br />

n(<br />

M ) <br />

(<br />

nM ) <br />

T<br />

<br />

1<br />

2<br />

P,<br />

n , n <br />

1<br />

2<br />

n(<br />

M ) <br />

dT <br />

P<br />

<br />

From the definition of partial molar properties<br />

1<br />

2<br />

T , n , n <br />

1<br />

2<br />

T , P1<br />

, n2.<br />

n3<br />

(<br />

nM ) <br />

dP <br />

ni<br />

<br />

T , P,<br />

n<br />

n(<br />

M ) n(<br />

M ) <br />

(<br />

nM ) <br />

<br />

dT <br />

<br />

dP M idni<br />

----------(3)<br />

T<br />

P<br />

<br />

P,<br />

n , n <br />

Subtracting equation (2) <strong>from</strong> equation (3)<br />

n(<br />

M ) <br />

<br />

T<br />

<br />

<br />

n(<br />

M ) <br />

dT <br />

<br />

P<br />

<br />

<br />

dP <br />

P,<br />

n , n <br />

1<br />

2<br />

1<br />

2<br />

T , n , n <br />

1<br />

2<br />

T , n , n <br />

1<br />

<br />

2<br />

n d M<br />

Equatin (4) <strong>is</strong> the fundamental f<strong>or</strong>m of Gibbs Duhem equation<br />

i<br />

i<br />

j<br />

dn<br />

0 ------------------(4)<br />

i<br />

T , P,<br />

n1<br />

, n3<br />

<br />

Special C<strong>as</strong>e<br />

At constant temperature and pressure dT and dP are equal to zero. The equation becomes<br />

i i 0 M d x<br />

<br />

F<strong>or</strong> binary solution at constant temperature and pressure the equation becomes<br />

x d M x d M 0<br />

1<br />

1<br />

2<br />

2<br />

x d M 1<br />

x ) d M 0 ------------(5)<br />

1<br />

1<br />

( 1 2<br />

dividing equation(5) by dx1<br />

x<br />

1<br />

d M<br />

dx<br />

1<br />

1<br />

<br />

( 1<br />

x<br />

1<br />

)<br />

d M<br />

dx<br />

1<br />

2<br />

0<br />

The above equation <strong>is</strong> Gibbs Duhem equation f<strong>or</strong> binary solution at constant temperature<br />

and pressure in terms of Partial molar properties.<br />

Any Data <strong>or</strong> equation on Partial molar properties must sat<strong>is</strong>fy Gibbs Duhem<br />

equation.<br />

2


Problem:<br />

Find weather the equation given below <strong>is</strong> thermodynamically cons<strong>is</strong>tent<br />

G x 150x<br />

x x ( 10x<br />

x )<br />

G<br />

G<br />

G<br />

G<br />

1<br />

2<br />

1<br />

2<br />

dG<br />

dx<br />

100 1<br />

2 1 2 1 2<br />

G<br />

G ( 1<br />

x1)<br />

x1<br />

G<br />

G x1<br />

x<br />

3<br />

1<br />

2<br />

18x1 35x1<br />

16x1<br />

101<br />

3<br />

1<br />

18x<br />

8x<br />

1<br />

1<br />

2<br />

2<br />

1<br />

150<br />

54x1 70x1<br />

16<br />

dG<br />

2<br />

2<br />

54x1 16x1<br />

dx1<br />

G D equation<br />

x<br />

1<br />

d M<br />

dx<br />

1<br />

1<br />

2<br />

<br />

( 1<br />

x<br />

1<br />

)<br />

d M<br />

dx<br />

1<br />

2<br />

0<br />

x1 ( 54 x1<br />

70 x1<br />

16 ) ( 1 x1<br />

)( 54 x1<br />

16 x1<br />

) 0<br />

<strong>It</strong> sat<strong>is</strong>fies the GD equation, the above equation <strong>is</strong> cons<strong>is</strong>tent.<br />

2


Ph<strong>as</strong>e Equilibrium<br />

Criteria f<strong>or</strong> ph<strong>as</strong>e equilibrium: If a system says to be in thermodynamic equilibrium,<br />

Temperature, pressure must be constant and there should not be any m<strong>as</strong>s transfer.<br />

The different criteria f<strong>or</strong> ph<strong>as</strong>e equilibrium are<br />

At Constant U and V: An <strong>is</strong>olated system do not exchange m<strong>as</strong>s <strong>or</strong> heat <strong>or</strong> w<strong>or</strong>k with<br />

surroundings. Theref<strong>or</strong>e dQ=0, dW=0 hence dU=0. A perfectly insulated vessel at<br />

constant volume dU=0 and dV=0.<br />

dS<br />

U , V <br />

0<br />

At Constant T and V: Helmoltz free energy <strong>is</strong> given by the expression<br />

A U TS<br />

U A TS<br />

on differentiating<br />

dU dA TdS SdT<br />

we know that<br />

dU TdS PdV<br />

TdS PdV dA TdS SdT<br />

dA PdV<br />

SdT<br />

Under the restriction of constant temperature and volume the equation simplifies to<br />

dAT<br />

, V <br />

0<br />

At Constant P and T<br />

The equation defines Gibbs free energy<br />

G H TS<br />

G U PV TS<br />

dG dU PdV VdP TdS SdT<br />

dU dG PdV VdP<br />

TdS SdT<br />

dU dG dG PdV TdS<br />

Under the restriction of constant Pressure and Temperature the equation simplifies to<br />

dGT<br />

, P <br />

0<br />

Th<strong>is</strong> means the Gibbs free energy decre<strong>as</strong>es <strong>or</strong> remains un altered depending on the<br />

reversibility and the irreversibility of the process. <strong>It</strong> implies that f<strong>or</strong> a system at<br />

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


Ph<strong>as</strong>e Equilibrium in single component system<br />

Consider a thermodynamic equilibrium system cons<strong>is</strong>ting of two <strong>or</strong> m<strong>or</strong>e ph<strong>as</strong>es of a<br />

single substance. Though the individual ph<strong>as</strong>es can exchange m<strong>as</strong>s with each other.<br />

Consider equilibrium between vap<strong>or</strong> and liquid ph<strong>as</strong>es f<strong>or</strong> a single substance at constant<br />

temperature and pressure. Appling the criteria of equilibrium<br />

dG 0<br />

a<br />

dG<br />

b<br />

dG 0<br />

a<br />

dG and<br />

b<br />

dG are chane in free energies of the ph<strong>as</strong>es a and b respectively.<br />

We know that<br />

i i dn G SdT VdP dG <br />

F<strong>or</strong> ph<strong>as</strong>e a<br />

a<br />

dG<br />

a a a a a a<br />

V dP S dT G dn<br />

F<strong>or</strong> ph<strong>as</strong>e b<br />

b<br />

dG<br />

b b b b b b<br />

V dP S dT G dn<br />

At constant temperature and pressure<br />

a<br />

dG<br />

a a<br />

G dn ,<br />

b<br />

dG<br />

b b<br />

G dn<br />

F<strong>or</strong> a whole system to be at equilibrium<br />

a<br />

dn<br />

b<br />

dn 0<br />

<strong>or</strong><br />

dn<br />

a<br />

dn<br />

a b a<br />

G G dn 0<br />

b<br />

a b<br />

G G<br />

When two ph<strong>as</strong>es are in equilibrium at constant temperature and pressure, Gibbs free<br />

energies must be same in each ph<strong>as</strong>e f<strong>or</strong> equilibrium.<br />

a<br />

b<br />

RT ln f C RT ln f C<br />

a<br />

f <br />

f<br />

b<br />

Clapeyron Clausius Equation<br />

Clapeyron Clausius Equation are developed two ph<strong>as</strong>es when they are in equilibrium<br />

(a) Solid- liquid<br />

(b) Liquid-Vap<strong>or</strong><br />

(c) Solid Vap<strong>or</strong>


dP<br />

dT<br />

dP<br />

dT<br />

Consider any two ph<strong>as</strong>es are in equilibrium with each other at given temperature and<br />

pressure. <strong>It</strong> <strong>is</strong> possible to transfer some amount of substance <strong>from</strong> one ph<strong>as</strong>e to other<br />

in a thermodynamically reversible manner(infinitely slow).The equal amount of<br />

substance will have same free energies at equilibrium .<br />

Consider GA <strong>is</strong> Gibbs free energy in ph<strong>as</strong>e A and GB <strong>is</strong> Gibbs free energy in ph<strong>as</strong>e B<br />

at equilibrium.<br />

GA=GB ------(1)<br />

G = GA -GB =0 ------(2)<br />

At new temperature and pressure the free energy / mole of substance in ph<strong>as</strong>e A <strong>is</strong><br />

G A dG A f<strong>or</strong> Ph<strong>as</strong>e B GB dGB<br />

From b<strong>as</strong>ic equation<br />

dG VdP SdT -----(3)<br />

dG A V AdP<br />

S AdT<br />

dGB VBdP<br />

S BdT<br />

VAdP S AdT<br />

VBdP<br />

S BdT<br />

V V<br />

dTS<br />

S <br />

dP <br />

B<br />

A<br />

B<br />

A<br />

dP S<br />

<br />

dT V<br />

V represents the change in volume when one mole of substance p<strong>as</strong>es <strong>from</strong><br />

ph<strong>as</strong>e A to Ph<strong>as</strong>e B<br />

Q<br />

S<br />

<br />

T<br />

Q<br />

<br />

TV<br />

Q<br />

----(4)<br />

T<br />

V V<br />

<br />

B<br />

A<br />

Th<strong>is</strong> <strong>is</strong> a b<strong>as</strong>ic equation of Clapey<strong>or</strong>n Cl<strong>as</strong>ius equation.<br />

B—Vap<strong>or</strong> state A- Liquid state<br />

Q=molar heat of vap<strong>or</strong>ization = V H <br />

VB <strong>is</strong> molar volume in vap<strong>or</strong> state, VA <strong>is</strong> molar volume in liquid state,


dP<br />

dT<br />

T<br />

<br />

Q<br />

V V <br />

g<br />

l<br />

Vg>>>Vl (G<strong>as</strong> volume <strong>is</strong> very high when compared to liquid volume)<br />

dP H<br />

<br />

dT TV<br />

V<br />

g<br />

H V (Latent heat of vap<strong>or</strong>ization),<br />

dP P<br />

<br />

dT TRT<br />

dP dT<br />

2<br />

P R T<br />

V g <br />

RT<br />

P<br />

Integrating the above equation <strong>from</strong> T1to T2 and pressure <strong>from</strong> P1 to P2.<br />

P<br />

2<br />

dP <br />

<br />

P R<br />

P<br />

1<br />

P<br />

ln<br />

P<br />

2<br />

1<br />

1<br />

<br />

R T1<br />

<br />

1<br />

T<br />

2<br />

<br />

<br />

<br />

Th<strong>is</strong> equation <strong>is</strong> used to calculate the vap<strong>or</strong> pressure at any desired temperature.<br />

Problem:<br />

The vap<strong>or</strong> pressure of water at 100 o C <strong>is</strong> 760mmHg. What will be the vap<strong>or</strong> pressure<br />

at 95 o C. The latent heat of vap<strong>or</strong>ization of water at th<strong>is</strong> temperature range <strong>is</strong><br />

41.27KJ/mole.<br />

P 760mmHg<br />

P<br />

1<br />

2<br />

T<br />

<br />

T<br />

2<br />

1<br />

<br />

dT<br />

T<br />

T<br />

T<br />

1<br />

2<br />

100 273 <br />

95 273 <br />

373K<br />

368K


P<br />

ln<br />

P<br />

2<br />

1<br />

P2<br />

<br />

ln<br />

<br />

760<br />

<br />

1<br />

<br />

R T1<br />

<br />

1<br />

T<br />

2<br />

<br />

<br />

<br />

41.<br />

27 10<br />

8.<br />

314<br />

P2 634.<br />

3mmHg<br />

3<br />

1<br />

<br />

373<br />

<br />

1 <br />

368 <br />

<br />

Ph<strong>as</strong>e Equilibrium in Multi component system<br />

An heterogeneous system contains two <strong>or</strong> m<strong>or</strong>e ph<strong>as</strong>es and each ph<strong>as</strong>e contains two <strong>or</strong><br />

m<strong>or</strong>e components in different prop<strong>or</strong>tions. Theref<strong>or</strong>e it <strong>is</strong> necessary to develop ph<strong>as</strong>e<br />

equilibrium f<strong>or</strong> multi component system in terms of chemical potential.<br />

The partial molar free energy <strong>or</strong> chemical potential <strong>is</strong> given <strong>as</strong><br />

G<br />

<br />

i<br />

Gi<br />

<br />

ni<br />

T , P,<br />

n j<br />

F<strong>or</strong> a system to be in equilibrium with respect to m<strong>as</strong>s transfer the driving f<strong>or</strong>ce f<strong>or</strong><br />

m<strong>as</strong>s transfer( Chemical potential) must have unif<strong>or</strong>m values f<strong>or</strong> each component in<br />

all ph<strong>as</strong>es.<br />

Consider a heterogeneous system cons<strong>is</strong>ts of ph<strong>as</strong>e , , and<br />

containing various components 1,2,3-------C , that constitutes the system.<br />

k<br />

i<br />

The symbol represents chemical potential of component “i”in ph<strong>as</strong>e k.<br />

At constant temperature and pressure, f<strong>or</strong> the criterion f<strong>or</strong> equilibrium <strong>is</strong><br />

dG=0 ----(1)<br />

Free energy f<strong>or</strong> multi component system given by the expression<br />

SdT VdP dG ------(2)<br />

<br />

i i dn<br />

At constant Temperature and pressure<br />

dG ----(3)<br />

i i dn<br />

Comparing equation 1 and 3<br />

0<br />

<br />

i i dn<br />

F<strong>or</strong> multi component system<br />

C<br />

<br />

<br />

i1 k <br />

<br />

k k<br />

i i dn<br />

0<br />

Expanding the above equation


1<br />

<br />

2<br />

<br />

c<br />

dn<br />

dn<br />

dn<br />

<br />

1<br />

<br />

2<br />

<br />

c<br />

<br />

<br />

<br />

1<br />

<br />

<br />

2<br />

<br />

c<br />

dn<br />

dn<br />

<br />

1<br />

dn<br />

<br />

2<br />

<br />

c<br />

<br />

<br />

1<br />

<br />

<br />

<br />

2<br />

<br />

c<br />

dn<br />

<br />

1<br />

dn<br />

dn<br />

<br />

<br />

2<br />

<br />

c<br />

<br />

<br />

<br />

<br />

1<br />

<br />

<br />

dn<br />

<br />

2<br />

<br />

c<br />

<br />

1<br />

dn<br />

dn<br />

<br />

<br />

2<br />

<br />

c<br />

<br />

0<br />

-----(4)<br />

Since the whole system <strong>is</strong> closed it should sat<strong>is</strong>fy the m<strong>as</strong>s conservation equation<br />

dn<br />

dn<br />

<br />

<br />

dn<br />

<br />

1<br />

<br />

2<br />

<br />

C<br />

dn<br />

dn<br />

<br />

1<br />

dn<br />

<br />

2<br />

<br />

C<br />

<br />

<br />

<br />

dn<br />

<br />

1<br />

dn<br />

<br />

2<br />

dn<br />

<br />

<br />

C<br />

<br />

0<br />

----(5)<br />

The variation in number of moles dni <strong>is</strong> independent of each other. However the sum of<br />

change in mole in all the ph<strong>as</strong>es must be zero. F<strong>or</strong> the criterion of equilibrium <strong>is</strong> that the<br />

chemical potential of each component must be equal in all ph<strong>as</strong>es.<br />

<br />

<br />

<br />

<br />

<br />

<br />

1<br />

<br />

2<br />

<br />

C<br />

1<br />

<br />

<br />

2<br />

<br />

C<br />

<br />

1<br />

<br />

<br />

2<br />

<br />

C<br />

<br />

At constant temperature and pressure the general criterion f<strong>or</strong> thermodynamic<br />

equilibrium in closed system f<strong>or</strong> heterogeneous multi component system<br />

At constant T and P<br />

<br />

i <br />

<br />

i<br />

i<br />

f<strong>or</strong> i=1,2,3--------C<br />

Since G RT ln f <br />

i i<br />

i<br />

The above equation <strong>is</strong> also sat<strong>is</strong>fied in terms of fugacity<br />

<br />

i<br />

<br />

<br />

f i f i f<strong>or</strong> i=1,2,3--------C<br />

f


Ph<strong>as</strong>e Diagram f<strong>or</strong> Binary solution<br />

Constant pressure equilibrium<br />

Consider a Binary system made up of component A and B. Where it <strong>is</strong> <strong>as</strong>sumed to be<br />

m<strong>or</strong>e volatile than B where vapour pressure ‘A’ <strong>is</strong> m<strong>or</strong>e than ‘B’. When the pressure <strong>is</strong><br />

fixed at the liquid composition can be changed the properities such <strong>as</strong> temperature and<br />

vapour compositions get quickly determined VLE at constant pressure <strong>is</strong> represented on<br />

T-xy diagram.<br />

Boiling point diagrams<br />

When temperature <strong>is</strong> plotted against liquid(x) and vapour(y) ph<strong>as</strong>e composition. The<br />

upper curve gives and lower curve gives. The lower curve <strong>is</strong> called <strong>as</strong> bubble point<br />

curve and upper curve <strong>is</strong> called <strong>as</strong> Dew point curve. The mixture below bubble point <strong>is</strong><br />

sub cooled liquid and above the Dew point <strong>is</strong> super heated vapour. The region between<br />

bubble point and Dew point <strong>is</strong> called mixture of liquid and vapour ph<strong>as</strong>e.<br />

Consider a liquid mixture whose composition and temperature <strong>is</strong> represented by<br />

point ‘A’. When the mixture <strong>is</strong> heated slowly temperature r<strong>is</strong>es and reach to point ‘B’,<br />

where the liquid starts boiling, temperature at that point <strong>is</strong> called boiling point of whether<br />

heating mixture reaches to point ‘G’ when all liquids converts to vapour the temperature<br />

at that point <strong>is</strong> called <strong>as</strong> Dew point. Further heating results in super heated vapour. The<br />

number of tic lies connects between vapour and liquid ph<strong>as</strong>e. F<strong>or</strong> a solution the term<br />

boiling point h<strong>as</strong> no meaning because temperature varies <strong>from</strong> boiling point to Dew point<br />

at constant pressure.


Effect of pressure on VLE<br />

The boiling point diagram <strong>is</strong> drawn <strong>from</strong> composition x=0 to x=1, boiling point of pure<br />

substances incre<strong>as</strong>es with incre<strong>as</strong>e in pressure. The variation of boiling point diagrams<br />

with pressure <strong>is</strong> <strong>as</strong> shown in diagram. The high pressure diagrams are above low<br />

pressure.


Equilibrium diagram<br />

Vap<strong>or</strong> composition <strong>is</strong> drawn against liquid comp at constant pressure. Vapour <strong>is</strong> always<br />

rich in m<strong>or</strong>e volatile component the curve lies above the diagonal line.<br />

Constant temperature Equilibrium<br />

VLE diagram <strong>is</strong> drawn against composition at constant temperature. The upper curve <strong>is</strong><br />

and lower curve <strong>is</strong> drawn at vapour comp(y). Consider a liquid at known pressure and<br />

composition at point ‘A’ <strong>as</strong> the pressure <strong>is</strong> decre<strong>as</strong>es and reaches to point ‘B’ where it<br />

starts boiling further decre<strong>as</strong>es in pressure reaches to point ‘C’ when all the liquid<br />

converts to vapour, further decre<strong>as</strong>es in pressure leads to f<strong>or</strong>mation of super heated<br />

vapour. In between B to C both liquid and vapour ex<strong>is</strong>ts together


Non ideal solution<br />

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

not obey Raoults law. The total pressure f<strong>or</strong> non ideal solutions may be greater <strong>or</strong> lower<br />

than that f<strong>or</strong> ideal solution. When total pressure <strong>is</strong> greater than pressure given by Raoults<br />

law, the system shows positive deviation <strong>from</strong> Raoults law.<br />

Eg: Ethanol-toluene<br />

When the total pressure at equilibrium <strong>is</strong> less than the pressure given by Roults law the<br />

system shows negative deviation <strong>from</strong> Raoults law.<br />

Eg: Tetrahydrofuron-ccl4<br />

Azeotropes<br />

Azeotropes are constant boiling mixtures. When the deviation <strong>from</strong> Raoults law <strong>is</strong> very<br />

large p-x and p-y curve meets at th<strong>is</strong> point y1=x1 and y2=x2<br />

A mixture of th<strong>is</strong> composition <strong>is</strong> known <strong>as</strong> Azeotrope. Azeotrope <strong>is</strong> a vapour liquid<br />

equilibrium mixture having the same composition in both the ph<strong>as</strong>es.<br />

Types of Azeotropes<br />

Azeotropes are cl<strong>as</strong>sified into two types<br />

1. Maximum curve pressure, minimum boiling azeotropes<br />

2. Minimum pressure maximum boiling azeotrope<br />

The azeotrope f<strong>or</strong>med when negative deviation <strong>is</strong> very large will exhibit minimum<br />

pressure <strong>or</strong> maximum boiling point, th<strong>is</strong> <strong>is</strong> known <strong>as</strong> minimum pressure Azeotrope.<br />

Eg : Ethanol –water, benzene-ethanol<br />

The azeotrope f<strong>or</strong>med when –ue deviation <strong>is</strong> very large will exhibit minimum pressure <strong>or</strong><br />

maximum boiling point, th<strong>is</strong> <strong>is</strong> known <strong>as</strong> minimum pressure azeotrope.<br />

Ph<strong>as</strong>e diagram f<strong>or</strong> both types of Azeotropes<br />

Minimum Temperature azeotropes


Maximum Tem – Min pressure Azeotropes<br />

Calculation of VLE f<strong>or</strong> ideal solution<br />

Ideal solution: Ideal solution <strong>is</strong> one which obeys Raoults law. Raoults law states that the<br />

partial pressure <strong>is</strong> equal to product of vap<strong>or</strong> pressure and mole fraction in liquid ph<strong>as</strong>e.<br />

PA PVAx<br />

A<br />

PA- Partial pressure<br />

PVA- Vap<strong>or</strong> pressure


xA- Mole fraction of component A<br />

F<strong>or</strong> binary solution (F<strong>or</strong> component A and B)<br />

We know that PT PA<br />

PB<br />

-------(1)<br />

PA PVAx<br />

A , PB PVB<br />

xB<br />

Substitute PA and PB in equation 1<br />

P P x P x<br />

T VA A VB B<br />

1 x A xB<br />

, 1<br />

x A xB<br />

PT PVAx<br />

A PVB<br />

( 1<br />

x A )<br />

P P P x P --------(2)<br />

x<br />

T<br />

A<br />

<br />

VA VB A VB<br />

P<br />

P<br />

T<br />

VA<br />

P<br />

VB<br />

P<br />

VB<br />

Assuming the vap<strong>or</strong> ph<strong>as</strong>e <strong>is</strong> also ideal<br />

PA<br />

PVAx<br />

A<br />

y A <br />

PT<br />

PT<br />

Substituting PT <strong>from</strong> equation (2)<br />

y<br />

A<br />

<br />

P<br />

VA<br />

x<br />

PVA PVB<br />

x A PVB<br />

A<br />

Dividing numerat<strong>or</strong> and denominat<strong>or</strong> by PVB<br />

y<br />

y<br />

A<br />

A<br />

<br />

P<br />

<br />

<br />

P<br />

<br />

VA<br />

VB<br />

P<br />

P<br />

VA<br />

VB<br />

x<br />

A<br />

<br />

1<br />

<br />

x<br />

A 1<br />

<br />

x<br />

A<br />

1x<br />

1<br />

A<br />

The above equation relates x and y<br />

<strong>is</strong> known <strong>as</strong> relative volatility of component A with respect to component B


Problem: The binary system acetone and acetone nitrile f<strong>or</strong>m an ideal solution. Using the<br />

following data prepare<br />

P-x-y diagram at 50 o C<br />

T-x-y diagram at 400mm Hg<br />

x-y diagram at 400 mm Hg<br />

T PV1 PV2<br />

38.45 400 159.4<br />

42 458.3 184.6<br />

46 532 216.8<br />

50 615 253.5<br />

54 707.9 295.2<br />

58 811.8 342.3<br />

62.3 937.4 400<br />

Solution: PT PV 1 PV<br />

2 x1 PV<br />

2<br />

,<br />

x1 PT y1<br />

0.0 253.5 0.0<br />

0.2 325.8 0.377<br />

0.4 398.1 0.6179<br />

0.6 470.4 0.784<br />

0.8 542.7 0.906<br />

1.0 615 1<br />

T-x-y diagram at 400 mm Hg , PT=400mmHg<br />

T PV1 PV2 x1 y1<br />

38.45 400 159.4 1 1<br />

42 458.3 184.6 0.7869 0.9015<br />

46 532 216.8 0.5812 0.7729<br />

50 615 253.5 0.405 0.6226<br />

54 707.9 295.2 0.2539 0.449<br />

58 811.8 342.3 0.122 0.2475<br />

62.3 937.4 400 0 0<br />

Calculation of VLE f<strong>or</strong> non Ideal solution<br />

F<strong>or</strong> non ideal solution<br />

Partial pressure <strong>is</strong> given by Pi i xi<br />

PVi<br />

i - Activity coefficient<br />

xi- Mole fraction<br />

PVi – vap<strong>or</strong> pressure<br />

y<br />

P<br />

x<br />

V1<br />

1<br />

A , PV1= 615, PV2 = 253.5<br />

PT<br />

x<br />

P<br />

P<br />

T V 2<br />

1 ,<br />

PV<br />

1 PV<br />

2<br />

y<br />

A<br />

<br />

P<br />

P<br />

T<br />

x<br />

V1<br />

1


F<strong>or</strong> binary mixture<br />

P x P <br />

1 1 1 V1<br />

P <br />

x<br />

P<br />

2 2 2 V 2<br />

P T<br />

T<br />

P P<br />

1<br />

P x P <br />

2<br />

x<br />

P<br />

1 1 V1<br />

2 2 V 2<br />

P1<br />

y1<br />

<br />

PT<br />

1x1P<br />

V1<br />

<br />

1x1P<br />

V1<br />

2 x2<br />

PV<br />

2<br />

1<br />

<br />

2 x2<br />

PV<br />

2<br />

1<br />

1x1P<br />

V1<br />

The above equation relates x and y f<strong>or</strong> non ideal solution<br />

Activity coefficients are functions of liquid composition x, many equations are available<br />

to estimate them. The imp<strong>or</strong>tant equations are<br />

Vanlaar equation<br />

Wilson equation<br />

Margules Equation<br />

Vanlaar Equation:<br />

Estimation of activity coefficient<br />

2<br />

Ax2<br />

ln<br />

1 <br />

2<br />

A <br />

<br />

x1<br />

x2<br />

<br />

B <br />

ln<br />

2<br />

2<br />

1<br />

Bx<br />

<br />

B <br />

<br />

x1<br />

x2<br />

<br />

A <br />

2<br />

Where A and b are known <strong>as</strong> vanlaar constants, if the constants are known the<br />

activity coefficients can be estimated.<br />

Estimation of Vanlaar constants<br />

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

constants A and B can be estimated by rearranging the equation<br />

A<br />

<br />

x<br />

<br />

ln<br />

<br />

2 2<br />

ln 1 1<br />

<br />

x1<br />

ln<br />

1 <br />

2


B <br />

<br />

x ln<br />

<br />

1 1<br />

ln 2 1<br />

<br />

x2<br />

ln<br />

2 <br />

2<br />

2 nd Method<br />

F<strong>or</strong> a systems f<strong>or</strong>ming azeotrope, if the temperature and pressure s are known at<br />

azeotropic composition then activity coefficients can be calculated <strong>as</strong> shown below.<br />

x P <br />

P<br />

1 1 1 V1<br />

P y x <br />

T<br />

P<br />

1 1 1 V1<br />

At azeotrpic composition x1=y1<br />

PT<br />

P<br />

1 ,<br />

PT<br />

P<br />

2<br />

V1<br />

V 2<br />

Van laar constants A and B can calculated using the above equations.<br />

Problem: The azeotope of ethanol and benzene h<strong>as</strong> composition of 44.8mol% C2H5OH at<br />

68 o C and 760mmHg.At 68 o C the vap<strong>or</strong> pressure of benzene and ethanol are 517 and 506<br />

mmHg.<br />

Calculate<br />

Vanlaar constants<br />

Prepare the graph of activity coefficients Vs composition<br />

Assuming the ratio of vap<strong>or</strong> pressure remains constant, prepare equilibrium<br />

diagram at 760mmHg.<br />

Solution:<br />

At azeotropic composition x1=y1=0.448, x2=y2=0.552, PV1=506mmHg,<br />

PV2 =517mmHg<br />

A <br />

P<br />

P<br />

T<br />

V1<br />

760<br />

506<br />

<br />

P<br />

<br />

T<br />

1 , 2<br />

PV<br />

2<br />

760<br />

1 1.<br />

501,<br />

2 1.<br />

47<br />

517<br />

x<br />

<br />

ln<br />

<br />

2 2<br />

ln 1 1<br />

<br />

x1<br />

ln<br />

1 <br />

( 0.<br />

552)<br />

ln( 1.<br />

47)<br />

<br />

A<br />

ln( 1.<br />

501)<br />

1<br />

<br />

( 0.<br />

448)<br />

ln( 1.<br />

501)<br />

<br />

<br />

<br />

2<br />

2


A=0.829<br />

B <br />

<br />

x ln<br />

<br />

1 1<br />

ln 2 1<br />

<br />

x2<br />

ln<br />

2 <br />

( 0.<br />

448)<br />

ln( 1.<br />

501)<br />

<br />

B ln( 1.<br />

47)<br />

1<br />

<br />

( 0.<br />

552)<br />

ln( 1.<br />

47)<br />

<br />

<br />

<br />

B=0.576<br />

2<br />

2<br />

Make use of the equations given below f<strong>or</strong> plotting graph of activity coefficients Vs<br />

composition.<br />

ln<br />

y<br />

1<br />

2<br />

2<br />

Ax2<br />

Bx1<br />

1 <br />

, ln<br />

2<br />

2 <br />

2<br />

T<br />

A<br />

<br />

x<br />

B<br />

1<br />

<br />

x2<br />

<br />

<br />

P1<br />

1x1P<br />

V1<br />

<br />

P x P x P<br />

x1<br />

1 1 V1<br />

x2<br />

2<br />

2<br />

V 2<br />

B <br />

<br />

x1<br />

x2<br />

<br />

A <br />

1<br />

<br />

2 x2<br />

P<br />

1<br />

x P<br />

1 2<br />

V 2<br />

1 1 V1<br />

0 1 6.745 1 0<br />

0.2 0.8 2.4 1.095 0.384<br />

0.4 0.6 1.644 1.374 0.4384<br />

0.6 0.4 1.21 1.721 0.5077<br />

0.8 0.2 1.0112 2.618 0.609<br />

1.0 0 1 3.723 1<br />

Margules equation<br />

Eetimation of activity coefficients<br />

ln x<br />

2<br />

A 2 B A x<br />

1<br />

2<br />

2<br />

2<br />

1<br />

<br />

B 2A<br />

B<br />

<br />

ln x x<br />

1<br />

2<br />

The constant A in the above equation <strong>is</strong> terminal value of ln1 at x1=0 and constant B <strong>is</strong><br />

the terminal value of ln2 at x2=0<br />

y1


When A=B<br />

2<br />

ln Ax ,<br />

1<br />

2<br />

ln Ax<br />

2<br />

2<br />

1<br />

The above equations are known <strong>as</strong> Margules suffix equation.<br />

Wilson Equation:<br />

Wilson proposed the following equation f<strong>or</strong> activity coefficients in binary solution<br />

12<br />

21 <br />

ln<br />

1 lnx1<br />

12<br />

x2<br />

x2<br />

<br />

x1<br />

12<br />

x2<br />

21x1<br />

x2<br />

<br />

ln<br />

2<br />

ln<br />

<br />

<br />

12<br />

21<br />

x2 21x1<br />

x1<br />

<br />

x1<br />

12<br />

x2<br />

21x1<br />

x2<br />

<br />

Wilson equation have two adjustable parameters 12<br />

component molar volumes.<br />

<br />

<br />

12<br />

21<br />

V2 11 2<br />

12<br />

<br />

V<br />

1<br />

<br />

V<br />

12<br />

<br />

exp<br />

RT<br />

12<br />

<br />

exp<br />

RT<br />

V<br />

<br />

V<br />

1<br />

V<br />

<br />

V<br />

a<br />

exp<br />

RT<br />

V1 22 1<br />

21<br />

2<br />

2<br />

<br />

<br />

<br />

a<br />

exp<br />

RT<br />

V1 and V2 – molar volumes of pure liquids<br />

- Energies of interaction between molecule<br />

<br />

<br />

<br />

<br />

<br />

and 21<br />

. These are related to pure<br />

Wilson equation suffers main d<strong>is</strong>advantages which <strong>is</strong> not suitable f<strong>or</strong> maxima <strong>or</strong> minima<br />

on ln versus x curves.<br />

Cons<strong>is</strong>tency of VLE data<br />

Gibbs duhem equation in terms of thermal cons<strong>is</strong>tency<br />

x<br />

1<br />

d ln <br />

dx<br />

1<br />

1<br />

<br />

( 1<br />

d ln <br />

x1)<br />

dx<br />

1<br />

2<br />

0<br />

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


Acc<strong>or</strong>ding to Gibbs Duhem equation both slopes must have oppsite sign then only the<br />

data <strong>is</strong> cons<strong>is</strong>tent, otherw<strong>is</strong>e it <strong>is</strong> incons<strong>is</strong>tent.<br />

F<strong>or</strong> the data to be cons<strong>is</strong>tent it h<strong>as</strong> sat<strong>is</strong>fy the following condition.<br />

1. If one of ln curves h<strong>as</strong> maximum at certain concentration and the other curve<br />

2.<br />

should be minimum at same composition.<br />

If there <strong>is</strong> no maximum <strong>or</strong> minimum point both must have + ue <strong>or</strong> –ue on entire<br />

range.<br />

Co-ex<strong>is</strong>tence equation<br />

The general f<strong>or</strong>m of Gibbs duhem equation at constant temperature and pressure<br />

0 d x ---(1)<br />

<br />

i i M<br />

In terms of partial molar free energies<br />

0 d x ------(2)<br />

<br />

i i G<br />

dividing equation (2) by dx1


x<br />

i<br />

dG<br />

dx<br />

1<br />

i<br />

0 -----(3)<br />

Gi RT ln f i <br />

at constant temperature<br />

dGi RT ln f i<br />

----(4)<br />

dx1<br />

dx1<br />

substituting equation (4) in equation (3)<br />

RT ln f<br />

xi<br />

dx<br />

d ln f<br />

xi<br />

dx<br />

1<br />

1<br />

i<br />

i<br />

0<br />

0------(5)<br />

F<strong>or</strong> binary system<br />

d ln f1<br />

d ln f 2<br />

x1 x2<br />

0 ----(6)<br />

dx dx<br />

1<br />

x d f x d ln f 0 -----(7)<br />

1<br />

ln 1 2 2<br />

1<br />

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

applicable f<strong>or</strong> both liquid and vap<strong>or</strong> ph<strong>as</strong>e<br />

F<strong>or</strong> liquids<br />

f x f<br />

i<br />

i<br />

i<br />

i<br />

i<br />

ln f ln<br />

ln x ln f<br />

i<br />

differentiating with respect to x1<br />

1<br />

1<br />

i<br />

d ln f i d ln<br />

i d ln xi<br />

d ln f i<br />

-------(8)<br />

dx dx dx dx<br />

1<br />

i<br />

fi <strong>is</strong> a pure component ffugacity it does not vary with x1<br />

Substituting equation (8) in equation (5)<br />

d ln<br />

i d ln x<br />

xi<br />

xi<br />

dx dx<br />

1<br />

1<br />

i<br />

=0<br />

0<br />

1<br />

=0


d ln<br />

xi<br />

dx<br />

1<br />

F<strong>or</strong> binary system<br />

d ln<br />

1 d ln<br />

x1<br />

x2<br />

dx dx<br />

1<br />

i<br />

0 ----------------(9)<br />

1<br />

2<br />

0 ---------(10)<br />

x d x d ln<br />

0 -----------(11)<br />

1<br />

ln 1 2 2<br />

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

F<strong>or</strong> ideal vap<strong>or</strong><br />

f i Pi<br />

From equation 5<br />

d ln Pi<br />

xi<br />

0 ------------(13)<br />

dx<br />

F<strong>or</strong> binary system<br />

1<br />

d ln P1<br />

d ln P2<br />

x1 x2<br />

0 ----------(14)<br />

dx dx<br />

1<br />

1<br />

x d P x d ln P 0 ---------------(15)<br />

1<br />

ln 1 2 2<br />

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

F<strong>or</strong> ideal vap<strong>or</strong><br />

P1 y1PT<br />

, P2 y2<br />

PT<br />

x1d<br />

ln( y1PT<br />

) x2d<br />

ln( y2<br />

PT<br />

) 0<br />

x1d<br />

ln y1<br />

x1d<br />

ln y2<br />

( x1<br />

x2<br />

) d ln PT<br />

0<br />

x x 1<br />

1<br />

x d<br />

1<br />

2<br />

ln y1<br />

2<br />

x1d<br />

ln y d ln PT<br />

0<br />

dP<br />

<br />

x1d<br />

ln y1<br />

x2d<br />

ln y<br />

P<br />

2


dP<br />

P<br />

x<br />

1<br />

dy<br />

y<br />

1<br />

1<br />

x<br />

2<br />

dy<br />

y<br />

2<br />

2<br />

y 2 y1<br />

0 , dy2 dy1<br />

on simplification<br />

dP <br />

x1<br />

( 1<br />

x1)<br />

<br />

dy1<br />

<br />

P y1<br />

( 1<br />

y1)<br />

<br />

Further simplification<br />

dP y1<br />

x1<br />

<br />

P<br />

<br />

dy1<br />

y(<br />

1<br />

y1)<br />

<br />

Th<strong>is</strong> <strong>is</strong> known <strong>as</strong> coex<strong>is</strong>tence equation. <strong>It</strong> relates P,x,y f<strong>or</strong> binary VLE system. Th<strong>is</strong><br />

equation <strong>is</strong> used to rest cons<strong>is</strong>tency of VLE data.<br />

Cons<strong>is</strong>tency test f<strong>or</strong> VLE data<br />

x1d<br />

ln 1 x2d<br />

ln<br />

2 0<br />

x1d<br />

ln<br />

d ln<br />

2 <br />

x<br />

d ln<br />

d ln<br />

2<br />

2<br />

2<br />

1<br />

x1d<br />

ln<br />

<br />

( 1<br />

x )<br />

1<br />

1<br />

x1d<br />

ln<br />

<br />

( 1<br />

x )<br />

Redlich K<strong>is</strong>ter Test<br />

1<br />

1<br />

The excess free energy of mixing f<strong>or</strong> a solution <strong>is</strong> given <strong>as</strong><br />

e<br />

G G G<br />

ideal<br />

<br />

G <br />

Nonideal<br />

e<br />

RT xi<br />

ln i<br />

F<strong>or</strong> binary system<br />

G RT x ln x ln<br />

e<br />

<br />

<br />

<br />

1<br />

1<br />

2<br />

2


differentiating with respect to x1<br />

dG e<br />

dx<br />

1<br />

<br />

RT x<br />

<br />

1<br />

d ln<br />

dx<br />

1<br />

1<br />

ln<br />

From Gibbs Duhem Equation<br />

d ln<br />

1 d ln<br />

2<br />

x1<br />

x2<br />

0<br />

dx dx<br />

dG e<br />

dx<br />

dx<br />

1<br />

dG e<br />

dx<br />

1<br />

dG e<br />

x1<br />

1<br />

<br />

x1<br />

0<br />

1<br />

dG<br />

1<br />

<br />

RT ln<br />

1 ln<br />

<br />

dx dx<br />

2<br />

RT<br />

1<br />

1<br />

2<br />

ln ln<br />

<br />

1<br />

<br />

RT ln<br />

<br />

e<br />

RT<br />

<br />

1<br />

2<br />

x1<br />

1<br />

x1<br />

0<br />

<br />

<br />

<br />

<br />

ln<br />

<br />

1<br />

2<br />

2<br />

dx<br />

dx<br />

<br />

dx<br />

<br />

1<br />

2<br />

1<br />

1<br />

x<br />

<br />

<br />

<br />

2<br />

d ln<br />

dx<br />

1<br />

2<br />

ln<br />

e<br />

The two limits indicates pure component f<strong>or</strong> which G 0 , where LHS <strong>is</strong> zero f<strong>or</strong> both<br />

limits.<br />

We can write<br />

x1<br />

1<br />

1 <br />

0 ln<br />

dx<br />

x 0<br />

2 <br />

1<br />

1<br />

Th<strong>is</strong> can be checked graphically. Net area should be equal to zero f<strong>or</strong><br />

cons<strong>is</strong>tency(Area=0).<br />

2<br />

dx<br />

dx<br />

2<br />

1


Problem<br />

Verify whether the following data <strong>is</strong> cons<strong>is</strong>tent<br />

X1 1 2<br />

0 0.576 1.00<br />

0.2 0.655 0.985<br />

0.4 0.748 0.930<br />

0.6 0.856 0.814<br />

0.8 0.950 0.626<br />

1.0 1.00 0.379<br />

Solution:<br />

We know that the Redlich k<strong>is</strong>ter test <strong>is</strong><br />

x1<br />

1<br />

1 <br />

0 ln<br />

dx1<br />

x 0<br />

2 <br />

1<br />

1<br />

2<br />

1<br />

ln<br />

2<br />

0.576 -0.552<br />

0.665 -0.408<br />

0.804 -0.218<br />

1.052 0.051<br />

1.518 0.417<br />

2.639 0.97<br />

Plot<br />

ln <br />

1<br />

Vs x1<br />

2<br />

Area under the curve <strong>is</strong> zero. Data <strong>is</strong> cons<strong>is</strong>tent.


CHEMICAL REACTION EQUILIBRIUM<br />

Energy change accompanies all chemical reactions. Because of th<strong>is</strong> energy<br />

change the temperature of the products may incre<strong>as</strong>e <strong>or</strong> decre<strong>as</strong>e depending on the<br />

exothermic <strong>or</strong> endothermic nature of the reaction. The energy change may be<br />

<strong>expressed</strong> in terms of heat of reaction, heat of combustion and heat of f<strong>or</strong>mation.<br />

Heat of reaction <strong>is</strong> the change in enthalpy of a reaction under pressure of 1.0<br />

atmosphere, starting & ending with all materials at a constant temperature T<br />

Heat of combustion <strong>is</strong> the heat of reaction of a combustion reaction.<br />

Heat of f<strong>or</strong>mation <strong>is</strong> the heat of reaction of a f<strong>or</strong>mation reaction. A f<strong>or</strong>mation<br />

reaction <strong>is</strong> one which results in the f<strong>or</strong>mation of one mole of a compound <strong>from</strong> the<br />

elements.<br />

Eq. H2 + ½ O2 H2O<br />

C +O2 CO2<br />

C + ½ H2 + 1/2 Cl2 CHCl3<br />

The standard heat of reaction, standard heat of f<strong>or</strong>mation, and standard heat of<br />

f<strong>or</strong>mation are respectively the heat of reaction, heat of combustion and heat of f<strong>or</strong>mation<br />

under 1 atmosphere starting and ending with all materials at constant temperature of<br />

25 0 C.<br />

In chemical industries, processes are carried out under <strong>is</strong>othermal conditions<br />

and th<strong>is</strong> requires the addition <strong>or</strong> removal of heat <strong>from</strong> the react<strong>or</strong>. Heat of reaction<br />

values will give the amount of heat to be removed <strong>or</strong> added. Knowing of th<strong>is</strong> heat<br />

value helps to design the heat exchange equipment.<br />

Standard heat of reaction<br />

Calculation of ∆H 298 f<strong>or</strong>m heats of f<strong>or</strong>mation data:<br />

The standard heat of reaction accompanying any chemical change <strong>is</strong> equal to<br />

the algebraic sum of the standard heats of f<strong>or</strong>mation of products minus the<br />

algebraic sum of the standard heat of f<strong>or</strong>mation of reactants.<br />

∆Hr = ∑∆H f 298 products - ∑∆Hf 298 reactants<br />

Heat of f<strong>or</strong>mation of any clement <strong>is</strong> zero


Heat effects of chemical Reaction:<br />

F<strong>or</strong> the reaction, aA + bB ↔ cC + dD<br />

Let heat capacity be Cp = α + βT + γT 2<br />

then we have CpA = αA + βAT + γAT 2<br />

CpB = αB + βBT + γBT 2<br />

CpC = αC + βCT + γCT 2<br />

CpD = αD + βDT + γDT 2<br />

And ΔH 0 298 be standard heat of reaction at 298 K and standard heat of reaction at any<br />

other temperature can be found by Kirchoff’s rule<br />

o <br />

d H<br />

dT<br />

C where<br />

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

γBT 2 )]<br />

<br />

C C C<br />

p p p<br />

products reac tan ts<br />

Cp <br />

cCpC dC pD <br />

aCpA bC pB then substituting the values of<br />

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

(a γA+ b γB )] T 2<br />

<strong>or</strong><br />

ΔCp = Δα + Δβ T + ΔγT 2<br />

Substituting in Kirchoff’s rule<br />

<br />

<br />

Pr oducts Re ac tan ts<br />

cC dC aC bC <br />

<br />

C D A B<br />

<br />

<br />

Pr oducts Re ac tan ts<br />

<br />

<br />

Pr oducts Re ac tan ts<br />

p<br />

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


H T T T<br />

0<br />

H<br />

T<br />

0 2<br />

<br />

<br />

H298<br />

0 2<br />

d<br />

H T T<br />

dT<br />

298<br />

<br />

<br />

2 3<br />

OR in general<br />

o<br />

2 3<br />

H T<br />

T T I<br />

P <br />

<br />

0 2<br />

H C dT T T dT<br />

<br />

= <br />

2 3<br />

2 3<br />

T T T I<br />

A chemical reaction proceeds in the direction of decre<strong>as</strong>ing free energy.The sum of the<br />

free energies of the reactants should be m<strong>or</strong>e than the sum of energies of products. F<strong>or</strong> a<br />

reaction to take place<br />

<br />

G G G <strong>or</strong> G 0<br />

Re action Re action<br />

products reaction<br />

Theref<strong>or</strong>e ΔG should be less than zero f<strong>or</strong> a reaction to occur. When equilibrium <strong>is</strong><br />

reached free energies of the reactants equal free energies of products,. Theref<strong>or</strong>e the<br />

criterion f<strong>or</strong> reaction equilibrium <strong>is</strong> ΔG = 0<br />

ΔGReaction Calculation: Consider the reaction aA + bB ↔ cC + dD<br />

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

reaction mixture.<br />

<br />

<br />

2 3<br />

o<br />

2 3<br />

H T<br />

T T I<br />

f f<br />

G G RT ln but a the activity of A<br />

0 A A<br />

A A 0<br />

fA 0<br />

fA<br />

A<br />

G G RT lna<br />

A<br />

0<br />

A A<br />

G G RT lna<br />

B<br />

0<br />

B B<br />

G G RT lna<br />

C<br />

0<br />

C C<br />

G G RT lna<br />

D<br />

0<br />

D D


G G G<br />

Re action<br />

<br />

At equilibrium ΔG = 0 and<br />

<br />

products Re ac tan ts<br />

cGC dGD aGA bGB<br />

<br />

0 GC RT<br />

0<br />

aC d GD RT aD<br />

<br />

0<br />

a GA 0<br />

RT ln aA b GB RT lnaB<br />

<br />

cGC dGD aGA bGB<br />

<br />

RT lnaC lnaD lnaA lnaB<br />

<br />

<br />

= c ln ln <br />

<br />

<br />

<br />

<br />

0<br />

GRe action G RT<br />

ln C<br />

c d<br />

a a <br />

D<br />

<br />

a b<br />

aA a <br />

B <br />

c d<br />

a a <br />

C D<br />

K K <strong>is</strong> equilibrium constant of the reaction<br />

a b<br />

aA a <br />

B <br />

a a <br />

G RT G RT K<br />

c d<br />

0 <br />

0<br />

<br />

C ln a<br />

aA D<br />

b<br />

aB<br />

<br />

<br />

0<br />

ln<br />

<br />

at equilibrium


EFFECT OF TEMPERATURE ON EQUILIBRIUM CONSTANT K<br />

Gibb’s - Helmoholtz equation at constant pressure <strong>is</strong> given<br />

o G<br />

<br />

d <br />

T H<br />

o<br />

but G RT<br />

ln K<br />

2<br />

dT T<br />

RT<br />

ln K <br />

d <br />

T<br />

<br />

H d R ln K<br />

H<br />

<strong>or</strong><br />

<br />

2 2<br />

dT T dT T<br />

d ln K<br />

H<br />

Van't Hoff equation.<br />

2<br />

dT RT<br />

Th<strong>is</strong> may be integrated to find the effect of temperature on K<br />

K2<br />

T2<br />

<br />

When H<br />

<strong>is</strong> contant d lnK <br />

K1<br />

T1<br />

K2 ln<br />

K1<br />

H<br />

1 1 <br />

<br />

R T2 T1<br />

<br />

if K <strong>is</strong> known at T ,then<br />

K can be calculated at other T<br />

1 1<br />

ln K<br />

H<br />

dT 2<br />

RT<br />

2 2<br />

When H varies with T<br />

d<br />

dT<br />

0<br />

H<br />

2<br />

RT<br />

<br />

T<br />

I <br />

d ln K<br />

dT 2<br />

RT 2R 3R<br />

<br />

RT <br />

0<br />

<br />

<br />

ln T <br />

ln K = <br />

R 2R 6R<br />

RT<br />

G RT ln K<br />

2<br />

T T I<br />

M<br />

Where M <strong>is</strong> a consatnt of integration<br />

2<br />

0 ln T T T<br />

I <br />

G RT M<br />

<br />

R 2R 6R<br />

RT <br />

T T<br />

G T I MRT<br />

2 6<br />

2 3<br />

<br />

0<br />

ln T


F<strong>or</strong> the re<strong>as</strong>ction C2H4 + ½ O2 ↔ C2H4O, develop equations f<strong>or</strong> ΔH 0 , K, and ΔG 0 , find<br />

ΔG 0 at 550 K<br />

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

Standard heat of reaction ΔH 0 298<br />

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

Re action 298<br />

In th<strong>is</strong> problem C<br />

p<br />

<br />

H H H<br />

Pr oducts Re ac tan ts<br />

T<br />

O2<br />

And ΔG 0 298 = -19070 Cal / mol<br />

1 <br />

= 1.59 3.68 6.39 5.285<br />

Pr oducts Re ac tan ts<br />

2 <br />

= 6.39 + 0.0021 T<br />

C2H4O = 1.59 + .00332 T<br />

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

1 <br />

= 0.0332 0.0224 0.0021 0.00975<br />

Pr oducts Re ac tan ts<br />

2 <br />

Then the variation of standard heat of reaction with temperature <strong>is</strong>


o 2 <br />

3 0.00975 2<br />

H T T T I 5.285 T T I<br />

2 3 2<br />

0 0.00975 2<br />

Given H298 =310 = 5.285 T T I then I = 1452<br />

2<br />

o<br />

0.00975 2 0<br />

H 5.285 T T 1452 Th<strong>is</strong> <strong>is</strong> variation of H298<br />

with temparature<br />

2<br />

2 I 5.285<br />

0.00975 1452<br />

lnK = ln T T T M = lnT<br />

T M<br />

R 2R 6R RT 2 2(2) 2T<br />

0<br />

0 G298 19070<br />

but G298 RT ln K298<br />

<strong>or</strong> lnK298 32<br />

RT 2(298)<br />

5.285<br />

0.00975 1452<br />

32 = ln298 298 M theref<strong>or</strong>e M = 48.763<br />

2 2(2) 2(298)<br />

5.285<br />

0.00975 1452<br />

lnK = lnT T 48.763<br />

2 4 2T<br />

0<br />

G RT lnK 2T<br />

lnK<br />

5.285 0.00975 1452 <br />

2T lnK 2T lnT T 48.763<br />

2 4 2T<br />

<br />

<br />

0 3<br />

2<br />

G 5.285T lnT 4.875(10)() T 1452 97.526() T<br />

G 5.285(550)ln550 4.875(10)(550) 1452 97<br />

.526(550)<br />

0<br />

550<br />

3<br />

2<br />

G 35320.62<br />

Cal / mol<br />

0<br />

550<br />

50 mol% each of SO2 and O2 <strong>is</strong> fed to a converter to f<strong>or</strong>m SO3.<br />

Show the variation of i) Standard heat of reaction with T ii) Equilibrium constant<br />

with T. The ΔHf and ΔGf at 298 K are<br />

Component<br />

ΔHf<br />

k Cal / k<br />

mol<br />

ΔGf<br />

k Cal / k<br />

mol<br />

SO2 -70960 -71680<br />

O2 0 0<br />

SO3 -94450 -88590


Feed contains 50 mole% SO2 and O2.Th<strong>is</strong> means f<strong>or</strong> every one mole SO2 there will be<br />

one mole O2 in the reactant side and ½ mole O2 in the product side, acc<strong>or</strong>ding to SO2 +<br />

½ O2 SO3<br />

Reactants: 1 mol SO2 + 1 mol O2 Products: 1 mol SO3 + ½ mol O2 (excess)<br />

1<br />

= 6.0377 + 6.148 7.116 6.148 - 4.152<br />

2<br />

Pr oducts reac tan ts<br />

-3<br />

= 23.53(10) +<br />

Pr oducts reac tan ts<br />

0 2 <br />

3<br />

H T T T I<br />

2 3<br />

-3<br />

12.474x10 2<br />

= -4.152T+<br />

T I<br />

2<br />

1<br />

3<br />

-3<br />

3.102(10) - 9.512 (10) + 3<br />

2<br />

<br />

-3<br />

.102 (10) <br />

=12.474<br />

x10<br />

-3<br />

0 0<br />

But H at 298 <strong>is</strong> given by H <br />

0<br />

H <br />

0<br />

H 94450 70960 23490<br />

k Cal / k mol<br />

298<br />

products Re ac tan ts<br />

-3 2<br />

-23490=-4.152 (298 ) + 6.237 x 10(298) <br />

<br />

I = -22818.19<br />

0<br />

3<br />

2<br />

H 4.152 T 6.237x10 T 22818.19<br />

Next f<strong>or</strong> K we have<br />

<br />

<br />

<br />

2 I<br />

lnK= lnT<br />

T T M<br />

R 2R 6R<br />

RT<br />

3<br />

4.152<br />

12.474x10 22818.19<br />

lnK= lnT<br />

T M<br />

2 2xR 2T<br />

I<br />

CP data:<br />

Component a b(10 3 )<br />

SO2 7.116 9.512<br />

O2 6.148 3.102<br />

SO3 6.0377 23.537


0<br />

0 G298<br />

We have G<br />

298= -RTln K298<br />

<strong>or</strong> lnK 298=<br />

2(298)<br />

G = G - G<br />

= - 88590 - (- 71680) = - 16910 k Cal / k mol<br />

0<br />

298<br />

<br />

Products Reactants<br />

16910<br />

lnK 298=<br />

28.372<br />

2(298)<br />

3<br />

4.152<br />

12474x10 22818.19<br />

28.372 ln298 <br />

298 M<br />

2 2x2<br />

2x298 M 0.9842<br />

11409.09<br />

3<br />

lnK 2.076lnT 3.1185x10 T 0.9842<br />

Derive the general equation f<strong>or</strong> the standard free energy f<strong>or</strong>mation f<strong>or</strong> ½ N2 + 3/2<br />

H2 NH3<br />

The absolute entropies in ideal g<strong>as</strong> state at 298 K and 1.0 atm are<br />

NH3 = 46.01 g Cal / g mol K<br />

N2 = 45.77<br />

H2 = 31.21<br />

ΔH 0 298 (g Cal / g mol) = - 11040<br />

Heat capacity Data <strong>is</strong> given by Cp= a + bT + cT 2 the values are<br />

Component a bx10 2<br />

T<br />

cx10 5<br />

NH3 6.505 0.613 0.236<br />

N2 6.903 - 0.038 0.193<br />

H2 6.952 - 0.46 0.096<br />

Compute the free energy change at 1500K. Is reaction fe<strong>as</strong>ible?


0 o 0<br />

We have G and<br />

0<br />

298<br />

H S S S S<br />

298<br />

<br />

Pr oducts Re ac tan ts<br />

G 11040 298( 23.69) 3980.38<br />

g Cal / g mol K<br />

1 3 <br />

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

2 2<br />

<br />

<br />

a a a<br />

Pr oducts Re ac tan ts<br />

1 6.505(6.903) <br />

2 3<br />

6. <br />

2<br />

<br />

952 7.3745<br />

<br />

b b b<br />

Pr oducts Re ac tan ts<br />

x<br />

1 <br />

2 x<br />

3<br />

<br />

2<br />

x<br />

<br />

<br />

<br />

c c c<br />

Pr oducts Re ac tan ts<br />

0 b 2 c<br />

3<br />

H298 aT T T I<br />

2 3<br />

1 0.236x10(0.193 10<br />

2 x<br />

3<br />

)(0.096 10) x<br />

2<br />

<br />

0.045 10<br />

<br />

x<br />

11040( 7.3745)298 <br />

3 7.01X10 298(<br />

2<br />

6<br />

2 0.045<br />

X10<br />

<br />

3<br />

3<br />

298) I OR I = -9153.26<br />

a b c<br />

2 I<br />

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

R 2R 6R<br />

RT<br />

0<br />

-3980.38<br />

G298 RT<br />

ln K298<br />

<strong>or</strong> ln K 298 = = 6.6784<br />

-(2x298)<br />

2 2<br />

2 3<br />

0.613 10( 0.038 10) ( 0.046 10 ) 7.01x10 5 5 5 6<br />

3 6<br />

7.3745 7.01x10 0.045x10<br />

2 9153.26<br />

6.6784 ln298 298(298) M<br />

2 2x2 6x2 2x298 M 11.7987<br />

The iation of G G<br />

0<br />

var with T <strong>is</strong> given by, = - RT lnK<br />

3 6<br />

0 -7.3745 7.01x10 0.045x10<br />

2 9153.26 <br />

G = -2T lnT T T 11.7987<br />

<br />

2 2x2 6x2 2T<br />

<br />

3 6<br />

0 -7.3745 7.01x10 0.045x10<br />

2 9153.26 <br />

G1500 = -2x1500 ln1500 1500 1500 11.7987<br />

<br />

2 2x2 6x2<br />

2x1500 <br />

G G G <br />

0<br />

1500<br />

<br />

Pr oducts Re ac tan ts<br />

g Cal<br />

28488.81 g mol<br />

The reaction <strong>is</strong> not fe<strong>as</strong>ible ΔG <strong>is</strong> highly +ve: F<strong>or</strong> fe<strong>as</strong>ibility it should equal to zero


The hydration of Ethylene to alcohol <strong>is</strong> given by C2H2 + H2O C2H5OH<br />

Temperature 0 C 145 6.8 x 10 -2<br />

Equilibrium Constant 320 1.9 x 10 -3<br />

The heat capacity data f<strong>or</strong> the component can be<br />

represented by Cp = a + bT where T <strong>is</strong> in kelvin<br />

and Cp <strong>is</strong> in cal / g mol 0 C<br />

Develop general expression f<strong>or</strong> the equilibrium constant and standard Gibb’s free<br />

energy change <strong>as</strong> function of temperature.<br />

<br />

a a a = 6.99(2.83 7.3) 3.14 <br />

Pr oducts Re ac tan ts<br />

<br />

b b b 39.741x10(28.6 10 x 2.46 10) x 8.68 10 x<br />

Pr oducts Re ac tan ts<br />

a b c<br />

I<br />

lnK = ln<br />

R 2R 6R<br />

RT<br />

2<br />

T T T M<br />

3 3 3 3<br />

3<br />

3.14<br />

8.68x10 I<br />

lnK = lnT<br />

T M<br />

2 2x2 2T<br />

3<br />

-2 3.14<br />

8.68x10 I <br />

At 418 K,<br />

ln 6.8x10 = ln 418 418 M <br />

2 2x2 2x418 <br />

Solving,<br />

I = -9655.807 & M = -5.667<br />

3<br />

-3 3.14<br />

8.68x10 I<br />

593 K, ln 1.9x10 ln593 593<br />

At M <br />

2 2x2 2x593 <br />

3<br />

3.14<br />

8.68x10 -9655.807<br />

lnK = lnT<br />

T <br />

-5.667<br />

2 2x2 2T<br />

3<br />

3.14 8.68x10 -9655.807 <br />

But G = -2T lnT T -5.667<br />

2 2x2 2T<br />

<br />

-3 2<br />

G = 3.14T ln T - 4.34x10 T 9655.807 11.334T<br />

Component a bx10 3<br />

C2H4 2.83 28.601<br />

H2O 7.3 2.46<br />

C2H5OH<br />

6.99 39.741


THERMODYNAMIC FEASIBILITY OF REACTIONS<br />

The equilibrium constant K <strong>is</strong> a me<strong>as</strong>ure of the concentration of the products<br />

f<strong>or</strong>med at equilibrium. <strong>It</strong> <strong>is</strong> related by the equation ΔG 0 = - RT ln K. The m<strong>or</strong>e the value<br />

of K, m<strong>or</strong>e will be the equilibrium conversion of products. When the value of ΔG < 0,<br />

means the value of K should be very large. Hence ΔG <strong>is</strong> the me<strong>as</strong>ure of fe<strong>as</strong>ibility of a<br />

chemical reaction.<br />

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

The m<strong>or</strong>e the –ve value, m<strong>or</strong>e will be the fe<strong>as</strong>ibility of the reaction<br />

ii) If ΔG 0 <strong>is</strong> +ve but less than 10 000 k Cal / mol, the reaction <strong>is</strong> not fe<strong>as</strong>ible,<br />

at atmosphere , may be fe<strong>as</strong>ible at any other pressure<br />

iii) If ΔG 0 <strong>is</strong> greater than 10 000 kCal / mol the reaction <strong>is</strong> not at all fe<strong>as</strong>ible<br />

under any condition.<br />

Equilibrium Calculations ( Homogeneous g<strong>as</strong> Ph<strong>as</strong>e reactions):<br />

aA (g) + bB (g) ↔ cC (g) + dD (g)<br />

c<br />

c y c c p<br />

<br />

We have<br />

f<br />

Activity a E f<br />

0<br />

f<br />

K<br />

c d<br />

a a <br />

C D<br />

<br />

a b<br />

a A a <br />

B <br />

f<strong>or</strong> g<strong>as</strong>es,<br />

0<br />

<strong>as</strong> = p = 1 atmosphere: a = f<br />

f = y but a f x p = y p<br />

c c c c c c<br />

c c c<br />

K =<br />

a y p<br />

D D D D<br />

a y p<br />

A A A A<br />

a y p<br />

B B B B<br />

<br />

<br />

<br />

<br />

<br />

<br />

<br />

<br />

<br />

<br />

<br />

<br />

A A A B B B A B A B A B<br />

<br />

<br />

D y DD p<br />

<br />

c D<br />

x<br />

y c yD x<br />

c D<br />

x<br />

p p<br />

y p y p y y p p<br />

c + d - (a + b) n<br />

()()()(P) y =()()()(P) where = activit y <br />

y co efficient f<strong>or</strong> g<strong>as</strong> ph<strong>as</strong>e<br />

<br />

<br />

<br />

<br />

d c d c d c d c d<br />

a b a b a b a b a b<br />

K k k k k k k<br />

y = mole fraction<br />

= <strong>Fugacity</strong> Co efficient f<strong>or</strong> g<strong>as</strong> ph<strong>as</strong>e<br />

P = Total pressure


Effect of variables on equilibrium :<br />

Effect of temperature:<br />

G RT lnK<br />

since ΔG depends only on temperature <strong>as</strong> the<br />

0<br />

298 298<br />

pressure <strong>is</strong> fixed at 1.0 atmosphere, K value varies with temperature. <strong>It</strong> <strong>is</strong> not affected by<br />

Pressure, Concentration, etc,. Variation of K with T <strong>is</strong> given by<br />

ln K<br />

d H<br />

2<br />

dT RT<br />

Van't Hoff equation.<br />

F<strong>or</strong> endothermic reactions ΔH 0 <strong>is</strong> + ve <strong>as</strong> T incre<strong>as</strong>es K also incre<strong>as</strong>es. Th<strong>is</strong><br />

means that the equilibrium conversion <strong>is</strong> m<strong>or</strong>e at higher pressure.<br />

F<strong>or</strong> exothermic reactions ΔH 0 <strong>is</strong> –ve. Theref<strong>or</strong>e K incre<strong>as</strong>es with T. Hence the<br />

equilibrium conversion decre<strong>as</strong>es <strong>as</strong> T incre<strong>as</strong>es. Eg. SO2 + 1/2 O2 ↔ SO3<br />

Effect of pressure: Consider a reaction of ideal g<strong>as</strong>es, then K = KyP Δn <strong>or</strong> Ky = K / P Δn<br />

When Δn > 0. An incre<strong>as</strong>e in pressure decre<strong>as</strong>es Ky. Hence equilibrium product<br />

yield <strong>is</strong> less at high pressures<br />

When Δn< 0. An incre<strong>as</strong>e in pressure incre<strong>as</strong>es Ky and equilibrium product<br />

yield<br />

pressure <strong>is</strong> required<br />

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

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

When Δn = 0 Here pressure h<strong>as</strong> no effect on reaction. F<strong>or</strong> ideal g<strong>as</strong>es the effect<br />

of pressure depends on the variation of γ and Φ with pressure<br />

Effect of Inert: Presence of inert h<strong>as</strong> the opposite effect of incre<strong>as</strong>e of pressure.<br />

Theref<strong>or</strong>e when Δn > 0, addition of inert incre<strong>as</strong>es Ky and equilibrium<br />

conversion<br />

And Δn < 0, addition of inert decre<strong>as</strong>es equilibrium conversion.<br />

Effect of excess reactants: Presence of excess reactants incre<strong>as</strong>es equilibrium<br />

conversion of the limiting reactant<br />

Presence of products in feed: decre<strong>as</strong>es the equilibrium conversion of reactants.<br />

Eg. CH3COOH + C2H5OH CH3COOC2H5 + H2O<br />

If water <strong>is</strong> added by 1.0 mole to the feed, equilibrium conversion of CH3COOH<br />

reduces <strong>from</strong> 30% to 15%


HCN <strong>is</strong> produced by the reaction N2 (g)+ C2H2 (g) 2HCN (g). The reactants are<br />

taken in stoichiometric ratio at 1.0 atmosphere and 300 0 C. At th<strong>is</strong> temperature ΔG 0 = -<br />

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

maximum conversion of C2H2<br />

30100<br />

8.314(573)<br />

0 3<br />

Δn=2– (1+1) = 0 we have G RT<br />

ln K <strong>or</strong> lnK = <strong>or</strong> K = 1.8029x10<br />

Component Moles<br />

in feed<br />

Moles<br />

reacted<br />

Moles<br />

present<br />

Mole<br />

fraction<br />

N2 1 X 1 – X (1 – X) / 2<br />

C2H2 1 X 1 – X (1 – X) / 2<br />

HCN -- 2X 2X<br />

K K K K P K K P P <br />

n n<br />

0<br />

y Assume 1 and 1<br />

y X<br />

K K <br />

Solving f<strong>or</strong> X<br />

2 2<br />

3<br />

1.8029(10) y<br />

HCN<br />

1 1<br />

yN y<br />

2 C2H2 1 X 1<br />

X <br />

, X = 0.0207<br />

Equilibrium Composition:<br />

Equilibrium Conversi on of C H<br />

<br />

2<br />

<br />

2<br />

<br />

<br />

1 X 1 0.0207 <br />

N2 100 100 48.965%<br />

2<br />

<br />

2<br />

<br />

<br />

1 X 1 0.0207 <br />

C2H2 100 100 48.965%<br />

2<br />

<br />

2<br />

<br />

<br />

HCN ()1 X 00 2.007%<br />

2 2<br />

Equilibrium conversion <strong>is</strong> max imum f<strong>or</strong> reversible reaction<br />

Moles of C H reacted X 0.0207<br />

2 2<br />

Max. Conversion of C2H2 100 100 100 2.07%<br />

Moles of C2H2 in feed 1 1<br />

(2X) / X =<br />

W<strong>or</strong>kout the problem, when the pressure <strong>is</strong> changed to 203 bar. <strong>Fugacity</strong> co efficients<br />

of N2,,C2H2 and HCN are 1.1, 0.928, and 0.54 Assume KΦ= 1<br />

X


y <br />

<br />

K K K K P <br />

<br />

c c<br />

y <br />

n<br />

C C<br />

0<br />

= (1)<br />

<br />

203 where a,b, and c are stoichiometric co efficients<br />

a b a b<br />

A <br />

B yA<br />

y <br />

B<br />

0.54 X<br />

<br />

(1.1)(0928) 1-X 1-X <br />

<br />

2<br />

<br />

2<br />

<br />

<br />

2 2<br />

-3<br />

and K <strong>is</strong> known already = 1.8029x10 = .(1)<br />

solving<br />

X = 0.0382<br />

then equilibrium Composion of C H <strong>is</strong> ( Max reversible reaction)<br />

2 2<br />

2 2<br />

moles C2H2reacted 0.0382<br />

= x 100 = 3.82%<br />

moles of C H in feed 1<br />

Calculate the K at 673K & 1.0 bar f<strong>or</strong> N2 (g) + 3 H2 (g) 2NH3 (g).<br />

Assume heat of reaction remains constant. Take standard heat of f<strong>or</strong>mation and<br />

standard free energy of f<strong>or</strong>mation of NH3 at 298 K to be - 46110 J / mol and – 16450 J<br />

/ mol respectively.<br />

G 2( 16450) 32900 J <br />

0<br />

298<br />

0<br />

298<br />

<br />

2 (-46110) = -92 220 J<br />

0<br />

G298 RT<br />

ln K298<br />

H <br />

G RT<br />

ln K<br />

0<br />

298 1<br />

- 32900 = - 8.314 (298) ln K<br />

K 564861.1<br />

0<br />

K2 H<br />

1 1 <br />

we have ln <br />

K1<br />

R T2 T1<br />

<br />

K2 92220 1 1 <br />

ln <br />

584861.1 8.314<br />

<br />

673 298<br />

<br />

<br />

K 5.75 10<br />

2<br />

4<br />

X <br />

Acetic acid <strong>is</strong> esterified in the liquid ph<strong>as</strong>e with ethanol at 373 K and 1.0 bar acc<strong>or</strong>ding<br />

to<br />

CH3COOH (L) + C2H5OH (L) CH3COOC2H5(L) + H2O (L)<br />

The feed cons<strong>is</strong>ts of 1.0 mol each of acetic acid & ethanol, estimate the mole fraction of<br />

ethyl acetate in the reacting mixture at equilibrium. The standard heat of f<strong>or</strong>mation<br />

and standard free energy of f<strong>or</strong>mation at 298 K are given below.<br />

Assume that the heat of reaction <strong>is</strong> independent of T and liquid mixture behaves <strong>as</strong><br />

ideal solution.<br />

1<br />

1


Δf 0 f<br />

(J)<br />

ΔG 0 f<br />

(J)<br />

ΔG 0 298 = – 318218 – 237130 + 389900 + 174780 = 9332 J<br />

We have, ΔG 0 298 = – RT ln K<br />

ΔH 0 298 = – 463250 – 285830 + 277690 + 484500 = 13110 J<br />

9332 = – 8.314 (298) ln K1 <strong>or</strong> K1 = 0.02313<br />

and f<strong>or</strong> K2 at 373K we have<br />

Component<br />

CH3COOH<br />

(L)<br />

Moles<br />

in feed<br />

C2H5OH<br />

(L)<br />

Moles<br />

reacted<br />

Moles<br />

present<br />

Mole<br />

fraction<br />

CH3COOH 1 X 1 – X (1 – X) / 2<br />

C2H5OH 1 X 1 – X (1 – X) / 2<br />

CH3COOC2H5 -- X X X / 2<br />

H2O -- X X X / 2<br />

2<br />

CH3COOC2H5(L) H2O (L)<br />

- 484500 - 277690 - 463250 - 285830<br />

- 389900 -174780 - 318218 - 237130<br />

K1 ln<br />

K2 H<br />

1 1 <br />

<br />

R T1 T2<br />

<br />

0.02313 12110 1 1 <br />

ln OR K 2<br />

K 8.314 298 373 <br />

0.067<br />

K2 0<br />

K y P<br />

x <br />

<br />

2<br />

<br />

<br />

0.067= <br />

2<br />

1 x <br />

<br />

2<br />

<br />

<br />

2<br />

x<br />

2<br />

(1) x<br />

solving, x = 0.252<br />

x 0.252<br />

mole fraction of CH3COOC2H 5 0.126<br />

2 2<br />

2<br />

2


The standard free energy change f<strong>or</strong> the reaction C4H8 (g) C4H6 (g) + H2 (g)<br />

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

/ mol and T in K<br />

a) Over what range of T, <strong>is</strong> the reaction prom<strong>is</strong>ing <strong>from</strong> thermodynamic view<br />

point?<br />

b) F<strong>or</strong> a reaction of pure butene at 800K, calculate equilibrium conversion at 1.0<br />

bar & 5.0 bar<br />

c) Repeat part (b) f<strong>or</strong> the feed with the 50 mol% butene and the rest inerts<br />

a. F<strong>or</strong> the reaction at ΔG = 0, T ≈ 812.4 K, above th<strong>is</strong> temperature ΔG <<br />

0, reaction <strong>is</strong> prom<strong>is</strong>ing below 550 K, ΔG > 0, reaction <strong>is</strong> unfav<strong>or</strong>able,<br />

b. At 800K, ΔG 0 = 1842.16 J / mol<br />

ΔG 0 = - RT ln K, <strong>or</strong> 1842.16 = - 8.314 ( 800 ) ln K <strong>or</strong><br />

K = 1.3191<br />

K = ky P Δn = ky P ( 2- 1 ) = ky P<br />

Theref<strong>or</strong>e at 1.0 bar K = kyP 1.3191 = ky ( 1 )<br />

ky = 1.31910At 5 bar, 1.3191 = ky ( 5 )<br />

ky = 0.26382<br />

Moles in<br />

feed<br />

C4H8 1 1 - X<br />

Moles in product stream yi<br />

(1 – X ) / (1 +<br />

C4H6 X X / (1 + X)<br />

H2 X X / (1 + X)<br />

1 + X<br />

X)


x x <br />

Y Y <br />

x<br />

k <strong>or</strong> 1.3191 = <strong>or</strong> x = 0.7541<br />

' ' 2<br />

C4H6 H2 1 x 1 x<br />

Y <br />

<br />

' 2<br />

Y 1 x<br />

C4H8 <br />

1 x<br />

<br />

1 x <br />

Conversion of betene = 75.41%<br />

2<br />

x<br />

At 5.0 bar, k y = 0.26382 = , x = 0.4569, Conversion of betene<br />

= 45.69 %<br />

2<br />

1 x<br />

C.<br />

Moles in<br />

feed<br />

Moles in product stream<br />

C4H8 1 1 - X (1 – X ) / (2 + X)<br />

C4H6 X X / (2 + X)<br />

H2 X X / (2 + X)<br />

Inerts 1.0 1 1 / (2 + X)<br />

2 + x<br />

2 x<br />

2 <br />

x<br />

yi<br />

2<br />

x 2 x<br />

Y 2<br />

<br />

at 1.0 bar, k K 1.3191 <br />

1 x <br />

x = 0.8194<br />

K<br />

2<br />

x 2 x<br />

Y 2<br />

<br />

at 5.0 bar k 0.26382 <br />

5 1 x <br />

x = 0.5501

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