Chemical Reaction Equilibria Chapter 13

The Reaction CoordinateApplication of Equilibrium Criteria to Chemical ReactionsThe Standard G and the Equilibrium ConstantEffect on Temperature on the Equilibrium ConstantEvaluation of the Equilibrium ConstantRelation of Equilibrium Constant to CompositionEquilibrium Conversion for Single Reactions

Chemical Engineering Thermodynamics II ChE 342 12-2

Introduction

Transformation of raw materials into products having commercial use and value is a major industry.

Raw Materials

Products{crude oil, minerals, etc..} {Gasoline, NaOH, H2

SO4

, ammonia, etc..}

This often involves chemical reactions and subsequent separation of the products.

The rate and maximum (or equilibrium) conversion of chemical reaction are of important concern in its commercial development. Both of these aspects depend on the T, P, and composition of the reactants.

We can often increase reaction rates by the use of CATALYST, but we can not change the equilibrium (or ultimate) conversion by using catalyst.

D

e

g

r

e

e

o

f

r

e

a

c

t

i

o

n

Time

With catalyst

Withou

t cataly

st

Uultimateconversion

Chemical Engineering Thermodynamics II ChE 342 12-3

Chemical Reaction Course concerns on study of reaction rate and

equilibrium, where the following tasks are often considered:

Design and operation of reaction equipment Effect of controllable variables upon conversion Extent of reaction as t

In general:Reaction Equilibrium is governed by THERMODYNAMICS

while,Reaction Rate is governed by REACTION KINETICS

Objectives:

Definitions of reaction stoichiometry

and criterion of reaction equilibrium. Dependence of equilibrium constant

on temperature and its evaluation.

Development of connection between equilibrium constant and composition.

Calculation of equilibrium conversion

for single reactions.

Chemical Engineering Thermodynamics II ChE 342 12-4

The Reaction CoordinateThe general chemical reaction can be written as:

44332211 +........++ AAAAwhere |i

| are stoichiometric

coefficients and Ai

denotes the chemical formulas of species. i

themselves are stoichiometric

numbers

Sign ConventionFor products: i

is positive (+) For reactants: i

is negative (-)

N.B. The reaction, as written above, does not represent any mechanisms. It represents only the overall macroscopic picture. It does not tell us what is happening at the microscopic level. Example:

CH4

+ H2

O

CO + 3H2CH4

= -1CO

= +1 H2

= +3

H2O

= -1

The stoichiomertric

number of an inert species

is zero.

(I)

Chemical Engineering Thermodynamics II ChE 342 12-5

In all reactions, the change in the number of moles of the species present are related to each other through the stoichiometric

number.

For example, for the reaction CH4

+ H2

O

CO + 3H2

, if 0.5 mole CH4

reacts, then 1.5 moles H2

and 0.5 mol CO generated, thats ratio of change in number of moles to the stoichiometric

number of CH4

is the same as that for H2

.Thus, in general for reaction (I), we can write:

1

1

2

2

=

dndn

1

1

3

3

=

dndn, , etc

or, in general:

.........=

=

= 3

3

2

2

1

1 dndndn

where d = the amount of reaction.d=

It will be observed that the sign of i

allows for the fact that:dni

> 0 for the product & dni

< 0 for the reactants The preceding equation provides the definition for , the Reaction Coordinate; a quantity characterizes the extent, or degree, to which chemical reaction has taken place.

is also called the Degree of Reaction

(II)

Chemical Engineering Thermodynamics II ChE 342 12-6

From II, we can write,ddn ii =

moles molesdimensionless

By convention,

= 0 at the initial state (before the reaction occurs). Integration yields:

0

=0

i

n

ni ddn

i

i

= f(t)

nn iii += 0Summing over all N species (reactant and products) yields: +==

0 iii nnnor, nn += 0 where, 0=0 inn = i&

The quantity

tells us whether there is a net increase or decrease in the total number of moles in the system. The mole fraction of species i

at any time, or stage of reaction, is by:

nn

y ii = nn ii

++

=0

0This expression is very useful, since it provides us with a relationship between species concentration and the progress of the reaction.

(III)

Chemical Engineering Thermodynamics II ChE 342 12-7

Example:

yCO 2+8+1

=

Consider the chemical reaction, CH4

+ H2

O

CO + 3H2Initially there present 2 moles CH4

, 1 mole H2

O, 1 mole CO and 4 moles H2

. Obtain yi

= f(). = i

= -1 -1

+ 1 + 3 = 2 n0

= n0i

= 2 + 1 + 1 + 4 = 8 moles

Application of the equation:

yCH 28-2

4 +=

nn

y iii ++

=0

0

y OH 28-1

2 +=

yH 2834

2 ++

=

Definition of a Mole of Reaction

This means that

has changed by a unit amount (i.e. 1 mol).

When

= 1 mole, the reaction proceeds to such an extent that the change in mole numbers for each reactant and product species is equal to its stoichiometric

numbers, thats: ni = i

Chemical Engineering Thermodynamics II ChE 342 12-8

Multireaction Stoichiometry

In this case, we allocate a reaction coordinate, j

, to each reaction

(j

= 1, 2, 3, R), where R

= number of separate reactions.

The stoichiometric

numbers are now identified as:

i,j Stoichiometric

number of species i

in reaction j

ni

may change because of the involvement of species i

in several reactions. The general equation analogous to Eq. (III) for single reactions is:

,=j

jjii ddn (i

= 1, 2, .N;

j = 1, 2, . , R)

Integration yields; )(+= ,0j

jjiii nn

Summing over all species: )(+== ,0i j

jjii nnn )(+= ,0j i

jji n

Similarly, ,=i

jij v )(+= 0j

jjnn

Mole fraction,nn

y ii =

+

)(+=

0

,0

jjj

jjjii

n

n(i

= 1, 2, . , N)

Chemical Engineering Thermodynamics II ChE 342 12-9

Example: Consider the following reactions,CH4

+ H2

O

CO + 3H2 (1) j

= 1CH4

+ 2H2

O

CO2

+ 4H2 (2) j

= 2

There are present 2 mol CH4

and 3 mol H2

O. Obtain expressions for yi

= f(1

, 2

).

Construct the following table for i,j

:

i

= CH4

H2

O CO CO2

H2j

12

-1-1

-1-2

10

01

34

j1

= 22

= 2

)(+= 0j

jjnn = 5 + 21

+ 22

21

21

225--2

4

yCH ++=

21

21

2252--3

2

y OH ++=

21

1

2+2+5=

yCO

21

2

2+2+5=

2

yCO21

21

2+2+54+3

=2

yH

Chemical Engineering Thermodynamics II ChE 342 12-10

Application of Equilibrium Criterion to Chemical ReactionsAs noted previously, in a closed system at equilibrium (constant

T

&P):

(dGt)T,P

= 0

Recalling that Gt

decreases during an irreversible process (eg. Mixing process, chemical rxn, etc) If a mixture is not in equilibrium, any reaction that occurs at

constant T

and P

must

lead to a decrease in the total energy of the system

is the single variable that characterizes the progress of the reaction.

Recall that

is related to the composition of the system

Criterion for equilibrium:1. (dGt)T,P

= 02. Gt()

is minimim

N.B. Equilibrium expression can also be used for OPEN SYSTEM at constant T

& P.

Chemical Engineering Thermodynamics II ChE 342 12-11

The Standard Gibbs Energy Change and the Equilibrium Constant

Recall the fundamental property relation for single-phase system:

)(-)()(i

iidndTnSdPnVnGd +=

ddn ii =If changes in ni

occur as a result of single reaction in a closed system, then:

)(-)()(i

ii ddTnSdPnVnGd +=

=]

)([ , iiPT

nG= 0.0 At equilibrium state, according to the

previous figure. The criterion for chemical reaction equilibrium can be expressed as:

0.0= iiSince ii G =Recall that

ln+)(= iii fRTT

Write this equation for pure species in its standard state at the same temperature

(A)

oii

oi fRTTG ln+)(= (B)

Chemical Engineering Thermodynamics II ChE 342 12-12

See section 4.3 (p. 133) for the definition of standard state:Gases: pure species in ideal gas state at 1 bar (or 1 atm)Liquids or solids: actual pure liquid or solid at 1 bar (or 1 atm)

Subtracting (B) from (A):o

iioii ffRTG /ln=_

ln= iaRTwhere is the activity of species i

in solution.ia

N.B. the activity coefficientia )/=(

iiii fxfln+= i

oii aRTGNow, & 0.0= ii

0=]ln+[ 0 iiii aRTG 0=)ln(+][ 0 i

iii aRTGor

RTG

a iivi i 0 -)(ln =or where product over all species

]/Exp[-)( 0 RTGa iivi i = = KGoi

= f(T) only, independent of pressure; K

= the equilibrium constant

= f(T) only

ln- ooii GGKRT == where G o

the Standard Gibbs Energy Change of Reaction.

Chemical Engineering Thermodynamics II ChE 342 12-13

Note: For a specific species the standard state represented by Gio

must be the same state represented by fio

upon which the activity coefficient is specified

iaCase of Gases

The standard state is the ideal gas state of pure species at a

pressure of 1 bar. fio

= 1 bar for each species in has phase reaction

= =

iii o

i

fa f

f = ( ) i viK f must be measured in bar

+ for product, -

for reactant

Example: A + 2B + 3C

D + 4 E

-2 -3-1 4= ( ) = ( ) ( ) ( ) ( ) ( )i vi A B C D EK f f f f f for

4

2 3

( ) ( )=

( ) ( ) ( )

D E

A B C

f fK

f f fCase of liquids and solids

The usual standard state is pure liquid or pure solid at 1 bar and T

of the system. for this state, fio

is not generally equal to 1 bar, thus =

ii o

i

fa

f

Chemical Engineering Thermodynamics II ChE 342 12-14

Other Standard Property Change of ReactionFor the general property, M, we can write

Mo

= i

Mio

For example: Ho

= i

Hio

CPo

= i

CPiof(T) only

Relationship between Hio and GioPreviously, we have derived:

Therefore, in the standard state for species i:

= PT

RTGT

RTH

])/(

- [

=dT

RTGdRTH

oio

i)/(

- 2 = f(T)Why we removed the partial derivative?

Summing over all species:

= ][- 2

RTG

dTd

RTHoiio

ii

Chemical Engineering Thermodynamics II ChE 342 12-15

= ][- 2RT

GdTd

RTHo

o Relationship between Ho

and Go

But, RTGK o /-ln = 2lnRT

HdT

Kd o= Highly important

relation

Note: If Ho

is negative (exothermic reactions) K

as T

Special case: Assuming Ho

is independent of T,

]1

-1

[

-)/ln(1

1 TTRH

KKo

= A plot of lnK

vs. 1/T

should be a straight line

See the next slide for plot of K

versus 1/T

for several reactions.

Chemical Engineering Thermodynamics II ChE 342 12-16

This figure allows determination of K at any temperature.

If K1

is known at T1

K

can be determined at any temperature for a given Ho.

Important Remarks:

Slope of the lines are ve

or +ve; -ve

for endothermic reactions

+ve

for exothermic reactions

Chemical Engineering Thermodynamics II ChE 342 12-17

Determination of Go

K

can be determined from the previous figure at a given temperature for certain

reactions if Ho

is constant (

f(T)) Another alternative method, when Ho

= f(T),

for determination of K

is by using:

RTG

Ko

- ln = = f(T)

If an estimate of Go

is available K

can be determined Recall that Go

= f(T)

Go

can be calculated by using the general relation: ooo STHG - =

Go

= i

GioHo

= i

HioSo

= i

SioCPo/R = i

GPio/RRecalling (from Chapter 4 and 5):

0

0

T

T

oPoo dT

RC

RHH += or: 0

0

T

T

oPoo

TdT

RC

RSS +=

Where H0o

and S0o is the standard enthalpy and entropy changes at a reference temperature T0

, respectively.

(A)

Substituting in (A) gives:

Chemical Engineering Thermodynamics II ChE 342 12-18

00

- -

00

T

T

oPo

T

T

oPoo

TdT

RC

RTSTdTRC

RHG +=

0

000

-

TGH

Soo

o =

According to (A):

00

-)-( -

000

0

T

T

oPoo

T

T

oPoo

TdT

RC

RTGHTT

dTRC

RHG +=whence:

00

- 1-

0

000T

T

oP

T

T

oP

oooo

TdT

RC

dTRC

TRTHG

RTH

RTG

++=

Division by RT:

CPio/R = f(T)

presented in Tables C.1-C.3 (pp. 657/658) Go

and Ho

values are tabulated for many formation reactions at standard conditions

Gof298

and Gof298

(G0o

and H0o) for a number of chemical compounds are listed in Table C.4 (pp. 569/660)

Go/RT

can be determined after determining the two integrals in the above equations

Hence, K can be determined

Chemical Engineering Thermodynamics II ChE 342 12-19

Example: calculate the equilibrium constant for the vapor-phase hydration of ethylene at 145 & 320oC from data given in Appendix C

C2

H4

(g) + H2

O(g)

C2

H5

OH(g)

Data: SpeciesC2

H4

(g)H2O(g)C2

H5

OH(g)

i-1-1+1

Ho29852510-241818-235100

Go29868460

-228572-168490

Ai1.4243.473.518

Bi

10314.3941.45

20.001

Ci

105-

4.392-----

-

6.002

where Cpi

/R = Ai

+ Bi

T

+ Ci

T2

298,o298 oii HH = = (-1)(52510) + (-1)(-241818)+(1)(-235100) = -

45792 J/mol

298,o298 oiiGG = = (-1)(52510) + (-1)(-241818)+(1)(-235100) = -

45792 J/mol2

,o )()()/(/ TBTBARCRC iiiiiio iPiP ++==

= [(-1)(1.424) + (-1)(3.47) + (1)(3.518)] + [(-1)(14.39) + (-1)(1.45) + (1)(20.0)]10-3T+ [(-1)(-4.392) + (-1)(0.0) + (1)(-6.002)]

10-5T2

= -1.376 + 4.157

10-3T -

1.61

10-6T2

Chemical Engineering Thermodynamics II ChE 342 12-20

418

298

26-3-

418

298

26-3-

]101.61-104.157-1.376[-

]101.61-104.157-1.376[418

1)298)(314.8(

457928378-)418)(314.8(

45792-

TdT

T

dTTRTG o

+

+++

+=

@ T

= 145oC = 418 K:

936.1

=RTG o@ T

= 145oC = 418 K:

RTG

Ko

- ln = K

= exp(-1.936) = 0.1443

Similarly @ T = 320oC = 593 K K

= 0.00294

K

with TExothermic rxn

Chemical Engineering Thermodynamics II ChE 342 12-21

Relationship between Equilibrium Constant and Composition

Gas-Phase ReactionsFor the gas phase reactions: ( where fio

= Po

= 1 bar)= =

iii o

i

fa f

f i

)( iifK =

),,( ii yPTff =K

= f(T) only

Reflects nonideality

of the equilibrium mixture

But

&This means that for a given T, the composition at eqm

must change

with P

in such away that remains constant iif )(

Now, Pyf iii = i

)( iii PyK =

Separate P

term, i ][)( ii PyK i= where = i

Or, i- )( iii

yPK =(recalling that yi

can be related to e

) Relates K

to e

for a given T

and P

Complicated problem because: ),,( ii yPTf = iterative solution is needed

Chemical Engineering Thermodynamics II ChE 342 12-22

Special cases: For equilibrium mixture as an ideal solution ii =

whence, i- )( iii yPK =

For sufficiently low pressure, equilibrium mixture behaves as an ideal gas

1 =i

whence, i- )( ii yPK = T, P, and composition-dependent terms are distinct are separate, thus solution for any one of

e

, T

or P

given the other two is straightforward Conclusion remarks

-

Dependence of T

can be seen through K: If Ho

+ve, ENDO

rxn, K

with T

(yi

)i

for a fixed P

e

rxn

shifts to the right

If Ho

-ve, EXO

rxn, K

with T

(yi

)i

for a fixed P

e

rxn

shifts to the left

P

@ constant T

(yi

)i

rxn

shifts to the right, e

If

(i

) -ve,-

P

@ constant T

(yi

)i

rxn

shifts to the left, e

If

(i

) +ve,

Chemical Engineering Thermodynamics II ChE 342 12-23

Example: Consider the water-gas shift reactionCO(g) + H2

O(g)

CO2

(g) + H2

(g)Assuming ideal gas mixture, what is the fraction of the steam reacted in each of the following cases: (Hint: fraction of steam converted = (ni0

nie

)/ni0

= -i

e

/ni0

)(a) n0,H2O

= 1, n0, CO

= 1 mol, T

= 1100 K, P

= 1 bar

n0

= 2 mol & = 0.0From Fig. 13.2: @ 1/T

= 1/(1100 K) = 9.09

10-4

for the above rxn

lnK

= 0 or K

= 1.0

1 = i i0 )( )1( iiyK = = 1

14/)-1(4 2

2

=e

e

,202

02

eeCO

y =

++

= ,2-1

yCO = ,22e

H

y =2-1

2

y OH =

Substituting:

1y

2

2CO =OHCO

H

yyy

e

= 0.5 fraction of steam reacts = 0.5

(b) n0,H2O

= n0, CO

= 1 mol, T

= 1100 K, P

= 10 bar Since

= 0 P

has no effect; e

= 0.5

(c) n0,H2O

= n0, CO

= 1 mol, n0,N2

= 2 mol, T

= 1100 K, P

= 1 bar

As an inert N2

= 0 Does not affect the calculations, e

remains the same.

Chemical Engineering Thermodynamics II ChE 342 12-24

(d) n0,H2O

= 2, n0, CO

= 1 mol, T

= 1100 K, P

= 1 bar

n0

= 3 mol

1)-1)(-2(

2

=ee

e

,32e

CO

y = ,3-1

yCO = ,32e

H

y =3-1

2

y OH =

Substituting: e

= 0.667 fraction of steam reacts = 0.667/2 = 0.33

(e) n0,H2O

= 1, n0, CO

= 2 mol, T

= 1100 K, P

= 1 bar

n0

= 3 mol

1)-1)(-2(

2

=ee

e

,22e

CO

y = ,3-2

yCO = ,32e

H

y =3-1

2

y OH =

Substituting: e

= 0.667 fraction of steam reacts = 0.667

(f) n0,H2O

= 2, n0, CO

= 1 mol, n0, CO2

= 1 mol, T

= 1100 K, P

= 1 bar

n0

= 3 mol

1)-1(

)1(2 =

+

e

ee

,

31

2

eCO

y

+= ,3

-1 yCO = ,32

eH

y =

3-1

2

y OH =

Substituting: e

= 0.0.33 fraction of steam reacts = 0.33(g) n0,H2O

= 1, n0, CO

= 1 mol, T

= 1650 K, P

= 1 bar From Fig. 13.2: @ 1/T

= 1/(1650 K)

lnK

= -1.15 or K

= 0.316

316.0)-1( 2

2

=e

e

e

= 0.36

Chemical Engineering Thermodynamics II ChE 342 12-25

Liquid-Phase ReactionsFor the liquid phase reactions: o

i

ii f

fa

= where fi

o

0

iviaK )(

=Then K

remains as

For liquid: iiii fxf =

oi

iiio

i

ii f

fx

ff

a ==

Determination of fi

/fio:

@ T

and P: iii fRTTG ln)( +=@ T

and Po: oii

oi fRTTG ln)( +=

Subtracting:

Integration at constant T

from Po

to P:

oii

oii ffRTGG /ln- =

But: dTSdPVdG ii i-=-P

Pi

oii

odPVGG =

(A)

(B)

Substitute (B) in (A): lnP

Pio

i

i

odPV

ff

RT =

For liquids Vi

is weak function of P

upon integration )-(ln oioi

i PPVff

RT =

or ]/)-(exp[ RTPPVff o

ioi

i =

Chemical Engineering Thermodynamics II ChE 342 12-26

]/)-(exp[ RTPPVxa oiiii =

ioiii RTPPVxK }]/)-(exp{[=and

io

ii RT

PPVii ex ][)(

)-(=

RTPPV

ii

oii

i ex)-(

)(=

])-(

exp[])([ i

ii

oi

ii VRTPPV

xK i=

Special cases At low P: exp[{Vi

(P-Po)/RT}{i

Vi

}]

1.0 (or when P

Po)i

iiixK )(= Requires determination i

, but i

= f(xi

) xi

can be found in this case iteratively.

For the case of ideal solution: i

= 1.0

iii

xK )(= Law of Mass Action

xi

= f(e

)

Chemical Engineering Thermodynamics II ChE 342 12-27

Example: Esterification

of acetic acid in the liquid phase at 100oC and 1 atm:

CH3

COOH(l) + C2

H5

OH(l)

CH3

COOC2

H5

(l) + H2

O(l)

Species

i

CH3

COOH(l), i=1C2

H5

OH(l), i=2CH3

COOC2

H5

(l), i=3H2

O(l), i=4

i-1-1+1+1

Hof298 (J)-484,500-277690-480,000-285,830

Gof298 (J)-389,900-174,780-332,200-273,129

n0,acetic acid

= n0,C2H5OH

= 1 molEstimate the mole fraction of ethyl acetate (x3

) at equilibrium. Assuming ideal solution and that Ho

does not change significantly with from 25oC to 100oC. i

iixK )(=

First, determine K: 298,o298 oii HH = = 484,500 + 277,690

480,00

285,830 = -

3640 J

298,o298 oiiGG = = 389,990 + 174,780 -332,200

273,129 = -

4650 J

lnK

= -Go/RT K298

= exp[-4650/(8.314)(298)] K298

= 6.527

Chemical Engineering Thermodynamics II ChE 342 12-28

For constant Ho, apply the following,]

1-

1[

-)/ln(

11 TTR

HKK

o

= ]2981

-373

1[

314.83640

-)527.6/ln( 373 =K

K373

= 4.859

21

34

xxxx

K =Now:

K

= 4.859, n0

= 2, = 0

0

0

nn

x eiii+

= ,2-1

1ex = ,23

ex = ,24ex =,

2-1

2ex =

859.44/)-1(

4/2

2

=e

e

e

= 0.6879

x3

= 0.344

(see other examples in the textbook)

Chemical Reaction EquilibriaChapter 13IntroductionSlide Number 3The Reaction CoordinateSlide Number 5Slide Number 6Example:Multireaction StoichiometryExample:Application of Equilibrium Criterion to Chemical ReactionsThe Standard Gibbs Energy Change and the Equilibrium ConstantSlide Number 12Slide Number 13Other Standard Property Change of ReactionSlide Number 15Slide Number 16Determination of GoSlide Number 18Slide Number 19Slide Number 20Relationship between Equilibrium Constant and CompositionSlide Number 22Slide Number 23Slide Number 24Slide Number 25Slide Number 26Slide Number 27Slide Number 28