A mechanism is a proposal from which you can work out a rate law that agrees with the observed rate laws. The fact that a mechanism explains the experimental results is not a proof that the mechanism is correct.
As a mechanism, no proof is required. Proposing a mechanism is an interesting academic exercise for a mature chemist. Students in general chemistry will not be required to propose a mechanism, but they are required to derive the rate law from a proposed mechanism.
Molecularity | Elementary step | Rate law |
---|---|---|
1 | A -> products | rate = k [A] |
2 | A + A -> products A + B -> products | rate = k [A]^{2} rate = k [A] [B] |
3 | A + A + A -> products A + 2 B -> products A + B + C -> products | rate = k [A]^{3} rate = k [A] [B]^{2} rate = k [A] [B] [C] |
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
Solution
Since step i is the rate-determining step, the rate law is
1 d[NO_{2}] - --- ------ = k [NO_{2}] [F_{2}] 2 dtSince both NO_{2} and F_{2} 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
i | Br_{2} | k_{1} ® ¬ k_{-1} | 2 Br | (both directions are fast) |
---|---|---|---|---|
ii | Br + H_{2} | k_{2} ® | HBr + H | (slow) |
iii | H + Br_{2} | k_{3} ® | HBr + Br | (fast) |
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] --- -------- = k_{2} [H_{2}] [Br] 2 dtThe 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,
k_{1} [Br_{2}] = k_{-1} [Br]^{2} or [Br] = ((k_{1}/k_{-1}) [Br_{2}])^{(1/2)}Substituting this in the rate expression results in,
The overall reaction order is 3/2, 1 with respect to [H_{2}] and 1/2 with respect to [Br_{2}].
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 k_{1}/k_{-1} is often written as K, and it is called the equilibrium constant for the reversible elementary steps.
Example 3
i. Cl_{2} + M = 2 Cl + M (fast equilibrium, K_{1}) ii. Cl + CO + M = ClCO + M (fast equilibrium, K_{2}) iii. ClCO + Cl_{2} = Cl_{2}CO + Cl (slow, k_{3})The overall reaction is
Solution
From the rate-determining (slow) step,
d[Cl_{2}CO] --------- = k_{3} [ClCO] [Cl_{2}] - - - (1) dtYou should express [ClCO] in terms of concentrations of Cl_{2} and CO. This is done by considering step ii.
You should express [Cl] in terms of [Cl_{2}]. For this, you may use step i.
where k = k_{1} K_{1}^{(1/2)} K_{2}, 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
Example 4
i. HNO_{2} + H^{+} = H_{2}O + NO^{+} (equilibrium, K_{1}) ii. NH_{4}^{+} = NH_{3} + H^{+} (equilibrium, K_{2}) iii. NO^{+} + NH_{3} -> NH_{3}NO^{+} (slow, k_{3}) iv. NH_{3}NO^{+} -> H_{2}O + H^{+} + N_{2} (fast, k_{4})Derive the rate law.
Solution
From the rate-determining step, you have
d[NH_{3}NO^{+}] ------------- = k_{3} [NO^{+}] [NH_{3}] - - - - (4) dtNeither NO^{+} nor NH_{3} is a reactant. You must express their concentrations in terms of [NH_{4}^{+}] and [HNO_{2}] from elementary processes i and ii.
From i, [NO^{+}] = K_{1} [HNO_{2}] [H^{+}] / [H_{2}O] - - (5)
From ii, [NH_{3}] = K_{2} [NH_{4}^{+}] / [H^{+}] - - - - (6)
Substituting (6) and (5) in (4) gives,
[HNO_{2}] [NH_{4}^{+}] Rate = k_{3} K_{1} K_{2} ----------------- [H_{2}O] = k [HNO_{2}] [NH_{4}^{+}]where k = k_{3} K_{1} K_{2} / [H_{2}O] is the overall rate constant.
Skill -
Figure out the order from a given mechanism.
What is the overall order?
Discussion -
If you got rate = k [A_{2}]^{(1/2)} [B_{2}],
congratulations.
Skill -
Figure out the order from a given mechanism.
You should get: rate = k [NO]^{2} [O_{2}]
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) (1/2)(d[F]/dt) = k[A] [C]^{4} (b) rate = k[A] [C]^{2} (c) -d[A]/dt = k[A]^{(1/2)} [C] (d) d[F]/dt = k[A] [C] (e) (d[F]/dt) = k[A]^{(1/2)}[C]^{2}
The method -
The rate determining step is ii. Express [B] in terms of [A] from i.