Fugacity-VK

7/29/2019 Fugacity-VK

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

to study extensively phase and chemical reaction equilibrium.We know that

SdTVdPdG -(1)

For isothermal conditionVdPdG For ideal gases

P

RTV

dPPRT

dG .

PRTddG ln

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

pressure. Which we call fugacity(f).

fRTddG ln -------(2)Applicable for all gases (ideal or real)

On differentiation

fRTG ln Is an constant depends on temperature and nature of gas.

fugacity has same units as pressure for an ideal gas.

For ideal GasesSdTVdPdG For Isothermal conditions

VdPdG from equation (2)

VdPfRTd ln

dPRT

Vfd ln

p

dpfd ln

pdfd lnln pf Fugacity = Pressure for Ideal gases


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

component to its pressure.

P

f

Is the measure of non ideal behavior of the gas.

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

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

state property and is denoted by subscripto.

oG = Standard Gibbs free energyo

f = Standard fugacity

Estimation of fugacity for gases

I method

SdTVdPdG For Isothermal conditions

VdPdG from equation (2)

VdPfRTd ln

dPRT

Vfd ln

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

P

dPRT

Vfd

1

ln

Lower limit is taken as P =1 atm

At 1 atm assuming the gases expected to behave ideally

P

VdPRT

f1

1ln

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


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II method

Using compressibility factor

VdPdG VdPfRTd ln --------------(2)

V in terms of compressibility terms it is given asP

ZRTV

Substitute V in equation 2

dPp

ZRTfRTd ln

pZdfd lnln subtracting both sides by dlnP

PdPZdPdfd lnlnlnln

Pdz

P

fd ln)1(ln

PdZd ln)1(ln Integrating the equation from 1 to and 0 to P

P

PdZd01

ln)1(ln

P

P

dPZ

0

)1(ln

Using generalized charts: using reduced properties a similar chart as compressibility chartis predicted for fugacity.

r

r

dPP

Z

P

f

1ln


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Fugacity is plotted against various reduced pressure at various reduced temperature

Using Residual Volume(): The residual volume is the difference between actualvolume (V) and the volume occupied by one mole of gas under same temperature and

pressureP

RTV ,

P

RTV

fRTddPP

RTdG ln

fRTd

P

dPdP

RT

RT ln

P

fddP

RTln

dPRTPf

ln

ProblemFor isopropanol vapor at 200oC the following equation is available

Z=1- 9.86 x 10-3P-11.45 x 10-5P2

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

253 1041.111086.91 PPRT

PVZ

)1041.111086.91( 253 PPP

RT

p

ZRTV

P

VdPRT

f1

1ln

P

dPPPP

RT

RTf

1

253 )1041.111086.91(1

ln

50

1

50

1

53

50

1

1041.111086.9ln PdPdPp

dpf


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2

)150(1041.111501086.9

1

50lnln

2253

f

f=26.744 bar

5348.050

744.26

Pf

Fugacity for liquids and solids

General expression for fugacity is

P

VdPRT

f1

1ln

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

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

constant. Integrating the above equation from condition 1 to 2

2

1

2

1

ln

P

P

f

f

dPRT

Vf

12

1

2ln PPRT

V

f

f

Problem:

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

volume 5.1x 10-2 m3/kg mole. What is the fugacity at 10 Mpa and 25oC

PaP

PaP

6

2

6

1

1010

1077.0

Paf6

1 107.0 KT 298Kgmole

JR 8314

121

2ln PPRT

V

f

f

f=0.846Mpa


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Activity(a):

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

of

fa

Effect of pressure on activity

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

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

fRTG ln oo fRTG ln

aRTf

fRTGGG

o

o lnln

at constant temperature VdPdG

G

G

P

PO o

dPVdG

)( oPPVG )(ln oPPVaRT

)(ln oPPRT

Va This equation predicts the effect of pressure on activity

Effect of Temperature on Activity

aRTGGG o ln

T

G

T

GaR

o

ln

Differentiating the above equation with T at constant P

P

o

P

P T

T

G

T

T

G

dT

adR

ln

22

ln

RT

H

RT

H

dT

adR

o

P


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2

ln

RT

HH

dT

adR

o

P

This equation predicts the effect of temperature on activity.


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Properties of solutionsThe relationships for pure component are not applicable to solutions. Which needs

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

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

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

Partial Molar properties:

The properties of a solution are not additive properties, it means volume of solution is notthe sum of pure components volume. When a substance becomes a part of a solution it

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

The term partial molar property is used to designate the component property when it isadmixture with one or more component solution.

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

got a set of properties associated with it like etcSV iP , . These properties are partially

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

It is defined as

ij

n

nMM

jnPTi

i

,,

iM = Partial molar property of component i.M= Any thermodynamic property of the solution

n = Total number moles in a solution

ni=Number of moles of component I in the solution

This equation defines how the solution property is distributed among the components.Thus the partial molar properties can be treated exactly as if they represented the molar

property of component in the solution.

The above expression is applicable only for an extensive property using

We can write

iiMnnM ii xn

n

iiMxM

xi=Mole fraction of component i in the solution.


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Measuring of partial molar propertiesTo understand the meaning of physical molar properties consider a open beaker

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

is 18x 10-6

m3

If the same amount of water is added to pure ethanol the volume increased

is approximately 14 x 10

-6

m

3

this is the partial molar volume of H2O in pure ethanol.The difference in volume can explain the volume applied by water molecules depending

on water molecules surrounding to them. When water is added to large amount ofethanol, ethanol molecules hence volume occupied surround all the water molecules will

be different in ethanol.

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

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

composition. The intermolecular forces also changes with change in thermodynamic

property.

Let wV = Partial molar volume of the water in ethanol water solution

wV = Molar volume of pure water at same temperature and pressuretV =Total volume of solution when water added to ethanol water mixture and allowed

for sufficient time so that the temperature remains constant

ww

t VnV

w

t

wn

VV

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

wV = property of solution for all infinitely small amount of water.

w

t

w

t

vwn

V

n

VV 0lim

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

water.

EnPTw

t

wn

VV

,,

nE- no of moles of ethanol

The partial molar volume of component i

inPTi

t

i

j

n

VV

,,


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Partial molar properties and properties of the solution

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

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

Let total property of the solutionnMMt

321 nnnn1,2,3 represents no of constituents

Thermodynamic property is a jnnnnPTf 321 ,,,,

For small change in the pressure and temperature and amount of various constituents canbe written as

i

inTP

t

nnpT

t

np

t

nT

tt dn

P

Mdn

P

MdT

P

MdP

P

MdM

j,,

1

,,,,, 32

At constant temperature and pressure dP and dTare equal to zero.The above equation reduces to

dnin

MdM

ijnTP

ni

i i

tt

,.1

dMtin terms of partial molar property

i

n

i

i

tdnMdM

1

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

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

property of the solution.

332211 dnMdnMdnMdMt

dnxMxMxMdMt 332211

332211332211nMnMnMnxMxMxMMt

iit MnM

Problem

A 30% mole by methanol water solution is to be prepared. How many m3

of pure

methanol (molar volume =40.7x10-3

m3/mol) and pure water (molar volume =

18.068x10-6m3/mol) are to be mixed to prepare 2m3 of desired solution. The partial molar

volume of methanol and water in 30% solution are 38.36x10-6

m3/mol and 17.765x10

-6

m3/mol respectively.


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Methanol =0.3 mole fraction

Water=0.7 mole fraction

Vt=0.3 x38.36x10

-6+0.7x17.765x10

-6

=24.025x10

-6

m

3

/molFor 2 m3soolution

mol36

10246.8310025.24

2

Number of moles of methanol in 2m3solution

=83.246x103x0.3= 24.97x10

3mol

Number of moles of water in 2m3solution

=83.246x103x07= 58.272x10

3mol

Volume of pure methanol to be taken

= 24.97x103 x 40.7x10-3 =1.0717 m3

Volume of pure water to be taken= 58.272x103 x 18.068x10-6 =1.0529 m3


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Estimation of Partial molar properties for a binary mixture :

Two methods for estimation

Analytical Method and Graphical Method(Tangent Intercept method)

Analytical Method: The general relation between partial molar property and molarproperty of the solution is given by

knPTK

kix

MxMM

,,

ik x1

For binary mixture

2,1 ki

PTx

MxMM

,2

21

112 xx , 12 1 xx , 12 xx

At constant Temperature and pressure

1

11 )1(x

MxMM ------(a)

1

2 1 x

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

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

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

Tangent Intercept method:

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

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


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

draw a tangent to this point to the curve. The intercept of the tangent with two axis x1=1and x1=0 are I1 and I2.

Slope of the tangent1

2

1 x

IM

dx

dM

1

12dx

dMxIM

112 dx

dM

xMI

Comparing I2 with equation (b) 22 MI

and also (right hand side)1

1

1 1 x

MI

dx

dM

1

11 1dx

dMxMI

1

11 1dx

dMxMI Comparing with equation(a) 11 MI

The intercept of the tangent gives the partial molar properties.


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

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

M and

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

M of component 2

2M

Problem

The Gibbs free energy of a binary solution is given by

mol

calxxxxxxG )10(150100 212121

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

dilution.

(b) Find the pure component properties

Sol:mol

calxxxxxxG )10(150100 212121

Substitute 12 1 xx and simplifying

1504989 12

1

3

1 xxxG

1

11 )1(x

GxGG

491627 12

1

1

xxx

G

101163518 12

1

3

11 xxxG

112 x

G

xGG

1508182

1

3

12 xxG

To find the partial molar properties of components 1 and 2

x2=0.8, x1=1-0.8 = 0.2


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101163518 12

1

3

11 xxxG

mol

calG 944.1021

150818 21312 xxG

mol

calG 824.1492

At infinite dilution

0111

atxGG

mol

calG 1011

1122

atxGG or x2=0

mol

calG 1602

To find the pure component property

1111 atxGG

mol

calG 1001

0122 atxGG

mol

calG 1501

Chemical Potential:


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

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

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

The chemical potential of component i

jnPTi

t

in

GGi

,,

21 ,,, nnTPfGt

ni

1i

i

n,Ti

t

n,P

t

n,T

tt dn

n

GdT

T

GdP

P

GdG

j

iinP

t

nT

tt dndT

T

GdP

P

GdG

,,

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

dTSdPVdG ttt

at constant temperature

t

T

VP

G

at constant pressure

t

P

ST

G

iittt dndTSdPVdG

At constant temperature and pressure

iiPTt dndG ,


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For binary solution 2211 xxG

Effect of temperature and pressure on chemical potential

Effect of temperature:

We know that

jnPTi

t

in

GGi

,,

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

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

inP

i

Tdn

G

T

2

,

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

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


ST

G

P

differentiating again w r t ni

i

nPi

t

i

Sn

S

nT

G

j

,

2

iS is partial molar entropy of component I

2

,

T

TT

T

Ti

i

nP

i

2T

ST ii

TSHG

In terms of partial molar properties


18/67

iii STHG

iii

iii

STH

STH

2

,T

H

T

i

nP

i

This equation represents the effect of temperature on chemical potential.

Effect of Pressure:

We know that

jnPTi

t

in

GGi

,,

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


inT

i

PdnG

P

2

,

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

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

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

VP

G

T


i

nTii

Vn

V

nP

G

j

,

2

,


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i

nT

i VP

,

This equation represents the effect of pressure on chemical potential

Fugacity in solutions

For pure fluids fugacity is explained as

fRTddG ln

1lim 0 P

fP

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

ii fRTdd ln

1P

flim

i

i

0P

i is Chemical potential

if is partial molar fugacity

For ideal gases Tii PyP

PT Total pressure.

Fugacity in Gaseous solutions

We know that

jnPTi

t

i n

G

Gi,,

------(1)


inT

i

Pdn

G

P

2

,

-----(2)


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SdTVdPdG -----(3)

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

VP

G

T


i

nTii

Vn

V

nP

G

j

,

2

,

i

nT

i V

P

,

PVii

dPVfRTd ii ln

Subtracting both sides by iPdln

ii ydPdPd lnlnln

Composition is constant dln yi=0

The above equation can be written as

P

dPPdPd i lnln

dPRT

Vfd ii ln

iiii PddPRT

VPdfd lnlnln

ii

i

i PddPRT

V

P

fd lnln


21/67

Modifying equation (a)

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

Ideal solutions

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

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

to its pure component volume at same temperature and pressure.

ii VV and iiii VxxVV --- Raoults lawIdeal solutins are formed when similar components or adjacent groups of group are

mixed ii VV Eg: Benzene-Toluene

Methanol- Ethanol

Hexane- HeptaneSolutin undergo change in volume due to mixing are known as non ideal solutions

ii VV Eg: Methanol-Water

Ethanol-water.

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

molecules are of the same magnitude.

Non-ideal solutions are formed when intermolecular forces between like molecules andunlike molecules of different magnitude.

For ideal solutions

ii

t VnV -------------------(1)

Vi is the molar volume of pure component I

ij

i

t

i Vn,P,Tn

VV

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

The residual volume for the pure component is

p

RTVi

Therefore we know from reduced properties

P

ii dP

P

RTV

RTP

f

0

1ln ----(3)

For component i in terms of partial molar properties

dPP

RTV

RTd ii

1

ln


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P

i

i

i dPP

RTV

RTP

f

0

1ln ----(4)

Subtracting equation (3) from (4)

P

ii

ii

i dPVVRTPf

Pf0

1ln ----(5)

we know that PyP ii equation (5) reduces to

P

ii

ii

i dPVVRTyf

f

0

1ln

comparing equation (2)

1ii

i

yf

f

iii yff --Lewis Randal ruleLewis Randal rule is applicable for evaluating fugacity of components in gas mixture.

Lewis Randal rule is valid for

1. At low pressure when gas behaves ideally.2. When Physical properties are nearly same.

3. At any pressure if component present in exess.

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

given as

iii kxf

iii kxP

ki- HenrysConstant, if -Partial molar fugacity,

iP -Partial pressure of component i.

Non Ideal solutions

For ideal solutions iii kxf ------------(a)


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For non ideal solutions iiii kxf ----------(b)

Where i is an activity coefficient of component i.

Comparing equation (a) and (b)

i

ii

ff Non ideal solution / Ideal solution

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

equation

i

i

ix

fxf lnln

iix lnln

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

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

Calculate the fugacity of the mixture.

Solution:

ccBBAA xxx lnlnlnln

)9.0ln(45.0)6.0ln(35.0)7.0ln(2.0ln

2975.0ln

7426.0

P

f , f =44.558atm

Gibbs Duhem Equation

Consider a multi component solution having ni moles of component I the property ofsolution be M in terms of partial molar properties

i

t MnnMM -----------------------(1).

Where n is the total no of moles of solution

Differentiating eq (1) we get

iiii dnMMdnnmd )( ----------------------------(2)We know that

21 ,,, nnPTfnM


24/67

2

,,,2

1

,,1,,,, 313.212121

)()()()()( dn

n

nMdn

n

nMdP

P

nMdT

T

nMnM

nnPTnnPTnnTnnP

i

nPTinnTnnP

dnn

nMdPPMndT

TMnnM

j,,,,,,

)()()()(

2121

From the definition of partial molar properties

ii

nnTnnP

dnMdPP

MndT

T

MnnM

2121 ,,,,

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

Subtracting equation (2) from equation (3)

0)()(

2121 ,,,,

ii

nnTnnP

MdndPP

MndT

T

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

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

Special Case

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

0ii Mdx

For binary solution at constant temperature and pressure the equation becomes02211 MdxMdx

0)1( 2111 MdxMdx ------------(5)

dividing equation(5) by dx1

0)1(1

21

1

11

dx

Mdx

dx

Mdx

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

and pressure in terms of Partial molar properties.

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

equation.


25/67

Problem:

Find weather the equation given below is thermodynamically consistent

)10(150100 212121 xxxxxxG

111

)1(x

GxGG

1

12x

GxGG

101163518 12

1

3

11 xxxG

1508182

1

3

12 xxG

167054 12

1

1

1 xxdx

Gd

1

2

1

1

2 1654 xxdx

Gd

G D equation

0)1(1

21

1

11

dx

Mdx

dx

Mdx

0)x16x54)(x1()16x70x54(x 12

111

2

11

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


26/67

Phase Equilibrium

Criteria for phase equilibrium: If a system says to be in thermodynamic equilibrium,Temperature, pressure must be constant and there should not be any mass transfer.

The different criteria for phase equilibrium are

At Constant U and V: An isolated system do not exchange mass or heat or work withsurroundings. Therefore dQ=0, dW=0 hence dU=0. A perfectly insulated vessel at

constant volume dU=0 and dV=0.

0, VUdSAt Constant T and V: Helmoltz free energy is given by the expression

TSUA TSAU

on differentiating

SdTTdSdAdU we know that

PdVTdSdU SdTTdSdAPdVTdS

SdTPdVdA

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

0, VTdAAt Constant P and T

The equation defines Gibbs free energyTSHG TSPVUG

SdTTdSVdPPdVdUdG

SdTTdSVdPPdVdGdU TdSPdVdGdGdU

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

0, PTdG

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

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

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


27/67

Phase Equilibrium in single component system

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

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

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

temperature and pressure. Appling the criteria of equilibrium

0dG

0 ba dGdGadG and bdG are chane in free energies of the phases a and b respectively.

We know that

iidnGSdTVdPdG For phase a

aaaaaaa dnGdTSdPVdG

For phase b bbbbbbb dnGdTSdPVdG At constant temperature and pressure

aaa dnGdG , bbb dnGdG For a whole system to be at equilibrium

ba

ba

dndn

or

dndn

0

0 aba dnGG

ba GG When two phases are in equilibrium at constant temperature and pressure, Gibbs freeenergies must be same in each phase for equilibrium.

CfRTCfRT ba lnln

baff

Clapeyron Clausius Equation

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

(a) Solid- liquid

(b) Liquid-Vapor(c) Solid Vapor


28/67

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

pressure. It is possible to transfer some amount of substance from one phase to otherin a thermodynamically reversible manner(infinitely slow).The equal amount of

substance will have same free energies at equilibrium .

Consider GA is Gibbs free energy in phase A and GB is Gibbs free energy in phase Bat equilibrium.

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

G = GA -GB =0 ------(2)At new temperature and pressure the free energy / mole of substance in phase A is

AA dGG for Phase B BB dGG

From basic equation

SdTVdPdG -----(3)

dTSdPVdG AAA

dTSdPVdG BBB

dTSdPVdTSdPV BBAA

ABAB SSdTVVdP

V

S

dT

dP

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

phase A to Phase B

T

QS

VT

Q

dT

dP

AB VVTQ

dTdP

----(4)

This is a basic equation of Clapeyorn Clasius equation.

BVapor state A- Liquid state

Q=molar heat of vaporization = VH

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


29/67

lg VVT

Q

dT

dP

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

g

V

TV

H

dT

dP

VH (Latent heat of vaporization),P

RTVg

TRT

P

dT

dP

2T

dT

RP

dP

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

211

2

T

1

T

1

RP

Pln

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

Problem:

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

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

41.27KJ/mole.

2

1

P

mmHg760P

K36827395T

K373273100T

2

1

2

1

2

1

T

T

P

PT

dT

RP

dP


30/67

211

2

T

1

T

1

RP

Pln

3681

3731

314.81027.41

760Pln

3

2

mmHg3.634P2

Phase Equilibrium in Multi component systemAn heterogeneous system contains two or more phases and each phase contains two or

more components in different proportions. Therefore it is necessary to develop phaseequilibrium for multi component system in terms of chemical potential.

The partial molar free energy or chemical potential is given as

jnPTi

iin

GG

,,

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

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

all phases.

Consider a heterogeneous system consists of phase ,, andcontaining various components 1,2,3-------C , that constitutes the system.

The symbolk

i represents chemical potential of component iin phase k.

At constant temperature and pressure, for the criterion for equilibrium isdG=0 ----(1)

Free energy for multi component system given by the expression

iidnSdTVdPdG ------(2)At constant Temperature and pressure

iidndG ----(3)

Comparing equation 1 and 3

ii dn0For multi component system

C

i k

k

i

k

i dn1

0

Expanding the above equation


31/67

0

22222222

11111111

cccccccc dndndndn

dndndndn

dndndndn

-----(4)

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

0

222

111

CCC dndndn

dndndn

dndndn

----(5)

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

change in mole in all the phases must be zero. For the criterion of equilibrium is that thechemical potential of each component must be equal in all phases.

CCC

222

111

At constant temperature and pressure the general criterion for thermodynamicequilibrium in closed system for heterogeneous multi component system

At constant T and P

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

Since iii fRTG ln

The above equation is also satisfied in terms of fugacity

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


32/67

Phase Diagram for Binary solution

Constant pressure equilibrium

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

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

fixed at the liquid composition can be changed the properities such as temperature andvapour compositions get quickly determined VLE at constant pressure is represented on

T-xy diagram.

Boiling point diagrams

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

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

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

sub cooled liquid and above the Dew point is super heated vapour. The region betweenbubble point and Dew point is called mixture of liquid and vapour phase.

Consider a liquid mixture whose composition and temperature is represented by

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

where the liquid starts boiling, temperature at that point is called boiling point of whetherheating mixture reaches to point G when all liquids converts to vapour the temperatureat that point is called as Dew point. Further heating results in super heated vapour. The

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

boiling point has no meaning because temperature varies from boiling point to Dew pointat constant pressure.


33/67

Effect of pressure on VLE

The boiling point diagram is drawn from composition x=0 to x=1, boiling point of puresubstances increases with increase in pressure. The variation of boiling point diagrams

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

pressure.


34/67

Equilibrium diagram

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

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

Constant temperature Equilibrium

VLE diagram is drawn against composition at constant temperature. The upper curve isand lower curve is drawn at vapour comp(y). Consider a liquid at known pressure and

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

starts boiling further decreases in pressure reaches to point C when all the liquidconverts to vapour, further decreases in pressure leads to formation of super heatedvapour. In between B to C both liquid and vapour exists together


35/67

Non ideal solution

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

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

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

law, the system shows positive deviation from Raoults law.Eg: Ethanol-toluene

When the total pressure at equilibrium is less than the pressure given by Roults law thesystem shows negative deviation from Raoults law.

Eg: Tetrahydrofuron-ccl4

Azeotropes

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

large p-x and p-y curve meets at this point y1=x1 and y2=x2A mixture of this composition is known as Azeotrope. Azeotrope is a vapour liquid

equilibrium mixture having the same composition in both the phases.

Types of Azeotropes

Azeotropes are classified into two types

1. Maximum curve pressure, minimum boiling azeotropes

2. Minimum pressure maximum boiling azeotropeThe azeotrope formed when negative deviation is very large will exhibit minimum

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

Eg : Ethanolwater, benzene-ethanol

The azeotrope formed whenue deviation is very large will exhibit minimum pressure or

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

Phase diagram for both types of AzeotropesMinimum Temperature azeotropes


36/67


37/67

Maximum TemMin pressure Azeotropes

Calculation of VLE for ideal solution

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

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

AVAA xPP PA- Partial pressurePVA- Vapor pressure


38/67

xA- Mole fraction of component A

For binary solution (For component A and B)

We know that BAT PPP -------(1)

AVAA xPP , BVBB xPP Substitute PA and PB in equation 1

BVBAVAT xPxPP

BA xx 1 , BA xx 1)1( AVBAVAT xPxPP

VBAVBVAT PxPPP --------(2)

VBVA

VBT

APP

PPx

Assuming the vapor phase is also ideal

T

AVA

T

AA

P

xP

P

Py

Substituting PT from equation (2)

VBAVBVAAVA

APxPP

xPy

Dividing numerator and denominator by PVB

11

A

VB

VA

A

VB

VA

A

xP

P

xP

P

y

11

A

AA

x

xy

The above equation relates x and y

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


39/67

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

following data prepare

P-x-y diagram at 50oC T-x-y diagram at 400mm Hg x-y diagram at 400 mm Hg

T PV1 PV238.45 400 159.4

42 458.3 184.6

46 532 216.8

50 615 253.5

54 707.9 295.2

58 811.8 342.3

62.3 937.4 400

Solution: 2121 VVVT PxPPP ,T

V

AP

xPy 11 , PV1= 615, PV2 = 253.5

x1 PT y10.0 253.5 0.0

0.2 325.8 0.377

0.4 398.1 0.6179

0.6 470.4 0.784

0.8 542.7 0.906

1.0 615 1

T-x-y diagram at 400 mm Hg , PT=400mmHg21

21

VV

VT

PP

PPx

,

T

V

AP

xPy 11

T PV1 PV2 x1 y138.45 400 159.4 1 1

42 458.3 184.6 0.7869 0.9015

46 532 216.8 0.5812 0.7729

50 615 253.5 0.405 0.6226

54 707.9 295.2 0.2539 0.449

58 811.8 342.3 0.122 0.2475

62.3 937.4 400 0 0Calculation of VLE for non Ideal solution

For non ideal solution

Partial pressure is given by Viiii PxP

i - Activity coefficient

xi- Mole fraction

PVivapor pressure


40/67

For binary mixture

1111 VPxP

2222 VPxP

21 PPPT

222111 VVT PxPxP

111

222222111

11111

1

1

V

VVV

V

T

Px

PxPxPx

Px

P

Py

The above equation relates x and y for non ideal solution

Activity coefficients are functions of liquid composition x, many equations are availableto estimate them. The important equations are

Vanlaar equation

Wilson equation

Margules Equation

Vanlaar Equation:Estimation of activity coefficient

2

21

2

21ln

xxBA

Ax

2

21

2

12ln

xA

Bx

Bx

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

activity coefficients can be estimated.

Estimation of Vanlaar constants

1st

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

constants A and B can be estimated by rearranging the equation2

11

221

ln

ln1ln

x

xA


41/67

2

22

112

ln

ln1ln

x

xB

2nd

Method

For a systems forming azeotrope, if the temperature and pressure s are known atazeotropic composition then activity coefficients can be calculated as shown below.

1111 VPxP

1111 VT PxyP

At azeotrpic composition x1=y1

1

1

V

T

P

P, 2

2

V

T

P

P

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

Problem: The azeotope of ethanol and benzene has composition of 44.8mol% C 2H5OH at

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

mmHg.

Calculate

Vanlaar constants Prepare the graph of activity coefficients Vs composition Assuming the ratio of vapor pressure remains constant, prepare equilibrium

diagram at 760mmHg.

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

PV2 =517mmHg

1

1

V

T

P

P, 2

2

V

T

P

P

501.1506

7601 , 47.1

517

7602

2

11

221

ln

ln1ln

x

xA

2

)501.1ln()448.0(

)47.1ln()552.0(1)501.1ln(

A


42/67

A=0.829

2

22

112

ln

ln1ln

x

xB

2

)47.1ln()552.0(

)501.1ln()448.0(1)47.1ln(

B

B=0.576

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

2

21

2

21ln

xxB

A

Ax ,

2

21

2

12ln

xA

Bx

Bx

111

222222111

11111

1

1

V

VVV

V

T

Px

PxPxPx

Px

P

Py

x1 x21 2 y1

0 1 6.745 1 00.2 0.8 2.4 1.095 0.384

0.4 0.6 1.644 1.374 0.4384

0.6 0.4 1.21 1.721 0.5077

0.8 0.2 1.0112 2.618 0.609

1.0 0 1 3.723 1

Margules equation

Eetimation of activity coefficients

12

21 2ln xABAx

22

12 2ln xBABx

The constant A in the above equation is terminal value of ln1 at x1=0 and constant B isthe terminal value of ln2 at x2=0


43/67

When A=B2

21ln Ax ,2

12ln Ax

The above equations are known as Margules suffix equation.

Wilson Equation:Wilson proposed the following equation for activity coefficients in binary solution

2121

21

2121

12221211 lnln

xxxxxxx

2121

21

2121

12112122 lnln

xxxxxxx

Wilson equation have two adjustable parameters 12 and 21 . These are related to pure

component molar volumes.

RT

a

V

V

RTV

V 12

1

21112

1

212 expexp

RT

aexp

V

V

RTexp

V

V 21

2

12212

2

1

21

V1 and V2molar volumes of pure liquids

- Energies of interaction between molecule

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

on ln versus x curves.Consistency of VLE data

Gibbs duhem equation in terms of thermal consistency

0ln

)1(ln

1

21

1

11

dx

dx

dx

dx

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


44/67

According to Gibbs Duhem equation both slopes must have oppsite sign then only thedata is consistent, otherwise it is inconsistent.For the data to be consistent it has satisfy the following condition.

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

should be minimum at same composition.

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

range.

Co-existence equation

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

0ii Mdx ---(1)In terms of partial molar free energies

0ii Gdx ------(2)dividing equation (2) by dx1


45/67

01dx

Gdx ii -----(3)

ii fRTG ln

at constant temperature

11

ln

dx

fRT

dx

Gd ii ----(4)

substituting equation (4) in equation (3)

0ln

1

dx

fRTx ii

0ln

1

dx

fdx ii ------(5)

For binary system

0lnln

1

22

1

11

dx

fdx

dx

fdx ----(6)

0lnln 2211 fdxfdx -----(7)

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

applicable for both liquid and vapor phase

For liquids

iiii fxf

iiii fxf lnlnlnln

differentiating with respect to x1

1111

lnlnlnln

dx

fd

dx

xd

dx

d

dx

fd iiii

-------(8)

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

Substituting equation (8) in equation (5)

0lnln

11

dx

xdx

dx

dx ii

i

i

=0

=


46/67

0ln

1

dx

dx ii

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

For binary system

0lnln

1

22

1

11

dx

dx

dx

dx

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

0lnln 2211 dxdx -----------(11)

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

For ideal vapor

ii Pf From equation 5

0ln

1

dx

Pdx ii ------------(13)

For binary system

0lnln

1

22

1

11

dx

Pdx

dx

Pdx ----------(14)

0lnln2211

PdxPdx ---------------(15)

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

For ideal vapor

TPyP 11 , TPyP 22

0)ln()ln( 2211 TT PydxPydx

0ln)(lnln 212111 TPdxxydxydx

121 xx

0lnlnln 2111 TPdydxydx

2211 lnln ydxydxP

dP


47/67

2

22

1

11

y

dyx

y

dyx

P

dP

012 yy , 12 dydy

on simplification

)1(

)1(

1

1

1

11

y

x

y

xdy

P

dP

Further simplification

)1( 1

11

1 yy

xyP

dy

dP

This is known as coexistence equation. It relates P,x,y for binary VLE system. Thisequation is used to rest consistency of VLE data.

Consistency test for VLE data

0lnln 2211 dxdx

2

112

lnln

x

dxd

)1(

lnln

1

112

x

dxd

)1(ln

ln1

112

x

dxd

Redlich Kister Test

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

Nonidealideal

e

GGG

iie xRTG ln

For binary system

2211 lnln xxRTGe


48/67

differentiating with respect to x1

1

22

1

221

1

11

1

lnln

lnln

dx

dx

dx

dx

dx

dxRT

dx

dG e

From Gibbs Duhem Equation

0lnln

1

22

1

11

dx

dx

dx

dx

1

221

1

lnlndx

dxRT

dx

dG e

12 dxdx

211

lnln RTdx

dG e

2

1

1

ln

RT

dx

dG e

1

0

1

2

1

1

0

1

1

1

1

ln

x

x

x

x

edxRTdG

The two limits indicates pure component for which 0eG , where LHS is zero for bothlimits.

We can write

1

0

1

2

11

1

ln0

x

x

dx

This can be checked graphically. Net area should be equal to zero forconsistency(Area=0).


49/67

Problem

Verify whether the following data is consistent

X1 1 20 0.576 1.00

0.2 0.655 0.985

0.4 0.748 0.930

0.6 0.856 0.814

0.8 0.950 0.6261.0 1.00 0.379

Solution:

We know that the Redlich kister test is

1

0

1

2

11

1

ln0

x

x

dx

2

1

2

1ln

0.576 -0.5520.665 -0.408

0.804 -0.218

1.052 0.051

1.518 0.417

2.639 0.97

Plot2

1ln

Vs x1

Area under the curve is zero. Data is consistent.


50/67

CHEMICAL REACTION EQUILIBRIUM

Energy change accompanies all chemical reactions. Because of this energy

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

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

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

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

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

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

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

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

elements.

Eq. H2 + O2 H2OC +O2 CO2

C + H2 + 1/2 Cl2 CHCl3

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

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

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

250C.

In chemical industries, processes are carried out under isothermal conditions

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

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

value helps to design the heat exchange equipment.

Standard heat of reaction

Calculation ofH298 form heats of formation data:

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

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

algebraic sum of the standard heat of formation of reactants.

Hr = H f 298 products - Hf 298 reactants

Heat of formation of any clement is zero


51/67

Heat effects of chemical Reaction:

For the reaction, aA + bB cC + dD

Let heat capacity be Cp = + T + T2

then we have CpA = A + AT + AT2

CpB = B + BT + BT2

CpC = C + CT + CT2

CpD = D + DT + DT2

And H0

298 be standard heat of reaction at 298 K and standard heat of reaction at any

other temperature can be found by Kirchoffs rule o

p

d HC

dT

where

= [c(C + CT + CT2) + d( D + DT + DT

2 ) ] - [a (A + AT + AT2

) + b (B + BT +

BT2 )]

= [(c C +d D ) (a A + b B ) ]+ [ (c C + d DT) (a A + b B)]T + [ c C + dD )

(a A+ b B )] T2

or

Pr Re tan

Pr Re tan

Pr Re tan

oducts ac ts

C D A B

oducts ac ts

oducts ac ts

cC dC aC bC

Substituting in Kirchoffs rule

, , ,pC pD pA pB

C C and C C

Cp = + T + T

tan

substituting the values of

p p p

products reac ts

p pC pD pA pB

C C C

C cC dC aC bC then


52/67

0

298

0 2

0 2

298

2 3

2 3

H T

H

o

H T T T

dH T T

dT

H T T T I

OR in general 0 2

2 3=2 3

PH C dT T T dT

T T T I

A chemical reaction proceeds in the direction of decreasing free energy.The sum of thefree energies of the reactants should be more than the sum of energies of products. For a

reaction to take place

Re Reor 0action actionproducts reaction

G G G G

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

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

criterion for reaction equilibrium is G = 0

GReaction Calculation: Consider the reaction aA + bB cC + dD

Let , , ,A B C DG G G and G be the partial molar free energies of A,B,C, and D in the

reaction mixture.

0 A A

0 0

0

0

0

0

f fln but the activ ity o f A

ln

ln

ln

ln

A A A

A A

A A A

B B B

C C C

D D D

G G R T af f

G G R T a

G G R T a

G G R T a

G G R T a

2 3

2 3

oH T T T I


53/67

ReRe tan

0 0

0 0

= c ln ln

ln ln

actionproducts ac ts

D BC A

C C D D

A A B B

C D A B

G G G

cG dG aG bG

G RT a d G RT a

a G RT a b G RT a

cG dG aG bG

0

Re

ln ln ln ln

ln

C

C D A B

c d

D

action a b

A B

RT a a a a

a a

G G RT a a

At equilibrium G = 0 and Cc d

D

a b

A B

a aK

a a

K is equilibrium constant of the reaction

0 00 ln lnCc d

D

a b

A B

a aG RT G RT K

a a

at equilibrium


54/67

EFFECT OF TEMPERATURE ON EQUILIBRIUM CONSTANT K

Gibbs - Helmoholtz equation at constant pressure is given

2

2 2

2

but ln K

ln K

ln KTor

ln KVan't Hoff equation.

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

When is

o

o

Gd

T HG RT

dT TRT

dd RH H

dT dT T T

d H

dT RT

H

2 2

1 1

K

2K

2

1 2 1

1 1

contant d lnK

K 1 1ln K

if K at T ,t

T

T

HdT

RT

H

R T T

is known

2 2

0

2

2

hen K can be calculated at other T

ln KWhen H varies with T

ln K2 3

d H

dT RT

T Id dT

RT R R RT

2ln Tln K =

2 6

T T IM

R R R RT

Where M is a consatnt of integration

0

20

2 30

ln

ln T

2 6

ln T2 6

G RT K

T T IG RT M

R R R RT

T TG T I MRT


55/67

For the reasction C2H4 + O2 C2H4O, develop equations for H0, K, and G0, find

G0at 550 K

Cp (Cal / mol K) Data: C 2H4 = 3.68 + 0.0224 T

O2 = 6.39 + 0.0021 T

C2H4O = 1.59 + .00332 T

Standard heat of reaction H0298

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

And G0298 = -19070 Cal / mol

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

p

Pr Re tan

In this problem C

1= 1.59 3.68 6.39 5.285

2oducts ac ts

T

Pr Re tan

1= 0.0332 0.0224 0.0021 0.00975

2oducts ac ts

Then the variation of standard heat of reaction with temperature is

Re 298

Pr Re tan

action

oducts ac ts

H H H


56/67

2 3 2

0 2

298

2 0

298

0.009755.285

2 3 2

0.00975Given =310 = 5.285 then I = 1452

2

0.00975

5.285 1452 This is variation of with temparature2

lnK =

o

o

H T T T I T T I

H T T I

H T T H

2

0

0 298

298 298 298

5.285 0.00975 1452ln = ln

2 6 2 2(2) 2

G 19070G ln or lnK 32

2(298)

5.285 0.00975 145232 = ln298 298 therefore M = 48.76

2 2(2) 2(298)

IT T T M T T M

R R R RT T

but RT K RT

M

0

0 3 2

0 3 2

550

3

5.285 0.00975 1452lnK = ln 48.763

2 4 2

G ln 2 ln5.285 0.00975 1452

2 ln 2 ln 48.7632 4 2

G 5.285 ln 4.875(10 )( ) 1452 97.526( )

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

T TT

RT K T K

T K T T T T

T T T T

0

550

.526(550)

G 35320.62 /Cal mol

50 mol% each of SO2and O2 is fed to a converter to form SO3.

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

with T. The Hfand Gfat 298 K are

Component

Hf

k Cal / k

mol

Gf

k Cal / k

mol

SO2 -70960 -71680

O2 0 0

SO3 -94450 -88590


57/67

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

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

O2 SO3

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

Pr tan

-3 3 -3 -3 -3

Pr tan

0 2 3

1= 6.0377 + 6.148 7.116 6.148 - 4.152

2

1= 23.53(10 ) + 3.102(10 ) - 9.512 (10 ) + 3.102 (10 ) =12.474 x10

2

2 3

=

oducts reac ts

oducts reac ts

H T T T I

-32

0 0 0 0

298Re tan

-3 2

0

12.474x10-4.152T+

2

298 is given by 94450 70960 23490 k Cal / k mol

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

I = -22818.19

4.152 6.237 1

products ac ts

T I

But H at H H H

I

H T x

3 2

2

3

0 22818.19

for K we have

lnK= lnR 2 6

4.152 12.474 10 22818.19lnK= ln

2 2 2

T

NextI

T T T M R R RT

xT T M

xR T

CPdata:

Component a b(10 )

SO2 7.116 9.512

O2 6.148 3.102

SO3 6.0377 23.537


58/67

00 298298 298 298

0

298Products Reactants

298

3

GWe have G = -RTln or lnK =

2(298)

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

16910lnK = 28.372

2(298)

4.152 12474 1028.372 ln298

2 2

K

x

3

22818.19298

2 2 298

0.9842

11409.09ln 2.076ln 3.1185 10 0.9842

Mx x

M

K T x T T

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

H2 NH3The absolute entropies in ideal gas state at 298 K and 1.0 atm are

NH3 = 46.01 g Cal / g mol K

N2 = 45.77

H2 = 31.21

H0

298 (g Cal / g mol) = - 11040

Heat capacity Data is given by Cp= a + bT + cT

2

the values areComponent a bx10

2cx10

5

NH3 6.505 0.613 0.236

N2 6.903 - 0.038 0.193

H2 6.952 - 0.46 0.096

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


59/67

0 0

298Pr Re tan

0

298

We have G and

1 3=46.01 (45.77) + (31.21) = 23.69 g Cal / g mol K

2 2

G 11040 298( 23.69) 39

o

oducts ac ts

H S S S S

Pr Re tan

2 2

Pr Re tan

80.38 g Cal / g mol K

1 36.505 (6.903) 6.952 7.3745

2 2

1 30.613 10 ( 0.038 10 ) ( 0.046 10

2 2

oducts ac ts

oducts ac ts

a a a

b b b x x x

2 3

5 5 5 6

Pr Re tan

0 2 3

298

3 62

) 7.01 10

1 30.236 10 (0.193 10 ) (0.096 10 ) 0.045 10

2 2

2 3

7.01 10 0.045 1011040 ( 7.3745)298 298 (

2 3

oducts ac ts

x

c c c x x x x

b cH aT T T I

X X

3

2

298) OR I = -9153.26

ln ln and R =2 gCal / g mol K2 6

I

a b c I K T T T M

R R R RT

0

298 298 298

3 62

-3980.38ln or ln K = = 6.6784

-(2x298)

7.3745 7.01 10 0.045 10 9153.266.6784 ln298 298 (298)

2 2 2 6 2 2 298

11.798

G RT K

x xM

x x x

M

0

3 60 2

3 60 2

1500

7

var with T is given by, = - RT lnK

-7.3745 7.01 10 0.045 10 9153.26= -2T ln 11.7987

2 2 2 6x2 2T

-7.3745 7.01 10 0.045 10= -2x1500 ln1500 1500 1500

2 2 2 6x2

The iation of G G

x xG T T T

x

x xG

x

0

1500Pr Re tan

9153.2611.7987

2x1500

g Cal28488.81

g moloducts ac tsG G G

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


60/67

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

The heat capacity data for the component can be

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

and Cp is in cal / g mol0C

Develop general expression for the equilibrium constant and standard Gibbs free

energy change as function of temperature.

Pr Re tan

3 3 3 3

Pr Re tan

2

3

= 6.99 (2.83 7.3) 3.14

39.741 10 (28.6 10 2.46 10 ) 8.68 10

lnK = lnR 2 6

3.14 8.68 10lnK = ln2 2 2 2

At 418 K

oducts ac ts

oducts ac ts

a a a

b b b x x x x

a b c I T T T M

R R RT

x IT T Mx T

3-2

3-3

3

3.14 8.68 10, ln 6.8x10 = ln 418 418

2 2 2 2 418

, I = -9655.807 & M = -5.667

3.14 8.68 10593 K, ln 1.9x10 ln593 593

2 2 2 2 593

3.14 8.68 10 -9655.80lnK = ln

2 2 2

x IM

x x

Solving

x IAt M

x x

xT T

x

3

-3 2

7-5.667

2

3.14 8.68 10 -9655.807But G = -2T ln -5.667

2 2 2 2

G = 3.14T ln T - 4.34x10 9655.807 11.334

T

xT T

x T

T T

Temperature 0C 145 6.8 x 10-2

Equilibrium Constant 320 1.9 x 10-

Component a bx103

C2H4 2.83 28.601

H2O 7.3 2.46

C2H5OH6.99 39.741


61/67

THERMODYNAMIC FEASIBILITY OF REACTIONS

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

formed at equilibrium. It is related by the equation G0

= - RT ln K. The more the value

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

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

chemical reaction.

i) If G0

< 0, there can be appreciable conversion of reactants in to products.

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

ii) If G0

is +ve but less than 10 000 k Cal / mol, the reaction is not feasible,

at atmosphere , may be feasible at any other pressure

iii) If G0 is greater than 10 000 kCal / mol the reaction is not at all feasible

under any condition.

Equilibrium Calculations ( Homogeneous gas Phase reactions):

aA (g) + bB (g) cC (g) + dD (g)

We have Cc d

D

a b

A B

a aK

a a

for gases,

0

0

c c c c c c c c

D D D D

A A A A

B B B B

c c c

as = p = 1 atmosphere: a = f

f = y but a p = y p

a y p

a y p

a y p

y pK =

c

c

fAc tivity a E f

ff x

D D D c D c D c D

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

c + d - (a + b) n

y p y y p px x

y p y p y y p p

( )( )( )(P ) =( )( )( )(P ) where = activity co efficient for gas phase

d c d c d c d c d

a b a b a b a b a b

y y

x

K k k k k k k

y = mole fraction

= Fugac ity Co efficient for gas phase

P = Total pressure


62/67

Effect of variables on equilibrium :

Effect of temperature: 0298 298

lnG RT K since G depends only on temperature as the

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

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

2

ln KVan't Hoff equation.

d H

dT RT

For endothermic reactions H0

is + ve as T increases K also increases. This

means that the equilibrium conversion is more at higher pressure.

For exothermic reactions H0 is ve. Therefore K increases with T. Hence the

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

Effect of pressure: Consider a reaction of ideal gases, then K = KyPn

or Ky = K / Pn

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

yield is less at high pressures

When n 0, addition of inert increases Ky and equilibrium

conversion

Andn < 0, addition of inert decreases equilibrium conversion.

Effect of excess reactants: Presence of excess reactants increases equilibrium

conversion of the limiting reactant

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

Eg. CH3COOH + C2H5OHCH3COOC2H5 + H2O

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

reduces from 30% to 15%


63/67

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

taken in stoichiometric ratio at 1.0 atmosphere and 3000C. At this temperature G0 = -

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

maximum conversion of C2H2

n=2(1+1) = 0 0 330100

we have ln or lnK = or K = 1.8029x108.314(573)

G RT K

ComponentMoles

in feed

Moles

reacted

Moles

present

Mole

fraction

N2 1 X 1X (1 X) / 2C2H2 1 X 1X (1 X) / 2

HCN -- 2X 2X(2X) / X =

X

2 2 2

0

2 23

1 1

Assume 1 and 1

1.8029(10) , X = 0.02071 1

2 2

Equilibrium Composition:

n n

y

HCN

y

N C H

K K K K P K K P P

y XK K Solving for X

X Xy y

2

2 2

1 1 0.0207N 100 100 48.965%

2 2

1 1 0.0207100 100 48.965%

2 2

( )100 2.007%

X

XC H

HCN X

Equilibrium Conversi

2 2

2 2

2 2

2 2

max for reversible reaction

of 0.0207Max. Conversion of 100 100 100 2.07%of in feed 1 1

on of C H

Equilibrium conversion is imum

Moles C H reacted XC H Moles C H

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

of N2,,C2H2and HCN are 1.1, 0.928, and 0.54 Assume K = 1


64/67

c c

0C C

a b a

A B A

2 2-3

y= (1) 203 where a,b, and c are stoichiometric co efficients

y

0.54 Xand K is known already = 1.8029x10 = .(1)

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

2 2

solving

n

y b

B

K K K K P y

2 2

2 2

2 2

X = 0.0382

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

moles C H 0.0382= x 100 = 3.82%

moles of C H in feed 1

reacted

Calculate the K at 673K & 1.0 bar for N2 (g) + 3 H2 (g) 2NH3 (g).Assume heat of reaction remains constant. Take standard heat of formation and

standard free energy of formation of NH3at 298 K to be - 46110 J / mol and 16450 J

/ mol respectively.0

298 0

298 2980

298

0

298 1

2( 16450) 32900 Jln

2 (-46110) = -92 220 J

ln

GG RT K

H

G RT K

1

1

02

1 2 1

2

- 32900 = - 8.314 (298) ln K

K 564861.1

K 1 1have ln

K

K 92220 1 1ln

584861.1 8.314 673 298

Hwe

R T T

4

2K 5.75 10X

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

to

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

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

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

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

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

ideal solution.


65/67

G0

298 = 318218237130 + 389900 + 174780 = 9332 J

H0

298 =463250285830 + 277690 + 484500 = 13110 J

We have, G0298 =RT ln K

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

and for K2 at 373K we have

ComponentMoles

in feed

Moles

reacted

Moles

present

Mole

fractionCH3COOH 1 X 1X (1X) / 2

C2H5OH 1 X 1X (1X) / 2

CH3COOC2H5 -- X X X / 2

H2O -- X X X / 2

2

2

20

2 2 2

3 2 5

x

20.067= solving, x = 0.252

(1 )1

2

0.252mole fraction of CH 0.126

2 2

y

xK K P

xx

xCOOC H

CH3COOH

(L)

C2H5OH

(L)CH3COOC2H5(L) H 2O (L)

f0f

(J)

- 484500 - 277690 - 463250 - 285830

G0f

(J)- 389900 -174780 - 318218 - 237130

1

2 1 2

2

2

1 1ln

0.02313 12110 1 1ln OR K 0.067

8.314 298 373

K H

K R T T

K


66/67

The standard free energy change for the reaction C4H8 (g) C4H6(g) + H2 (g)is given by GT

0=1.03665 X 10

5 20.9759T ln T 12.9372 T, where range of GT0

in J

/ mol and T in K

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

point?

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

bar & 5.0 bar

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

a. For the reaction at G = 0, T 812.4 K, above this temperature G 0, reaction is unfavorable,

b. At 800K, G0 = 1842.16 J / mol

G0 = - RT ln K, or 1842.16 = - 8.314 ( 800 ) ln K or

K = 1.3191

K = ky Pn

= ky P( 2- 1 ) = ky P

Therefore at 1.0 bar K = kyP 1.3191 = k y ( 1 )

ky

= 1.31910At 5 bar, 1.3191 = k y

( 5 )

ky = 0.26382

Moles in

feed

Moles in product streamyi

C4H8 1 1 - X(1 X ) / (1 +

X)

C4H6 X X / (1 + X)

H2 X X / (1 + X)

1 + X


67/67

4 6

4

' ' 2C H H2

Y ' 2

C H8

2

y 2

x x

Y Y x1 x 1 xk or 1.3191 = or x = 0.7541

1 xY 1 x

1 x

Conversion of betene = 75.41%

xAt 5.0 bar, k = 0.26382 = , x = 0.4569, Conversion of b

1 x

etene = 45.69 %

C.

Moles in

feed

Moles in product streamyi

C4H8 1 1 - X (1X ) / (2 + X)

C4H6 X X / (2 + X)

H2 X X / (2 + X)

Inerts 1.0 1 1 / (2 + X)

2 + x

2

Y 2

2

Y 2

x 2 xat 1.0 bar, k K 1.3191

1 x2 x

x = 0.8194

K x 2 x

at 5.0 bar k 0.263825 1 x2 x

x = 0.5501

Documents

Fugacity-VK