What is finding orders from initial-rate data?

Compare experiments in which only one concentration changes: - rate unchanged when a concentration doubles, order 0; - rate doubles, order 1; - rate quadruples (× 2^2), order 2.

Pearson Edexcel Edexcel 2019-style practice: The reaction between A and B was studied. In experiment 1, [A] = 0.10, [B] = 0.10, rate = 2.0 × 10^-3. In experiment 2, [A] = 0.20, [B] = 0.10, rate = 4.0 × 10^-3. In experiment 3, [A] = 0.20, [B] = 0.20, rate = 1.6 × 10^-2\ mol dm^-3\,s^-1. Determine the order with respect to A and B, write the rate equation, and calculate k with units.

Compare experiments where one concentration changes at a time. Experiments 1 and 2: [A] doubles, [B] constant, rate doubles, so order in A is **1** (1). Experiments 2 and 3: [B] doubles, [A] constant, rate goes from 4.0 × 10^-3 to 1.6 × 10^-2, a factor of 4 = 2^2, so order in B is **2** (1). Rate equation: rate = k[A][B]^2; overall order 3 (1). k = rate[A][B]^2 = 2.0 × 10^-30.10 × (0.10)^2 = 2.0 × 10^-31.0 × 10^-3 = 2.0 (2). Units: mol dm^-3\,s^-1(mol dm^-3)^3 = mol^-2\,dm^6\,s^-1 (1).

EnglandChemistrySyllabus dot point

How do we turn observations about rate into a mathematical rate equation?

Rate equations and orders of reaction, the rate constant, finding orders from initial-rate and concentration-time data, the rate-determining step, and the Arrhenius equation.

An Edexcel 9CH0 Topic 15 answer covering rate equations and orders, the rate constant, finding orders from rate data, the rate-determining step, and the Arrhenius equation.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking

Edexcel Topic 15 wants you to write rate equations, determine orders of reaction from initial-rate and concentration-time data, find and give units for the rate constant, link the rate-determining step to a mechanism, and use the Arrhenius equation to find an activation energy.

The answer

Rate equations and orders

A zero-order reactant ( $[\text{X}]^0 = 1$ ) does not appear in the rate equation, so changing its concentration has no effect on the rate. A first-order reactant doubles the rate when its concentration doubles; a second-order reactant quadruples it.

Finding orders from initial-rate data

Compare experiments in which only one concentration changes:

rate unchanged when a concentration doubles, order $0$ ;
rate doubles, order $1$ ;
rate quadruples ( $\times 2^2$ ), order $2$ .

Finding orders from concentration-time graphs

For a first-order reaction the half-life is constant (independent of concentration), which is the quickest way to spot first order. A first-order rate constant can also be found from $k = \ln 2 / t_{1/2}$ . For a zero-order reaction the concentration falls linearly with time; for second order the half-life increases as the reaction proceeds.

The rate-determining step

If the rate equation is $\text{rate} = k[\text{A}][\text{B}]$ but the overall equation involves two moles of A, the mechanism must have only one A and one B reacting in (or before) the slow step. Proposed mechanisms must be consistent with the experimentally found rate equation.

The Arrhenius equation

The rate constant increases with temperature according to the Arrhenius equation, $k = Ae^{-E_a/RT}$ , where $A$ is the pre-exponential (frequency) factor. Taking natural logs gives the linear form:

Finding the order, rate constant and units

A reaction has $\text{rate} = k[\text{A}]^2$ . When $[\text{A}] = 0.050\ \text{mol dm}^{-3}$ the rate is $3.0 \times 10^{-4}\ \text{mol dm}^{-3}\,\text{s}^{-1}$ . Find $k$ and its units, then the rate when $[\text{A}] = 0.10\ \text{mol dm}^{-3}$ .

step 1: Rearrange for k

$k = \dfrac{\text{rate}}{[\text{A}]^2}$ .

step 2: Substitute

$k = \dfrac{3.0 \times 10^{-4}}{(0.050)^2} = \dfrac{3.0 \times 10^{-4}}{2.5 \times 10^{-3}} = 0.12$ .

step 3: Units

$\dfrac{\text{mol dm}^{-3}\,\text{s}^{-1}}{(\text{mol dm}^{-3})^2} = \text{mol}^{-1}\,\text{dm}^3\,\text{s}^{-1}$ , so $k = 0.12\ \text{mol}^{-1}\,\text{dm}^3\,\text{s}^{-1}$ .

step 4: New rate

Doubling $[\text{A}]$ multiplies the rate by $2^2 = 4$ : rate $= 4 \times 3.0 \times 10^{-4} = 1.2 \times 10^{-3}\ \text{mol dm}^{-3}\,\text{s}^{-1}$ .

Examples in context

Example 1. Iodine clock reactions. In the iodine clock, the time for a fixed small amount of product to form is measured as concentrations are varied. Because the time is inversely proportional to the initial rate, plotting $1/t$ against concentration reveals the order in each reactant directly. This is the standard Edexcel practical for determining orders by the initial-rate method described above.

Example 2. Why food keeps longer in a fridge. The Arrhenius equation explains refrigeration: lowering the temperature reduces $k$ exponentially, so the spoilage reactions (and microbial enzyme reactions) run far more slowly. A rough rule that reaction rate halves for every $10\ ^\circ\text{C}$ drop is the Arrhenius relationship in everyday form, and it is why a chilled environment, not a chemical change, preserves food.

Try this

Q1. A reaction obeys $\text{rate} = k[A][B]^2$ . State the overall order and the effect of doubling $[B]$ . [2 marks]

Cue. Overall order $3$ ; doubling $[B]$ multiplies the rate by $2^2 = 4$ .

Q2. Explain how a constant half-life on a concentration-time graph identifies the order. [1 mark]

Cue. A constant half-life indicates a first-order reaction.

Exam-style practice questions

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

Edexcel 20196 marksThe reaction between A and B was studied. In experiment 1,

[\text{A}] = 0.10

[\text{B}] = 0.10

, rate

= 2.0 \times 10^{-3}

. In experiment 2,

[\text{A}] = 0.20

[\text{B}] = 0.10

, rate

= 4.0 \times 10^{-3}

. In experiment 3,

[\text{A}] = 0.20

[\text{B}] = 0.20

, rate

= 1.6 \times 10^{-2}\ \text{mol dm}^{-3}\,\text{s}^{-1}

. Determine the order with respect to A and B, write the rate equation, and calculate

k

with units.

Show worked answer →

Compare experiments where one concentration changes at a time.

Experiments 1 and 2: $[\text{A}]$ doubles, $[\text{B}]$ constant, rate doubles, so order in A is 1 (1).

Experiments 2 and 3: $[\text{B}]$ doubles, $[\text{A}]$ constant, rate goes from $4.0 \times 10^{-3}$ to $1.6 \times 10^{-2}$ , a factor of $4 = 2^2$ , so order in B is 2 (1).

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

$k = \dfrac{\text{rate}}{[\text{A}][\text{B}]^2} = \dfrac{2.0 \times 10^{-3}}{0.10 \times (0.10)^2} = \dfrac{2.0 \times 10^{-3}}{1.0 \times 10^{-3}} = 2.0$ (2). Units: $\dfrac{\text{mol dm}^{-3}\,\text{s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{mol}^{-2}\,\text{dm}^6\,\text{s}^{-1}$ (1).

Edexcel 20224 marksValues of the rate constant

k

were measured at several temperatures and a graph of

\ln k

against

1/T

gave a straight line of gradient

-6.05 \times 10^3\ \text{K}

. Use the Arrhenius equation to calculate the activation energy. (

R = 8.31\ \text{J K}^{-1}\,\text{mol}^{-1}

Show worked answer →

Use the logarithmic form of the Arrhenius equation.

Taking logs: $\ln k = \ln A - \dfrac{E_a}{R}\cdot\dfrac{1}{T}$ , so the gradient is $-\dfrac{E_a}{R}$ (1).

Therefore $E_a = -\text{gradient} \times R = -(-6.05 \times 10^3) \times 8.31$ (1) $= 5.03 \times 10^4\ \text{J mol}^{-1}$ (1) $= 50.3\ \text{kJ mol}^{-1}$ (1).

Markers reward identifying gradient $= -E_a/R$ , the substitution, and converting to $\text{kJ mol}^{-1}$ .

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Sources & how we know this

Pearson Edexcel A-Level Chemistry (9CH0) specification — Pearson Edexcel (2015)