Edexcel International A Level Chemistry: Introduction to kinetics and equilibria (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 9: Introduction to Kinetics and Equilibria of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, outcomes 9.1 to 9.11. This IAS topic comes from Unit 2: Energetics, Group Chemistry, Halogenoalkanes and Alcohols. Read the study guide and revision notes before you try this set. The checklist, course hub and diagnostics help you track gaps.

Relative atomic masses to use: C = 12.0, O = 16.0, H = 1.0.

Questions

1. An effervescent vitamin tablet fizzes in water as an acid and a carbonate in it react. Explain, using collision theory, why the fizzing is faster if:

(a) the tablet is crushed into powder first [2] (b) the water is warm instead of cold. [2]

2. A pile of dry wood in a shed does not burn at room temperature, but it keeps burning once it has been lit with a match. Explain these observations in terms of activation energy. [2]

3. Equal lengths of magnesium ribbon are added to excess hydrochloric acid of different concentrations at one temperature. The time for the ribbon to disappear is recorded.

[HCl] / mol dm⁻³	0.50	1.00	1.50	2.00
Time / s	160	82	54	41

(a) Calculate the rate, as 1/time, for each run. Give units. [2] (b) Use your answers to deduce how the rate depends on the acid concentration. [1]

4. Hydrogen peroxide solution decomposes when manganese(IV) oxide is added: 2H₂O₂(aq) → 2H₂O(l) + O₂(g). The volume of oxygen is plotted against time. The tangent at t = 0 goes through (0 s, 0 cm³) and (20 s, 36 cm³). The tangent at t = 60 s goes through (0 s, 30 cm³) and (120 s, 78 cm³).

(a) Calculate the initial rate of reaction, with units. [2] (b) Calculate the rate at 60 s. [1] (c) Explain why the rate at 60 s is lower than the initial rate. [2]

5. A sample of gas is warmed from T₁ to a higher temperature T₂. Describe how the Maxwell-Boltzmann distribution at T₂ differs from that at T₁, and use it to explain why the rate of reaction increases. [4]

6. A gas reaction has an uncatalysed activation energy of 150 kJ mol⁻¹ and ΔH = −60 kJ mol⁻¹. With a catalyst the reaction takes place in two steps. Step 1 has a barrier 52 kJ mol⁻¹ above the reactants and forms an intermediate 20 kJ mol⁻¹ above the reactants. Step 2 has a barrier 45 kJ mol⁻¹ above the intermediate.

(a) Deduce the activation energy of the catalysed route. [2] (b) Calculate how much lower this is than the uncatalysed activation energy. [1] (c) Describe the reaction profile for the catalysed route, stating the energy of each feature relative to the reactants. [2]

7. Peroxodisulfate ions oxidise iodide ions slowly in aqueous solution: S₂O₈²⁻(aq) + 2I⁻(aq) → 2SO₄²⁻(aq) + I₂(aq). A few drops of iron(III) solution make the reaction much faster. Use the Maxwell-Boltzmann distribution to explain how the iron ions increase the rate at constant temperature. [3]

8. Methanol can be made over a copper-based catalyst by two routes.

Route A, from carbon dioxide: CO₂ + 3H₂ → CH₃OH + H₂O Route B, from carbon monoxide: CO + 2H₂ → CH₃OH

(a) Calculate the atom economy of route A for making methanol. [2] (b) State the atom economy of route B, and give a reason. [1] (c) Explain two ways in which catalysts help make industrial processes more sustainable. [2]

9. A sealed glass bulb at 450 °C holds an equilibrium mixture of hydrogen, iodine vapour and hydrogen iodide: H₂(g) + I₂(g) ⇌ 2HI(g). Iodine vapour is purple; the other two gases are colourless.

(a) The purple colour does not change over time. Explain, in terms of rates and concentrations, why the system is in dynamic equilibrium. [2] (b) More hydrogen is injected at constant temperature and volume. Predict and explain the change in colour. [2]

10. Phenylethene (styrene) is made by removing hydrogen from ethylbenzene over a catalyst. The forward reaction is endothermic.

C₆H₅CH₂CH₃(g) ⇌ C₆H₅CH=CH₂(g) + H₂(g)

Predict and justify the effect on the equilibrium yield of phenylethene of:

(a) raising the temperature [2] (b) raising the pressure [2] (c) removing hydrogen from the mixture as it forms. [2]

11. Gas Z is made over a catalyst: X(g) + 2Y(g) ⇌ Z(g). The forward reaction is exothermic. The data are illustrative.

Temperature / °C	200	250	300	350
Yield of Z at 5 MPa / %	38	24	13	6
Yield of Z at 20 MPa / %	62	47	31	18
Time to reach equilibrium / min	240	45	10	2

(a) How does temperature affect the yield of Z? Explain your answer. [2] (b) Explain the effect of pressure on the yield of Z. [2] (c) A manufacturer chooses 300 °C. Use the data to justify this as a compromise. [2] (d) Suggest one reason why the manufacturer might choose 5 MPa, even though 20 MPa gives a higher yield. [1] (e) State the effect of the catalyst on the yield of Z and on the time taken to reach equilibrium. [1]

Answers

1. (a) Crushing gives a larger surface area, so more acid and carbonate particles are exposed to the water [1]. Collisions between the reacting particles become more frequent, so the rate increases [1]. (b) In warm water the particles have more kinetic energy, so a larger fraction of collisions reach the activation energy [1]. Successful collisions happen more often [1]. Examiner insight: For (b), collision frequency alone usually scores less than the fraction of particles with at least Ea.

2. At room temperature very few molecules have energy at or above the activation energy, so the rate is negligible [1]. The match supplies energy to overcome Ea; the exothermic combustion then releases enough heat to keep nearby molecules above Ea [1]. Examiner insight: “The wood needs energy”, with no mention of activation energy, tends to miss the first mark.

3. (a) 1/160 = 0.00625; 1/82 = 0.0122; 1/54 = 0.0185; 1/41 = 0.0244 [1]; unit s⁻¹ [1]. (b) Rate ÷ concentration is roughly constant (about 0.0122 to 0.0125), so rate is directly proportional to [HCl] [1]. Examiner insight: “Rate increases with concentration” is too vague for the deduction mark; show the constant ratio or that doubling concentration doubles the rate.

4. (a) Gradient = (36 − 0) ÷ (20 − 0) [1] = 1.8 cm³ s⁻¹ [1]. (b) (78 − 30) ÷ (120 − 0) = 0.40 cm³ s⁻¹ [1]. (c) Hydrogen peroxide is used up, so its concentration falls [1]. Collisions between H₂O₂ molecules and the catalyst surface become less frequent, so the rate falls [1]. Examiner insight: Take both points from the tangent; points on the curve give an average rate, not the rate at that instant.

5. At T₂ the peak is lower [1] and at a higher energy [1]. The area under the curve is unchanged, as the number of molecules is the same [1]. A much greater proportion of molecules have energy at or above Ea, so more collisions succeed and the rate increases [1]. Examiner insight: Mark Ea on the energy axis; without it, the curve is not linked to the rate.

6. (a) Second barrier = 20 + 45 = 65 kJ mol⁻¹ above the reactants [1]. This is higher than the first barrier (52), so catalysed Ea = 65 kJ mol⁻¹ [1]. (b) 150 − 65 = 85 kJ mol⁻¹ [1]. (c) First hump at +52, dip for the intermediate at +20 and second, higher hump at +65 [1]; products at −60, the same level as for the uncatalysed route [1]. Examiner insight: Ea is taken to the highest point on the catalysed route; choosing the first barrier gives 52, which is wrong here.

7. The iron ions provide an alternative route with a lower activation energy [1]. The distribution is unchanged at constant temperature, but the lower Ea lies further left, so more ions have energy at or above it [1]. A greater proportion of collisions succeed, giving more successful collisions per unit time [1]. Examiner insight: Saying the catalyst “gives the ions more energy” loses credit.

8. (a) Mr(CH₃OH) = 32.0; Mr(H₂O) = 18.0; sum of products = 32.0 + 18.0 = 50.0 [1]. Atom economy = 32.0 ÷ 50.0 × 100 = 64.0% [1]. (b) 100%, because methanol is the only product [1]. (c) Lower Ea allows a useful rate at a lower temperature, so less fuel is used [1]. A catalyst can allow a route with fewer by-products, so atom economy is higher and there is less waste [1]. Examiner insight: Include water in the denominator; omitting it wrongly gives 100% for route A.

9. (a) The forward and backward reactions continue at equal rates [1], so the concentrations of H₂, I₂ and HI stay constant and the colour does not change [1]. (b) The purple colour becomes paler [1]. The added hydrogen moves the position to the right to use it up, and this also uses up iodine [1]. Examiner insight: Writing that the concentrations are “equal” loses the mark; they are constant.

10. (a) The forward reaction is endothermic [1], so raising the temperature moves the position to the right to absorb the heat; the yield of phenylethene increases [1]. (b) 1 gas molecule on the left and 2 on the right [1]. Raising the pressure moves the position to the left, the side with fewer gas molecules; the yield decreases [1]. (c) Removing a product moves the position right to replace it [1]; the yield increases [1]. Examiner insight: Count gas molecules from the coefficients; “more molecules on the right” without numbers can lose the justification mark.

11. (a) As temperature rises the yield falls, for example from 38% to 6% at 5 MPa [1]. The forward reaction is exothermic, so higher temperature moves the position to the left, in the endothermic direction [1]. (b) Higher pressure gives a higher yield at every temperature [1], because the right side has fewer gas molecules (1 compared with 3) [1]. (c) At 300 °C equilibrium takes 10 minutes, against 240 minutes at 200 °C, so the rate is much faster [1]. The yield (13% or 31%) is still higher than at 350 °C, so 300 °C balances yield against rate [1]. (d) Any one: high pressure needs stronger, more expensive equipment; more energy is used to compress the gases; higher pressure is a greater safety risk [1]. (e) No effect on the yield; equilibrium is reached in a shorter time [1]. Examiner insight: “Use the data” means quote figures from the table; general statements without numbers usually score only part of the marks.

Where marks are usually lost

• Writing “more collisions” instead of “more frequent collisions”.
• Explaining a temperature effect by collision frequency only.
• Taking tangent readings from the curve itself, or leaving out units.
• Measuring the catalysed Ea to the first hump or the intermediate, not the highest point.
• Drawing a new Maxwell-Boltzmann curve for a catalyst.
• Leaving a by-product such as water out of an atom economy calculation.
• Saying concentrations are “equal” at equilibrium.
• Predicting a shift with no reason.
• Claiming a catalyst increases the yield.
• Not quoting figures in a “use the data” question.

Next steps

• Revise with the kinetics and equilibria revision notes.
• Reread any weak area in the kinetics and equilibria study guide.
• Browse the Edexcel International A Level Chemistry hub.
• Tick off outcomes on the printable checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Topic 9: Introduction to Kinetics and Equilibria (outcomes 9.1 to 9.11). Published by Pearson Education Limited.