1. 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:
d[E] d[F ]
(a) Rate = − = = k 2 [ D][ E ] ............(1)
dt dt
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.
d[D]
= k1 [ A][ B] − k −1 [C ][ D] − k 2 [ D][ E ] = 0
dt
k1 [ A][ B] = k −1 [C ][ D] + k 2 [ D][ E ] = (k −1 [C ] + k 2 [ E ])[ D]
k1 [ A][ B]
[ D] = ...................(2)
(k −1 [C ] + k 2 [ E ])
Substitute equation (2) in equation (1)
d[ E ] d[ F ]  k1 [ A][ B]  k1k 2 [ A][ B][ E ]
Rate = − = = k 2 [ E ] =
dt dt  (k −1[C ] + k 2 [ E ])  k −1[C ] + k 2 [ E ]
(b) Used the second step as the rate-determining step
d[E] d[F ]
Rate = − = = k 2 [ D][ E ] ............(3)
dt dt
From the first step which is in near-equilibrium,
k1 [ A][ B] = k −1 [C ][ D]
[C ][ D] k1
= = K eq
[ A][ B] k −1
k1 [ A][ B]
[ D] = ..........(4)
k −1 [C ]
Substitute equation (4) in equation (3)
d[E] d[F ]  k [ A][ B]  k1 k 2 [ A][ B][ E ]
Rate = − = = k 2 [ E ] 1 =
dt dt  k −1[C ]  k −1 [C ]
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2. 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 k1 2CH3
→
Chin transfer CH3 + C2H6 k 2 → CH4 + C2H5
k
Propagation C2H5  31 C2H4 + H
→
H + C2H6 k 4
→ H2 + C2H5
Termination H + C2H5 k5 → C2H6
What is the rate law for the above reaction, using the usual assumption?
Answer:
d [C 2 H 6 ]
Rate = - = k1 [C 2 H 6 ] + k 2 [CH 3 ][C 2 H 6 ] + k 4 [ H ][C 2 H 6 ] .......(1)
dt
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;
d [CH 3 ]
= 2k1 [C 2 H 6 ] − k 2 [CH 3 ][C 2 H 6 ] = 0 ....... (2)
dt
d [C 2 H 5 ]
= k 2 [CH 3 ][C 2 H 6 ] + k 4 [ H ][C 2 H 6 ] − k 3 [C 2 H 5 ] − k 5 [ H ][C 2 H 5 ] = 0 .......(3)
dt
d[H ]
= k 3 [C 2 H 5 ] − k 4 [ H ][C 2 H 6 ] − k 5 [ H ][C 2 H 5 ] = 0 …… .(4)
dt
2 k1
From (2) yields [CH3]= …………..(5)
k2
(2k1 + k 4 [ H ])[C 2 H 6 ]
Substituting (5) in (3) yield [C2H5]= ……(6)
k3 + k5 [H ]
Substituting (5) and (6) in (4) yields
 (2k + k 4 [ H ])[C 2 H 6 ]   (2k + k 4 [ H ])[C 2 H 6 ] 

k3  1  − k 4 [ H ][C 2 H 6 ] − k 5 [ H ] 1
  =0

 k3 + k5 [H ]   k3 + k5 [H ] 
2k1 k 3 [C 2 H 6 ] + k 3 k 4 [ H ][C 2 H 6 ] − k 3 k 4 [ H ][C 2 H 6 ]
− k 4 k 5 [ H ]2 [C 2 H 6 ] − 2k1 k 5 [ H ][C 2 H 6 ] − k 4 k 5 [ H ] 2 [C 2 H 6 ] = 0
2k1 k 3 + k 3 k 4 [ H ] − k 3 k 4 [ H ] − k 4 k 5 [ H ] 2 − 2k1 k 5 [ H ] − k 4 k 5 [ H ]2 = 0
− 2k 4 k 5 [ H ] 2 − 2k1 k 5 [ H ] + 2k1k 3 = 0
k 4 k 5 [ H ] 2 + k1 k 5 [ H ] − k1 k 3 = 0
− k1 k 5 ± (k1k 5 )2 − 4(k 4 k 5 )(−k1k 3 )
Solve quadratic equation gives: [H]=
2k 4 k 5
1/ 2
k k 
In general, k1 is small so that [H]=  1 3 
k k  ………(7)
 4 5
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3. 1/ 2
d [C 2 H 6 ]  2k  kk 
Rate = − = k1 [C 2 H 6 ] + k 2  1 [C 2 H 6 ] + k 4  1 3 
k  k k  [C 2 H 6 ]
dt  2   4 5
1/ 2
k k k 
= k1 [C 2 H 6 ] + 2k1 [C 2 H 6 ] +  1 3 4
 k 
 [C 2 H 6 ]
 5 
d [C 2 H 6 ]  
1/ 2
k k k 
Rate = − = 3k1 +  1 3 4
 k 
 [C 2 H 6 ]
dt 
  5  

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 − + H+ 2
Answer:
Lets: CH3CO2H = A, CH3CO − = B, and H+ = C
2
d [ A] [ B]eq [C ]eq k d
Rate = − = k d [ A] − k a [ B][C ] At equilibrium, Keq= = =1.73×10-5
dt [ A]eq ka
After temperature jump, and a new equilibrium, [A]= [A]eq+ x, [B] = [ B]eq-x and
[C]= [C]eq- x,
d [ A] d ([ A]eq + x)
= = − k d ([ A]eq + x) + k a ([ B]eq − x)([C ]eq − x)
dt dt
dx
= − k d [ A]eq − k d x + k a [ B]eq [C ]eq − k a [ B]eq x − k a [C ]eq x + k a x 2
dt
k d [ A]eq = k a [ B]eq [C ]eq and x 2 ≈ small = 0
dx
dt
[
= −(k d ) x − (k a [ B]eq + k a [C ]eq ) x = − k d + k a ([ B]eq + [C ]eq ) x ]
1
= k d + k a ([ B]eq + [C ]eq )
τ
[ B ]eq [C ]eq kd −5 x2 x2
K eq = = = 1.73 × 10 = ≈
[ A]eq ka 0.10 − x 0.10
x = 1.73 × 10 −5 × 0.10 = 1.32 × 10 −3 = [ B]eq = [C ]eq
1 1
= −9
= k d + k a (1.32 × 10 −3 + 1.32 × 10 −3 )
τ 8.5 × 10
= 1.73 × 10 −5 k a + 2.64 × 10 −3 k a = 2.6573 × 10 −3 k a
k a = 4.42 × 1010 L mol −1 s −1
k d = 1.73 × 10 −5 k a = 7.65 × 10 5 s −1
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4. 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
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5. 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×106 /s-1
20.0 7.62
37.0 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. 1.04 × 1010 s −1 B. 7.66 × 10 6 s −1 C. 7.66 × 10 −6 s −1 D. 1.04 × 10 −10 s −1
5. For the mechanism
Which of the following is the rate law of the mechanism?
d [ D] d [ B]
A. Rate = = k 2 [ B] B. Rate = = k1 [ A] − k −1[ B][C ]
dt dt
d [ B] d [ A] d [ D] k1 k 2
C. Rate = = 0 D. Rate = = −k1 [ A] + k −1[ B ][C ] E. Rate = = [ A][C ] −1
dt dt dt k −1
6. The mechanism for decomposition of ozone, 2O3→ 3O3 has been proposed as follows:
O3 + M k1 O2+ O + M
→
O + O2 k 2 → O3
O + O3 k 3 → 2O2
Where M is any molecule. Which of the following is NOT TRUE?
d [O3 ] d [O]
A. = − k1 [ M ][O3 ] +k2[O][O3]-k3[O][O3] B. =0
dt dt
d [O2 ] d [O2 ]
C. = k1 [O3 ][M ] − k 2 [O][O2 ] − 2k 2 [ M ][O3 ] D. =0
dt dt
E. None of the above
7. Consider the opposing reaction:
Both of which are first order reactions. Which of the equations below is TRUE ?
d [ A] d [ A]
A. = −k1 [ A] + k −1 [ B] B. = k1 [ A] − k −1 [ B]
dt dt
d [ A] [ A]eq k
C. - = k1 [ A] + k −1 [ B] D. K eq = E. K eq = −1
dt [ B]eq k1
42