Chapter 6- Chemical Reactian Equilibria

8/8/2019 Chapter 6- Chemical Reactian Equilibria

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Chemical EngineeringThemodynamics

Chapter 6

Chemical-reactionEquilibria


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In a chemical reaction, both the rate and the

equilibrium conversion are considered for

commercial purposes. Equilibrium conversion

represent the maximum possible conversion

regardless the reaction rate or catalyst.

This chapter focus on the effect of

temperature, pressure, and the initial

composition on the equilibrium conversions of

chemical reactions.


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Chemical-reaction

Equilibria

6.1 The reaction

Coordinate

6.2 Application ofEquilibrium Criteria to

Chemical Reactions

6.3 The Standard

Gibbs-Energy change and

The Equilibrium Constant

6.4 Effect of Temperature

on The Equilibrium

Constant

6.5 Evaluation of

Equilibrium Constants

6.6 Relation of Equilibrium

Constant to Composition

6.7 Equilibrium Conversionsfor Single Reactions

6.8 Phase Rule and DuhemsTheorem For Reacting System

6.9 Multireaction Equilibria


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The general chemical reaction:

...... 44332211 p AvAvAvAv

where |vi| is a stoichiometric coefficient

andAistands for a chemical formula.

Forvi :Positive ( + ) for product

Negative ( - ) for reactants

6.1 The Reaction Coordinate


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For example,

CH4 + H2O CO + 3H2

The stoichiometric numbers are :

14 !CHv 12 !OHv 1!Cv 32 !Hv

The stoichiometric number for aninert

species is zero.

Since:Id

v

dn

v

dn

v

dn

v

dn!!!!! ...

4

4

3

3

2

2

1

1


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The differential change dni in the number ofmoles of a reacting species andd is :

dni

= vi

d (i = 1 , 2 ,.N)

This variable called the reaction

coordinate,characterizes the extent or

degree to which a reaction has taken place.


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!I

I00

dvdn in

n

i

i

i

givesIiii vnn ! 0 (i = 1 , 2 ,.N)

Summation over all species yields : !!

i

i

i

i

i

i vnnn I0

or n = n0 + v

where |i

inn |i

inn 00 |i

ivv


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Thus, the mole fractions yiof the speciespresent are related to by :

I

I

vn

vn

n

ny

iii

i

!!

0

0


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Example 13.2

Consider a vessel which initially contains

only n0 mol ofwatervapor. Ifdecomposition

occurs according to the reaction,

2222

1OHOH p

find expressions which relate the number of

moles and the mole fraction of each chemicalspecies to the reaction coordinate.


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Solution 13.2

For the given reaction2

1

2

111 !!v

Application of Eqs.(13.4) and (13.5) yields:

I! 02 nnH

I

I

2

10

0

2

!

n

nyH

I!2H

n

I

I

2

10

2

!

n

yH

I2

12!On

I

I

2

12

1

0

2

!

n

yO


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The fractional

decompositio

nofw

atervaporis:

00

00

0

0 2

nn

nn

n

nn OH II!

!

Thus whenno= 1, is identifiedwith the

fractional decomposition of the watervapor.


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!j

jjii dvdn I,

After integration:

jj

jiiiv

nn I!,0

Summing over all species yields :

j

j i

ji

i

j

j

ji

i

i vnvnn II

!! ,0,0

For the mole fraction of

the species presents in

particular reaction:

!

j

jj

j

jjii

ivn

vn

yI

I

0

,0

(i = 1 , 2 ,.N)


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At the equilibrium state :

(dGt)T, P= 0

The total Gibbs energy,Gt is a minimum.

Its differential is zero.

6.2 Application of Equilibrium Criteria To

Chemical Reactions


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The fundamental property relation for

single-phase systems, provides an

expression for the total differential of theGibbs energy:

!i

iidndTSdPnVnGd Q

6.3 The Standard Gibbs-Energy Change

and The Equilibrium Constant


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Since nGis a state function, the right side

of this an exact differential expression:

PT

t

PTi

ii

GnGv

,,

-

xx

!

-

x

x!

IIQ

If changes in mole numbers nioccur as the

result of a single chemical reaction in aclosed system , then each dn

imay be

replaced by the product vid.

! i ii dvdTSdPnVnGd IQ


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For chemical-reaction equilibrium :

0!iiiv Q

Recall the definition of the fugacity of a

species in solution:

iii fRTT ln+!QFor a pure species iin its standard start

at the same temperature :

QQ

iii fRTTG ln+!

The difference between these equations:

Q

Q

i

i

ii

f

fRTG

ln!Q


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For the equilibrium state of a chemical reaction:

? A 0ln !i iiii ffTGvQQ

0ln ! i

v

ii

i

ii

i

ffRTGv QQ

RTGvff i

ii

i

v

ii

i

!

Q

Qln

!i

v

ii KffiQln

where

(|

RT

GK

Q

exp


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Alternative expression for K :

RT

GKQ(|ln

where

|(

i ii

GG QQ R

K is dependence on temperature.In spite of its dependence on temperature,

K is called the equilibrium constant for the

reaction.

Called as standard Gibbs-energy change

of reaction.


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The dependence of G onT :

2RT

H

dT

RTGd QQ (!

(

Then become:

2

ln

RT

H

dT

Kd Q(!

Equation above gives the effect of

temperature on equilibrium constant and

hence on the equilibrium conversion.

6.4 Effect of Temperature on the

Equilibrium Constant


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WhenH

negative =exothermic reaction

and the equilibrium constant,K

decreases as the temp. increases.

positive =endothermic reaction

and the equilibrium constant,Kincreases as the temp. increases.

d

(!

d TTR

H

K

K 11ln

Q

IfHis assumed independent of T,

integration from a reference T (at 298K)to random T, then:


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In Figure 13.2, a plot

of lnKvs 1/T for anumber of common

reactions, illustrates

this near linearity.


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The effect of temperature on the equilibrium

constant is based on the definition of thestandard Gibbs energy:

Gi = Hi -TSi

Multiplication by viand summation over all

species gives:

!i

ii

i

ii

i

ii SvTHvGvQQ


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Hence, the standard property change of

reaction;

G=H - TS

where the standard heat of reaction andstandard entropy change is related to

temperature:

dTR

C

RHHT

T(

(!(0

0

Q

QQ

(

(!(T

T

P

T

dT

R

CRSS

00

Q

QQ


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However

0

000

T

GH

S

QQ

Q ((

!(

whence

(((((!(T

T

PT

T

P

TdT

RCdT

RC

TRTH

RTHG

RTG

00

10

00

QQQQQQ

and recall,

RTGKQ

(!ln


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As an alternative, the preceding equation

may be reorganized so as to factorK intothree terms , each representing a basic

contribution to its value :K = K0 K1 K2

K0 represents the equilibrium constant atreference temperature T

0:

(|

0

0

0RT

G

K

Q

-

(! T

T

RT

HK 0

0

01 1exp

Q

(

(

(

-

(!

2

2

2

0

22

0

2

02

1

2

121

6

11

2

11lnexp

X

X

X

XX

X

X

X

X

X

T

DCTBTAK


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Example 13.4

Calculate the equilibrium constant for the

vapor-phase hydration of ethylene at 145and at 320C from data given inApp. C.

Solution 13.4

First determine values forA,B,C, and

D for the reaction:

C2H4 (g) + H2O(g) C2H5OH(g)

6.5 Evaluation of Equilibrium Constant


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The meaning of is indicated by:A = (C2H5OH) - (C2H4) - (H2O). Thus, from the heat-capacity data of Table C.1:

A = 3.518 - 1.424 - 3.470 = - 1.376B = (20.001 - 14.394 - 1.450) x 10-3

= 4.157 x l0-3

C= (-6.002 + 4.392 - 0.000) x 10-6

= -1.610 x 10-6

D = (-0.000 - 0.000 - 0.121) x 105

= -0.121 x 105


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Values ofH298 andG298 at 298.15K for the

hydration reaction are found from the heat-of-formation and Gibbs-energy-of-formationdata

of Table C.4:

H298=

-235,100 - 52

,510 - (-241

,818)= -45,792 J mol-1

G298 = -168,490 - 68,460 - (-228,572)

= -8,378 J mol-1

ForT = 145 + 273.15 = 418.15 K,values

of the integrals in Eq. (13.18) are:

121.230

!(

T

T

PdT

R

CQ

06924.00

!

(

T

T

T

dT

R

CQ


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At 418.15K,

: lnK= ln -1.9356 andK= 1.443 x 10-1

At 593.15K,

: lnK= ln -5.8286 andK= 2.942x 10-3

Application of Eqs. (13.21), (13.22), and (13.24)provides an alternative solution to this example.

By Eq. (13.21),

366.2915.298314.8378,8

exp0!!K

473.18

15.298314.8

792,45

0

0!

!

(

RT

HQ


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The following results are obtained:

T/K X K0

K1

K2

K

298.15 1 29.366 1 1 29.366

418.15 1.4025 29.366 4.985x10-3 0.9860 1.443x10-1

593.15 1.9894 29.366 1.023x10-4 0.9794 2.942x10-3

Clearly, the influence ofK1, is far greater

than that ofK

2. This is a typical result, andaccounts for the fact that the lines on Fig.

13.2 are nearly linear.


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The standard state for gas is the ideal

gas-state of the pure gas at the standard-

state pressure P of 1 bar.

Since forideal gas : fi = P

Thus : QQ Pf

ff i

i

i ! and KPf

i

v

i

i

!

Q

where the constant K is a function of temp..

6.6 Relation of Equilibrium Constant

to Composition and Pressure


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For a fixed temperature the composition at

equilibrium must change with pressure insuch a way that KPf

i

v

i

i

!Q

Hence anequilibrium expressiondisplaying

pressure and composition;

KP

Py

v

i

v

ii

i

! QJ

P is the standard state pressure of 1 bar.

constant .

| i ivv

reactants

products

i

i

y

y

reactants

products

i

i

R

R


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Forpressure sufficiently low or temperature

sufficiently high, the equilibrium behaves

essentially as an ideal gas :

For equilibrium mixture assumed to be

inideal solution,each is assumed to bethe fugacity of pure species i :

KP

Py

v

i

v

iii

! QJ

iJ

KP

Py

v

i

v

ii

! Q

1!iJ

iJ

| i ivv

reactants

products

i

i

y

y

reactants

products

i

i

R

R


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Example 13.5

The water-gas-shift reaction,

CO(g) + H2O(g) CO2(g) + H2(g)is carried out under the different sets of

conditions described below. Calculate the

fraction of steam reacted in each case.

Assume the mixture behaves as an ideal gas.

6.7 Equilibrium Conversions for Single

Reactions


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Calculate if the conditions:

(a)The reactants consist of 1 mol ofH2O

vapor and 1 mol of CO. The temperature

is 1,100 K and the pressure is 1 bar.

(b) Same as (a) except that the pressure

is 10 bar.

(c) Same as (a) except that 2 mol of N2 is

included in the reactants.


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(d)The reactants are 2 mol ofH2O and 1 mol of

CO. Other conditions are the same as in (a).

(e) The reactants are 1 mol ofH2O and 2 mol of

CO. Other conditions are the same as in (a).

(f)The initial mixture consists of 1 mol ofH2O, 1

mol of CO, and 1 mol of CO2. Other

conditions are the same as in (a)

(g) Same as (a) except that the temperature

is 1,650K

.


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Solution 13.5

(a) For the given reaction at 1,100 K,104/T = 9.05, andFig. 13.2 provides the

value. lnK0

orK= 1. For this reaction01111 !!!

ii

vv

. Since the reactio

nmixture

)(12

22

AK

yy

yy

OHCO

COH!!

By Eq. (13.5):

2

1e

COyI

!

2

12

e

OHyI

!2

2

e

COyI

!

2

2

e

HyI

!

is an ideal gas,Eq. (13.28) applies, and

here becomes:

KP

P

yy

yyv

v

OHCO

COH

i

!

Q2

22


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Substitution of these values into Eq. (A) gives:

1

12

2

!

e

e

II

Therefore the fraction of the steam that

reacts is 0.5.

(b) Since v= 0 , the increase in pressure

has no effect on the ideal-gas reaction,

ande is still 0.5.

or e

= 0.5


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(c) The N2

does not take part in the reaction,

and serves only as a diluent. It does increase

the initial number of moles no

from 2 to 4, and

the mole fractions are all reduced by a factor

of 2. However, Eq. (A) is unchanged andreduces to the same expression as before.

Therefore,e

is again 0.5.


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(d)In this case, the mole fractions at

equilibrium are:

3

1e

COyI

!3

22

e

OHyI

!

32e

COyI

!

32e

HyI

!

and Eq. (A) becomes:

1

21

2

!

ee

e

II

I

The fraction of steam that reacts isthen 0.667/2 = 0.333

or e

= 0.667


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(e)Here the expressions foryCO andyH2O are

interchanged, but this leaves the equilibrium

equation the same as in (d). Therefore e=

0.667, and the fraction of steam that reacts is

0.667.

(f) In this case Eq. (A) becomes:

111

2!

e

ee

I

II

The fraction of steam reacted is 0.333.

or e = 0.333


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(g) At 1,650 K, 104/T= 6.06, and from Fig.

13.2, in K = -1.15or K = 0.316.

Therefore Eq. (A) becomes:

316.01

2

!

e

e

I

I

The reaction is exothermic, and conversion

decreasesw

ith increasing temperature.

or e

= 0.36

(Must try: Examples 13.6, 13.7,13.8)


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This is phase rule for reacting systems.

F= 2 +

Nr

where is number of phases ,N number of

chemical species andris number of

independent chemical reactions atequilibrium within the system .

6.8 Phase Rule and Duhems Theorem

for Reacting System


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j

i

v

i

i Kf

fji

!

,

Q

where j is the reaction index.

For gas phase reaction:

j

i

v

i KP

fji

!

,

Q

For the equilibrium mixture is an ideal-gas,

jv

i

v

i KP

Py

j

ji

! Q

,

6.9 Multireaction Equilibria

(Must try: Examples 13.12, 13.13)


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Chapter 6- Chemical Reactian Equilibria