The short answer
A rate equation, rate = k[A]ᵐ[B]ⁿ, links the rate of a reaction to reactant concentrations. The orders m and n are found by experiment, not from the balanced equation, and at A Level are 0, 1 or 2. The rate constant k changes with temperature.
A rate equation tells you how the rate of a reaction depends on the concentrations of its reactants. It has the form rate = k[A]ᵐ[B]ⁿ, where m and n are the orders of reaction with respect to A and B, and k is the rate constant. The orders are found by experiment, not read from the balanced equation, and at A Level they are 0, 1 or 2.
AQA (7405) and Cambridge International (9701) both teach rate equations in exactly this form, both restrict the orders to 0, 1 and 2, and both treat the topic as A Level content rather than AS. They part company on temperature, as the section on the Arrhenius equation shows, so keep your own specification to hand. If you sit Pearson Edexcel or OCR, check the matching section of your specification for its exact wording.
What each part of the rate equation means
- Rate is the change in concentration per unit time, usually in mol dm⁻³ s⁻¹.
- [A] and [B] are the concentrations of the reactants, in mol dm⁻³.
- m and n are the orders with respect to each reactant. The overall order is their sum, m + n.
- k is the rate constant. It stays the same at a fixed temperature and changes when the temperature changes.
A reactant of order zero does not appear in the rate equation at all, because any concentration raised to the power zero equals 1. That does not mean it takes no part in the reaction. It still reacts; in the mechanisms you meet at A Level, it does so after the step that controls the rate.
Why the orders come from experiments
AQA's specification puts it plainly: the rate equation is an experimentally determined relationship. The balanced equation describes the overall change, not how fast each step happens, so a reaction such as NO₂ + CO → NO + CO₂ can have a rate equation with no CO in it at all. The orders reflect the mechanism, and in particular its slowest step, which is why only data can reveal them.
What zero, first and second order look like
Many questions on this topic come down to recognising three patterns.
| Order with respect to A | Effect of doubling [A] on the rate | Concentration–time graph | Rate–concentration graph |
|---|---|---|---|
| 0 | No change | Straight line sloping down at a constant gradient | Horizontal line |
| 1 | Rate doubles | Curve with a constant half-life | Straight line through the origin |
| 2 | Rate increases four times | Curve whose successive half-lives get longer | Upward curve, with rate proportional to [A]² |
For a zero-order reaction, the gradient of the concentration–time line gives the rate, and because rate = k, the size of that gradient is the rate constant. AQA lists this calculation by name.
Method 1: initial rates
The initial-rates method runs the reaction several times, changing one concentration at a time, and measures the rate at the very start of each run. Take this illustrative data for a reaction A + B → products:
| Experiment | [A] / mol dm⁻³ | [B] / mol dm⁻³ | Initial rate / mol dm⁻³ s⁻¹ |
|---|---|---|---|
| 1 | 0.010 | 0.010 | 2.0 × 10⁻⁴ |
| 2 | 0.020 | 0.010 | 4.0 × 10⁻⁴ |
| 3 | 0.010 | 0.020 | 8.0 × 10⁻⁴ |
- Compare experiments 1 and 2. [A] doubles, [B] stays the same and the rate doubles, so the reaction is first order with respect to A.
- Compare experiments 1 and 3. [B] doubles, [A] stays the same and the rate increases four times, so the reaction is second order with respect to B.
- Write the rate equation: rate = k[A][B]². The overall order is 1 + 2 = 3.
- Rearrange for k and substitute experiment 1: k = rate ÷ ([A][B]²) = (2.0 × 10⁻⁴) ÷ (0.010 × 0.010²) = (2.0 × 10⁻⁴) ÷ (1.0 × 10⁻⁶) = 200.
- Find the units the same way: (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³)³ = dm⁶ mol⁻² s⁻¹. So k = 200 dm⁶ mol⁻² s⁻¹.
Check the result against a run you did not use. When two concentrations change at once, multiply their effects: doubling [A] and tripling [B] should raise the rate 2 × 3² = 18 times.
In the laboratory, AQA's required practical 7 measures the rate of reaction both by an initial-rate method and by continuous monitoring, and AQA suggests the iodine clock reaction as one way to find an order.
Method 2: concentration–time graphs and half-life
Continuous monitoring follows a single run and plots concentration against time. Drawing tangents to the curve at several points gives the rate at each concentration, which you can then plot as a rate–concentration graph and match against the patterns in the first table.
Half-life is the shortcut for first order. A first-order reaction has a half-life that does not depend on concentration: the reactant takes as long to fall from 0.80 to 0.40 mol dm⁻³ as from 0.40 to 0.20 mol dm⁻³. Cambridge's syllabus names the half-life method and the relationship k = 0.693 ÷ t½ explicitly. A first-order reaction with a half-life of 50 s has k = 0.693 ÷ 50 = 0.0139 s⁻¹.
Units of the rate constant
The units of k change with the overall order, which is why examiners ask for them. Derive them by rearranging the rate equation, as in step 5 above, and use this table as a check:
| Overall order | Units of k |
|---|---|
| 0 | mol dm⁻³ s⁻¹ |
| 1 | s⁻¹ |
| 2 | dm³ mol⁻¹ s⁻¹ |
| 3 | dm⁶ mol⁻² s⁻¹ |
The rate equation and the rate-determining step
In a multi-step reaction, the slowest step limits the overall rate. Species that take part in that step, or in a step before it, appear in the rate equation; species that only join in afterwards do not. Go back to NO₂ + CO → NO + CO₂, which has the experimentally found rate equation rate = k[NO₂]². A mechanism consistent with it is:
- NO₂ + NO₂ → NO₃ + NO (slow, rate-determining)
- NO₃ + CO → NO₂ + CO₂ (fast)
Two molecules of NO₂ react in the slow step, which accounts for the second order in NO₂. CO enters only in the fast second step, so it is zero order. NO₃ is made in step 1 and used up in step 2, which makes it an intermediate. Adding the two steps and cancelling what appears on both sides gives back the overall equation, the first check on any mechanism you propose.
Cambridge expects you to work in both directions: suggest a mechanism that fits a given rate equation, and deduce the rate equation from a given mechanism and rate-determining step. AQA asks you to use the orders to say what they reveal about the rate-determining step.
Temperature and the Arrhenius equation
Raising the temperature increases k, so the rate rises even when the concentrations are unchanged. This is where the two boards differ most.
- AQA 7405 requires calculations with the Arrhenius equation, k = A × e^(−Eₐ/RT), where A is the Arrhenius constant, Eₐ the activation energy and T the temperature in kelvin. You also need the rearranged form, ln k = −Eₐ/RT + ln A: a graph of ln k against 1/T is a straight line with gradient −Eₐ/R. AQA gives the equations and the gas constant R when a question needs them.
- Cambridge 9701, in its 2025–2027 syllabus, asks you to describe qualitatively how a change in temperature affects the rate constant and hence the rate. The syllabus does not name the Arrhenius equation.
For AQA, a typical graph question gives you the gradient and asks for Eₐ. If the gradient of ln k against 1/T is −6000 K, then Eₐ = 6000 × 8.31 = 49 900 J mol⁻¹, or about 49.9 kJ mol⁻¹. The minus sign belongs to the gradient, not to the activation energy, and the conversion from J to kJ comes at the end.
Mistakes that cost marks
- Taking the orders from the balanced equation instead of the data.
- Comparing two runs in which more than one concentration changed, and crediting the whole effect to one reactant.
- Leaving the units off k, or using the units for a different overall order.
- Losing a power of ten when squaring: 0.010² is 1.0 × 10⁻⁴, not 1.0 × 10⁻².
- Writing a zero-order reactant into the rate equation for completeness.
- Proposing a mechanism whose steps do not add up to the overall equation.
Where to go from here
How far each board takes this topic, from Cambridge's half-life calculations to AQA's Arrhenius graphs, is set out in your own specification, and our A Level Chemistry tutoring page shows how AQA, OCR, Pearson Edexcel and Cambridge International each examine the subject, paper by paper. If kinetics is one of several weak spots, one-to-one A Level tutoring matched to your specification starts from your marked work and the questions where the marks are being lost.