18.3. Reaction Mechanisms http://www.ck12.org
A bimolecular elementary reaction could be one of two types. Either a molecule of A could react with a molecule of
B, or two molecules of A could react with each other. In either case, the rate of reaction depends on how frequently
the collisions between reactant molecules occur.
A+B→products rate=k[A][B]
2A→products rate=k[A]^2The reaction order for each reactant in an elementary step is equal to its stoichiometric coefficient in the equation
for that step. In the first equation above, each coefficient is a 1, and so the reaction is first-order with respect to A
and first-order with respect to B. In the second equation, the coefficient of 2 means the reaction is second-order with
respect to A.
The determination of a reaction mechanism can only be made in the laboratory. When a reaction occurs in a sequence
of elementary steps, the overall reaction rate is governed by whichever one of those steps is the slowest. Therate-
determining stepis the slowest step in the reaction mechanism. To get an idea of how one step is rate determining,
imagine driving on a one-lane road where it is not possible to pass another vehicle. The rate of flow of traffic on such
a road would be dictated by whichever car is traveling at the lowest speed. The decomposition of hydrogen peroxide
is discussed below and illustrates how reaction mechanisms can be determined through experimental studies.
Decomposition of Hydrogen Peroxide
Recall that a catalyst is a substance that increases the rate of a chemical reaction without being consumed. Catalysts
lower the overall activation energy for a reaction by providing an alternative mechanism for the reaction to follow.
One such catalyst for the decomposition of hydrogen peroxide is the iodide ion (I−).
2H 2 O 2 (aq)I−
→2H 2 O(l)+O 2 (g)By experiment, the rate of this reaction is found to be first-order with respect to both H 2 O 2 and I−, which makes it
second-order overall.
rate = k[H 2 O 2 ][I−]Based on this experimental rate law, we know that the reaction cannot occur in a single elementary step. If it did,
the reaction would be second-order with respect to H 2 O 2 , since the coefficient of H 2 O 2 in the balanced equation is
- A reaction mechanism can be constructed which accounts for the rate law and for the detection of the IO−ion as
an intermediate. It consists of two bimolecular elementary steps.
Step 1: H 2 O 2 (aq)+I−(aq)→H 2 O(l)+IO−(aq)
Step 2: H 2 O 2 (aq)+IO−(aq)→H 2 O(l)+O 2 (g)+I−(aq)If step 1 is the rate-determining step, then the rate law for that step will be the rate law for the overall reaction.
rate = k[H 2 O 2 ][I−]The rate law for the slow step of the proposed mechanism agrees with the overall experimentally determined rate
law. IO−is present as an intermediate in the reaction. The iodide ion catalyst also appears in the mechanism. It is
a reactant in the first elementary step and a product in the second step. By definition, catalysts are not used up in a
reaction, so if they are consumed by one step in a mechanism, they must be regenerated by a subsequent step.