The Rate-Determining Step and Intermediates
The rate-determining step is the slowest step in a reaction mechanism, and it controls the overall rate; an intermediate is a species formed in one step and consumed in a later step, so it does not appear in the overall equation. In brief: in the decomposition of hydrogen peroxide catalyzed by iodide, the slow first step H2O2 + I− → H2O + IO− sets the rate law rate = k[H2O2][I−], and IO− is the intermediate.
When a reaction takes place in several steps, the overall rate cannot be faster than the slowest one. The idea is like traffic on a road with a narrow section: no matter how wide the rest of the road is, the flow is limited by the bottleneck. This article explains how to find the rate law from a mechanism, how to handle a fast equilibrium before a slow step, how to tell intermediates from catalysts and how the energy profile looks.
What is the rate-determining step?
In a multistep mechanism, each step has its own rate constant. If one step is much slower than the others, it limits the overall rate, and is called the rate-determining step (or rate-limiting step). The steps after it are fast, and do not influence the rate. When the slow step is the first one, the rate law of the overall reaction is the rate law of that elementary step, using the coefficients of the step as the orders.
The steps before the slow step can matter if they produce species that are used in it. Then the concentrations of such intermediates must be expressed in terms of the concentrations of the reactants, because the rate law of the overall reaction cannot contain intermediates.
What are intermediates?
An intermediate is formed in an early step and used up in a later one. It has a real, if short, existence, and in some cases it can be detected spectroscopically or even isolated at low temperature. Typically, intermediates are very reactive atoms, radicals or ions. They are not part of the overall equation, since they cancel when the steps are added.
Catalyst, intermediate and transition state compared
| Species | Formed | Used up | In the overall equation? | Can it be isolated? |
|---|---|---|---|---|
| Intermediate | In an early step | In a later step | No | Sometimes |
| Catalyst | Consumed first, regenerated later | Regenerated | No (on both sides it cancels) | Yes |
| Transition state | At the top of an energy barrier | Immediately | No | No |
The simplest way to tell a catalyst from an intermediate is by the order in which they appear: a catalyst is used in an early step and regenerated afterwards, while an intermediate is produced first and used afterwards.
Example 1: iodide-catalyzed decomposition
The overall reaction is 2H2O2 → 2H2O + O2, and the experimental rate law is rate = k[H2O2][I−]. The mechanism is:
- H2O2 + I− → H2O + IO− (slow)
- H2O2 + IO− → H2O + O2 + I− (fast)
Adding the steps gives 2H2O2 → 2H2O + O2, and I− and IO− cancel. The intermediate is IO− (formed in step 1 and used in step 2), and the catalyst is I− (used in step 1 and regenerated in step 2). The slow step is bimolecular, which gives rate = k1[H2O2][I−], in agreement with the experiment. This is the classic “elephant toothpaste” demonstration, in which the oxygen foams a soap solution.
Example 2: nitrogen dioxide and carbon monoxide
The overall reaction below about 225 °C is NO2 + CO → NO + CO2, with the rate law rate = k[NO2]2. The mechanism is:
- NO2 + NO2 → NO3 + NO (slow)
- NO3 + CO → NO2 + CO2 (fast)
The sum is NO2 + CO → NO + CO2 (after cancelling one NO2 and the intermediate NO3). The slow step involves two NO2 molecules, so rate = k[NO2]2. Carbon monoxide takes part only after the slow step, so the rate is zero order in CO.
Example 3: a fast step before the slow step
For 2NO + O2 → 2NO2 the experimental rate law is rate = k[NO]2[O2]. A mechanism that explains it is:
- NO + NO ⇌ N2O2 (fast, reversible)
- N2O2 + O2 → 2NO2 (slow)
The rate of the slow step is rate = k2[N2O2][O2], which includes an intermediate. Since step 1 is a fast equilibrium, K = [N2O2]/[NO]2, so [N2O2] = K[NO]2. Substituting gives rate = k2K[NO]2[O2], which is the experimental rate law. The observed rate constant is k = k2K.
What does the energy profile look like?
For a mechanism with two steps, the energy profile has two peaks, with a valley between them that corresponds to the intermediate. The step with the higher peak (the highest barrier measured from the preceding minimum) is the rate-determining step. The valley is an energy minimum, so the intermediate is a real species, unlike the transition states at the peaks. A catalyst provides a different set of steps with lower barriers.
Who developed these ideas?
The idea of a rate-determining step in a series of reactions was used in the early studies of chain reactions. Max Bodenstein developed in 1913 the approach for very reactive intermediates, which are assumed to be present at a low steady concentration, an approach later called the steady-state approximation. It is widely used when the rate-determining step is not obvious, in enzyme kinetics and in atmospheric chemistry.
Which mistakes are common?
- Having an intermediate in the rate law: express it in terms of reactants.
- Assuming that the first step is always the slow one: it can be any step.
- Mixing up a catalyst and an intermediate: check which one appears first.
- Forgetting that steps must add up to the overall equation: check the sum.
- Thinking that a faster step after the slow one speeds up the reaction: it does not.