A rate law for an overall reaction must be measured. There is one exception: a reaction that happens in a single collision, an elementary step. For it, the rate law can be written straight from the equation.
Key idea
The molecularity of an elementary step is the number of particles that collide in it: unimolecular (one), bimolecular (two) or termolecular (three, rare). For an elementary step only, the rate law uses the coefficients as orders: $\mathrm{A + B \to}$ products gives $\text{rate} = k[\mathrm{A}][\mathrm{B}]$, and $\mathrm{2A \to}$ products gives $\text{rate} = k[\mathrm{A}]^2$.
Why it works
In a bimolecular step, every reaction event is a collision between one A and one B. The number of such collisions per second in a liter is proportional to $[\mathrm{A}]$ (more A to be hit) and to $[\mathrm{B}]$ (more B doing the hitting), so the rate is proportional to $[\mathrm{A}][\mathrm{B}]$. For a unimolecular step, each molecule falls apart or rearranges on its own with a fixed chance per second, so the rate is $k[\mathrm{A}]$. Three particles rarely meet at the same instant, which is why termolecular steps are uncommon.
Worked examples
Worked example: A bimolecular step
The elementary step $\mathrm{NO + O_3 \to NO_2 + O_2}$ occurs in polluted air. Give its molecularity and rate law.
- Two particles collide: bimolecular.
- Because it is elementary, the coefficients are the orders: $\text{rate} = k[\mathrm{NO}][\mathrm{O_3}]$.
Answer: Bimolecular; $\text{rate} = k[\mathrm{NO}][\mathrm{O_3}]$.
Worked example: Two of the same molecule
Write the rate law of the elementary step $\mathrm{2NO_2 \to NO_3 + NO}$.
- Two $\mathrm{NO_2}$ molecules collide: bimolecular.
- $\text{rate} = k[\mathrm{NO_2}]^2$: one factor for each colliding molecule.
Answer: $\text{rate} = k[\mathrm{NO_2}]^2$.
| Elementary step | Molecularity | Rate law |
|---|---|---|
| $\mathrm{A \to}$ products | unimolecular | $k[\mathrm{A}]$ |
| $\mathrm{A + B \to}$ products | bimolecular | $k[\mathrm{A}][\mathrm{B}]$ |
| $\mathrm{2A \to}$ products | bimolecular | $k[\mathrm{A}]^2$ |
| $\mathrm{2A + B \to}$ products | termolecular | $k[\mathrm{A}]^2[\mathrm{B}]$ |
Common mistake
Writing the rate law of an overall reaction from its coefficients as if it were one elementary step. $\mathrm{2N_2O_5 \to 4NO_2 + O_2}$ is not a collision of two molecules, and its measured rate law is first order. Check: the problem must say the step is elementary (or be part of a mechanism) before the coefficients become orders.
Molecularity is a whole number by its nature: you cannot have half a molecule colliding. Orders of overall reactions can be fractions, which is one sign that they come from several steps.