Thermodyn Compiled Gmmadhu

8/6/2019 Thermodyn Compiled Gmmadhu

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e-Notes by G.M.Madhu, RVCE, Bangalore

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 thatSdTVdPdG = -(1)

For isothermal conditionVdPdG =

For ideal gases

P

RTV =

= dPP

RTdG .

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 conditionsVdPdG =

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 acomponent 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 standardstate property and is denoted by subscripto.

oG = Standard Gibbs free energy

of = Standard fugacity

Estimation of fugacity for gases

I methodSdTVdPdG =

For Isothermal conditionsVdPdG =

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 factorVdPdG =

VdPfRTd =ln --------------(2)

V in terms of compressibility terms it is given as

P

ZRTV =

Substitute V in equation 2

dPp

ZRTfRTd =ln

pZdfd lnln =subtracting both sides by dlnP

PdPZdPdfd lnlnlnln =

PdzP

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 chart

is predicted for fugacity.

r

r

dPP

ZPf = 1ln

Fugacity is plotted against various reduced pressure at various reduced temperature


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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=

+=

fRTdP

dPdP

RTRT ln=

+

=

P

fddP

RTln

= dPRTPf

ln

Problem

For 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(

1ln

=50

1

50

1

53

50

1

1041.111086.9ln PdPdPpdpf

( )2

)150(1041.111501086.9

1

50lnln

2253

= f

f=26.744 bar


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5348.050

744.26 ===P

f

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

lnP

P

f

f

dPRT

Vf

( )121

2ln PPRT

V

f

f=

Problem:

Liquid chlorine at 25oC has a vapour pressure of 0.77Mpa, fugacity 0.7Mpa and Molarvolume 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 298= KgmoleJ

R 8314=

( )12

1

2ln PPRT

V

f

f=

f=0.846Mpa

Activity(a):

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

of

fa =

Effect of pressure on activity


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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=

aRT

f

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

+=

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:


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

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

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

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

admixture with one or more component solution.

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

got a set of properties associated with it 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 solutionn = Total number moles in a solutionni=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 molarproperty of component in the solution.

The above expression is applicable only for an extensive property using

We can write

= ii MnnM ii xn

n=

= ii MxM

xi=Mole fraction of component i in the solution.

Measuring of partial molar properties

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

containing huge volume of water in one mole of water is added to it, the volume increaseis 18x 10-6m3 If the same amount of water is added to pure ethanol the volume increased

is approximately 14 x 10-6m3 this is the partial molar volume of H2Oin pure ethanol. The

difference in volume can explain the volume applied by water molecules depending onwater molecules surrounding to them. When water is added to large amount of ethanol,

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

different in ethanol.


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If same quantity of water is added to an equimolar mixture of H 2O and ethanol, the

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

composition. The intermolecular forces also changes with change in thermodynamicproperty.

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

wV = Molar volume of pure water at same temperature and pressuret

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

allowedfor 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

=

,,

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 solution

nMMt =+++= 321 nnnn

1,2,3 represents no of constituents

Thermodynamic property is a )jnnnnPTf 321 ,,,,For small change in the pressure and temperature and amount of various constituents can

be written as


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+

+

+

+

=

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 dPand dTare equal to zero.The above equation reduces to

dnin

MdM

ijnTP

ni

i i

tt

=

=

=,.1

dMt in terms of partial molar property

i

n

i

i

t dnMdM =

=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.

+++= 332211dnMdnMdnMdM

t

( )dnxMxMxMdM t +++= 332211( )

332211332211nMnMnMnxMxMxMM t ++=+++=

= iit MnM

ProblemA 30% mole by methanol water solution is to be prepared. How many m 3 of puremethanol (molar volume =40.7x10-3m3/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.

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= 24.97x103mol


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Number of moles of water in 2m3solution

=83.246x103x07= 58.272x103mol

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

Estimation of Partial molar properties for a binary mixture :Two methods for estimation

Analytical Method and Graphical Method(Tangent Intercept method)

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

property of the solution is given by

knPTK

kixMxMM

,, = ik x1

For binary mixture2,1 == 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).


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Tangent Intercept method:

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

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

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

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

and x1=0 are I1 and I2.

Slope of the tangent 1

2

1 x

IM

dx

dM =

1

12dx

dMxIM =

1

12dx

dMxMI =

Comparing I2 with equation (b) 22 MI =

and also (right hand side)1

1

1 1 x

MI

dx

dM

=

1

11 1dx

dMxMI =


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1

11 1dx

dMxMI += Comparing with equation(a) 11 MI =

The intercept of the tangent gives the partial molar properties.

Limiting cases: For infinite dilution of component when at x1=0 a tangent is drawn atx1=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

2

M

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 infinitedilution.

(b) Find the pure component properties

Sol:mol

calxxxxxxG )10(150100

212121+++=

Substitute 12 1 xx = and simplifying

15049891

2

1

3

1++= xxxG

+=1

11 )1(x

GxGG

491627 12

1

1

=

xxx

G

101163518 12

13

11 ++= xxxG

=1

12x

GxGG

1508182

1

3

12+= xxG


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To find the partial molar properties of components 1 and 2

x2=0.8, x1=1-0.8 = 0.2

1011635181

2

1

3

11++= xxxG

mol

calG 944.1021 =

1508182

1

3

12+= xxG

mol

calG 824.1492 =

At infinite dilution

0111 ==

atxGG

mol

calG 1011 =

1122 ==

atxGG orx2=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

dnn

GdT

T

GdP

P

GdG

j

+

+

= ii

nP

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=

+= ii

tttdndTSdPVdG

At constant temperature and pressure

= iiPTt dndG ,

For binary solution 2211 xxG +=


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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

iii STHG =


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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

Pdn

G

P

=

2

,

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

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

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

VP

G

T

=


i

nTii

V

n

V

nP

G

j

=

=

,

2

,

i

nT

i VP

=

,

This equation represents the effect of pressure on chemical potential


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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

in

GGi

,,

== ------(1)


inT

i

Pdn

G

P

=

2

,

-----(2)

SdTVdPdG = -----(3)

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

VP

G

T

=



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i

nTii

Vn

V

nP

G

j

=

=

,

2

,

i

nT

i VP

=

,

PVii =

dPVfRTd ii =ln

Subtracting both sides by iPd ln

ii ydPdPd lnlnln +=

Composition is constant dlnyi=0

The above equation can be written as

P

dP

PdPd i == lnln

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.

dPRT

Vfd ii =ln

ii

ii PddPRT

VPdfd lnlnln =

ii

i

i PddPRT

V

P

fd lnln =

[ ] dPP

RTV

RTd ii

= 1ln


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iiVV = and == iiii VxxVV --- Raoults law

Ideal 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

iiVV

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

iVn,P,T

n

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

=

P

i

i

i dPP

RTV

RTP

f

0

1ln ----(4)

Subtracting equation (3) from (4)

( ) =P

ii

ii

i dPVVRTPf

Pf

0

1ln ----(5)

we know that PyP ii =equation (5) reduces to

( ) =P

ii

ii

i dPVVRTyf

f

0

1ln


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comparing equation (2)

1=ii

i

yf

f

iii yff = --Lewis Randal rule

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

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

3. At any pressure if component present in exess.

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

Where i is an activity coefficient of component i.

Comparing equation (a) and (b)

==i

ii

f

f Non ideal solution / Ideal solution

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

equation

=i

i

i

x

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.


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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 of

solution 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

+

+

+

= ,,,2

1

,,1,,,, 313.212121

)()()()()( dn

n

nMdn

n

nMdP

P

nMdT

T

nMnM

nnPTnnPTnnTnnP

+

+

=

i

nPTinnTnnP

dnn

nMdP

P

MndT

T

MnnM

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


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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 becomes

02211 =+ 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 Duhemequation.

Problem:

Find weather the equation given below is thermodynamically consistent

)10(150100 212121 xxxxxxG +++=

1

11 )1(

x

GxGG

+=

1

12x

GxGG

=

1011635181

2

1

3

11+++= xxxG

1508182

1

3

12+= xxG

167054 12

1

1

1 += xxdx

Gd

1

2

1

1

2 1654 xx

dx

Gd=

G D equation

0)1(1

21

1

11 =+

dx

Mdx

dx

Mdx

0)x16x54)(x1()16x70x54(x1

2

111

2

11=


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

Phase Equilibrium

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

The different criteria for phase equilibrium are

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

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

constant volume dU=0 and dV=0.0

,VUdS

At Constant T and V: Helmoltz free energy is given by the expressionTSUA =TSAU +=

on differentiatingSdTTdSdAdU ++=

we know thatPdVTdSdU =

SdTTdSdAPdVTdS ++=

SdTPdVdA =

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

0, VTdA

At Constant P and T


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The equation defines Gibbs free energy TSHG =

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.

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 constanttemperature and pressure. Appling the criteria of equilibrium

0=dG

0=+ ba dGdG adG and bdG are chane in free energies of the phases a and b respectively.

We know that

ii dnGSdTVdPdG +=For phase a

aaaaaaa dnGdTSdPVdG +=For phase b

bbbbbbbdnGdTSdPVdG +=

At constant temperature and pressureaaa

dnGdG = , bbb dnGdG =

For a whole system to be at equilibrium

ba

ba

dndn

or

dndn

=

=+ 0

[ ] 0= aba dnGG


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

energies must be same in each phase for equilibrium.CfRTCfRT ba +=+ lnln

ba

ff =

Clapeyron Clausius Equation

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

(a) Solid- liquid

(b) Liquid-Vapor

(c) Solid Vapor

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 isAA dGG + for Phase B BB dGG +

From basic equationSdTVdPdG -----(3)

dTSdPVdG AAA dTSdPVdG BBB

dTSdPVdTSdPV BBAA [ ] [ ]

ABAB SSdTVVdP

V

S

dT

dP

=


27/66

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

phase A to Phase B

T

QS=

VTQ

dTdP =

[ ]AB VVT

Q

dT

dP

----(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,

[ ]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.

= 21

2

1

T

T

P

PT

dT

RP

dP


28/66

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 ==

211

2

T

1

T

1

RP

Pln

= 36813731314.8 1027.41760Pln3

2

mmHg3.634P2 =

Phase Equilibrium in Multi component system

An heterogeneous system contains two or more phases and each phase contains two ormore components in different proportions. Therefore it is necessary to develop phase

equilibrium for multi component system in terms of chemical potential.

The partial molar free energy or chemical potential is given as

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 inall 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 is

dG=0 ----(1)

Free energy for multi component system given by the expression

+= ii dnSdTVdPdG ------(2)At constant Temperature and pressure

ii dndG ----(3)


29/66

Comparing equation 1 and 3

= ii dn0For multi component system

= =

=C

i k

k

i

k

i dn1

0

Expanding the above equation

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 dni is independent of each other. However the sum of

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

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

CCC ==

==

==

222

111

At constant temperature and pressure the general criterion for thermodynamic

equilibrium 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


30/66

Phase Diagram for Binary solution

Constant pressure equilibrium

Consider a Binary system made up of component A and B. Where it is assumed to bemore volatile than B where vapour pressure A is more than B. When the pressure is

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

vapour compositions get quickly determined VLE at constant pressure is represented onT-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 bypoint A. When the mixture is heated slowly temperature rises and reach to point B,

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

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

at that point is called as Dew point. Further heating results in super heated vapour. Thenumber of tic lies connects between vapour and liquid phase. For a solution the term

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

at constant pressure.


31/66

Effect of pressure on VLEThe boiling point diagram is drawn from composition x=0 to x=1, boiling point of pure

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

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


32/66

Equilibrium diagram

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

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

Constant temperature Equilibrium

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

and lower curve is drawn at vapour comp(y). Consider a liquid at known pressure andcomposition at point A as the pressure is decreases and reaches to point B where it

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

converts to vapour, further decreases in pressure leads to formation of super heatedvapour. In between B to C both liquid and vapour exists together


33/66

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 Raoultslaw, 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 verylarge 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 azeotrope

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

pressure or maximum boiling point, this is known as minimum pressure Azeotrope.Eg : Ethanol water, benzene-ethanol

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

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

Phase diagram for both types of AzeotropesMinimum Temperature azeotropes


34/66


35/66

Maximum Tem Min pressure Azeotropes

Calculation of VLE for ideal solution

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

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

AVAA xPP =

PA- Partial pressurePVA- Vapor pressure


36/66

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

VBTA

PP

PPx

=

Assuming the vapor phase is also ideal

T

AVA

T

AA

P

xP

P

Py ==

Substituting PT from equation (2)

( ) VBAVBVA

AVA

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


37/66

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 PV238.4

5

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

VA

P

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

VA

P

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 0.44958 811.8 342.3 0.122 0.2475

62.3 937.4 400 0 0

Calculation of VLE for non Ideal solution

For non ideal solution

Partial pressure is given by Viiii PxP =

i- Activity coefficient


38/66

xi- Mole fractionPVi vapor pressure

For binary mixture

1111 VPxP =2222 VPxP =

21PPPT +=

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 available

to estimate them. The important equations areVanlaar equation

Wilson equation

Margules Equation

Vanlaar Equation:Estimation of activity coefficient

2

21

2

2

1ln

+

=

xxB

A

Ax

2

21

2

1

2ln

+

=

xA

Bx

Bx

Where A and b are known as vanlaar constants, if the constants are known theactivity coefficients can be estimated.

Estimation of Vanlaar constants1st Method: If the activity coefficients are known at any one composition, then vanlaar

constants A and B can be estimated by rearranging the equation


39/66

2

11

22

1ln

ln1ln

+=

x

xA

2

22

11

2

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 compositionx1=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 506mmHg.

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

11

=V

T

P

P, 2

2

=V

T

P

P

501.1506

7601 == , 47.1

517

7602 ==

2

11

22

1ln

ln1ln

+=

x

xA


40/66

2

)501.1ln()448.0(

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

+=A

A=0.829

2

22

11

2ln

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 Vs

composition.

2

21

2

2

1ln

+

=

xxB

A

Ax

,2

21

2

1

2ln

+

=

xA

Bx

Bx

111

222222111

1111

1

1

1

V

VVV

V

T

Px

PxPxPx

Px

P

Py

+=

+==

x1 x2 1 2 y1

0 1 6.745 1 0

0.2 0.8 2.4 1.095 0.384

0.4 0.6 1.644 1.374 0.4384

0.6 0.4 1.21 1.721 0.5077

0.8 0.2 1.0112 2.618 0.609

1.0 0 1 3.723 1

Margules equation

Eetimation of activity coefficients

[ ][ ]12

21

2lnxABAx +=

[ ][ ]2

2

122ln xBABx +=

The constant A in the above equation is terminal value of ln 1 at x1=0 and constant B isthe terminal value of ln 2 at x2=0


41/66

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

2

12 expexp

=

RT

aexp

V

V

RTexp

V

V 21

2

12212

2

1

21

V1 and V2 molar volumes of pure liquids - Energies of interaction between molecule

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

onln

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


42/66

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 curveshould 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

=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


43/66

0ln

1

=dx

fdx ii ------(5)

For binary system

0

lnln

1

2

21

1

1 =+ dxfd

xdx

fd

x ----(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

fdiiii ++=

-------(8)

fi is a pure component ffugacity it does not vary withx1

Substituting equation (8) in equation (5)

0lnln

11

=+dx

xdx

dx

dx ii

i

i

0ln

1

=dx

dx ii

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

For binary system

0lnln

1

2

2

1

1

1=+

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

=0

=0


44/66

0ln

1

=dx

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

For binary system

0lnln

1

22

1

11 =+

dx

Pdx

dx

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

0lnln 2211 =+ 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 =+++ TPdxxydxydx121 =+ xx

0lnlnln 2111 =++ TPdydxydx

2211 lnln ydxydxP

dP=

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. This

equation is used to rest consistency of VLE data.


45/66

Consistency test for VLE data

0lnln 2211 =+ dxdx

2

112

lnlnx

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 +=

differentiating with respect tox1

+++=

1

2

2

1

2

21

1

1

1

1

lnln

lnln

dx

dx

dx

dx

dx

dxRT

dx

dG e

From Gibbs Duhem Equation

0lnln

1

2

2

1

1

1=+

dx

dx

dx

dx

+=

1

2

21

1

lnlndx

dxRT

dx

dG e

12 dxdx =

[ ]211

lnln = RTdx

dG e


46/66

=

2

1

1

ln

RT

dx

dGe

=

=

=

=

=

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 0=eG , where LHS is zero for bothlimits.

We can write

=

==

1

0

1

2

1

1

1

ln0

x

x

dx

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

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 0.626

1.0 1.00 0.379Solution:

We know that the Redlich kister test is

=

=

=

1

0

1

2

1

1

1

ln0

x

x

dx


47/66

2

1

2

1ln

0.57

6

-0.552

0.66

5

-0.408

0.80

4

-0.218

1.05

2

0.051

1.518

0.417

2.639

0.97

Plot2

1

ln

Vs x1

Area under the curve is zero. Data is consistent.


48/66

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 of H298 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 = Hf 298 products - Hf 298 reactants


49/66

Heat of formation of any clement is zero


50/66

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 + AT

2

CpB = B + BT + BT2

CpC = C + CT + CT2

CpD = D + DT + DT2

And H0298 be standard heat of reaction at 298 K and standard heat of reaction at any

other temperature can be found byKirchoffs rule( )o

p

d H

CdT

= where

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

2 ) ] - [a (A + AT +

AT2 ) + b (B + BT + BT

2 )]

= [(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

( )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

= + +

= + +

= + + +

, , , pC pD pA pBC C and C C

Cp = + T + T2

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

=

= + +


51/66

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 the

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

reaction to take place

Re Reor 0action action products reaction

G G G G =


52/66

( ) ( )

( ) ( )

( ) ( )

( )

ReRe tan

0 0

0 0

= c ln ln

ln ln

action products 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

actiona b

A B

RT a a a a

a aG G RT

a a

+ +

= +

At equilibrium G = 0 andC

c d

D

a b

A B

a aK

a a

=

K is equilibrium constant of the reaction

[ ]0 0

0 ln ln

C

c d

D

a b

A B

a aG RT G RT K

a a

= + = +

at equilibrium

EFFECT OF TEMPERATURE ON EQUILIBRIUM CONSTANT K


53/66

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

2

K

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

0

2

2

hen K can be calculated at other

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


54/66

0

20

2 30

ln

ln T

2 6

ln T 2 6

=

= + + +

= +

G RT K

T T IG RT M

R R R RT

T T

G T I MRT

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

G

0

at 550 KCp (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

actionoducts ac ts

H H H =


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( )

( )

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 variationof with temparatu2

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 = ln 298 298 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) ln 550 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 m ol =

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

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

with T. The Hfand Gf at 298 K are

Componen

t

Hf

k Cal / k

mol

Gf

k Cal / k

mol

SO2 -70960 -71680

O2 0 0

SO3 -94450 -88590

CPdata:

Componen

ta b(103)

SO2 7.116 9.512

O2 6.148 3.102

SO36.037

7

23.53

7


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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

= +

=

= + + +

[ ]

-3 2

0 0 0 0

298

Re 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= ln2 2 2

T

Next

IT T T M

R R RT

xT T M xR T

+ + +

+ + +

00 298

298 298 298

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 ln2982 2

K

x

=

= +

3

22818.192982 2 298

0.9842

11409.09ln 2.076ln 3.1185 10 0.9842

Mx x

M

K T x T T

+ + =

= + + +


57/66

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

H0298 (g Cal / g mol) = - 11040

Heat capacity Data is given by Cp= a + bT + cT2 the values are

Component a bx102 cx105

NH3 6.505 0.613 0.236

N2 6.903 - 0.038 0.193

H2 6.952 - 0.46 0.096

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

0 0

298

Pr 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 3

6.505 (6.903) 6.952 7.37452 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

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

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

+

= + + +


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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 ln 298 298 (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

1500

Pr 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

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.

Temperature0

C 145 6.8 x 10-2

Equilibrium Constant 320 1.9 x 10-3

Componen

ta bx103

C2H4 2.83 28.601

H2O 7.3 2.46

C2H5OH 6.99 39.741


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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 = ln

2 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 M

x 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 -9655.80

lnK = ln2 2 2

x IM

x x

Solving

x I At M

x x

x

T Tx

+ + = + +

+ 3

-3 2

7

-5.6672

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

+

+ +

+

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):


60/66

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

We haveC

c 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

f Activity a E f

f

f 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 xy p y p y y p p

( )( )( )(P ) =( )( )( )(P ) where = activit y co efficient for gas phase

d c d c d c d c d

a b a b a b a b a b

y y

x

K k k k k k k

=

=

y = mole fraction

= Fugacity Co efficient for gas phase

P = Total p ressure

Effect of variables on equilibrium :Effect of temperature:

0

298 298lnG 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 H0is 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 / P

n

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

yield is less at high pressures


61/66

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

yield

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

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

pressure is required

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

of pressure depends on the variation of and with pressure

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

Therefore whenn > 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%

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) = 00 330100we have ln or lnK = or K = 1.8029x10

8.314(573)G RT K

=

Component

Moles

in feed

Moles

reacted

Moles

present

Mole

fraction

N2 1 X 1 X (1 X) / 2

C2H2 1 X 1 X (1 X) / 2

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

X


62/66

2 2 2

0

2 23

1 1

Assume 1 and 1

1.8029(10) , X = 0.01 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

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,,C2H2 and HCN are 1.1, 0.928, and 0.54 Assume K= 1

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 = 0.0382then 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 NH3 at 298 K to be - 46110 J / mol and 16450 J

/ mol respectively.


63/66

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

10

2

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

2

K 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.

CH3COOH

(L)

C2H5OH

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

f0f

(J)- 484500 - 277690 - 463250 - 285830

G0f

(J)- 389900 -174780 - 318218 - 237130


64/66

G0298 = 318218 237130 + 389900 + 174780 = 9332 J

H0298 = 463250 285830 + 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

fraction

CH3COOH 1 X 1 X (1 X) / 2

C2H5OH 1 X 1 X (1 X) / 2

CH3COOC2H5 -- X X X / 2

H2O -- X X X / 2

2

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

= =

= = =

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

=

= =


65/66

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

0 =1.03665 X 105 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 = ky ( 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


66/66

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 (1 X ) / (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 xat 5.0 bar k 0.26382

5 1 x2 x

x = 0.5501

+ = = == +

+ = = = +

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