Chapter X

Kinetics of Complex Reactions

Levine: p.559 17.9

In this section we are to consider some examples of reactions more complex than A + B P, and see how the integrated rate laws are modified.

complex reactions:

reaction contains more than one elementary reaction

typical complex reactions

1) Opposing Reaction

2) Parallel Reaction

3) Consecutive Reaction

§1. Typical complex reactions

1) Opposing Reaction / reversible reaction

majority of the reactions are reversible, i.e., the forward and the backward / reverse reaction take place simultaneously.

(1) kinetic equilibrium constant

for opposing reaction consisting of elementary reactions:

[A] [B]a br k [G] [H]g hr k

As reaction proceeds, r+ increases while r decreases. When r+ becomes equal to r, equilibrium is reached.

+

-

A B G Hk

ka b g h

[A] [B] [G] [H]a b g hk k

[G] [H]

[A] [B]

g h

ca b

kK

k

k

kKc

In this way we arrive at a very important connection

between the equilibrium constant and the rate coefficients of

simple reactions. This relation, named as kinetic equilibrium

constant, is correct only for elementary reactions.

therefore

For first-first order opposing reaction:

(2) rate equation

t = 0 a 0

t = t a-x x

t = te a-xe xe

The change rate of [A] has two contributions: A is depleted by the forward reaction at a rate k+[A], but is replenished by

the reverse reaction at a rate k-[B]. The total rate of change of

the concentration of A is therefore

xkxakdt

dx )(

A Bk

k

e e( )k a x k x e

e

( )k a xk

x

Under equilibrium conditions

e

e

( )( )

a xdxk a x k x

dt x

e

e

( )x xdxk a

dt x

e e

e

ln( )

x xk

at x x

e e

e

ln( )

x xk

at x x

e e

e

ln( )

a x xk

at x x

e e

e

ln( )

a x xk

at x x

which suggests that k+ and k can be determined by measuring x at t and equilibrium concentration.

e

e

1ln

( )

xk k

t x x e

e

1ln

( )

xk k

t x x

e

e

ln ( )( )

xk k t kt

x x e

e

ln ( )( )

xk k t kt

x x

Similar to the rate equation of first-order reaction

e e

e

ln( )

x xk

at x x

e e

e

ln( )

x xk

at x x

e e

e

ln( )

a x xk

at x x

e e

e

ln( )

a x xk

at x x

1-2 opposing reaction

2-2 opposing reaction

A B Ck

k

A + B C + Dk

k

Principle of relaxation method for studying fast reaction

2) parallel reaction / Competing reaction

1

[B][A]

dk

dt

2

[C][A]

dk

dt

))(()()( 2121 xakkxakxakdt

dx

21 kk When )(1 xakdt

dx

When 21 kk )(2 xakdt

dx

The rate of parallel reaction is determined mainly by the faster one.

))(( 21 xakkdt

dx

Integration of the equation yields:

tkkxa

a)(ln 21

])(exp[)( 21 tkkaxa

A B C

a 0 0

a-x y z

x = y + z

)(1 xakdt

dy

)(2 xakdt

dz

a

t

For production of B and C:

)(1 xakdt

dy ])(exp[)( 21 tkkaxa ])(exp[)( 21 tkkaxa

])(exp[ 211 tkkakdt

dy ]})(exp[1{ 21

21

1 tkkkk

aky

]})(exp[1{ 21

21

1 tkkkk

aky

2

1

k

k

z

y

2

1

k

k

z

y

The composition of the final products is fixed.

selectivity of the reaction.

A

B C

t

c]})(exp[1{ 21

21

2 tkkkk

akz

]})(exp[1{ 21

21

2 tkkkk

akz

Optimum temperature for better selectivity

Example

A B A1 Ea, 1

A C A2 Ea, 2

When A1>A2, Ea,1>Ea,1, to in

crease the ratio of B in the products, should higher temperature or lower temperature be chosen?

logA2

1/T

logA1

log k

B

C

When A1> A2, Ea,1<Ea,2, to

increase the ratio of B in the

products, should higher

temperature or lower

temperature be chosen?

logA2

1/T

logA1

logk

B

C

The selectivity of the parallel reaction can be improved by adoption of appropriate catalyst.

Using catalyst to better selectivity

Main reaction and Side reaction:

reaction with higher k is taken as the main reaction, while others side reactions.

Reaction that produces the demanded product is the main reaction.

product

consumed

nS

nSelectivity:

3 Consecutive reaction

Some reactions proceed through the formation of intermediate.

CH4 + Cl2 CH3Cl CH2Cl2 CHCl3 CCl4

A B C

t = 0 a 0 0

t = t x y z

a = x + y + z

General reaction 1 2A B Ck k

xkdt

dx1

tkx

a1ln )exp( 1tkax

ykxkdt

dy21

)exp()exp( 2112

1 tktkkk

aky

)exp()exp( 21

12

1 tktkkk

aky

ykdt

dz2

C

tmax t

A

C

B

)exp(

)exp(1

212

1

112

2

tkkk

k

tkkk

k

az

)exp(

)exp(1

212

1

112

2

tkkk

k

tkkk

k

az

shows that the intermediate’s concentration rises from zero to a maximum and then drops back to zero as A is depleted and C dominates in the mixture. ��

)exp()exp( 2112

1 tktkkk

aky

)exp()exp( 21

12

1 tktkkk

aky

If C is the demanded product, the reaction time should be prolonged. If B is the demanded product, the reaction should be interrupted at optimum time, i.e., tmax.

0dt

dy

21

21max

)/ln(

kk

kkt

At tmax, the concentration of B = ?

C

tmax t

A

C

B

21

21max

)/ln(

kk

kkt

)exp()exp( 21

12

1 tktkkk

aky

)/ln

exp()/ln

exp(21

212

21

211

12

1max kk

kkk

kk

kkk

kk

aky

21

2

21

1

)/exp(ln)/ln(exp 212112

1max

kk

k

kk

k

kkkkkk

aky

21

2

21

1

)()(2

1

2

1

12

1max

kk

k

kk

k

k

k

k

k

kk

aky

1)()( 1

2

1

2

1

12

1max

21

2

k

k

k

k

kk

aky kk

k12

2

)(2

1max

kk

k

k

kay 12

2

)(2

1max

kk

k

k

kay

k2/k11/5 5 10 100 103 108

tmax2.01 0.40 0.25 0.047 710-3 10-7

ymax/a 0.67 0.13 0.08 7 10-3 10-3 0

Ea,1Ea,2-0.4 4.0 5.7 11.5 17.2 46.1

12

2

)(2

1max

kk

k

k

kay 12

2

)(2

1max

kk

k

k

kay

When k2 >> k1, ymax would

be very small, and the tmax

would be very short.

2 1

2 1max

ln( k / k )t

k k

2 1

2 1max

ln( k / k )t

k k

t

y

0

k1/k2

0dt

dy 0dt

dy

Physical meaning of k2 >> k1

B is a active intermediate (Such as active atom: Cl, H, etc., ra

dicals: CH3•, H2C:, C+, C-, etc., activated molecules: A*), it is d

ifficult to form but easy to decompose to product.

For consecutive reaction with large k2/k1 ratio, once the

reaction take place, the active intermediate (B) rapidly attains

its maximum concentration and its concentration keeps nearly

unchanged during the whole reaction.

0dt

dy 0dt

dy Steady-state approximation

1 2A B Ck k

)exp()exp(1 2

12

11

12

2 tkkk

ktk

kk

kaz

1 21 2

2 1

Zdc ak kexp( k t ) exp( k t )

dt k k

1 2

1 22 1

Zdc ak kexp( k t ) exp( k t )

dt k k

When k2 >> k1 1 1Zdc ak exp( k t )dt

1 1Zdc ak exp( k t )dt

The total rate is determined mainly by k1

When k2 << k12 2

Zdc ak exp( k t )dt

2 2Zdc ak exp( k t )dt

The total rate is determined mainly by k2

The rate of the overall consecutive reaction depends only on the smaller rate constant (rate-determining step).

rate-determining step (r. d. s.): the step with the slowest rate.

?? !! It’s a r.d.s

patient !

The rate of the elementary step with the lowest rate constant, i.e., r.d.s., can be used to express the actual rate of the overall reaction.

Its activation energy should be 10 kJmol-1 more than that of other steps.

What is a eligible r. d. s.?

Rate-determining step approximation

Procedure for synthesis of ammonia:

1) diffusion; 2) absorption; 3) activation; 4) reduction; 5)

protonation; 6) desorption; 7) diffusion.

which step is the r.d.s?

Key step for optimization of the reaction conditions

0 30 60 90 120 150 180 210 2400.0

0.2

0.4

0.6

0.8

1.0

5

4

3

2

1

C(mmol/L)

Ti me( mi n)

( 1) p- ni t r openol ( 2) hydr oqui none ( 3) pyocat echol ( 4) benzoqui none ( 5) mal ei c aci d

Electrocatalytic degradation of p-nitrophenol