WJEC WJEC 2021-style practice: For the reaction A + B → products, doubling [A] at constant [B] doubles the rate, while doubling [B] at constant [A] quadruples the rate. Deduce the rate equation, the overall order, and calculate the rate constant if the rate is 2.0 × 10^-3\ mol dm^-3 s^-1 when [A] = 0.10 and [B] = 0.10\ mol dm^-3.

Doubling [A] doubles the rate, so the reaction is first order in A. Doubling [B] quadruples the rate, so it is second order in B. Rate equation: rate = k[A][B]^2; overall order = 1 + 2 = 3. Rearrange: k = rate[A][B]^2 = 2.0 × 10^-30.10 × 0.10^2 = 2.0 × 10^-31.0 × 10^-3 = 2.0 mol-2 dm6 s-1. Markers reward each order, the overall order, the rearrangement, and the rate constant with its units.

WalesChemistrySyllabus dot point

How do we express rate mathematically and deduce a mechanism?

Rate equations, orders of reaction, the rate constant, half-life, determining orders from data, the rate-determining step and the Arrhenius equation.

A focused answer to WJEC A-Level Chemistry Unit 3, covering rate equations and orders of reaction, the rate constant and its units, half-life, determining orders from initial-rate data, the rate-determining step and the Arrhenius equation.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

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What this dot point is asking

WJEC wants you to write rate equations, determine orders of reaction and the rate constant from data, use half-life, identify the rate-determining step, and apply the Arrhenius equation linking rate constant to activation energy and temperature.

The answer

Rate equations and orders

Determining orders

Use initial-rate data: see how the rate changes when each concentration is varied in turn. No change means zero order; proportional means first order; quadrupling for doubling means second order.

Rate-determining step and Arrhenius

The rate-determining step is the slowest step; only species involved up to and including it appear in the rate equation, so the orders reveal the mechanism. The Arrhenius equation $k = A e^{-E_a/RT}$ shows $k$ rising with temperature and falling with activation energy.

Finding the rate constant and its units

A reaction is first order in $\text{X}$ and zero order in $\text{Y}$ . When $[\text{X}] = 0.20$ mol dm $^{-3}$ the rate is $4.0 \times 10^{-4}$ mol dm $^{-3}$ s $^{-1}$ . Find $k$ .

step 1 Write the rate equation

$\text{rate} = k[\text{X}]$ (zero order in $\text{Y}$ , so $\text{Y}$ is absent).

step 2 Rearrange for k

$k = \dfrac{\text{rate}}{[\text{X}]}$ .

step 3 Substitute

$k = \dfrac{4.0 \times 10^{-4}}{0.20} = 2.0 \times 10^{-3}$ .

step 4 Units

$\dfrac{\text{mol dm}^{-3}\,\text{s}^{-1}}{\text{mol dm}^{-3}} = \text{s}^{-1}$ , so $k = 2.0 \times 10^{-3}$ s $^{-1}$ .

Determining orders from initial-rate data

The cleanest way to find an order is the initial-rates method: run several experiments changing one concentration at a time and compare the initial rates. If doubling a concentration leaves the rate unchanged, that reactant is zero order; if the rate doubles, first order; if it quadruples, second order. Comparing two experiments where only one concentration differs isolates the effect of that reactant. Once all orders are known, substitute one full set of data into the rate equation to find the rate constant $k$ , then derive its units from the overall order.

From orders to mechanism

The orders reveal the slowest, rate-determining step, because only species involved up to and including that step appear in the rate equation. If a reactant in the overall equation is absent from the rate equation (zero order), it must react after the rate-determining step. If a species appears to a higher power than its stoichiometric coefficient, more than one of its molecules is involved in or before the slow step. This is how kinetics, an experimental method, lets chemists deduce a mechanism they cannot watch directly.

Examples in context

Mechanism of nucleophilic substitution. Whether a halogenoalkane reacts by a one-step or two-step pathway is revealed by its order: first order overall points to a rate-determining ionisation, second order to a single concerted step. Enzyme kinetics. At high substrate concentration enzyme reactions become zero order because the enzyme is saturated, an everyday example of order changing with conditions.

Try this

Q1. State what is meant by the order of a reaction with respect to a reactant. [1 mark]

Cue. The power to which that reactant's concentration is raised in the rate equation.

Q2. Give the units of the rate constant for a second-order reaction. [1 mark]

Cue. mol $^{-1}$ dm $^3$ s $^{-1}$ .

Q3. State what a constant half-life tells you about the order. [1 mark]

Cue. The reaction is first order.

Q4. A reaction is first order in X and second order in Y. State the overall order. [1 mark]

Cue. Third order ( $1 + 2$ ).

Q5. Explain why a zero-order reactant does not appear in the rate equation. [1 mark]

Cue. It reacts after the rate-determining step, so its concentration does not affect the rate.

Exam-style practice questions

Practice questions written in the style of WJEC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WJEC 20215 marksFor the reaction

\text{A} + \text{B} \rightarrow \text{products}

, doubling

[\text{A}]

at constant

[\text{B}]

doubles the rate, while doubling

[\text{B}]

at constant

[\text{A}]

quadruples the rate. Deduce the rate equation, the overall order, and calculate the rate constant if the rate is

2.0 \times 10^{-3}\ \text{mol dm}^{-3}\text{ s}^{-1}

when

[\text{A}] = 0.10

and

[\text{B}] = 0.10\ \text{mol dm}^{-3}

Show worked answer →

Doubling $[\text{A}]$ doubles the rate, so the reaction is first order in $\text{A}$ . Doubling $[\text{B}]$ quadruples the rate, so it is second order in $\text{B}$ .

Rate equation: $\text{rate} = k[\text{A}][\text{B}]^2$ ; overall order $= 1 + 2 = 3$ .

Rearrange: $k = \dfrac{\text{rate}}{[\text{A}][\text{B}]^2} = \dfrac{2.0 \times 10^{-3}}{0.10 \times 0.10^2} = \dfrac{2.0 \times 10^{-3}}{1.0 \times 10^{-3}} = 2.0$ mol-2 dm6 s-1.

Markers reward each order, the overall order, the rearrangement, and the rate constant with its units.

WJEC 20193 marksA first-order reaction has a constant half-life of 120 s. Explain what is meant by a constant half-life and calculate the fraction of reactant remaining after 360 s.

Show worked answer →

A constant half-life means the time for the concentration to halve is the same regardless of the starting concentration, which is the signature of a first-order reaction.

$360$ s is $360/120 = 3$ half-lives.

After $3$ half-lives the fraction remaining is $(1/2)^3 = 1/8 = 0.125$ .

Markers reward the meaning of constant half-life as evidence of first order, the number of half-lives, and the fraction $1/8$ .

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Sources & how we know this

WJEC A-level Chemistry specification — WJEC (2015)