Rate Law

Expresses reaction rate in terms of reactant concentrations and their experimentally determined orders: rate = k[A]^m[B]^n.

\text{rate} = k[A]^m[B]^n

rate = k[A]^m[B]^n

What it means

The rate law is the experimentally discovered link between concentration and speed: it says the rate equals a temperature-dependent constant k times each reactant's concentration raised to its order. The orders m and n reveal the mechanism's sensitivity — a second-order dependence on A means doubling A quadruples the rate. Critically, these exponents come from measured data, never from the balanced equation; a stoichiometric coefficient of 2 does not make a reactant second order. When to use it: interpreting initial-rate data to find orders and k, predicting how concentration changes alter speed, and feeding k into integrated rate laws for time-dependent predictions.

Variables

Symbol	Name	Unit	Meaning
$rate$	Reaction rate	M/s	How fast the reaction proceeds, usually measured as the change in concentration per unit time.
$k$	Rate constant	varies (e.g. M⁻¹s⁻¹)	The rate constant at a given temperature; its units depend on the overall order.
$[A]$	Concentration of A	M (mol/L)	Molar concentration of reactant A.
$[B]$	Concentration of B	M (mol/L)	Molar concentration of reactant B.
$m$	Order in A	unitless	The experimentally determined exponent for A.
$n$	Order in B	unitless	The experimentally determined exponent for B.

Assumptions

The orders m and n are determined experimentally, not from stoichiometric coefficients.
The temperature is constant — k changes with temperature.
The equation describes a single elementary step or an empirically fitted overall rate.

Worked example

Problem

Determine the rate law and the rate constant for A + B → products from the data given.

Given:

Exp 1: [A]=0.100 M, [B]=0.100 M, rate=2.00×10⁻³ M/s
Exp 2: [A]=0.200 M, [B]=0.100 M, rate=8.00×10⁻³ M/s
Exp 3: [A]=0.100 M, [B]=0.200 M, rate=4.00×10⁻³ M/s

Compare Exp 1 and Exp 2 ([B] constant): [A] doubles (0.100 → 0.200 M) and the rate quadruples (2.00×10⁻³ → 8.00×10⁻³). A factor of 4 = 2², so m = 2.
Compare Exp 1 and Exp 3 ([A] constant): [B] doubles (0.100 → 0.200 M) and the rate doubles (2.00×10⁻³ → 4.00×10⁻³). A factor of 2 = 2¹, so n = 1.
The rate law is rate = k[A]²[B]; overall order = 2 + 1 = 3.
Solve for k from Exp 1: k = rate / ([A]²[B]) = (2.00×10⁻³) / ((0.100)²(0.100)).
Calculate: (0.100)²(0.100) = 0.00100 M³; 2.00×10⁻³ / 1.00×10⁻³ = 2.00 M⁻²s⁻¹.
Verify with Exp 2: 2.00 × (0.200)²(0.100) = 2.00 × 0.00400 = 8.00×10⁻³ M/s, which matches.

rate = (2.00 M⁻²s⁻¹)[A]²[B]; overall order 3

Doubling A quadruples the rate (second order in A); doubling B doubles the rate (first order in B) — the rate law is empirically pinned down.