Introduction to Rate Equations Revision Notes | A Level AQA Chemistry

1. Why Rate Equations Matter

Here is a question that should bother you: two reactions both involve the same reactants, yet one is ten times faster than the other at the same temperature. How do you describe that difference mathematically? The answer lies in the rate equation — one of the most powerful tools in physical chemistry.

Rate equations connect something you can measure in a lab (how fast a reaction goes) to something you can control (the concentration of reactants). Once you have a rate equation, you can predict how changing concentration will affect speed, identify the slowest step in a multi-step mechanism, and calculate a numerical constant that characterises the reaction at a given temperature.

This topic is A-level only — it does not appear at GCSE — so the ideas here are genuinely new. Build them carefully from the ground up and the rest of the kinetics topic will fall into place.

2. The Rate Equation: Form and Meaning

Consider a reaction between two reactants A and B. Experiments show that the rate depends on the concentrations of A and B, but not necessarily in the same way for each. The AQA rate equation captures this:

The exponents $m$ and $n$ are called the orders of reaction with respect to A and B. They are not taken from the balanced equation — they are found only by experiment. This is a point the AQA mark scheme tests repeatedly, so fix it in your mind now.

The overall order of the reaction is the sum of all individual orders: $m + n$ for a two-reactant system. A reaction with $m = 2$ and $n = 1$ is third order overall. AQA questions often ask for both the individual orders and the overall order, so always state both.

3. What Each Order Actually Means

The numbers 0, 1, and 2 are not abstract — each one tells you something concrete about how that reactant's concentration controls the rate. Picture a series of experiments where you double the concentration of one reactant while keeping everything else constant.

Notice the pattern: doubling the concentration multiplies the rate by $2^m$. For zero order, $2^0 = 1$ (no change). For first order, $2^1 = 2$ (doubles). For second order, $2^2 = 4$ (quadruples). This relationship is the core tool for reading orders from experimental data tables.

4. Determining Orders from Experimental Data

In practice, you are given a table of initial rate experiments and asked to deduce the orders. The method is always the same: find two experiments where only one concentration changes, then compare the rate ratio to the concentration ratio.

Worked example: Determining Orders from an Initial Rates Table

The table below shows results from three experiments for the reaction A + B → products. Experiment 1: [A] = 0.10 mol dm⁻³, [B] = 0.10 mol dm⁻³, rate = 2.0 × 10⁻³ mol dm⁻³ s⁻¹ Experiment 2: [A] = 0.20 mol dm⁻³, [B] = 0.10 mol dm⁻³, rate = 4.0 × 10⁻³ mol dm⁻³ s⁻¹ Experiment 3: [A] = 0.20 mol dm⁻³, [B] = 0.30 mol dm⁻³, rate = 3.6 × 10⁻² mol dm⁻³ s⁻¹ Determine the order with respect to A, the order with respect to B, and write the rate equation.

1. Experiments 1 and 2: [B] is constant at 0.10 mol dm⁻³. [A] doubles: 0.10 → 0.20 mol dm⁻³ (factor of 2) Rate ratio: (4.0 × 10⁻³) ÷ (2.0 × 10⁻³) = 2 — When [A] doubles, rate doubles. So rate ∝ [A]^1 — first order in A.
2. Experiments 2 and 3: [A] is constant at 0.20 mol dm⁻³. [B] triples: 0.10 → 0.30 mol dm⁻³ (factor of 3) Rate ratio: (3.6 × 10⁻²) ÷ (4.0 × 10⁻³) = 9 — When [B] triples, rate increases by a factor of 9. Check: 3^2 = 9. So rate ∝ [B]^2 — second order in B.
3. rate = k[A]^1[B]^2 Overall order = 1 + 2 = 3 — Combine the individual orders. The overall order is third order.
4. 2.0 × 10⁻³ = k × (0.10)^1 × (0.10)^2 2.0 × 10⁻³ = k × 0.10 × 0.010 2.0 × 10⁻³ = k × 1.0 × 10⁻³ k = (2.0 × 10⁻³) ÷ (1.0 × 10⁻³) = 2.0 — Rearrange rate = k[A][B]² for k and substitute known values. Always verify: 0.10 × 0.10 × 0.10 = 0.001 = 1.0 × 10⁻³.
5. rate = k[A][B]² mol dm⁻³ s⁻¹ = k × (mol dm⁻³) × (mol dm⁻³)² mol dm⁻³ s⁻¹ = k × mol³ dm⁻⁹ k = mol dm⁻³ s⁻¹ ÷ mol³ dm⁻⁹ = mol⁻² dm⁶ s⁻¹ — Cancel units algebraically. For a third-order reaction, k always has units of mol⁻² dm⁶ s⁻¹.

Answer: Order in A: first order. Order in B: second order. Rate equation: rate = k[A][B]². Overall order: third. k = 2.0 mol⁻² dm⁶ s⁻¹.

That five-step method works for every initial rates question AQA sets. The key discipline is always comparing experiments where only one concentration changes — if two concentrations change simultaneously, you cannot isolate the individual orders.

5. The Rate Constant k: Definition, Units, and Temperature Dependence

The rate constant $k$ is the proportionality constant that makes the rate equation an equality rather than just a proportionality. Its value encodes the activation energy of the reaction and the frequency and geometry of effective collisions — factors that are formalised in the Arrhenius equation.

The units of $k$ are not fixed — they vary with the overall order of the reaction. You derive them by rearranging the rate equation for $k$ and cancelling units algebraically. This is a standard AQA calculation, so the method below is worth practising until it is automatic.

Worked example: Deriving the Units of k for Any Overall Order

Derive the units of k for a reaction that is first order in A and first order in B (second order overall).

1. rate = k[A]^1[B]^1 — Start from the rate equation with the known orders.
2. $k = \frac{\text{rate}}{[A][B]}$ — Divide both sides by [A][B] to isolate k.
3. $k = \frac{\text{mol dm}^{-3} \text{ s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})}$ — Rate has units mol dm⁻³ s⁻¹; each concentration term has units mol dm⁻³.
4. $k = \frac{\text{mol dm}^{-3} \text{ s}^{-1}}{\text{mol}^2 \text{ dm}^{-6}} = \text{mol}^{-1} \text{ dm}^{3} \text{ s}^{-1}$ — mol dm⁻³ ÷ mol² dm⁻⁶ = mol^(1-2) × dm^(-3-(-6)) × s⁻¹ = mol⁻¹ dm³ s⁻¹.

Answer: For a second-order reaction, k has units of mol⁻¹ dm³ s⁻¹.

There is a useful pattern here: for an overall order of $n$, the units of $k$ are mol$^{1-n}$ dm$^{3(n-1)}$ s$^{-1}$. You do not need to memorise this formula — deriving it from scratch each time is faster and less error-prone — but recognising the pattern helps you sense-check your answer.

6. Worked Calculation: Finding k and Predicting Rate

Now that the theory is in place, here is a complete exam-style calculation that brings together order determination, rate equation writing, and k calculation in one question. This mirrors the style of AQA Paper 1 and Paper 2 multi-part questions.

Worked example: Complete Rate Equation Calculation

The reaction between P and Q was studied at constant temperature. The following initial rate data were collected. Experiment 1: [P] = 0.20 mol dm⁻³, [Q] = 0.10 mol dm⁻³, rate = 1.6 × 10⁻⁴ mol dm⁻³ s⁻¹ Experiment 2: [P] = 0.40 mol dm⁻³, [Q] = 0.10 mol dm⁻³, rate = 3.2 × 10⁻⁴ mol dm⁻³ s⁻¹ Experiment 3: [P] = 0.40 mol dm⁻³, [Q] = 0.20 mol dm⁻³, rate = 3.2 × 10⁻⁴ mol dm⁻³ s⁻¹ (a) Determine the order with respect to P and with respect to Q. (b) Write the rate equation. (c) Calculate k, including units. (d) Predict the rate when [P] = 0.30 mol dm⁻³ and [Q] = 0.50 mol dm⁻³.

1. [Q] constant at 0.10 mol dm⁻³. [P] doubles: 0.20 → 0.40 (factor 2). Rate ratio: (3.2 × 10⁻⁴) ÷ (1.6 × 10⁻⁴) = 2. — Doubling [P] doubles the rate. Rate factor = 2 = 2^1, so order in P is 1 (first order).
2. [P] constant at 0.40 mol dm⁻³. [Q] doubles: 0.10 → 0.20 (factor 2). Rate ratio: (3.2 × 10⁻⁴) ÷ (3.2 × 10⁻⁴) = 1. — Doubling [Q] has no effect on rate. Rate factor = 1 = 2^0, so order in Q is 0 (zero order).
3. rate = k[P]^1[Q]^0 = k[P] — Since [Q]^0 = 1, Q does not appear in the rate equation. Overall order = 1 + 0 = 1.
4. 1.6 × 10⁻⁴ = k × 0.20 k = (1.6 × 10⁻⁴) ÷ 0.20 k = 8.0 × 10⁻⁴ — Rearrange rate = k[P] for k. Division: 1.6 × 10⁻⁴ ÷ 0.20 = 1.6 × 10⁻⁴ ÷ (2.0 × 10⁻¹) = 0.80 × 10⁻³ = 8.0 × 10⁻⁴.
5. k = rate ÷ [P] k = (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³) = s⁻¹ — For a first-order reaction, k always has units of s⁻¹. The mol dm⁻³ terms cancel.
6. rate = k[P] = 8.0 × 10⁻⁴ × 0.30 rate = 2.4 × 10⁻⁴ mol dm⁻³ s⁻¹ — Q is zero order, so [Q] = 0.50 mol dm⁻³ has no effect. Only [P] matters. Multiply: 8.0 × 10⁻⁴ × 0.30 = 8.0 × 3.0 × 10⁻⁴ × 10⁻¹ = 24.0 × 10⁻⁵ = 2.4 × 10⁻⁴.

Answer: (a) First order in P; zero order in Q. (b) rate = k[P]. (c) k = 8.0 × 10⁻⁴ s⁻¹. (d) rate = 2.4 × 10⁻⁴ mol dm⁻³ s⁻¹.

Step 6 is where many students drop marks: they substitute [Q] into the rate equation even though Q is zero order. Because $[Q]^0 = 1$, the concentration of Q is irrelevant to the rate. The rate equation itself tells you this — Q does not appear in it.

7. Concentration–Rate Graphs for Each Order

AQA questions frequently present a graph of rate against concentration and ask you to identify the order. Each order produces a characteristic shape, so recognising these patterns is a quick route to marks.

Zero order gives a horizontal line — rate is independent of concentration. First order gives a straight line through the origin — rate is directly proportional to concentration. Second order gives an upward-curving parabola — rate increases more steeply as concentration rises.

8. Exam Technique and Common Mistakes

Rate equations questions are highly structured, which means the marks are predictable — and so are the mistakes. The guidance below is drawn from the patterns that appear most frequently in AQA mark schemes.

Rate equations sit at the heart of AQA kinetics and connect directly to later topics including the Arrhenius equation and reaction mechanisms. The core skills — determining orders from data, writing the rate equation, calculating k with correct units, and predicting rates — are examined in almost every series. Master the method here and those later topics will build naturally on a solid foundation.