Orders of Reaction and Rate Equations Revision Notes | A Level AQA Chemistry

1. Why Rate Equations Matter

Here is a question that trips up almost every student at first: can you look at a balanced equation and predict how the rate depends on concentration? The answer is no — and understanding why not is the foundation of this entire topic.

Consider the reaction $2NO_{(g)} + O_{2(g)} \rightarrow 2NO_{2(g)}$. You might expect the rate to depend on $[NO]^2$ and $[O_2]^1$ from the stoichiometry. Experimentally, that happens to be true for this reaction — but for many others, the experimental orders bear no resemblance to the stoichiometric coefficients at all. The rate equation is an experimentally determined relationship, not a theoretical one derived from the equation.

This matters enormously for AQA exams: you will never be asked to write a rate equation from a balanced equation alone. Every rate equation must come from data.

2. The Rate Equation and Orders of Reaction

At A-level, orders are restricted to 0, 1, or 2. Each order tells you something specific about how concentration controls rate. A zero-order reactant has no effect on rate — doubling its concentration changes nothing. A first-order reactant doubles the rate when its concentration doubles. A second-order reactant quadruples the rate when its concentration doubles.

The graph shapes are a reliable diagnostic tool. A straight line through the origin on a rate vs concentration plot confirms first order. A curve that becomes more steeply upward-curving than a straight-line proportionality indicates second order. On concentration-time plots, second-order decay becomes less steep with time and the half-life increases as the reaction proceeds.

3. Deducing Orders from Concentration–Rate Data

The most common exam task is a table of initial rate data. The method is always the same: compare two experiments where only one concentration changes, and work out the ratio of rates.

The key relationship is: if $[A]$ is multiplied by a factor $f$, and the rate is multiplied by a factor $r$, then the order $m$ satisfies $f^m = r$. For the values you will encounter at A-level, this reduces to three cases: $r = 1$ (zero order), $r = f$ (first order), $r = f^2$ (second order).

Worked example: Deducing Orders and Writing the Rate Equation from a Data Table

The table below shows initial rate data for the reaction A + B + C → products. Exp 1: [A] = 0.10 mol dm⁻³, [B] = 0.10 mol dm⁻³, [C] = 0.10 mol dm⁻³, rate = 2.0 × 10⁻³ mol dm⁻³ s⁻¹ Exp 2: [A] = 0.20 mol dm⁻³, [B] = 0.10 mol dm⁻³, [C] = 0.10 mol dm⁻³, rate = 4.0 × 10⁻³ mol dm⁻³ s⁻¹ Exp 3: [A] = 0.10 mol dm⁻³, [B] = 0.20 mol dm⁻³, [C] = 0.10 mol dm⁻³, rate = 2.0 × 10⁻³ mol dm⁻³ s⁻¹ Exp 4: [A] = 0.10 mol dm⁻³, [B] = 0.10 mol dm⁻³, [C] = 0.20 mol dm⁻³, rate = 8.0 × 10⁻³ mol dm⁻³ s⁻¹ Deduce the order with respect to each reactant and write the rate equation.

1. Compare Exp 1 and Exp 2: [A] doubles (0.10 → 0.20), [B] and [C] unchanged. Rate ratio = (4.0 × 10⁻³) ÷ (2.0 × 10⁻³) = 2 Concentration factor = 2; rate factor = 2 2^m = 2, so m = 1 — Only [A] changes between these two experiments. The rate doubles when [A] doubles, confirming first order with respect to A. [M1]
2. Compare Exp 1 and Exp 3: [B] doubles (0.10 → 0.20), [A] and [C] unchanged. Rate ratio = (2.0 × 10⁻³) ÷ (2.0 × 10⁻³) = 1 Concentration factor = 2; rate factor = 1 2^n = 1, so n = 0 — Doubling [B] has no effect on rate. B is zero order — it will not appear in the rate equation. [M1]
3. Compare Exp 1 and Exp 4: [C] doubles (0.10 → 0.20), [A] and [B] unchanged. Rate ratio = (8.0 × 10⁻³) ÷ (2.0 × 10⁻³) = 4 Concentration factor = 2; rate factor = 4 2^p = 4, so p = 2 — Doubling [C] quadruples the rate. C is second order. [M1]
4. rate = k[A]^1[C]^2 Or equivalently: rate = k[A][C]² — B is zero order so it is omitted. The overall order is 1 + 0 + 2 = 3. [A1]

Answer: rate = k[A][C]² (first order in A, zero order in B, second order in C, third order overall)

4. The Initial Rate Method

The initial rate method is the experimental strategy behind the data tables you have been analysing. The rate is measured at the very start of each experiment — before significant concentration changes occur — so the initial concentrations are known precisely.

Each experiment uses a different set of initial concentrations, with typically only one concentration changed at a time. By comparing the initial rates across experiments, you can isolate the effect of each reactant and deduce its order. This is why the method is so powerful: it gives clean, interpretable data.

One important limitation of the initial rate method is that it only gives you the rate at the very start of the reaction. If the mechanism changes at higher concentrations, or if a product inhibits the reaction, the initial rate data will not capture that behaviour. This is why the rate equation is described as experimentally determined — it reflects the kinetics under the specific conditions tested.

5. Calculating the Rate Constant k

Once you have the rate equation and the orders, you can calculate the rate constant k using any one set of experimental data. Rearrange the rate equation to make k the subject, substitute in the values, and carry the units through carefully.

The units of k depend on the overall order of reaction. You must derive them each time rather than memorising a list — the method is always the same: substitute units into the rearranged rate equation and simplify.

Worked example: Calculating the Rate Constant k and Its Units

For the reaction with rate equation rate = k[A][C]², use Experiment 1 from the earlier data table to calculate k and determine its units. Exp 1: [A] = 0.10 mol dm⁻³, [C] = 0.10 mol dm⁻³, rate = 2.0 × 10⁻³ mol dm⁻³ s⁻¹

1. $k = \frac{\text{rate}}{[A][C]^2}$ — Divide both sides of rate = k[A][C]² by [A][C]². [M1]
2. $k = \frac{2.0 \times 10^{-3}}{(0.10)(0.10)^2}$ — Insert the values from Experiment 1. [M1]
3. $(0.10)^2 = 0.10 \times 0.10 = 0.010$ $(0.10)(0.010) = 0.0010 = 1.0 \times 10^{-3}$ — Expand the square explicitly to avoid arithmetic errors. 0.10 × 0.10 = 0.010, then multiply by the remaining 0.10.
4. $k = \frac{2.0 \times 10^{-3}}{1.0 \times 10^{-3}} = 2.0$ — Dividing the numerator by the denominator gives 2.0. [A1]
5. $k = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^2} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{\text{mol}^3\text{ dm}^{-9}}$ $= \text{mol}^{-2}\text{ dm}^{6}\text{ s}^{-1}$ — Substitute units into the expression for k and cancel. The overall order is 3 (1 + 2), so the units of k are mol⁻² dm⁶ s⁻¹. [B1]

Answer: k = 2.0 mol⁻² dm⁶ s⁻¹

6. Reading Orders from Concentration–Time and Rate–Concentration Graphs

Experimental data often arrives as a graph rather than a table. Two graph types appear regularly in AQA questions: concentration–time graphs and rate–concentration graphs. Each gives you different information, and you need to be fluent with both.

On a concentration–time graph, the gradient at any point gives the rate at that moment. To find the initial rate, draw a tangent at t = 0 and calculate its gradient. The shape of the curve also carries information: a straight line (constant gradient) indicates zero order; an exponential decay indicates first order; a curve that decreases more steeply than exponential indicates second order.

On a rate–concentration graph, the pattern is direct: a horizontal line means zero order, a straight line through the origin means first order, and an upward-curving parabola means second order. If you plot rate against [A]² and get a straight line through the origin, that confirms second order in A — this is a useful check when the curve shape is ambiguous.

7. Exam Technique and Common Mistakes

Rate equation questions are highly structured, which means the marks are predictable — and so are the mistakes. The guidance below covers the errors that cost students marks most frequently, along with the precise language that earns them.

Bringing this all together: the rate equation encodes the kinetic behaviour of a reaction in a compact, quantitative form. Every element of it — the orders, the rate constant, and its units — must be grounded in experimental evidence. That is the central principle of this topic, and it is the idea the examiner will test from multiple angles across a paper.