If the slow step is second, the predicted
rate law will usually
involve an intermediate.
Possible mechanismFor reaction A + B

C
Step 1: fast equilibrium
A + B

intermediate
Step 2: slow
intermediate

C
Predicted overall rate lawrate =
k[intermediate]
It is not possible to test this rate law by varying [intermediate] because the intermediate is consumed as soon as it is formed, and is thus never present in significant concentration.
To test such a pathway,
[intermediate] must be
expressed in terms of
concentrations of stable species (reactants and/or products).This can be done
using the
equilibrium constant (
K) for the step forming the intermediate.
[intermediate] =
K[A][B]
Substitution of the expression for the concentration of the intermediate into the predicted rate law above gives a rate law that can be experimentally verified.
rate =
k ×
K[A][B]
The predicted rate constant for the overall reaction of A and B to give C would be the product of the rate constant (
k - lower case) for the second step and the equilibrium constant (
K - upper case) for the first step.