OCR A Level Chemistry A: How fast? (H432)

This study guide teaches section 5.1.1, How fast?, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, covering outcomes (a) to (k). H432 is the full A level. OCR’s assessment overview places Module 5 in Periodic table, elements and physical chemistry (01) and in Unified chemistry (03).

For collision theory and Boltzmann curves (3.2.2), read our guide to Reaction rates first. This page pairs with the How fast? revision notes and the How fast? practice set. Use our H432 course page, the printable outcome list and a free diagnostic to plan.

What this unit covers

Statement	Learning outcome in brief
5.1.1(a)	Use the key terms below
5.1.1(b)	Orders from initial rates; rate = k[A]^m[B]^n (PAG10)
5.1.1(c)	k and related quantities, with units
5.1.1(d)	Concentration–time graphs: order 0 or 1; rates from gradients
5.1.1(e)	Constant half-life of a first-order reaction
5.1.1(f)	Find k from k = ln 2/t½
5.1.1(g)	Rate–concentration graphs: order 0, 1 or 2; k from a first-order gradient
5.1.1(h)	Initial rates and continuous monitoring, including colorimetry (PAG9, PAG10)
5.1.1(i)	Rate-determining step and mechanism steps
5.1.1(j)	Effect of temperature on rate and k
5.1.1(k)	Arrhenius equation; Ea and A from ln k against 1/T

Key terms (a)

• Rate of reaction: how quickly a reactant’s concentration falls or a product’s concentration rises, per unit time, usually in mol dm⁻³ s⁻¹.
• Order with respect to a reactant: the exponent on [A] in the rate equation, for reactant A.
• Overall order: add the individual orders together.
• Rate constant, k: the number linking the rate to the concentration terms. It is fixed at one temperature only.
• Half-life, t½: how long it takes for [A] to halve, starting from any chosen value.
• Rate-determining step: in a mechanism with more than one step, the step with the lowest rate; the overall reaction cannot go faster than it.

Orders and rate equations from initial rates (b)

In rate = k[A]^m[B]^n each order is 0, 1 or 2, found by experiment, not from the balanced equation. Compare two experiments where only one concentration changes.

• Rate unchanged: order 0.
• Rate changes by the same factor as the concentration: order 1.
• Rate changes by the square of the factor: order 2.

Worked example 1. Chlorine dioxide reacts with hydroxide ions in water:

2ClO₂(aq) + 2OH⁻(aq) → ClO₃⁻(aq) + ClO₂⁻(aq) + H₂O(l)

Initial rates at constant temperature:

Experiment	[ClO₂] / mol dm⁻³	[OH⁻] / mol dm⁻³	Rate at t = 0 / mol dm⁻³ s⁻¹
1	0.0150	0.0200	9.00 × 10⁻⁴
2	0.0300	0.0200	3.60 × 10⁻³
3	0.0150	0.0500	2.25 × 10⁻³

1. Experiments 1 and 2: [ClO₂] doubles with [OH⁻] fixed. The rate rises by 3.60 × 10⁻³ ÷ 9.00 × 10⁻⁴ = 4 = 2². Second order in ClO₂.
2. Experiments 1 and 3: [OH⁻] rises by 0.0500 ÷ 0.0200 = 2.5 with [ClO₂] fixed. The rate rises by 2.25 × 10⁻³ ÷ 9.00 × 10⁻⁴ = 2.5. First order in OH⁻.
3. Rate equation: rate = k[ClO₂]²[OH⁻]. Overall order = 3.

Calculating k and its units (c)

Rearrange the rate equation and substitute one experiment’s values.

Worked example 1, continued. From experiment 1:

k = rate ÷ ([ClO₂]²[OH⁻]) = 9.00 × 10⁻⁴ ÷ (0.0150² × 0.0200) = 200 dm⁶ mol⁻² s⁻¹

Units: mol dm⁻³ s⁻¹ divided by (mol dm⁻³)³ leaves dm⁶ mol⁻² s⁻¹.

Related quantities follow from the same equation. For [ClO₂] = 0.0450 and [OH⁻] = 0.0100 mol dm⁻³:

rate = 200 × 0.0450² × 0.0100 = 4.05 × 10⁻³ mol dm⁻³ s⁻¹

Each extra order adds dm³ mol⁻¹ to the units: overall orders 0, 1, 2 and 3 give mol dm⁻³ s⁻¹, s⁻¹, dm³ mol⁻¹ s⁻¹ and dm⁶ mol⁻² s⁻¹.

Concentration–time graphs (d)

Continuous monitoring of one mixture gives a concentration–time graph. Its shape shows the order for the reactant followed.

• Zero order: a straight line sloping down. The gradient, and so the rate, is constant until the reactant runs out.
• First order: a curve whose gradient falls as the concentration falls, with a constant half-life.

The rate at any time is the gradient of the tangent there (see 3.2.2(b)), quoted as a positive value.

Worked example 2 (zero order). Ammonia decomposes on a hot tungsten wire. [NH₃] falls in a straight line from 2.40 × 10⁻³ mol dm⁻³ at 0 s to 0.60 × 10⁻³ mol dm⁻³ at 300 s.

Rate = (2.40 − 0.60) × 10⁻³ ÷ 300 = 6.0 × 10⁻⁶ mol dm⁻³ s⁻¹. Since rate = k[NH₃]⁰ = k, k = 6.0 × 10⁻⁶ mol dm⁻³ s⁻¹.

Half-life and k for first-order reactions (e) and (f)

A first-order reaction has a constant half-life: it takes the same time to fall from any concentration to half that value. Read two successive half-lives to show this.

The specification gives k = ln 2/t½ (ln 2 = 0.693). You do not need to derive it from [A] = [A]₀e^(–kt), and integrated rate equations are not required.

Worked example 3. Cyclopropane isomerises to propene in the gas phase at a fixed temperature:

t / s	0	420	840	1260	1680	2520
[C₃H₆] / mol dm⁻³	0.0480	0.0339	0.0240	0.0170	0.0120	0.0060

1. 0.0480 → 0.0240 takes 840 s. 0.0240 → 0.0120 takes 1680 − 840 = 840 s. 0.0120 → 0.0060 takes 840 s.
2. The half-life is constant at 840 s, so the reaction is first order.
3. k = ln 2 ÷ 840 = 8.25 × 10⁻⁴ s⁻¹.

Check with a tangent at 840 s through (240 s, 0.0359) and (1440 s, 0.0121): rate = 0.0238 ÷ 1200 = 1.98 × 10⁻⁵ mol dm⁻³ s⁻¹, and k = 1.98 × 10⁻⁵ ÷ 0.0240 = 8.25 × 10⁻⁴ s⁻¹. The methods agree.

Rate–concentration graphs (g)

These data usually come from separate initial-rate experiments.

Order	Shape of rate against concentration
0	Flat line: the rate is the same at every concentration
1	Line through (0, 0): doubling [A] doubles the rate
2	Curve through (0, 0) that gets steeper (rate ∝ [A]²)

For a first-order reactant, rate = k[A], so the gradient equals k.

Worked example 4. Initial rates for the decomposition of sulfuryl chloride, SO₂Cl₂(g) → SO₂(g) + Cl₂(g):

[SO₂Cl₂] / mol dm⁻³	0.050	0.100	0.150	0.200
Rate / mol dm⁻³ s⁻¹	1.2 × 10⁻⁶	2.4 × 10⁻⁶	3.6 × 10⁻⁶	4.8 × 10⁻⁶

The points lie on a straight line through the origin: first order. Gradient = 4.8 × 10⁻⁶ ÷ 0.200 = 2.4 × 10⁻⁵ s⁻¹ = k.

Practical techniques (h)

Continuous monitoring (PAG9). Follow one mixture over time by measuring gas volume, mass loss or colour. In colorimetry, choose a filter of the colour the coloured species absorbs. Absorbance is proportional to concentration, so a calibration curve made with standard solutions converts absorbance to concentration.

Initial rates (PAG10). Run separate experiments, changing one concentration at a time and keeping temperature and the other concentrations fixed. Find each initial rate from a tangent at t = 0, or use a clock reaction. In a clock reaction you measure the time until a set, small quantity of product has appeared. Because so little reaction has happened, the rate is almost constant over that time, so the initial rate is proportional to 1/t. This approximation fails if the reaction has gone too far.

Worked example 5. A clock reaction is timed at three concentrations of reactant X:

[X] / mol dm⁻³	0.010	0.020	0.030
t / s	72	18	8
1/t / s⁻¹	0.0139	0.0556	0.125

Relative rates are 1 : 4 : 9 when [X] is in the ratio 1 : 2 : 3. The rate depends on [X]², so the reaction is second order in X.

Rate-determining step and mechanisms (i)

Species in the rate equation react in, or before, the rate-determining step; the orders give how many of each. A zero-order reactant joins later.

To propose a mechanism:

1. Use the rate equation to write a slow step with the right particles.
2. Add fast steps so that all the steps add up to the overall equation.
3. Check that any intermediate formed is used up later.

Worked example 6. For H₂(g) + 2ICl(g) → I₂(g) + 2HCl(g), rate = k[H₂][ICl].

The rate-determining step involves one H₂ and one ICl:

• Step 1 (slow): H₂ + ICl → HI + HCl
• Step 2 (fast): HI + ICl → I₂ + HCl

The steps add to the overall equation; HI is the intermediate. The second ICl reacts after the slow step, so it is not in the rate equation.

Temperature and the rate constant (j)

Heating shifts the Boltzmann distribution to higher energies (3.2.2(f) and (g)), so more of the molecules now have at least the activation energy, and more collisions per second succeed. The concentrations are unchanged by heating, so the faster rate means k has increased. k is constant only at one temperature.

The Arrhenius equation (k)

k = Ae^(–Ea/RT)

Ea is in J mol⁻¹, T in kelvin, R = 8.314 J mol⁻¹ K⁻¹ and A is the pre-exponential factor (no explanation of A is needed). Taking natural logs:

ln k = –Ea/RT + ln A

Matching it to y = mx + c, the y-values are ln k and the x-values are 1/T. The graph is a straight line whose gradient is –Ea/R and whose intercept is ln A. OCR’s additional guidance says R and these equations are provided on the Data Sheet.

Worked example 7. k for a second-order reaction at five temperatures:

T / K	300	310	320	330	340
k / dm³ mol⁻¹ s⁻¹	5.36 × 10⁻³	1.21 × 10⁻²	2.60 × 10⁻²	5.33 × 10⁻²	0.105
1/T / K⁻¹	3.333 × 10⁻³	3.226 × 10⁻³	3.125 × 10⁻³	3.030 × 10⁻³	2.941 × 10⁻³
ln k	–5.229	–4.415	–3.650	–2.932	–2.254

1. Plot ln k against 1/T and draw the best-fit straight line.
2. Gradient, using points on the line at the two ends: (–2.254 − (–5.229)) ÷ (2.941 × 10⁻³ − 3.333 × 10⁻³) = –7.59 × 10³ K.
3. Ea = –gradient × R = 7.59 × 10³ × 8.314 = 6.31 × 10⁴ J mol⁻¹ = 63.1 kJ mol⁻¹.
4. ln A = ln k – gradient × (1/T) = –5.229 + 7.59 × 10³ × 3.333 × 10⁻³ = 20.07, so A = e^20.07 = 5.2 × 10⁸ dm³ mol⁻¹ s⁻¹. A has the same units as k.

Using k = Ae^(–Ea/RT) directly: with Ea = 63.0 kJ mol⁻¹ and A = 5.0 × 10⁸ dm³ mol⁻¹ s⁻¹ at 300 K, Ea/RT = 25.26 and k = 5.0 × 10⁸ × e^(–25.26) = 5.36 × 10⁻³ dm³ mol⁻¹ s⁻¹.

Common errors

• Reading orders from the balanced equation instead of from data.
• Assuming every change in concentration is a doubling, as in experiment 3 of worked example 1.
• Using k = ln 2/t½ for a reaction that is not first order.
• Leaving Ea in J mol⁻¹ when kJ mol⁻¹ is asked for, or putting T in °C.
• Writing a slow step that contains a zero-order reactant.

Next steps

Once the ideas make sense, drill them with the condensed How fast? notes and test them on the How fast? questions. Tangent and gradient skills are revisited in our page on written-paper practical skills.

Official syllabus

Source: section 5.1.1 How fast?, learning outcomes (a) to (k), in the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, published by OCR.