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Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Rate equations & orders · Core · Paper 4 style · 6 marks
(d) Rate eq matches the stoichiometry 2NO + O₂ (a termolecular collision); a three-particle collision with correct orientation and energy is very improbable.
B1 orders match equation coefficients; B1 simultaneous collision of three particles very unlikely.
Q6Rate equations & orders · Challenge · Paper 5 style · 8 marks
A student investigates the reaction S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂ using an iodine clock. Each mixture contains 10.0 cm³ of K₂S₂O₈(aq), 5.0 cm³ of Na₂S₂O₃(aq), 1 cm³ of starch, a volume V of KI(aq) and enough water to make the total volume 26.0 cm³. The time t for the blue-black colour to appear is recorded.V / cm³: 10, 8, 6, 4, 2t / s: 22.0, 27.5, 36.7, 55.0, 110.0
(a)Explain why the total volume of each mixture is kept the same.[1]
(b)Calculate 1000/t for each experiment, giving values to 3 significant figures.[2]
(c)Explain why 1/t can be used as a measure of the initial rate.[1]
(d)Use your values to deduce the order of reaction with respect to iodide ions. Explain your answer.[2]
(e)State two variables, other than total volume, that must be kept constant.[2]
Show answer
(a) So that the volume of KI solution is proportional to [I⁻] in the mixture.
B1.
(b) 1000/t = 45.5, 36.4, 27.2, 18.2, 9.09 s⁻¹
B2 all five correct (B1 for three or four).
(c) Each run stops when the same fixed amount of iodine has been formed (just enough to use up all the thiosulfate), so rate ∝ amount/t ∝ 1/t.
B1 same (fixed) amount of I₂ formed/reacted before the blue-black colour appears in every run.
(d) First order
M1 1/t is directly proportional to volume of KI (e.g. halving volume halves 1/t; graph straight line through origin); A1 first order.
(e) Temperature; concentration/volume of peroxodisulfate (and of thiosulfate and starch).
1 mark each.
Q7Half-life & first-order kinetics · Core · Paper 4 style · 3 marks
The decomposition of dinitrogen pentoxide, 2N₂O₅(g) → 4NO₂(g) + O₂(g), is first order with a constant half-life of 24 minutes at a certain temperature.
(a)Calculate the fraction of the original N₂O₅ remaining after 96 minutes.[1]
(b)Calculate the rate constant, in min⁻¹, using k = 0.693/t½.[1]
(c)State how the half-life would change if the initial concentration of N₂O₅ were doubled. Explain.[1]
Show answer
(a) 1/16
B1 96/24 = 4 half-lives; (½)⁴.
(b) 0.0289 min⁻¹
B1 0.693/24.
(c) No change – the half-life of a first-order reaction is independent of the starting concentration.
B1.
Q8Half-life & first-order kinetics · Standard · Paper 4 style · 5 marks
Hydrogen peroxide decomposes in the presence of a catalyst. The concentration of H₂O₂ is measured over time.t / s: 0, 100, 200, 300, 400, 600[H₂O₂] / mol dm⁻³: 0.800, 0.566, 0.400, 0.283, 0.200, 0.100
(a)Use the data to show that the reaction is first order with respect to H₂O₂.[2]
(b)Calculate the rate constant, k, using k = 0.693/t½, and give its units.[2]
(c)Calculate the initial rate of reaction.[1]
Show answer
(a) Successive half-lives are constant: 0.800→0.400 in 200 s, 0.400→0.200 in 200 s, 0.200→0.100 in 200 s.
M1 two or more half-lives found from data; A1 they are equal (200 s), so first order.
(b) 3.47 × 10⁻³ s⁻¹ (accept 3.46 × 10⁻³ s⁻¹)
M1 k = 0.693/200 = 0.003465; A1 3.47 × 10⁻³ (0.00346–0.00347 accepted) with units s⁻¹.
Q9Half-life & first-order kinetics · Standard · Paper 4 style · 4 marks
2-chloro-2-methylpropane is hydrolysed by water in a first-order reaction: rate = k[(CH₃)₃CCl], with k = 4.62 × 10⁻⁴ s⁻¹ at 25 °C.
(a)Calculate the half-life of the reaction.[1]
(b)Calculate the time taken for the concentration of 2-chloro-2-methylpropane to fall to 12.5 % of its initial value.[1]
(c)Explain why the concentration of water does not appear in the rate equation, even though water is a reactant.[2]
Show answer
(a) 1500 s
B1 0.693 ÷ 4.62 × 10⁻⁴.
(b) 4500 s
B1 three half-lives.
(c) Water is the solvent and in large excess, so its concentration is effectively constant; and/or it is not involved in the slow step (SN1 – RDS is ionisation of the C–Cl bond).
B1 water in large excess/constant; B1 RDS involves only the halogenoalkane.
Q10Half-life & first-order kinetics · Challenge · Paper 4 style · 7 marks
Ammonia decomposes on the surface of a hot tungsten wire: 2NH₃(g) → N₂(g) + 3H₂(g).t / s: 0, 100, 200, 300, 400[NH₃] / 10⁻³ mol dm⁻³: 2.00, 1.60, 1.20, 0.80, 0.40
(a)Deduce the order of reaction with respect to ammonia. Explain your answer.[2]
(b)Calculate the rate constant and give its units.[2]
(c)Calculate the time for the first half-life and the time for the next half-life (from 1.00 × 10⁻³ to 0.50 × 10⁻³ mol dm⁻³).[2]
(d)Suggest why the rate does not depend on the concentration of ammonia in this experiment.[1]
Show answer
(a) Zero order
M1 concentration falls by the same amount (0.40 × 10⁻³) in each 100 s – constant rate/straight line; A1 zero order.
(d) The tungsten surface is fully covered with adsorbed ammonia (all active sites occupied), so extra NH₃ cannot react faster.
B1.
Q11Half-life & first-order kinetics · Challenge · Paper 4 style · 5 marks
A medicine is removed from the body by a first-order process with a half-life of 6.0 hours.
(a)Calculate the mass of the drug remaining in the body 24 hours after a 400 mg dose.[2]
(b)Calculate the first-order rate constant, in h⁻¹.[1]
(c)The drug is effective only while at least 60 mg remains. A doctor says a second dose is needed after 12 hours. Use a calculation to show whether this advice is sensible.[2]
Show answer
(a) 25 mg
M1 24/6.0 = 4 half-lives; A1 400 × (½)⁴ = 25 mg.
(b) 0.116 h⁻¹ (accept 0.115 h⁻¹)
B1 0.693/6.0 = 0.1155 → 0.116 h⁻¹.
(c) Sensible: after 12 h (2 half-lives) 100 mg remains, still above 60 mg; after 18 h only 50 mg would remain (< 60 mg), so the level falls below 60 mg between 12 h and 18 h (at about 16.4 h) and a second dose at 12 h keeps the drug effective.
M1 amounts at 12 h (100 mg) and 18 h (50 mg); A1 conclusion.
Q12Homogeneous & heterogeneous catalysis · Core · Paper 4 style · 6 marks
(a)Explain the difference between homogeneous and heterogeneous catalysis.[2]
(b)Classify each of the following as homogeneous or heterogeneous: (i) iron in the Haber process (ii) Fe²⁺ ions in the reaction of S₂O₈²⁻ with I⁻ (iii) platinum/rhodium in a catalytic converter (iv) NO₂ in the oxidation of atmospheric SO₂.[4]
Show answer
(a) Homogeneous: catalyst in the same phase as the reactants; heterogeneous: catalyst in a different phase.
B1 each.
(b) (i) heterogeneous (ii) homogeneous (iii) heterogeneous (iv) homogeneous
1 mark each.
Q13Homogeneous & heterogeneous catalysis · Standard · Paper 4 style · 5 marks
Peroxodisulfate ions oxidise iodide ions: S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂. The reaction is catalysed by Fe²⁺(aq).
(a)Explain why the uncatalysed reaction between S₂O₈²⁻ and I⁻ ions is slow.[1]
(b)Write two equations to show how Fe²⁺ ions catalyse the reaction.[2]
(c)Explain why Fe³⁺ ions can catalyse this reaction equally well.[1]
(d)State the property of iron that allows it to act as this catalyst.[1]
Show answer
(a) Both ions are negatively charged, so they repel each other and collisions with enough energy are rare (high Ea).
(c) The two steps occur in reverse order: Fe³⁺ first oxidises I⁻ to make Fe²⁺, which then reduces S₂O₈²⁻ – the catalyst is regenerated either way.
B1.
(d) It has variable oxidation states (+2 and +3).
B1.
Q14Homogeneous & heterogeneous catalysis · Standard · Paper 4 style · 5 marks
Cars on the roads of Lahore are fitted with catalytic converters containing platinum and rhodium.
(a)Write an equation for the reaction between carbon monoxide and nitrogen monoxide in a catalytic converter.[1]
(b)Describe the steps by which the solid catalyst speeds up this reaction.[3]
(c)Explain why the catalyst is spread as a thin coating on a ceramic honeycomb.[1]
Show answer
(a) 2CO + 2NO → 2CO₂ + N₂
B1.
(b) Reactant molecules adsorb onto active sites on the metal surface; bonds in reactants weaken and molecules are held close together in suitable orientation, so reaction occurs; products desorb, freeing the sites.
(c) Gives a large surface area so more active sites, and uses less expensive metal.
B1.
Q15Homogeneous & heterogeneous catalysis · Challenge · Paper 4 style · 7 marks
Over industrial cities, sulfur dioxide from burning fuel is oxidised in the air faster when nitrogen dioxide from vehicle engines is present.
(a)Write two equations to show how nitrogen dioxide catalyses the oxidation of sulfur dioxide to sulfur trioxide in the atmosphere.[2]
(b)Give the oxidation numbers of nitrogen in NO₂ and in NO, and of sulfur in SO₂ and in SO₃.[2]
(c)Explain why this is described as homogeneous catalysis and how the overall equation shows that NO₂ is a catalyst.[2]
(d)State why this reaction contributes to acid rain.[1]
Show answer
(a) SO₂ + NO₂ → SO₃ + NO; NO + ½O₂ → NO₂
B1 each.
(b) N: +4 → +2; S: +4 → +6
B1 N; B1 S.
(c) All species are gases (same phase); NO₂ is used in the first step and regenerated in the second, so it does not appear in the overall equation SO₂ + ½O₂ → SO₃.
B1 same phase; B1 regenerated/cancels.
(d) SO₃ reacts with water to form sulfuric acid, H₂SO₄.
B1.
Q16Mechanisms & rate-determining step · Core · Paper 4 style · 4 marks
The reaction between nitrogen dioxide and carbon monoxide occurs in two steps:Step 1 (slow): NO₂ + NO₂ → NO₃ + NOStep 2 (fast): NO₃ + CO → NO₂ + CO₂
(a)Write the overall equation for the reaction.[1]
(b)Deduce the rate equation. Explain your answer.[2]
(c)Identify the intermediate in this mechanism.[1]
Show answer
(a) NO₂ + CO → NO + CO₂
B1 add steps and cancel NO₃ and one NO₂.
(b) rate = k[NO₂]²
M1 step 1 is slow/rate-determining; A1 involves two NO₂, CO not in RDS.
(c) NO₃
B1.
Q17Mechanisms & rate-determining step · Standard · Paper 4 style · 5 marks
2-bromo-2-methylpropane reacts with aqueous sodium hydroxide. The rate equation is rate = k[(CH₃)₃CBr].
(a)State the order of reaction with respect to hydroxide ions and explain what this shows about the mechanism.[2]
(b)Describe the two-step mechanism, naming the intermediate.[2]
(c)Predict the effect on the initial rate of doubling [(CH₃)₃CBr] and halving [OH⁻] at the same time.[1]
Show answer
(a) Zero order; OH⁻ is not involved in the rate-determining step, which is the slow ionisation of the C–Br bond.
B1 slow step contains one H₂ and one ICl; B1 fast step uses second ICl, steps add to overall equation; B1 HI intermediate.
(b) Orders come from the rate-determining step, which need not match the stoichiometry; they must be found by experiment.
B1.
(c) No change
B1 3 × ⅓ = 1.
Q19Temperature & Boltzmann distribution · Core · Paper 2 style · 6 marks
A mixture of gases reacts with activation energy Ea.
(a)Describe the Boltzmann distribution curve for a gas at temperature T₁. Refer to the axes and the start of the curve.[2]
(b)Describe how the curve changes at a higher temperature T₂.[2]
(c)Use the Boltzmann distribution to explain why the rate of reaction increases with temperature.[2]
Show answer
(a) y-axis: number/proportion of molecules; x-axis: energy. Curve starts at the origin, rises to a peak (most probable energy) and falls gradually, never touching the energy axis.
B1 axes labelled; B1 starts at origin, single peak, asymptotic to x-axis.
(b) Peak moves to higher energy and is lower; curve is broader with more molecules at high energy; area under curve unchanged.
B1 peak lower and to the right; B1 same area / more molecules beyond Ea.
(c) A larger proportion of molecules have energy ≥ Ea, so more frequent successful collisions (also slightly more frequent collisions).
B1 more molecules with E ≥ Ea; B1 more successful/effective collisions per unit time.
Q20Temperature & Boltzmann distribution · Standard · Paper 1 style · 1 mark
(a)The temperature of a sample of gas is increased. Which statement about its Boltzmann distribution is correct?[1]
Show answer
(a) C The peak becomes lower and moves to higher energy.
At higher temperature the most probable energy increases and the peak falls; the area (total number of molecules) stays the same, so A is wrong; Ea is unchanged by temperature.
Q21Rate measurement & graphs · Core · Paper 3 style · 6 marks
Marble chips (calcium carbonate, in excess) are added to hydrochloric acid in a conical flask on a balance. The loss in mass, caused by carbon dioxide escaping, is recorded.t / s: 0, 30, 60, 90, 120, 150, 180, 210, 240loss in mass / g: 0.00, 0.44, 0.78, 1.02, 1.18, 1.26, 1.30, 1.32, 1.32
(a)Calculate the amount, in mol, of carbon dioxide produced in total.[1]
(b)Calculate the mean rate of reaction, in g s⁻¹, over the first 30 s and over the period 60 s to 90 s.[2]
(c)Explain, in terms of particles, why the rate decreases during the reaction.[2]
(d)Suggest why a loose plug of cotton wool is placed in the neck of the flask.[1]
Show answer
(a) 0.0300 mol
B1 1.32/44.0.
(b) 0.0147 g s⁻¹ and 0.00800 g s⁻¹
B1 0.44/30; B1 0.24/30.
(c) Concentration of acid falls (and chip surface shrinks), so fewer particles per unit volume and fewer collisions per second.
B1 concentration decreases; B1 collision frequency falls.
(d) Lets CO₂ escape but stops acid spray being lost (which would add to the mass loss).
B1.
Q22Rate measurement & graphs · Standard · Paper 5 style · 7 marks
A student plans to find how the concentration of hydrochloric acid affects the rate of its reaction with magnesium ribbon, collecting the hydrogen in a 100 cm³ gas syringe.
(a)Identify the independent variable, the dependent variable and two control variables.[3]
(b)A tangent drawn to the curve at t = 20 s passes through the points (0 s, 12 cm³) and (40 s, 52 cm³). Calculate the rate at 20 s.[2]
(c)Suggest one source of error in measuring the volume of gas at the start, and an improvement.[2]
Show answer
(a) Independent: concentration of HCl; dependent: volume of H₂ (with time); control: mass/length of Mg, volume of acid, temperature.
B1 independent; B1 dependent; B1 two control variables.
(b) 1.0 cm³ s⁻¹
M1 gradient = (52 − 12)/(40 − 0); A1 1.0 cm³ s⁻¹.
(c) Gas lost while inserting the bung after adding Mg; place Mg in a small tube inside the flask and tip it over after sealing (or use a divided flask).
B1 error; B1 matching improvement.
Q23Catalysis · Core · Paper 1 style · 1 mark
(a)Which statement about the effect of a catalyst on a reversible reaction is correct?[1]
Show answer
(a) C It increases the rates of the forward and backward reactions equally.
A catalyst provides an alternative route with lower Ea for both directions, so equilibrium is reached faster but its position is unchanged. B is wrong because the reverse Ea is lowered by the same amount.
Q24Catalysis · Standard · Paper 2 style · 6 marks
Hydrogen peroxide decomposes: 2H₂O₂(aq) → 2H₂O(l) + O₂(g), ΔH = −98 kJ per mol of H₂O₂. The activation energy is 75 kJ mol⁻¹ without a catalyst and 23 kJ mol⁻¹ with the enzyme catalase (found in potato and liver).
(a)Calculate the activation energy of the reverse reaction for the uncatalysed and the catalase-catalysed reactions.[2]
(b)Use the Boltzmann distribution to explain how a catalyst increases the rate of a reaction.[3]
(c)State how the catalyst affects the enthalpy change of the reaction.[1]
Show answer
(a) 173 kJ mol⁻¹ and 121 kJ mol⁻¹
B1 75 + 98; B1 23 + 98.
(b) Catalyst provides an alternative pathway with lower Ea; on the Boltzmann curve the lower Ea lies further to the left; so a greater proportion of molecules have E ≥ Ea and more collisions are successful per unit time.
B1 alternative route lower Ea; B1 more molecules have energy ≥ Ea (shaded area larger); B1 more effective collisions per unit time.
(c) No effect – ΔH stays −98 kJ mol⁻¹.
B1.
Q25Rate constant & temperature · Standard · Paper 4 style · 5 marks
For a reaction, k = 2.0 × 10⁻³ s⁻¹ at 300 K and k = 8.0 × 10⁻³ s⁻¹ at 320 K.
(a)State the Arrhenius equation and explain what happens to k as temperature increases.[2]
(b)Using ln k = ln A − Ea/RT, calculate the activation energy, in kJ mol⁻¹. [R = 8.31 J K⁻¹ mol⁻¹][3]
Show answer
(a) k = Ae^(−Ea/RT); as T increases, −Ea/RT becomes less negative so k increases (exponentially).
(d) Extrapolate to find the y-intercept, which equals ln A; A = e^(intercept).
B1.
Q27Collision theory (concentration, pressure, surface area) · Core · Paper 2 style · 5 marks
In a fertiliser factory near Multan, the rates of several gas and solution reactions are controlled carefully.
(a)Explain why increasing the pressure increases the rate of a reaction between two gases.[2]
(b)Explain why powdered zinc reacts faster with dilute sulfuric acid than the same mass of zinc granules.[2]
(c)State why increasing concentration does not change the proportion of collisions that are successful.[1]
Show answer
(a) Same number of molecules in a smaller volume / more molecules per unit volume; so collisions are more frequent (more successful collisions per second).
B1 each.
(b) Powder has a larger surface area exposed to acid; so more frequent collisions between acid particles and zinc atoms.
B1 each.
(c) Concentration does not change the energy of the particles, so the fraction with E ≥ Ea is unchanged.
B1.
Q28Collision theory (concentration, pressure, surface area) · Standard · Paper 1 style · 1 mark
(a)Which change increases the frequency of collisions between reacting particles but does NOT change the fraction of collisions that have energy greater than the activation energy?[1]
Show answer
(a) C increasing the concentration of a reactant in solution
Higher concentration means more particles per unit volume, so more frequent collisions, but the energy distribution is unchanged. B raises both the collision frequency and the fraction with E ≥ Ea.
Q29Activation energy & rate · Core · Paper 2 style · 4 marks
Milk turns sour because of reactions caused by bacterial enzymes.
(a)Define the term activation energy.[1]
(b)A rough rule says the rate of a reaction doubles for every 10 °C rise. Calculate the factor by which the rate increases when the temperature of milk rises from 5 °C (in a fridge) to 35 °C (on a hot day).[1]
(c)Explain why a small rise in temperature causes a large increase in rate, even though the mean kinetic energy of the particles rises only a little.[2]
Show answer
(a) The minimum energy that colliding particles must have for a reaction to occur.
B1.
(b) 8 times
B1 three 10 °C rises: 2³ = 8.
(c) The Boltzmann distribution shifts so that the number of particles with energy ≥ Ea increases by a large proportion; only these particles react, so the number of successful collisions per second increases greatly.
B1 many more particles with E ≥ Ea; B1 more successful collisions per unit time.
Q30Rate & equilibrium (rate of forward/back reactions) · Standard · Paper 2 style · 8 marks
Hydrogen and iodine are sealed in a flask at 450 °C: H₂(g) + I₂(g) ⇌ 2HI(g).
(a)Describe how the rates of the forward and backward reactions change from the moment H₂ and I₂ are mixed until equilibrium is reached.[3]
(b)State the effect of adding a catalyst on (i) the rates of the forward and backward reactions (ii) the position of equilibrium (iii) the time taken to reach equilibrium.[3]
(c)The forward reaction is exothermic. Explain why raising the temperature decreases the equilibrium yield of HI even though both rates increase.[2]
Show answer
(a) Forward rate is greatest at the start and decreases as [H₂] and [I₂] fall; backward rate starts at zero and increases as [HI] rises; at equilibrium the two rates are equal (and non-zero).
B1 forward decreases; B1 backward increases from zero; B1 equal at equilibrium.
(b) (i) both increase by the same factor (ii) no change (iii) decreases
1 mark each.
(c) The backward (endothermic) reaction has the larger Ea, so its rate increases by a greater proportion; equilibrium shifts left.
B1 endothermic direction favoured/its rate increases more; B1 equilibrium shifts to the left.
Similar questions
Questions that test the same subtopic in different years, grouped so you can work on one skill at a time.
References only: find each question in its paper (official link where available, otherwise through your school).
