Deriving Rate laws from Reaction Mechanisms Involving Equilibrium Elementary Steps
You all have experienced that an accident or road construction on a free way slows all the traffic on the
road, because limitations imposed by the accident and constructions apply to all cars on that road. The
narrow passing stretch limits the speed of the traffic.
If several steps are involved in an overall chemical reaction, the slowest step limits the rate of the
reaction. Thus, a slow step is called a rate determining step. The following examples illustrate the
method of deriving rate laws from the proposed mechanism. Please learn the technique.
Example 1
If the reaction
2 NO2 + F2  2 NO2F
follows the mechanism,
i. NO2 + F2  NO2F + F (slow)
ii. NO2 + F  NO2F (fast)
What is the rate law?
Solution
Since step i is the rate-determining step, the rate law is
1 d[NO2]
- --- ------ = k [NO2] [F2]
2 dt
Since both NO2 and F2 are reactants, this is the rate law for the reaction.
Discussion:
Addition of i. and ii. gives the overall reaction, but the step ii does not affect the rate law. Note
that the rate law is not derived from the overall equation either.
Example 2
For the reaction
H2 + Br2 = 2 HBr,
the following mechanism has been proposed
Br2
k1
i
2 Br
(both directions are fast)
k-1
k2
ii Br + H2  HBr + H (slow)
k3
iii H + Br2  HBr + Br (fast)
Derive the rate law that is consistent with this mechanism.
Solution
For a problem of this type, you should give the rate law according to the rate-determining (slow)
elementary process. In this case, step ii is the rate-determining step, and the rate law is,
1 d[HBr]
--- -------- = k2 [H2] [Br]
2 dt
The factor 1/2 results from the 2 HBr formed every time, one in step ii and one in step iii. Since
[Br] is not one of the reactants, its relationship with the concentration of the reactants must be
sought. The rapid reaction in both direction of step i implies the following relationship,
k1 [Br2] = k-1 [Br]2
or
[Br] = ((k1/k-1) [Br2])(1/2)
Substituting this in the rate expression results in,
rate = k2 (k1/k-1)(1/2) [H2] [Br2](1/2)
The overall reaction order is 3/2, 1 with respect to [H2] and 1/2 with respect to [Br2].
Discussion:
The important point in this example is that the rapid equilibrium in step i allows you to express
the concentration of an intermediate, ([Br]) in terms of concentrations of reactants ([Br 2]) so that
the rate law can be expressed by concentrations of the reactants.
The ratio k1/k-1 is often written as K, and it is called the equilibrium constant for the reversible
elementary steps.
Example 3
Derive the rate law that is consistent with the proposed mechanism in the formation of
phosgene from Cl2 and CO. (K1 = k1/k-1 and K2 = k2/k-2 may be considered as equilibrium
constants of the elementary processes, and M is any inert molecule.)
i. Cl2 + M
2 Cl + M (fast equilibrium, K1)
ii. Cl + CO + M
ClCO + M (fast equilibrium, K2)
iii. ClCO + Cl2  Cl2CO + Cl (slow, k3)
The overall reaction is
Cl2 + CO  Cl2CO
Solution
From the rate-determining (slow) step,
d[Cl2CO]
--------- = k3 [ClCO] [Cl2] - - - (1)
dt
You should express [ClCO] in terms of concentrations of Cl2 and CO. This is done by
considering step ii.
[ClCO] = K2 [Cl] [CO] - - - (2)
You should express [Cl] in terms of [Cl2]. For this, you may use step i.
[Cl] = K1(1/2) [Cl2](1/2) - - - (3)
Substituting (3) in (2) and then in (1) gives the Rate,
Rate = k3 K1(1/2) K2 [CO] [Cl2](3/2)
= k [CO] [Cl2](3/2)
where k = k1 K1(1/2) K2, the observed rate constant. The overall order of the reaction is 5/2, strange
but that is the observed rate law.
Discussion:
This example shows how the concentrations of intermediates are related to those of the reactants
in a two-step equilibrium.
If the third step is
iii. ClCO + Cl  Cl2CO (slow, k3)
the rate law will be different from the result derived above.
As an exercise, derive the rate law using this alternate step.
Example 4
In an acid solution, the mechanism for the reaction
NH4+ + HNO2 -> N2 + 2 H2O + H+
is:
i. HNO2 + H+ H2O + NO+ (equilibrium, K1)
ii. NH4+
NH3 + H+ (equilibrium, K2)
+
iii. NO + NH3  NH3NO+
(slow, k3)
iv. NH3NO+  H2O + H+ + N2 (fast, k4)
Derive the rate law.
Solution
From the rate-determining step, you have
d[NH3NO+]
------------- = k3 [NO+] [NH3] - - - - (4)
dt
Neither NO+ nor NH3 is a reactant. You must express their concentrations in terms of [NH4+] and
[HNO2] from elementary processes i and ii.
From i, [NO+] = K1 [HNO2] [H+] / [H2O] - - (5)
From ii, [NH3] = K2 [NH4+] / [H+] - - - - (6)
Substituting (6) and (5) in (4) gives,
[HNO2] [NH4+]
Rate = k3 K1 K2 ----------------[H2O]
= k [HNO2] [NH4+]
where k = k3 K1 K2 / [H2O] is the overall rate constant.
Confidence Building Questions
1. If the reaction
2 A + B2 = 2 AB
follows the mechanism,
i. A + B2  AB + B (slow)
ii. A + B  AB (fast)
What is the order with respect to [B2]? ____________________
2. Figure out the order from a given mechanism.
What is the overall order?
Kinetic studies of the reaction
A2 + B2  2 AB
suggested the mechanism as follows,
i. A2 2 A (fast equilibrium)
ii. A + B2  AB + B (slow)
iii. B + A2  AB + A (fast)
What is the order of the reaction with respect to [A2]? ___________________
3. If the mechanism for the reaction
2 NO + O2 -> 2 NO2
is
i. NO + NO = N2O2 (fast equilibrium)
ii. N2O2 + O2 -> 2 NO2 (slow)
What is the power of [NO] in the differential rate law? ____________________
4. If the reaction mechanism consists of these elementary processes,
i.
A
2 B (fast, equilibrium)
ii. B + 2 C  E (slow)
iii.
E  F (fast)
Choose the correct differential rate law for the reaction
A+4C2F
(a)
(b)
(c)
(d)
(e)
(1/2)(d[F]/dt) = k[A] [C]4
rate = k[A] [C]2
-d[A]/dt = k[A](1/2) [C]
d[F]/dt = k[A] [C]
(d[F]/dt) = k[A](1/2)[C]2