As an example, consider the gas-phase reaction + CO → NO + . If this reaction occurred in a single step, its
reaction rate (
r) would be proportional to the rate of
collisions between and CO molecules:
r =
k[][CO], where
k is the reaction
rate constant, and square brackets indicate a
molar concentration. Another typical example is the
Zel'dovich mechanism.
First step rate-determining In fact, however, the observed reaction rate is
second-order in and zero-order in CO, with rate equation
r =
k[]2. This suggests that the rate is determined by a step in which two molecules react, with the CO molecule entering at another, faster, step. A possible mechanism in two elementary steps that explains the rate equation is: • + → NO +
(slow step, rate-determining) • + CO → +
(fast step) In this mechanism the
reactive intermediate species is formed in the first step with rate
r1 and reacts with CO in the second step with rate
r2. However, can also react with NO if the first step occurs in the
reverse direction (NO + → 2 ) with rate
r−1, where the minus sign indicates the rate of a reverse reaction. The concentration of a reactive intermediate such as [] remains low and almost constant. It may therefore be estimated by the
steady-state approximation, which specifies that the rate at which it is formed equals the (total) rate at which it is consumed. In this example is formed in one step and reacts in two, so that : \frac{d\ce{[NO3]}}{dt} = r_1 - r_2 - r_{-1} \approx 0. The statement that the first step is the slow step actually means that the first step
in the reverse direction is slower than the second step in the forward direction, so that almost all is consumed by reaction with CO and not with NO. That is,
r−1 ≪
r2, so that
r1 −
r2 ≈ 0. But the overall rate of reaction is the rate of formation of final product (here ), so that
r =
r2 ≈
r1. That is, the overall rate is determined by the rate of the first step, and (almost) all molecules that react at the first step continue to the fast second step.
Pre-equilibrium: if the second step were rate-determining The other possible case would be that the second step is slow and rate-determining, meaning that it is slower than the first step in the reverse direction:
r2 ≪
r−1. In this hypothesis,
r1 − r−1 ≈ 0, so that the first step is (almost) at
equilibrium. The overall rate is determined by the second step:
r =
r2 ≪
r1, as very few molecules that react at the first step continue to the second step, which is much slower. Such a situation in which an intermediate (here ) forms an equilibrium with reactants
prior to the rate-determining step is described as a
pre-equilibrium For the reaction of and CO, this hypothesis can be rejected, since it implies a rate equation that disagrees with experiment. • + → NO +
(fast step) • + CO → +
(slow step, rate-determining) If the first step were at equilibrium, then its
equilibrium constant expression permits calculation of the concentration of the intermediate in terms of more stable (and more easily measured) reactant and product species: :K_1 = \frac{\ce{[NO][NO3]}}{\ce{[NO2]^2}}, :[\ce{NO3}] = K_1 \frac{\ce{[NO2]^2}}{\ce{[NO]}}. The overall reaction rate would then be :r = r_2 = k_2 \ce{[NO3][CO]} = k_2 K_1 \frac{\ce{[NO2]^2 [CO]}}{\ce{[NO]}}, which disagrees with the experimental rate law given above, and so disproves the hypothesis that the second step is rate-determining for this reaction. However, some other reactions are believed to involve rapid pre-equilibria prior to the rate-determining step,
as shown below. ==Nucleophilic substitution==