Edexcel International A Level Chemistry: Kinetics (YCH11) – 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.

This set covers Topic 11: Kinetics of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, outcomes 11.1 to 11.13, including Core Practicals 9a, 9b and 10. Topic 11 belongs to Unit 4: Rates, Equilibria and Further Organic Chemistry, an International A2 (IA2) unit. Work through the study guide and revision notes before you start; the checklist and course hub list other topics.

Use R = 8.31 J mol⁻¹ K⁻¹ wherever needed. The Arrhenius equation is k = Ae^(−Ea/RT).

Questions

1. Define each term.

(a) Order of reaction with respect to a substance [1] (b) Rate constant [1] (c) Half-life [1]

2. For a reaction, rate = k[C][D]².

(a) State the overall order. [1] (b) Deduce the units of k. [1] (c) At one temperature, k = 160 in these units. Calculate the rate when [C] = 0.025 mol dm⁻³ and [D] = 0.040 mol dm⁻³. [2] (d) [C] is multiplied by 1.5 and [D] is doubled. State the factor by which the rate changes. [1]

3. Methyl methanoate is hydrolysed in aqueous acid: HCOOCH₃ + H₂O → HCOOH + CH₃OH. Water is the solvent. Use these data.

Experiment	[HCOOCH₃] / mol dm⁻³	[H⁺] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.40	0.10	2.2 × 10⁻⁵
2	0.80	0.10	4.4 × 10⁻⁵
3	0.80	0.25	1.1 × 10⁻⁴
4	0.20	0.30	to find

(a) Deduce the order with respect to HCOOCH₃ and to H⁺, explaining your reasoning. [2] (b) Give the rate equation for the hydrolysis. [1] (c) Use experiment 1 to find the value and units of k. [2] (d) Predict the initial rate for experiment 4. [1] (e) Suggest why the order with respect to water cannot be found from these experiments. [1]

4. Hydrogen peroxide decomposes in the presence of a catalyst: 2H₂O₂ → 2H₂O + O₂. Use these data.

t / min	0	3	6	9	12	18
[H₂O₂] / mol dm⁻³	0.800	0.566	0.400	0.283	0.200	0.100

(a) Use two half-lives to show that the reaction is first order in H₂O₂. [2] (b) A tangent at t = 6 min gives a rate of 0.046 mol dm⁻³ min⁻¹. Calculate k, with units. [2] (c) Suggest a technique for collecting these data continuously. [1]

5. Initial rates were measured at three concentrations of reactant K.

[K] / mol dm⁻³	0.10	0.20	0.30
Rate / mol dm⁻³ s⁻¹	1.5 × 10⁻⁴	6.0 × 10⁻⁴	1.35 × 10⁻³

(a) Deduce the order with respect to K and describe the shape of the rate-concentration graph. [2] (b) Another reactant, L, is zero order. Describe its concentration-time graph and how its half-life changes as the reaction proceeds. [2]

6. Select and justify a technique to obtain rate data for each reaction.

(a) MgCO₃(s) + 2HNO₃(aq) → Mg(NO₃)₂(aq) + H₂O(l) + CO₂(g) [2] (b) 2MnO₄⁻(aq) + 5C₂O₄²⁻(aq) + 16H⁺(aq) → 2Mn²⁺(aq) + 10CO₂(g) + 8H₂O(l), with excess ethanedioate [2] (c) CH₃COOC₂H₅(aq) + OH⁻(aq) → CH₃COO⁻(aq) + C₂H₅OH(aq) [2]

7. In Core Practical 9a, propanone and sulfuric acid are in large excess. 5.00 cm³ samples are run into sodium hydrogencarbonate solution and titrated with 0.00500 mol dm⁻³ sodium thiosulfate. I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻.

Time / min	2	6	10	14
Titre / cm³	21.60	19.20	16.80	14.40

(a) Explain why the samples are run into sodium hydrogencarbonate. [1] (b) Use the titres to deduce the order with respect to iodine. [2] (c) Calculate the rate of reaction of iodine in mol dm⁻³ min⁻¹. [3] (d) Rate = k[CH₃COCH₃][H⁺]. State what this shows about the rate-determining step and explain why H⁺ is described as a catalyst. [2]

8. Two halogenoalkanes are hydrolysed by aqueous hydroxide ions. For 2-chloro-2-methylbutane, rate = k[C₅H₁₁Cl]. For 1-bromopropane, rate = k[C₃H₇Br][OH⁻].

(a) Name the mechanism for each and explain how each rate equation supports it. [4] (b) State the effect on each rate of doubling [OH⁻]. [1]

9. Hydrogen reacts with iodine monochloride: H₂(g) + 2ICl(g) → I₂(g) + 2HCl(g). Rate = k[H₂][ICl].

(a) State which species, and how many of each, are involved up to and including the rate-determining step. [1] (b) Propose a two-step mechanism consistent with the rate equation and the overall equation, labelling the slow step. [3]

10. In Core Practical 10, a student times a clock reaction at four temperatures, using the same mixture each time.

Temperature / °C	25	35	45	55
t / s	96	51	28	16

(a) Explain why 1/t can be used in place of k. [1] (b) Calculate 1/T (in K⁻¹) and ln(1/t) for each run. [2] (c) Use the gradient of a graph of ln(1/t) against 1/T to calculate Ea in kJ mol⁻¹. [3] (d) Suggest one way to keep the temperature of each run steady. [1]

11. This question is about catalysts.

(a) Classify each as homogeneous or heterogeneous: H⁺(aq) in the iodination of propanone; solid iron in the reaction of nitrogen and hydrogen gases. [2] (b) Explain how a solid catalyst increases the rate of a gas-phase reaction. [3]

12. In a Harcourt-Esson clock, a student varies the volume of potassium iodide solution, adding water so the total volume is 50 cm³ every time.

Volume of KI / cm³	8	12	16	20
t / s	75	50	38	30

(a) Explain why water is added. [1] (b) Deduce the order with respect to iodide ions. [2] (c) State the role of the sodium thiosulfate in the mixture. [1]

Answers

1. (a) The exponent on that substance’s concentration term in the rate equation [1]. (b) The k linking rate to the concentration terms, constant at a fixed temperature [1]. (c) The time for a reactant’s concentration to drop to 50% of a chosen value [1]. Examiner insight: Link order to the rate equation, not to the balanced equation.

2. (a) 3 [1] (b) mol⁻² dm⁶ s⁻¹ [1] (c) Rate = 160 × 0.025 × 0.040² [1] = 6.4 × 10⁻³ mol dm⁻³ s⁻¹ [1]. (d) 1.5 × 2² = 6 [1] Examiner insight: Square the factor for [D] before multiplying; adding factors is a common slip.

3. (a) Experiments 1 and 2: [ester] doubles, rate doubles, so first order in ester [1]; experiments 2 and 3: [H⁺] × 2.5, rate × 2.5, so first order in H⁺ [1]. (b) Rate = k[HCOOCH₃][H⁺] [1] (c) k = 2.2 × 10⁻⁵ ÷ (0.40 × 0.10) = 5.5 × 10⁻⁴ [1]; mol⁻¹ dm³ s⁻¹ [1]. (d) 5.5 × 10⁻⁴ × 0.20 × 0.30 = 3.3 × 10⁻⁵ mol dm⁻³ s⁻¹ [1] (e) Water is the solvent, in such large excess that its concentration stays effectively constant [1]. Examiner insight: Quote the experiments you compared.

4. (a) 0.800 → 0.400 takes 6 min [1]; 0.400 → 0.200 takes another 6 min (or 0.200 → 0.100 takes 6 min), so t½ is constant and the reaction is first order [1]. (b) k = rate ÷ [H₂O₂] = 0.046 ÷ 0.400 [1] = 0.115 min⁻¹ [1]. (c) Measure the volume of oxygen given off with a gas syringe [1]. Examiner insight: Show two separate half-lives; one alone cannot show they are constant.

5. (a) [K] doubles, rate × 4; [K] triples, rate × 9, so second order [1]; the graph is an upward curve through the origin [1]. (b) A straight line with a negative gradient [1]; the half-life gets shorter as [L] falls [1]. Examiner insight: Show the numerical factor for each change.

6. (a) Mass change on a balance [1]; CO₂ is dense, so the mass lost is large enough to measure [1]. (b) Colorimetry [1]; purple MnO₄⁻ is the only strongly coloured species, so absorbance falls as it reacts [1]. (c) Conductivity [1]; mobile OH⁻ ions are replaced by less conductive ethanoate ions, so conductivity falls [1]. (Titrating quenched samples with standard acid is also acceptable.) Examiner insight: Each justification needs the property that changes.

7. (a) It neutralises the acid catalyst, stopping the reaction at that time [1]. (b) The titre falls by 2.40 cm³ in each 4 min interval [1]; the graph is a straight line, so the rate is constant and independent of [I₂] [1]. (c) Gradient = 2.40 ÷ 4 = 0.600 cm³ min⁻¹ [1]; moles of I₂ per minute = 0.600 × 10⁻³ × 0.00500 ÷ 2 = 1.50 × 10⁻⁶ mol min⁻¹ [1]; rate = 1.50 × 10⁻⁶ ÷ 0.00500 dm³ = 3.00 × 10⁻⁴ mol dm⁻³ min⁻¹ [1]. (d) The rate-determining step involves one propanone and one H⁺; iodine reacts after it [1]. H⁺ is used in one step and regenerated later [1]. Examiner insight: Remember the 1 : 2 ratio between iodine and thiosulfate, and divide by the sample volume in dm³.

8. (a) Tertiary chloride: SN1 [1]; only the halogenoalkane is in the slow step, forming a carbocation [1]. 1-bromopropane: SN2 [1]; both the halogenoalkane and OH⁻ are in the single step, via one transition state [1]. (b) Tertiary unchanged; primary doubles [1]. Examiner insight: Tie each mechanism to the species in its rate equation.

9. (a) One H₂ and one ICl [1]. (b) Step 1 (slow): H₂ + ICl → HI + HCl [1][1]; step 2 (fast): HI + ICl → I₂ + HCl [1]. Examiner insight: Add your steps to check they give the overall equation; other mechanisms that fit both equations are equally valid.

10. (a) The same amount of reaction happens in each run, so rate ∝ 1/t, and k ∝ rate at fixed concentrations [1]. (b) 1/T: 3.356, 3.247, 3.145, 3.049 (× 10⁻³ K⁻¹) [1]; ln(1/t): −4.56, −3.93, −3.33, −2.77 [1]. (c) Gradient = (−2.77 − (−4.56)) ÷ ((3.049 − 3.356) × 10⁻³) [1] = −5830 K [1]; Ea = 5830 × 8.31 = 48 400 J mol⁻¹ ≈ 48 kJ mol⁻¹ [1]. (d) Warm the solutions separately in a thermostatic water bath before mixing [1]. Examiner insight: Take the gradient from points far apart on your best-fit line, and convert J to kJ at the end.

11. (a) H⁺(aq): homogeneous [1]; iron: heterogeneous [1]. (b) Reactant gases bond weakly (adsorb) at active sites on the solid [1]; bonds weaken and molecules are held close together, lowering the activation energy [1]; products desorb, freeing the sites [1]. Examiner insight: Use the words adsorb and desorb.

12. (a) To keep the total volume constant, so [I⁻] is proportional to the volume of KI used [1]. (b) 1000/t ÷ volume ≈ 1.67, 1.67, 1.64, 1.67: rate is proportional to [I⁻] [1]; first order in iodide [1]. (c) It removes the iodine formed, delaying the blue-black colour until a fixed amount of reaction has occurred [1]. Examiner insight: Use all four runs, not just one pair.

Where marks are usually lost

• Reading orders from the balanced equation.
• Leaving out the units of k, or using units for the wrong overall order.
• Forgetting the 1 : 2 iodine to thiosulfate ratio in Core Practical 9a.
• Using °C instead of K when calculating 1/T.
• Missing the negative sign of the gradient, or quoting J mol⁻¹ as kJ mol⁻¹.
• Proposing mechanism steps that do not add up to the overall equation.

Next steps

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Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Pearson Education Limited – Unit 4, Topic 11: Kinetics, outcomes 11.1 to 11.13.