12.4 Reaction Mechanisms and Rate Laws
553
Exercise 12.16
a. Show that the proposed mechanism in the previous example leads to the correct rate law.
b. If H 2 O and NO 3 are the products of step 2, and if no further HNO 3 enters in later steps,
propose steps 3 and 4 to complete the mechanism and give the correct stoichiometry.
Fractional orders can occur if one of the substances in the rate-limiting step is
produced in a previous step with a stoichiometric coefficient greater than unity, as in
Eq. (12.4-28).
The Temperature Dependence of Rates of Nonelementary
Reactions
The rate law of a reaction corresponding to an assumed mechanism contains the rate
constants of some or all of the steps of the mechanism. If the temperature dependence
of these rate constants is known, the temperature dependence of the overall rate can
be deduced. The rate law for the ozone decomposition of Eq. (12.4-4) is given by
Eq. (12.4-7):
rate
1
3
d[O 2 ]
dt
k app
[O 3 ] 2
[O 2 ]
k 2
k 1 [O 3 ] 2
k
1 [O 2 ]
(12.4-45)
If each of the elementary rate constants is governed by the Arrhenius formula,
Eq. (12.3-2), then the temperature dependence of the overall rate constant, k app , is
given by
k app
A 2 A 1
A 1
e −E a2 /RT e −E a1 /RT
e −E a1 /RT
A app exp
−
E a2 + E a1 − E a1
RT
A app exp
−
E a,app
RT
(12.4-46)
where the apparent activation energy E a,app is
E a,app E a2 + E a1 − E a1
Since the ratio k 1 /k
1 is equal to an equilibrium constant for the first step, we can rewrite
Eq. (12.4-46) as
k app A 2 exp
−
E a2 + ∆G ◦
1
RT
(12.4-47)
It is possible that the apparent activation energy of a reaction with a multistep mechanism is negative. This occurs in the recombination of iodine atoms to form I 2 , as
described in Problem 12.41.
Exercise 12.17
The temperature dependence of a rate law corresponding to a steady-state approximation is more
complicated than that of the previous example. Write the temperature dependence of the apparent
first-order rate constant for the decomposition of N 2 O 5 given in Eq. (12.4-34).
553
Exercise 12.16
a. Show that the proposed mechanism in the previous example leads to the correct rate law.
b. If H 2 O and NO 3 are the products of step 2, and if no further HNO 3 enters in later steps,
propose steps 3 and 4 to complete the mechanism and give the correct stoichiometry.
Fractional orders can occur if one of the substances in the rate-limiting step is
produced in a previous step with a stoichiometric coefficient greater than unity, as in
Eq. (12.4-28).
The Temperature Dependence of Rates of Nonelementary
Reactions
The rate law of a reaction corresponding to an assumed mechanism contains the rate
constants of some or all of the steps of the mechanism. If the temperature dependence
of these rate constants is known, the temperature dependence of the overall rate can
be deduced. The rate law for the ozone decomposition of Eq. (12.4-4) is given by
Eq. (12.4-7):
rate
1
3
d[O 2 ]
dt
k app
[O 3 ] 2
[O 2 ]
k 2
k 1 [O 3 ] 2
k
1 [O 2 ]
(12.4-45)
If each of the elementary rate constants is governed by the Arrhenius formula,
Eq. (12.3-2), then the temperature dependence of the overall rate constant, k app , is
given by
k app
A 2 A 1
A 1
e −E a2 /RT e −E a1 /RT
e −E a1 /RT
A app exp
−
E a2 + E a1 − E a1
RT
A app exp
−
E a,app
RT
(12.4-46)
where the apparent activation energy E a,app is
E a,app E a2 + E a1 − E a1
Since the ratio k 1 /k
1 is equal to an equilibrium constant for the first step, we can rewrite
Eq. (12.4-46) as
k app A 2 exp
−
E a2 + ∆G ◦
1
RT
(12.4-47)
It is possible that the apparent activation energy of a reaction with a multistep mechanism is negative. This occurs in the recombination of iodine atoms to form I 2 , as
described in Problem 12.41.
Exercise 12.17
The temperature dependence of a rate law corresponding to a steady-state approximation is more
complicated than that of the previous example. Write the temperature dependence of the apparent
first-order rate constant for the decomposition of N 2 O 5 given in Eq. (12.4-34).