When the slow step comes second, its rate law usually contains an intermediate. Intermediates are hard to measure and do not appear in a measured rate law, so we need a way to replace them. The trick uses dynamic equilibrium.
Key idea
If a fast, reversible first step comes before the slow step, its forward and reverse rates are equal: $k_1[\text{reactants}] = k_{-1}[\text{intermediate}]$. Solve for the intermediate and substitute into the slow step's rate law. The final rate law contains only reactants (and sometimes products).
Why it works
The first step goes back and forth many times while the slow step barely runs. So it has time to settle into equilibrium, and only a trickle of intermediate leaks away through the slow step. Setting the two rates equal is the same as writing the equilibrium constant of the first step, $K_1 = k_1/k_{-1}$.
Worked examples
Worked example: Nitrogen monoxide and oxygen
Mechanism for $\mathrm{2NO + O_2 \to 2NO_2}$:
Step 1 (fast, reversible): $\mathrm{2NO \rightleftharpoons N_2O_2}$
Step 2 (slow): $\mathrm{N_2O_2 + O_2 \to 2NO_2}$
Derive the rate law.
Step 1 (fast, reversible): $\mathrm{2NO \rightleftharpoons N_2O_2}$
Step 2 (slow): $\mathrm{N_2O_2 + O_2 \to 2NO_2}$
Derive the rate law.
- Slow step: $\text{rate} = k_2[\mathrm{N_2O_2}][\mathrm{O_2}]$. $\mathrm{N_2O_2}$ is an intermediate, so it must go.
- Fast equilibrium: $k_1[\mathrm{NO}]^2 = k_{-1}[\mathrm{N_2O_2}]$, so $[\mathrm{N_2O_2}] = \dfrac{k_1}{k_{-1}}[\mathrm{NO}]^2$.
- Substitute: $\text{rate} = \dfrac{k_2k_1}{k_{-1}}[\mathrm{NO}]^2[\mathrm{O_2}] = k[\mathrm{NO}]^2[\mathrm{O_2}]$.
Answer: $\text{rate} = k[\mathrm{NO}]^2[\mathrm{O_2}]$, with $k = k_2k_1/k_{-1}$.
Worked example: Hydrogen and iodine
A mechanism for $\mathrm{H_2 + I_2 \to 2HI}$:
Step 1 (fast, reversible): $\mathrm{I_2 \rightleftharpoons 2I}$
Step 2 (slow): $\mathrm{H_2 + 2I \to 2HI}$
Derive the rate law.
Step 1 (fast, reversible): $\mathrm{I_2 \rightleftharpoons 2I}$
Step 2 (slow): $\mathrm{H_2 + 2I \to 2HI}$
Derive the rate law.
- Slow step: $\text{rate} = k_2[\mathrm{H_2}][\mathrm{I}]^2$.
- Fast equilibrium: $k_1[\mathrm{I_2}] = k_{-1}[\mathrm{I}]^2$, so $[\mathrm{I}]^2 = \dfrac{k_1}{k_{-1}}[\mathrm{I_2}]$.
- $\text{rate} = \dfrac{k_2k_1}{k_{-1}}[\mathrm{H_2}][\mathrm{I_2}]$. Notice: this rate law looks just like one elementary collision of $\mathrm{H_2}$ with $\mathrm{I_2}$, yet the mechanism is different. A rate law alone cannot prove a mechanism.
Answer: $\text{rate} = k[\mathrm{H_2}][\mathrm{I_2}]$.
Common mistake
Leaving the intermediate's concentration in the final rate law. A measured rate law contains only species you can add to the flask. Check: after substituting, scan the rate law for anything that is made in one step and used in another; if you find it, substitute again.
An overall order of 3 (as in the NO example) is one hint of a pre-equilibrium: three-body collisions are rare, but a fast equilibrium followed by a two-body collision gives the same rate law.