OCR A Level Chemistry A: How fast? (H432) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions practise 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. They test every outcome from 5.1.1(a) to (k), including the PAG9 and PAG10 techniques. H432 is the full A level. Questions 4 and 10 are longer problems.

Stuck? Use the How fast? study guide or the How fast? revision notes. Use R = 8.314 J mol⁻¹ K⁻¹.

Questions

1. Explain what is meant by each term.

(a) The order of reaction with respect to a reactant. [1] (b) The half-life of a reactant. [1] (c) The rate-determining step. [1]

2. For each rate equation, state the overall order and deduce the units of k.

(a) rate = k[R][S] [2] (b) rate = k[T]²[V] [2]

3. Substances M, N and Z react in solution. Initial rates at a fixed temperature:

Run	[M] / mol dm⁻³	[N] / mol dm⁻³	[Z] / mol dm⁻³	Rate, t = 0 / mol dm⁻³ s⁻¹
1	0.100	0.200	0.050	9.0 × 10⁻⁵
2	0.300	0.200	0.050	2.7 × 10⁻⁴
3	0.100	0.400	0.050	9.0 × 10⁻⁵
4	0.100	0.200	0.150	8.1 × 10⁻⁴

(a) Deduce the order with respect to M, N and Z. Explain your reasoning. [3] (b) Give the rate equation for the reaction. [1] (c) Calculate k, with units. [2] (d) Work out the starting rate when [M] = 0.250, [N] = 0.100 and [Z] = 0.080 mol dm⁻³. [1]

4. Purple crystal violet, CV⁺, fades as it reacts with excess hydroxide ions. A colorimeter follows the reaction. Starting [CV⁺] = 2.00 × 10⁻⁵ mol dm⁻³, absorbance 0.880.

t / s	0	80	160	240	320	480
Absorbance	0.880	0.622	0.440	0.311	0.220	0.110

(a) Describe how absorbance readings can be converted into concentrations of CV⁺. [2] (b) Use the data to show that the reaction is first order with respect to CV⁺. [2] (c) Calculate the rate constant from the half-life. [2] (d) A tangent to the curve at t = 160 s passes through (0 s, 0.745) and (391 s, 0.000). Calculate the rate at 160 s in mol dm⁻³ s⁻¹. [3]

5. Compound W decomposes on a hot metal surface. [W] falls in a straight line from 0.0500 mol dm⁻³ at t = 0 to 0.0180 mol dm⁻³ at t = 160 s.

(a) State the order with respect to W and give a reason. [2] (b) Calculate the rate constant and give its units. [2] (c) Calculate when W would be used up. [1]

6. Initial rates for 2N₂O₅(g) → 4NO₂(g) + O₂(g) at one temperature:

[N₂O₅] / mol dm⁻³	0.30	0.60	0.90	1.20
Initial rate / mol dm⁻³ s⁻¹	1.5 × 10⁻⁴	3.0 × 10⁻⁴	4.5 × 10⁻⁴	6.0 × 10⁻⁴

(a) Describe the shape of the rate–concentration graph and deduce the order. [2] (b) Determine k from the gradient. [2] (c) Calculate the half-life of N₂O₅. [2]

7. A gas-phase reaction has the overall equation A + 2B → AB₂ and the rate equation rate = k[B]².

(a) Suggest a two-step mechanism for this reaction, showing which step is rate-determining. [3] (b) Explain why A does not appear in the rate equation. [1]

8. A clock reaction measures the time, t, for a fixed small amount of product to form at different [Y].

[Y] / mol dm⁻³	0.040	0.080	0.100
t / s	60	30	24

(a) Explain why the reciprocal of t is proportional to the initial rate. [2] (b) Deduce the order with respect to Y, showing your working. [2] (c) State two variables to keep constant. [2]

9. A first-order reaction has Ea = 96.0 kJ mol⁻¹ and A = 2.5 × 10¹² s⁻¹.

(a) Use k = Ae^(–Ea/RT) to calculate k at 300 K. [2] (b) Calculate k at 310 K. [1] (c) Calculate the factor by which k increases between 300 K and 310 K. [1] (d) Use the Boltzmann distribution to explain why heating increases k. [3]

10. Rate constants for a second-order reaction:

T / K	293	303	313	323	333
k / dm³ mol⁻¹ s⁻¹	0.0595	0.131	0.276	0.554	1.07
(1/T) / 10⁻³ K⁻¹	3.413	3.300		3.096
ln k	–2.82	–2.03		–0.59

(a) Complete the two missing columns. [2] (b) Show how the Arrhenius equation predicts a linear plot of ln k versus 1/T. [1] (c) All five points fit one line. From the end points, calculate Ea in kJ mol⁻¹. [3] (d) Calculate the pre-exponential factor, A, with units. [3]

Answers

1. (a) The index n in [A]^n within the rate equation, for reactant A [1]. (b) The time taken for its concentration to fall to half of its starting value [1]. (c) The slowest step in a mechanism, which controls the overall rate [1]. [3] Examiner insight: “Time for half the reaction” is too vague; the definition needs concentration and half.

2. (a) Overall order 2 [1]; k = rate ÷ [R][S] gives dm³ mol⁻¹ s⁻¹ [1]. (b) Overall order 3 [1]; k = rate ÷ [T]²[V] gives dm⁶ mol⁻² s⁻¹ [1]. [4] Examiner insight: Show the units cancelling; inverted powers, such as mol² dm⁻⁶ s⁻¹, lose the mark.

3. (a) Runs 1 and 2: [M] × 3, rate × 3, so first order in M [1]. Runs 1 and 3: [N] × 2, rate unchanged, so zero order in N [1]. Runs 1 and 4: [Z] × 3, rate × 9 = 3², so second order in Z [1]. (b) rate = k[M][Z]² [1] (c) k = 9.0 × 10⁻⁵ ÷ (0.100 × 0.050²) = 0.36 [1] dm⁶ mol⁻² s⁻¹ [1] (d) rate = 0.36 × 0.250 × 0.080² = 5.8 × 10⁻⁴ mol dm⁻³ s⁻¹ [1] [7] Examiner insight: Name the runs you compare; bare orders with no reasoning earn little if one is wrong.

4. (a) Measure the absorbance of standard CV⁺ solutions of known concentration, using the same filter [1]. Plot a calibration graph and read concentrations from it [1]. (b) 0.880 → 0.440 takes 160 s; 0.440 → 0.220 takes 160 s; 0.220 → 0.110 takes 160 s [1]. The half-life is constant, so the reaction is first order in CV⁺ [1]. (c) k = 0.693 ÷ 160 [1] = 4.33 × 10⁻³ s⁻¹ [1] (d) Gradient = 0.745 ÷ 391 = 1.91 × 10⁻³ absorbance units s⁻¹ [1]. Each unit of absorbance is 2.00 × 10⁻⁵ ÷ 0.880 = 2.27 × 10⁻⁵ mol dm⁻³ [1]. Rate = 1.91 × 10⁻³ × 2.27 × 10⁻⁵ = 4.33 × 10⁻⁸ mol dm⁻³ s⁻¹ [1]. [9] Examiner insight: Quote at least two successive half-lives with values; “the half-life is constant” alone is not evidence.

5. (a) Zero order [1]: a straight line means the rate stays constant as [W] falls [1]. (b) rate = (0.0500 − 0.0180) ÷ 160 = 2.0 × 10⁻⁴; rate = k, so k = 2.0 × 10⁻⁴ [1] mol dm⁻³ s⁻¹ [1]. (c) t = 0.0500 ÷ 2.0 × 10⁻⁴ = 250 s [1] [5] Examiner insight: For zero order, k has the units of rate; s⁻¹ here loses the units mark.

6. (a) A straight line through the origin [1], so the reaction is first order in N₂O₅ [1]. (b) Gradient = 6.0 × 10⁻⁴ ÷ 1.20 [1] = k = 5.0 × 10⁻⁴ s⁻¹ [1] (c) t½ = 0.693 ÷ 5.0 × 10⁻⁴ [1] = 1.39 × 10³ s (about 23 minutes) [1] [6] Examiner insight: Take the gradient from the whole line, not one point, and give k with units.

7. (a) Step 1 (slow, rate-determining): B + B → B₂ [1]. Step 2 (fast): A + B₂ → AB₂ [1]. The steps add up to A + 2B → AB₂ and the intermediate B₂ cancels [1]. (b) A reacts only in a fast step after the rate-determining step, so changing [A] does not change the rate (zero order) [1]. [4] Examiner insight: Check that your steps add up to the overall equation; a correct slow step alone may not earn full credit.

8. (a) So little product forms in time t that the rate is almost constant [1]. Rate ≈ fixed amount ÷ t, so rate ∝ 1/t [1]. (b) 1/t = 0.0167, 0.0333 and 0.0417 s⁻¹, a ratio of 1 : 2 : 2.5 [1]. [Y] is in the ratio 1 : 2 : 2.5, so the rate is proportional to [Y]: first order [1]. (c) Any two of: temperature [1]; concentrations of the other reactants [1]; total volume of the mixture. [6] Examiner insight: “Keep the amounts the same” is too vague; name each variable precisely.

9. (a) Ea/RT = 96 000 ÷ (8.314 × 300) = 38.49 [1]. k = 2.5 × 10¹² × e^(–38.49) = 4.81 × 10⁻⁵ s⁻¹ [1] (b) Ea/RT = 37.25; k = 1.67 × 10⁻⁴ s⁻¹ [1] (c) Factor = 1.67 × 10⁻⁴ ÷ 4.81 × 10⁻⁵ = 3.46 (3.47 from the rounded values) [1] (d) The Boltzmann distribution shifts to higher energies [1]. More molecules have energy ≥ Ea [1]. More collisions per second succeed; concentrations are unchanged, so k must increase [1]. [7] Examiner insight: Convert Ea to J mol⁻¹ before using R; 96.0 in the exponent gives k close to A.

10. (a) 313 K: 1/T = 3.195 × 10⁻³ K⁻¹, ln k = –1.29 [1]. 333 K: 1/T = 3.003 × 10⁻³ K⁻¹, ln k = 0.07 [1]. (b) ln k = –Ea/RT + ln A has the form y = mx + c, with –Ea/R as the constant gradient and ln A as the intercept [1]. (c) Gradient = (0.07 − (–2.82)) ÷ (3.003 × 10⁻³ − 3.413 × 10⁻³) = –7.05 × 10³ K [1]. Ea = –gradient × R = 7.05 × 10³ × 8.314 [1] = 58.6 kJ mol⁻¹ [1]. (d) ln A = ln k + (Ea/R)(1/T) = –2.82 + 7049 × 3.413 × 10⁻³ [1] = 21.238 [1]. A = e^21.238 = 1.67 × 10⁹ dm³ mol⁻¹ s⁻¹ [1]. [9] Examiner insight: Carry the unrounded gradient into (d); small rounding in ln A changes A noticeably.

Where marks are usually lost

• Taking orders from equation coefficients instead of from the data.
• Treating a factor of 3 or 2.5 as a doubling.
• Units of k with the powers of mol and dm inverted.
• Forgetting to convert absorbance to concentration.
• Claiming first order from a single half-life.
• A slow step containing a zero-order reactant.
• Using T in °C, or kJ inside Ea/RT.

Next steps

• Recap the methods in the How fast? revision notes.
• Reread the worked examples in the How fast? study guide.
• Find related topics on the OCR A Level chemistry course page and track progress with the printable outcome list.
• Try one of our free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Source: the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, published by OCR. Section 5.1.1, How fast?, learning outcomes (a) to (k), in Module 5, Physical chemistry and transition elements.