38

TOPIC 9- Mechanism of chemical reactions

Examples of Solved Problems

1. (a) Apply the steady-state approximation to the mechanism;

To obtain the rate law. (b) Now apply the rate-determining-step approximation, assuming that

the first step is in near-equilibrium and the second step is rate-determining.

Answer:

(a) )1..(..........]][[][][

2 EDkdt

Fd

dt

EdRate ==−=

Choose the simplest or shortest equation of rate law. B is catalyst , D is an

intermediate ( species that occurs in the reaction steps but do not appear in the

overall reaction). Write expression for its net rate of formation, then change of the

concentration of intermediate is set equal to zero.

)2.........(..........])[][(

]][[][

]])[[][(]][[]][[]][[

0]][[]][[]][[dt

d[D]

21

1

21211

211

EkCk

BAkD

DEkCkEDkDCkBAk

EDkDCkBAk

+=

+=+=

=−−=

−

−−

−

Substitute equation (2) in equation (1)

][][

]][][[

])[][(

]][[][

][][

21

21

21

12

EkCk

EBAkk

EkCk

BAkEk

dt

Fd

dt

EdRate

+=

+==−=

−−

(b) Used the second step as the rate-determining step

)3..(..........]][[][][

2 EDkdt

Fd

dt

EdRate ==−=

From the first step which is in near-equilibrium,

)4(..........][

]][[][

]][[

]][[

]][[]][[

1

1

1

1

11

Ck

BAkD

Kk

k

BA

DC

DCkBAk

eq

−

−

−

=

==

=

Substitute equation (4) in equation (3)

][

]][][[

][

]][[][

][][

1

21

1

12

Ck

EBAkk

Ck

BAkEk

dt

Fd

dt

EdRate

−−

=

==−=

39

2. Consider the pyrolysis of ethane to ethylene: C2H6 = C2H4 + H2

At temperature of 700 to 900 K and pressures above about 0.2 bar, this reaction has the

following mechanism:

Initiation C2H6 → 1k 2CH3

Chin transfer CH3 + C2H6 → 2k CH4 + C2H5

Propagation C2H5 → 13kC2H4 + H

H + C2H6 → 4k H2 + C2H5

Termination H + C2H5 → 5k C2H6

What is the rate law for the above reaction, using the usual assumption?

Answer:

Rate = - )1.......(]][[]][[][][

6246232621

62 HCHkHCCHkHCkdt

HCd++=

The rate of formation of ethane in the last step is ignored because its occurs only after

long chains producing the products ( as Termination step).

CH3, C2H5 and H are intermediates, apply steady-state approximations;

(4)0]][[]][[][][

)3.......(0]][[][]][[]][[][

)2(.......0]][[][2][

525624523

525523624623252

6232621

3

.……=−−=

=−−+=

=−=

HCHkHCHkHCkdt

Hd

HCHkHCkHCHkHCCHkdt

HCd

HCCHkHCkdt

CHd

From (2) yields [CH3]= 2

12

k

k …………..(5)

Substituting (5) in (3) yield [C2H5]= ( )

][

][][2

53

6241

Hkk

HCHkk

+

+ ……(6)

Substituting (5) and (6) in (4) yields

( ) ( )

0][][2][][][2

0][][]][[2][][

]][[]][[][2

0][

][][2][]][[

][

][][2

2

5451

2

54434331

62

2

54625162

2

54

624362436231

53

6241

5624

53

6241

3

=−−−−+

=−−−

−+

=

+

+−−

+

+

HkkHkkHkkHkkHkkkk

HCHkkHCHkkHCHkk

HCHkkHCHkkHCkk

Hkk

HCHkkHkHCHk

Hkk

HCHkkk

0][][

02][2][2

3151

2

54

3151

2

54

=−+

=+−−

kkHkkHkk

kkHkkHkk

Solve quadratic equation gives: [H]= ( )

54

3154

2

5151

2

))((4

kk

kkkkkkkk −−±−

In general, k1 is small so that [H]=

2/1

54

31

kk

kk………(7)

40

][][2

][][

62

2/1

54

31

462

2

1

2621

62 HCkk

kkkHC

k

kkHCk

dt

HCdRate

+

+=−=

][3][

][][2][

62

2/1

5

4311

62

62

2/1

5

431

621621

HCk

kkkk

dt

HCdRate

HCk

kkkHCkHCk

+=−=

++=

3. For acetic acid in dilute aqueous solution at 25

o C , K= 1.73×10

-5 and the relaxation

time is 8.5×10-9

s for a 0.1 M solution. Calculate the rate constants for the forward and

reversed step in CH3CO2H = CH3CO −

2 + H+

Answer:

Lets: CH3CO2H = A, CH3CO −

2 = B, and H+

= C

]][[][][

CBkAkdt

AdRate ad −=−= At equilibrium, Keq=

a

d

eq

eqeq

k

k

A

CB=

][

][][=1.73×10

-5

After temperature jump, and a new equilibrium, [A]= [A]eq+ x, [B] = [ B]eq-x and

[C]= [C]eq- x,

2][][][][][

)])([]([)]([)]([][

xkxCkxBkCBkxkAkdt

dx

xCxBkxAkdt

xAd

dt

Ad

aeqaeqaeqeqadeqd

eqeqaeqd

eq

+−−+−−=

−−++−=+

=

[ ]

)][]([1

)][]([)][][()(

0][][][ 2

eqeqad

eqeqadeqaeqad

eqeqaeqd

CBkk

xCBkkxCkBkxkdt

dx

smallxandCBkAk

++=

++−=+−−=

=≈=

τ

eqeq

a

d

eq

eqeq

eq

CBx

x

x

x

k

k

A

CBK

][][1032.110.01073.1

10.010.01073.1

][

][][

35

225

==×=××=

≈−

=×===

−−

−

155

1110

335

33

9

1065.71073.1

1042.4

106573.21064.21073.1

)1032.11032.1(105.8

11

−−

−−

−−−

−−

−

×=×=

×=

×=×+×=

×+×+=×

=

skk

smolLk

kkk

kk

ad

a

aaa

adτ

41

Exercise 9a

1. Consider the following mechanism:

(a) Derive the rate using the steady-state approximation to eliminate [C].

(b) Assuming that k2 << k-1 , express the pre-exponential factor A and Ea for the

apparent second-order rate constant.

2. Derive the steady-state rate equation for the following mechanism for a trimolecular

reaction:

3. The reaction mechanism for the decomposition of A2

Involves an intermediate A. Deduce the rate law for the reaction in two ways by (a) assuming

that the first step is in near-equilibrium and the second step is rate-determining and (b) making

a steady-state approximation.

4. Derive the expression for the relaxation times τ for the following reactions:

Exercise 9b (Objective questions)

1. Which of the statements below is NOT TRUE about pre-exponential factor, A?

A. The pre-exponential factor, A has the same units for all reaction.

B. The unit of the pre-exponential factor, A depends on the order of the reaction

C. The pre-exponential factor, A has the same units as those of k.

D. As in general, the value of the pre-exponential factor, A may depends on T.

E. None of the above.

2. Which of the statements below is TRUE about activation energy, Ea?

A. Activation energies are never negative.

B. Activation energies are always positive.

C. Activation energies have the same units as the pre-exponential factor.

D. The value of the activation energies independent on T.

E. For reaction with Ea > 0, the greater Ea, the more rapidly the rate constant

increases with T

42

For the following data, answer question 3-4:

A substance decomposes acceding to first-order kinetics; A→ P .The rate constants at

two difference temperatures are as follows;

Temperature/ oC Rate constant, k×10

6 /s

-1

20.0

37.0

7.62

51.5

3. The activation energy of the reaction above is...

A.-84.89 J mol-1

B.- 84.89 J mol-1

C. 84.89 J mol-1

D.84.89 kJ mol-1

4. The pre-exponential factor A of the reaction above is ..

A. 1101004.1 −× s B. 161066.7 −

× s C. 161066.7 −−× s D. 1101004.1 −−

× s

5. For the mechanism

Which of the following is the rate law of the mechanism?

A. Rate = ][][

2 Bkdt

Dd= B. Rate = ]][[][

][11 CBkAk

dt

Bd−

−=

C. Rate = 0][

=dt

Bd D. Rate = ]][[][

][11 CBkAk

dt

Ad−

+−= E. Rate = 1

1

21 ]][[][ −

−

= CAk

kk

dt

Dd

6. The mechanism for decomposition of ozone, 2O3→ 3O3 has been proposed as follows:

O3 + M →1k O2+ O + M

O + O2 →2k O3

O + O3 →3k 2O2

Where M is any molecule. Which of the following is NOT TRUE?

A. ]][[][

313 OMk

dt

Od−= +k2[O][O3]-k3[O][O3] B. 0

][=

dt

Od

C. ]][[2]][[]][[][

3222312 OMkOOkMOk

dt

Od−−= D. 0

][ 2 =dt

Od

E. None of the above

7. Consider the opposing reaction:

Both of which are first order reactions. Which of the equations below is TRUE ?

A. ][][][

11 BkAkdt

Ad−

+−= B. ][][][

11 BkAkdt

Ad−

−=

C. - ][][][

11 BkAkdt

Ad−

+= D. eq

eq

eqB

AK

][

][= E.

1

1

k

kK eq

−=