In the rate law rate = k[A]^m[B]^n, the powers m and n are found:

by experiment. Orders are experimental — they reflect the mechanism, not the stoichiometric coefficients.

Doubling [A] doubles the rate. The reaction is ___ order in A.

first. Rate proportional to concentration means first order.

Doubling [B] makes the rate four times larger. The order in B is:

second. ×4 = 2², so the reaction is second order in B.

For rate = k[A][B]², the overall order is:

3. The overall order is the sum of the individual orders: first order in A (power 1) plus second order in B (power 2) gives 1 + 2 = 3, so the reaction is third order overall.

The rate constant k increases most when you raise the:

temperature. k depends strongly on temperature (Arrhenius); concentration changes the rate but not k.

The rate-determining step of a reaction is the:

slowest step. The slowest step is the bottleneck that controls the overall rate.

A reactant's concentration halves every 30 s regardless of its starting value. The reaction is:

first order. A constant half-life is the signature of a first-order reaction.

The order of a reaction with respect to a reactant is found by:

experiment. Orders reflect the mechanism, so they must be determined experimentally, not read from the equation.

Reaction Kinetics

Rate laws, order of reaction, and what the rate constant depends on

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Practise the full question bank:

The rate law

Collision theory told you which way rate changes; kinetics puts it in an equation. The rate law relates rate to concentration: for a reaction of A and B, rate = k[A]^m[B]^n, where k is the rate constant and the powers m and n are the orders with respect to each reactant.

Crucially, the orders are found by experiment, not from the balanced equation — they reflect the actual mechanism, not the stoichiometry.

Order of reaction

The order with respect to a reactant tells you how its concentration affects rate. Zero order: changing it has no effect. First order: double the concentration, double the rate. Second order: double it, and the rate quadruples (×2²). The overall order is the sum of the individual orders (m + n).

Finding the order from data

The standard method is initial rates: change one reactant's concentration while holding the others fixed, and see how the initial rate responds. If doubling [A] doubles the rate, the reaction is first order in A; if it quadruples the rate, second order; if the rate is unchanged, zero order. Repeat for each reactant to build the full rate law.

The rate constant and mechanisms

The rate constant k is fixed for a reaction at a given temperature — but it rises sharply with temperature, which is why heating speeds reactions so effectively (the Arrhenius relationship links k to temperature and activation energy). Its units depend on the overall order.

Many reactions happen in several steps — a mechanism. The slowest step, the rate-determining step, controls the overall rate, which is why the experimental rate law often involves only some of the reactants: the ones taking part up to and including that slow step.

Orders come from experiment, not the equation

You cannot read the order off the balanced equation. For 2A + B → products the reaction might be first order in A, or second, or zero — only experiment (initial rates) tells you, because the order reflects the mechanism.

A worked example

Worked example

Doubling [A] doubles the rate; doubling [B] quadruples it. Write the rate law and give the overall order.

Doubling [A] doubles the rate → first order in A (m = 1).Rate ∝ [A]¹.
Doubling [B] quadruples the rate (×4 = 2²) → second order in B (n = 2).Rate ∝ [B]².
Rate law: rate = k[A][B]². Overall order = 1 + 2 = 3.Overall order is the sum of the individual orders.

rate = k[A][B]², overall third order — determined from how the rate responds to each concentration, not from any equation.

Half-life and first-order reactions

The half-life of a reaction is the time for a reactant's concentration to fall to half its value. A telling feature of a first-order reaction is that its half-life is constant — every successive halving takes the same time, regardless of the starting concentration. Spotting a constant half-life on a concentration–time graph is a quick way to identify first-order behaviour.

For a zero-order reactant the rate does not depend on its concentration at all, so a concentration–time graph is a straight line.

Catalysis and the rate-determining step

Most reactions happen in several steps, and the slowest step — the rate-determining step — controls the overall rate. Only species involved up to and including that step appear in the rate law, which is why the rate equation is found by experiment, not from the overall balanced equation.

A catalyst offers a different mechanism with a lower activation energy. A homogeneous catalyst is in the same phase as the reactants (e.g. an acid catalyst in solution); a heterogeneous catalyst is in a different phase (e.g. solid iron in the gaseous Haber process), where reactants adsorb onto its surface, react, then leave.

Why the rate law comes from experiment

The balanced equation tells you nothing about the mechanism. The orders in the rate law reflect the rate-determining step, so they must be measured — for example by seeing how the initial rate changes when you double one concentration at a time.

Practice questions

6 questions · 21 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

4 · 9 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]

Define the term rate of reaction.

Model answer

The change in concentration of a reactant or product per unit time.

Examiner tip. Units are usually mol dm⁻³ s⁻¹.

SQ2[2 marks]

What is meant by the order of a reaction with respect to a reactant?

Model answer

The power to which the concentration of that reactant is raised in the rate equation.

Examiner tip. Overall order = the sum of the individual orders in the rate equation.

SQ3[3 marks]

A reaction has rate equation rate = k[A][B]². State the order with respect to each reactant, the overall order, and the effect on rate of doubling [B].

Model answer

First order in A, second order in B, overall third order. Doubling [B] multiplies the rate by 2² = 4.

Examiner tip. The exponent tells you the factor: doubling a second-order reactant quadruples the rate.

SQ4[2 marks]

The rate equation for a reaction is rate = k[A]²[B]. State the overall order and the effect on the rate of doubling [A].

Model answer

Overall order = 2 + 1 = 3 (third order). Doubling [A] multiplies the rate by 2² = 4.

Examiner tip. Add the individual orders for the overall order; the exponent gives the multiplying factor.

Long questions

2 · 12 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[6 marks]

The rate of a reaction between A and B was studied by varying concentrations.

Explain how the rate constant k is affected by increasing temperature, and why. [3]
Explain what is meant by the rate-determining step. [2]
State the units of k for a first-order reaction. [1]

Mark scheme

k increases with temperature[1]
Particles have more energy and collide more often[1]
and a greater fraction exceed the activation energy, so more collisions are successful[1]
The rate-determining step is the slowest step in a multi-step reaction[1]
It controls the overall rate[1]
s⁻¹[1]

k rises with temperature (more successful collisions); rate-determining step = the slowest step, which limits overall rate; first-order k has units s⁻¹.

Examiner tip. Only temperature (and a catalyst) change k — not concentration.

LQ2[6 marks]

A reaction between P and Q was studied by measuring the initial rate at different concentrations.

Explain what is meant by the rate-determining step. [2]
Explain how a catalyst increases the rate constant k. [2]
Explain why increasing temperature increases k. [2]

Mark scheme

The slowest step in a multi-step reaction[1]
which controls (limits) the overall rate[1]
A catalyst provides an alternative pathway with lower activation energy[1]
so a greater fraction of collisions are successful, raising k[1]
Higher temperature gives molecules more energy[1]
so more exceed the activation energy — more successful collisions and a larger k[1]

Rate-determining step = slowest step controlling the rate; catalyst and higher temperature both raise k by increasing the fraction of successful (Ea-exceeding) collisions.

Examiner tip. k depends on temperature and catalyst, never on concentration.