A-level Chemistry: Rate Equations and the Arrhenius Equation

Specification focus — AQA 7405, 3.1.9. This is A-level only content. In an examination, marks are awarded for using experimental data accurately, applying the correct rate relationship, and communicating a conclusion with units.

Overview

Rate equations turn collision theory into a usable mathematical model. A rate equation shows how a reaction rate depends on reactant concentrations and therefore gives evidence about a mechanism that may proceed in several steps. The core form is Rate = k[A]^m[B]^n, where the powers m and n are experimentally determined orders and k is the rate constant. AQA limits orders in this topic to 0, 1 or 2.[1]

Do not confuse a rate equation with a balanced symbol equation or a Kc expression. A balanced equation shows overall stoichiometry; a rate equation is found from data. A Kc expression includes equilibrium concentrations and may contain products; a rate equation for this course contains the relevant reactant concentrations. Candidates who state that an order was “read from the balanced equation” receive no credit.

1. The rate equation: orders and their meaning

For a reaction involving A and B:

A reliable method is factor → factor → power. Compare two experiments where only one concentration changes. Calculate the concentration factor, calculate the rate factor, and select the power that produces it. For example, if [A] doubles and the initial rate becomes four times greater, A is second order because 2² = 4. The memory hook is “Square the Second.”

Worked mini-example: deducing an order from initial rates

Only [A] changes. It doubles (×2), while the rate rises from 1.50 × 10⁻³ to 6.00 × 10⁻³ (×4). Therefore, rate ∝ [A]², so the reaction is second order with respect to A. In a 2-mark question, one mark is typically awarded for the numerical comparison and one for the correct order. Simply writing “second order” without evidence may lose the explanation mark.

2. Deriving and using a complete rate equation

Repeat the comparison for B using a pair of experiments in which [A] is constant. Then assemble the equation. If A is second order and B is first order, write:

**Rate = k[A]²[B]**The overall order is the sum of the separate orders: 2 + 1 = 3. This is useful when deriving the units of k, but it does not mean that three particles necessarily collide in a single overall reaction. Use careful mechanism wording: an elementary proposed slow step must be consistent with the experimentally determined orders. The slowest step is the rate-determining (or rate-limiting) step.

Examiner wording: “The order with respect to a reactant provides information about the rate-determining step.” Avoid the absolute claim that the overall balanced equation directly gives the mechanism.

Calculating k and its units

Rearrange before substituting. For Rate = k[A]²[B],

**k = Rate / ([A]²[B])**For a rate measured in mol dm⁻³ s⁻¹, the unit calculation is:

(mol dm⁻³ s⁻¹) / ((mol dm⁻³)² × (mol dm⁻³)) = dm⁶ mol⁻² s⁻¹.

Write units with the final answer. They are often a separate mark. Remember: “Overall order controls the units of k.”

3. Rate constant, temperature and activation energy

At a fixed temperature, k is constant for a particular reaction. When temperature increases, k increases because a greater proportion of particles have energy equal to or greater than the activation energy, so a greater fraction of collisions is successful. For a 2-mark qualitative explanation, candidates should mention both more particles with E ≥ Eₐ and more successful collisions.

The Arrhenius equation is:

**k = Ae^(−Eₐ/RT)**Here, A is the Arrhenius constant, Eₐ is activation energy in J mol⁻¹, R is 8.31 J K⁻¹ mol⁻¹, and T is temperature in K. The non-negotiable conversion is kJ mol⁻¹ → J mol⁻¹: multiply by 1000 before substitution. Memory hook: **“Eₐ must be J.”**Rearranging produces a straight-line relationship:

ln k = (−Eₐ/R)(1/T) + ln AThis matches y = mx + c. When plotting ln k against 1/T, the gradient is −Eₐ/R and the y-intercept is ln A. A negative gradient is expected because 1/T falls as temperature rises while k rises.

4. Formula and conversion checklist

5. Required Practical 7: measuring reaction rate

AQA Required Practical 7 requires measuring rate by an initial-rate method and a continuous-monitoring method.[2] Your centre may use a different approved reaction system, so learn the principles rather than a single recipe.

For an initial-rate method, candidates should state what is changed, what is controlled, what is measured, and how rate is calculated. “Repeat the experiment” earns limited credit unless the answer explains that repeats improve reliability and a mean can be calculated. For continuous monitoring, the word tangent is essential when asked for the rate at one instant; a chord gives an average rate over an interval.

Common errors are starting the timer late, changing total volume while changing concentration, allowing temperature to drift, reading a gas syringe at an angle, choosing an inconsistent visual endpoint, or using an absorbance range outside the colorimeter’s reliable range. Examiners may test these through a method-evaluation question, a graph interpretation task, or a request to explain why only one variable must change.

6. Graph and data skills

A concentration–time graph needs named axes and units. The gradient of a tangent is Δconcentration/Δtime, normally negative for reactant concentration. State the rate as a positive magnitude unless the question specifically asks for the gradient. For a zero-order reaction, [reactant] decreases linearly, so the magnitude of the gradient equals k.

On an Arrhenius graph, plot ln k on the y-axis and 1/T in K⁻¹ on the x-axis. Use a large triangle across the best-fit line when calculating gradient; do not use adjacent raw points unless instructed. Preserve the negative sign in the gradient, then use Eₐ = −gradient × R. Convert the final answer to kJ mol⁻¹ if requested.

7. Exam technique: command words and timing

Allow about one minute per mark. For a calculation, show the formula, the substitution, an unrounded intermediate value where helpful, and a final answer with units and appropriate significant figures. Method marks can be credited even if the final arithmetic is wrong, provided the working is clear.

8. Synoptic links and retrieval practice

This topic connects directly to 3.1.5 Kinetics, because collision theory explains why temperature raises the rate constant. It connects to 3.1.8 Thermodynamics, where activation energy appears on enthalpy-profile diagrams. It also connects to 3.3.3 Halogenoalkanes: kinetic evidence can distinguish a unimolecular pathway from a bimolecular one. Finally, contrast it with 3.1.6 Kc: kinetics describes the speed of change, whereas Kc describes the equilibrium composition at a fixed temperature.

Cover and recall: Cover the answers and say: (1) What factor does the rate change by when a second-order concentration is tripled? (2) What are the units of k for a first-order reaction? (3) What does the gradient represent on a ln k against 1/T plot? (4) Why must Eₐ be in joules? (5) What graph construction is required for an instantaneous rate?

Elaborate: Why can a rate equation not be obtained from the overall symbol equation? What would happen to the Arrhenius plot if a catalyst lowered Eₐ? How could a systematic gas leak alter the apparent rate–concentration relationship?

References

[1] AQA A-level Chemistry specification, section 3.1.9

[2] AQA Required practical 7 specification wording