Advanced Thermodynamics

Note 12Chemical-Reaction Equilibria

Lecturer: 郭修伯

Chemical reaction

• Both the rate and equilibrium conversion of a chemical reaction depend on the temperature, pressure, and composition of reactants.– Although reaction rates are not susceptible to

thermodynamic treatment, equilibrium conversions are.

– The purpose of this lecture is to determine the effect of temperature, pressure, and initial composition on the equilibrium conversions of chemical reactions.

• A general chemical reaction:

– A differential change in the number of moles of a reacting species:

...... 44332211 AvAvAvAv

dv

dn

v

dn

v

dn

v

dn ...

4

4

3

3

2

2

1

1

Reaction coordinate, which characterizes the extent or degree to which a reaction has taken place.

)...,,2,1( Nidvdn ii )...,,2,1(00

Nidvdn i

n

n i

i

i

Reaction coordinate

)...,,2,1(0 Nivnn iii vnnn

ii 0

vn

vn

n

ny iii

i

0

0

For a system in which the following reaction occurs,assume there are present initially 2 mol CH4, 1 mol H2O, 1 mol CO and 4 mol H2. Determine expressions for the mole fractions yi as functions of ε.

23111 i

i

224 3HCOOHCH

vn

vn

n

ny iii

i

0

0

8411200 i

inn

28

24

CHy

28

1

COy

28

12

OHy

28

342

Hy

Consider a vessel which initially contains only n0 mol of water vapor. If decomposition occurs according to the reaction,

find expressions which relate the number of moles and the mole fraction of each chemical species to the reaction coordinate ε.

2

1

2

111

ii

222 2

1OHOH

vn

vn

n

ny iii

i

0

0

000 nnni

i

21

0

02

n

ny OH

21

0

2

nyH

21

21

0

2

nyO

02nn OH

2

12On

2Hn

Multireaction

• Two or more independent reactions proceed simultaneously– νi,j : the stoichiometric number of species i in reaction j.

– the change of the moles of a species ni: jj

jii dvdn ,

i

jij ,

total stoichiometric number:

jj

jiii vnn ,0

integration

summation

jj i

jivnn

,0

jj

jvnn 0jj

j

jj

jii

i vn

vn

y

0

,0

Consider a system in which the following reactions occur,

if there are present initially 2 mol CH4 and 3 mol H2O, determine expressions for the yi as functions of ε1 and ε2 .

224 3HCOOHCH 2224 42 HCOOHCH

i CH4 H2O CO CO2 H2

j νj

1 -1 -1 1 0 3 22 -1 -2 0 1 4 2

jj

j

jj

jii

i vn

vn

y

0

,021

21

225

24

CHy

21

21

225

232

OHy

21

1

225

COy

21

2

2252

COy21

21

225

432

Hy

Equilibrium criteria to chemical reactions

• ε is the single variable that characterizes the progress of the reaction– the total Gibbs energy at constant T and P is

determined by ε

– if not in chemical equilibrium, any reaction leads to a decrease in the total Gibbs energy of the system

• The condition for equilibrium:– the total Gibbs energy is a minimum

– Its differential is zero 0, PTtdG

)(fG t

Fig 13.1

Standard Gibbs energy change and the equilibrium constant

i

iidndTnSdPnVnGd )()()(

dvdn ii

i

ii ddTnSdPnVnGd )()()(

0)()(

,,

mequilibriuat

PT

t

PTiii

GnG

The fugacity of a species in solution: iii fRTT ˆln)(

For pure species i in its standard state:o

iioi fRTTG ln)(

oi

ioii f

fRTG

ˆln

0i

ii

0ˆ

ln

io

i

ioii f

fRTG

0ˆ

ln

i i

v

oi

ioii

i

f

fRTG

RT

G

f

f i

oii

i

v

oi

i

i

ˆln

RT

GK

o

exp

Kf

f

i

v

oi

i

i

ˆ

i

oii

o GG

RT

GK

o

expThe equilibrium constant for the reaction, f(T)

i

oii

o GG The standard Gibbs energy change of reaction, f(T)

i

oii

o MM Other standard property changes of reaction :

dT

RTGd

RTH

oi

oi

2

dT

RT

Gd

RTH

i

oii

i

oii

2

dT

RTGd

RTH

o

o

2

dT

Kd

RT

H o ln2

oi

oi

oi TSHG

i

oiii

oiii

oii STHG

For a chemical species in its standard state:

ooo STHG

summation

T

T

oPoo dT

R

CRHH

00

T

T

oPoo

T

dT

R

CRSS

00

0

000 T

GHS

ooo

T

T

oP

T

T

oPoooo

T

dT

R

CRTdT

R

CRGH

T

THG

0000

00

T

T

oP

T

T

oPoooo

T

dT

R

CRTdT

R

CRGH

T

THG

0000

00

RT

GK

oln

Readily calculated at any temperature from the standard heat of reaction and the standard Gibbs energy change of reaction at a reference temperature

210 KKKK

0

00 exp

RT

GK

o

T

T

RT

HK

o0

0

01 1exp

T

T

oP

T

T

oP

T

dT

R

CdT

R

C

TK

00

1exp2

Fig 13.2

Calculation the equilibrium constant for the vapor-phase hydration of ethylene at 145 and at 320 °C from data given in App.c

OHHCOHHC 52242

The heat capacity data:

210 KKKK

OHHCOHHC 24252

0

00 exp

RT

GK

o

T

T

RT

HK

o0

0

01 1exp

T

T

oP

T

T

oP

T

dT

R

CdT

R

C

TK

00

1exp2

From Table C.4

molJHH oo 457922980

molJGG oo 83782980

T K0 K1 K2 K

298.15 29.366 1 1 29.366

418.15 29.366 4.985x10-3 0.986 1.443x10-1

593.15 29.366 1.023x10-4 0.9794 2.942x10-3

• Gas phase reaction:– Ideal-gas state: – Standard-state pressure Po of 1 bar

ooi Pf

Kf

f

i

v

oi

i

i

ˆ ooi Pf

KP

f

i

v

oi

i

ˆ Pyf iii ̂ˆ KP

Py

oi

v

ii

i

ˆ

i

i

An ideal solution: ii ˆ KP

Py

oi

vii

i

An ideal gas: 1ˆ i KP

Py

oi

vi

i

f (T)f (P)f (composition)

• Liquid phase reaction:

Kf

f

i

v

oi

i

i

ˆ iiii fxf ˆ

Kf

fx

i

v

oi

iii

i

oi

iP

P ioii f

fRTdPVGG

oln

RT

PPV

f

f oi

oi

i ln

iii

o

i

vii V

RT

PPKx i exp

Except at high pressures

Kxi

vii

i Ideal solution Kxi

vi

i Law of mass action

Equilibrium conversions for single reactions

• Single reaction in a homogeneous system:– assuming an ideal gas and the equilibrium constant is

known:

– assuming an ideal solution and the equilibrium constant is known:

– the phase composition at equilibrium can be obtained

KP

Py

oi

vi

i

Kxi

vi

i

The water-gas-shift reaction, is carried out under the different sets of conditions described below. Calculate the fraction of steam reacted in each case. Assume the mixture behave as an ideal gas.

222 HCOOHCO

(a) the reactant consists of 1 mol of H2O vapor and 1 mol of CO, T = 1100 K, P = 1 bar

From Fig 13.2, K = 1 K

P

Py

oi

vi

i

0i

i 1

2

22 OHCO

COH

i

vi yy

yyy i

vn

vn

n

ny iii

i

0

0

2

1 eCOy

22

eCOy

2

12

eOHy

22

eHy

5.0e

(b) the same as (a) except that the pressure is 10 bar

Since v = 0, the increase in pressure has no effect: 5.0e

(c) the same as (a) except that 2 mol of N2 is included in the reactants

Since N2 does not take part in the reaction and serves as a diluent: 5.0e

(d) the reactants are 2 mol of H2O and 1 mol of CO. other conditions are the same as (a)

vn

vn

n

ny iii

i

0

0

3

1 eCOy

32

eCOy

3

22

eOHy

32

eHy

667.0e

The fraction of steam that reacts is then 0.667 / 2 = 0.333

(e) the reactants are 1 mol of H2O and 2 mol of CO. other conditions are the same as (a)

vn

vn

n

ny iii

i

0

0

3

2 eCOy

32

eCOy

3

12

eOHy

32

eHy

667.0e

The fraction of steam that reacts is then 0.667

(f) the initial mixture consists of 1 mol of H2O, 1 mol of CO and 1 mol of CO2 . other conditions are the same as (a)

vn

vn

n

ny iii

i

0

0

3

1 eCOy

3

12

eCOy

3

12

eOHy

32

eHy

333.0eThe fraction of steam that reacts is then 0.333

(g) same as (a) except that the temperature is 1650 K

From Fig 13.2, K = 0.316 K

P

Py

oi

vi

i

0i

i316.0

2

22 OHCO

COH

yy

yy

vn

vn

n

ny iii

i

0

0

2

1 eCOy

22

eCOy

2

12

eOHy

22

eHy

36.0e

The reaction is exothermic, and conversion decreases with increasing temperature.

Estimate the maximum conversion of ethylene to ethanol by vapor phase hydration at 250°C and 35 bars for an initial steam-to-ethylene ratio of 5.

For a temperature of 250°C, K = 10.02 x 10-3

KP

Py

oi

v

ii

i

ˆ

Assuming the reaction mixture is an ideal solution.ii ˆ

OHHCOHHC 52242

1i

i

)1002.10( 3

224242

oOHOHHCHC

EtOHEtOH

P

P

yy

y

Tc /K Pc /bar ω Tri Pri B0 B1φ i

C2H4 282.3 50.40 0.087 1.853 0.694 -0.074 0.126 0.977H2O 647.1 220.55 0.345 0.808 0.159 -0.511 -0.281 0.887

EtOH 513.9 61.48 0.645 1.018 0.569 -0.327 -0.021 0.827

)1002.10(1

35

)887.0()977.0(

)827.0( 3

242224242

OHHC

EtOH

OHOHHCHC

EtOHEtOH

yy

y

yy

y

e

eHCy

6

142

vn

vn

n

ny iii

i

0

0

e

eOHy

6

52

e

eEtOHy

6

233.0e

The equilibrium conversion is a function of temperature, pressure, and the steam-to-ethylene ratio in the feed:

Fig 13.4

In a laboratory investigation, acetylene is catalytically hydrogenated to ethylene at 1120 °C and 1 bar. If the feed is an equilmolar mixture of acetylene and hydrogen, what is the composition of the product stream at equilibrium?

222 2 HCHC 42222 HCHC

42222 HCHHC

oII

oI

o GGG

RT

GK

oln

III KRTKRTKRT lnlnln

1105.2104 65 III KKK

KP

Py

oi

vi

i

Ideal gas1

222

42 HCH

HC

yy

y

e

eHy

2

12

vn

vn

n

ny iii

i

0

0

e

eHCy

2

122

e

eHCy

242

293.0e

414.0222 HCH yy172.0

42HCy

Acetic acid is esterified in the liquid phase with ethanol at 100 °C and atmospheric pressure to produce ethyl acetate and water according to the reaction:

If initially there is 1 mol of each of acetic acid and ethanol, estimate the mole fraction of ethyl acetate in the reacting mixture at equilibrium.

OHHCOOCCHOHHCCOOHCH 2523523

2

1 eEtOHAcH xx

vn

vn

n

nx iii

i

0

0

22

eOHEtAc xx

6879.0e

344.02/6879.0 EtAcx

For the reaction at standard state (298 K):

JG

JHo

o

4650

3640

298

298

RT

GK

oln

8759.115.298314.8

4650ln 298

RT

GK

o dT

Kd

RT

H o ln2

8586.4373 K

Assuming ideal solution:

Kxi

vi

i Kxx

xx

EtOHAcH

OHEtAc 2

The gas phase oxidation of SO2 to SO3 is carried out at a pressure of 1 bar with 20% excess air in an adiabatic reactor. Assuming that the reactants enter at 25°C and that equilibrium is attained at the exit, determine the composition and temperature of the product stream from the reactor

322 2

1SOOSO For the reaction at standard state (298 K):

JG

JHo

o

70866

98890

298

298

Assuming 1 mol of SO2 entering the reactor,O2: 0.5 x (1.2) = 0.6 mol enteringN2: 0.6 x (79/12) = 2.257 mol entering

In the product stream: SO2: 1 - εe

O2: 0.6 - 0.5εe

SO3: εe

N2: 2.257total moles: 3.857-0.5εe

Energy balance: 0298 HHH oPe

o

The enthalpy change of the products at T

)15.298()15.298( TCnTCHi

H

oPiiH

oP

oP

15.298298

H

oP

eo

C

HT

KP

Py

oi

vi

i

5.0i

i

Ke

e

e

e

5.0

5.06.0

5.0857.3

1

Assuming ideal gas:

210 KKKK

0

00 exp

RT

GK

o

T

T

RT

HK

o0

0

01 1exp

T

T

oP

T

T

oP

T

dT

R

CdT

R

C

TK

00

1exp2

IDCPHT

IDCPST

K14.11894

3054.11ln

Assume T ln K e T Converges at T = 855.7 K, εe = 0.77

0062.05.0857.3

12

e

eSOy

...

• Reactions in heterogeneous systems– a criterion of vapor/liquid equilibrium, must be

satisfied along with the equation of chemical-reaction equilibrium

Estimate the compositions of the liquid and vapor phases when ethylene reacts with water to form ethanol at 200°C and 34.5 bar, conditions which assure the presence of both liquid and vapor phases. The reaction vessel is maintained at 34.5 bar by connection to a source of ethylene at this pressure. Assume no other reactions.

According to phase rule (see later): F = 2 (say, P, T)the material balance equations do not enter into the solution of this problem.

Regarding the reaction occurring in the vapor phase: )(52)(2)(22 ggg OHHCOHHC

Assuming ideal gas :

KP

f

i

v

oi

i

ˆ o

OHHC

EtOH Pff

fK

242

ˆˆ

ˆ

barPo 1

KT 15.473031.015.473 KK

Phase equilibrium:

li

vi ff ˆˆ

o

OHHC

EtOH Pff

fK

242

ˆˆ

ˆ o

ll

lo

vv

v

Pff

fP

ff

fK

OHHC

EtOH

OHHC

EtOH

242242

ˆˆ

ˆ

ˆˆ

ˆ

For vapor phase: Pyf iiv

i ̂ˆ

For liquid phase: liii

li fxf ˆ

o

lOHOHOHHCHC

lEtOHEtOHEtOH P

fxPy

fxK

2224242

ˆ

sati

sati

sati

li Pff ~

ii ˆIdeal solution (vapor):

o

satOH

satOHOHOHHCHC

satEtOH

satEtOHEtOHEtOH P

PxPy

PxK

22224242

P

Pxy

i

sati

satiii

i

P

Px

P

Pxy

OH

satOH

satOHOHOH

EtOH

satEtOH

satEtOHEtOHEtOH

HC

2

2222

421

OHEtOHHC yyy242

1

o

satOH

satOHOHOHHCHC

satEtOH

satEtOHEtOHEtOH P

PxPy

PxK

22224242

4221 HCEtOHOH xxx

volatile

Known values: P, T, , , , ,

satOH2

satEtOH

42HC satEtOHP

satOHP

2

OHOHHC

EtOHEtOH

xy

xK

2242

0493.0

OHOHEtOHEtOHHC xxy2242

493.0907.01

EtOHOH xx 12

OHOHHC

EtOHEtOH

xy

xK

2242

0493.0

OHOHEtOHEtOHHC xxy2242

493.0907.01

Values of , are determined experimentallyOHEtOH 2

Assume xEtOH xH2O yC2H4 K

check 031.015.473 KK

Converged at:xi yi

EtOH 0.06 0.180H2O 0.94 0.464C2H4 0.00 0.356

Σ =1 Σ =1

Phase rule and Duhem’s theorem for reacting systems

• For non-reacting systems– π phases and N chemical species:

• For reacting systems– Phase rule variables in each phase: temperature, pressure

and N-1 mole fraction. Total number of these variable: 2 + (N-1) π

– Phase-equilibrium equations: (π-1) N– r independent chemical reactions at equilibrium within

the system: r equations– F = [2+(N-1) π ] - [(π-1) N] – [r] :– With special constraints s:

NF 2

rNF 2

srNF 2

Determine the number of degree of freedom F for each of the following systems:(a) A system of two miscible non-reacting species which exists as an azeotrope in

vapor/liquid equilibrium.(b) A system prepared by partially decomposing CaCO3 into an evacuated space.(c) A system prepared by partially decomposing NH4Cl into an evacuated space.(d) A system consisting of the gases CO, CO2, H2, H2O, and CH4 in chemical

equilibrium(a) Two non-reacting species in two phases

With no azeotrope: With azeotrope (x1=y1, one constraint):

202222 rNF 12 srNF

(b) Single reaction:Three chemical species and three phases:

)(2)()(3 gss COCaOCaCO )(2)()(3 gss COCaOCaCO

113322 rNF

(c) Single reaction:Three chemical species and two phases:One constraint: gas phase is equimolar

)()(3)(4 ggs HClNHClNH

1113222 srNF

)()(3)(4 ggs HClNHClNH

(d) 5 species, a single gas phase, and no special constraints: srNF 2???The formation reactions:

COOC 22

1

22 COOC

OHOH 222 2

1

422 CHHC

222

1COOCO

2224 2 COHOCH

Eliminating C

Eliminating C

OHOH 222 2

1

222

1COOCO

2224 2 COHOCH

OHCOHCO 222 Eliminating O2

2224 42 COHOHCH Eliminating O2

r = 2

4025122 srNF

Duhem’s theorem

• For non-reacting system :– For any closed system formed initially from given masses

of particular chemical species, the equilibrium state is completely determined by specification of any two independent variables.

–

• For reacting system:– A variable is introduced for each independent reaction: εj – A equilibrium relation can be written for each independent

reaction–

2])1[(])1(2[ NNNF

211])1[(])1(2[ NNNF

Multireaction equilibria

• A separate equilibrium constant is evaluated for each reaction:

– gas phase reaction

ji

v

oi

i Kf

fji

,ˆ

ji

v

oi K

P

fji

,ˆ

Ideal gas

joi

vi K

P

Py

j

ji

,

A feed stock of pure-n-butane is cracked at 750K and 1.2 bar to produce olefins. Only two reactions have favorable equilibrium conversions at these conditions:

If these reactions reach equilibrium, what is the product composition?

6242104 HCHCHC 463104 CHHCHC

With a basis of 1 mol of n-butane feed:

III

IIIHCy

1

1104

III

IHCHC yy

16242

III

IICHHC yy

1463

joi

vi K

P

Py

j

ji

,

Assuming ideal gas:Io

HC

HCHC KP

P

y

yy 1

104

6242

IIoHC

CHHC KP

P

y

yy 1

104

463

1068.0I 8914.0II 001.0104HCy 0534.0

6242 HCHC yy 4461.0

463 CHHC yy

A bed of coal in a coal gasifier is fed with steam and air, and produce a gas stream containing H2, CO, O2, H2O, CO2, and N2. If the feed to the gasifier consists of 1 mol of steam and 2.38 mol of air, calculate the equilibrium composition of the gas stream at P = 20 bar for temperature 1000, 1100, 1200, 1300, 1400, and 1500 K.

The feed stream:1 mol C, O2 = (0.21)(2.38) = 0.5 mol, N2 = (0.79)(2.38) = 1.88 mol

The formation reactions for the compounds are:

OHOH 222 2

1

COOC 22

1

22 COOC

Δ Gºf J/molT (K) H2O CO CO2

1000 -192424 -200240 -3957901100 -187000 -209110 -3959601200 -181380 -217830 -3960201300 -175720 -226530 -3960801400 -170020 -235130 -3961301500 -164310 -243740 -396160

RT

GK

oln

K = ...

joi

vi K

P

Py

j

ji

,

Assuming ideal gases for the remaining species:

21

22

2

oHO

OHI P

P

yy

yK

21

2

oO

COII P

P

y

yK

2

2

O

COIII y

yK

With a basis of 1 mol of C feed:

238.3

2III

IHy

238.3 III

IICOy

238.3

121

2III

IIIIII

Oy

238.3

12

III

IOHy

238.3

2III

IIICOy

238.3

88.12

IIINy

Carbon is a pure solid phase: 1)1(@

)20(@ˆ

barf

barf

f

fo

C

Co

C

C

Three equations and three unknowns … can be solved.

Non- stoichiometric method

• Alternative method to solve chemical reaction equilibrium problems– base on the fact that the total Gibbs energy of the

system has it minimum value

– the basis for a general scheme of computer solution

–

– Find the set {ni} which minimizes Gt for specified T and P, subject to the constraints of the material balances.

NPTt nnnngG ...,,, 321,

Lagrange’s undetermined multiplier method

• The material balance on each element k :

ki

iki Aan

0

k

iikik Aan

multiply the Lagrange multipliers

0

kk

iikik Aan

summed over k

kk

iikik

t AanGF

kk

iikik

t AanGF

F (and Gt) minimum value?

),...,2,1(0,,,,

Nian

G

n

F

kikk

nPTi

t

nPTijj

),...,2,1(0 Niak

ikki

oi

ioii f

fRTG

ˆln gas phase reaction

pure ideal gas

oio

ii P

fRTG

ˆln

oiio

fii P

PyRTG

ˆ

ln

),...,2,1(0ˆ

ln NiaP

PyRTG

kikko

iiofi

Calculate the equilibrium compositions at 1000 K and 1 bar of a gas-phase system containing the species CH4, H2O, CO, CO2, and H2. In the initial unreacted state, there are present 2 mol of CH4 and 3 mol of H2O. Values of ΔG°f at 1000 K of each species are given.

Element kC O H

Ak = no. of atomic masses of k in the system2 3 14

species aik = no. of atoms of k per molecule of iCH4 1 0 4H2O 0 1 2CO 1 1 0CO2 1 2 0H2 0 0 2

),...,2,1(0ˆ

ln NiaP

PyRTG

kikko

iiofi

Ideal gas: 1ˆ i 1oP

P

),...,2,1(0ln NiaRTn

n

RT

G

kik

k

i i

i

ofi

),...,2,1(0ln NiaRTn

n

RT

G

kik

k

i i

i

ofi

For the 5 species:

CH4 : 04

ln19720

4

RTRTn

n

RTHC

i i

CH

...

5 equations

For the 3 elements:

C : 224 COCOCH nnn

H : 14224224 HOHCH nnn

O : 3222 COCOOH nnn

3 equations

2224 HCOCOOHCHi i nnnnnn 1 equations

Solve simultaneously