2022 May/June · Paper 13 · Q101 markHess's law cycles
2022 May/June · Paper 21 · Q1 (f(ii))3 marksReaction pathway diagrams & activation energy
2022 May/June · Paper 22 · Q1 (b(iii), b(iv))5 marksEnthalpy calculations from experimental data · Standard enthalpy changes (definitions)
2022 May/June · Paper 22 · Q3 (b)2 marksBond energies
2022 May/June · Paper 23 · Q3 (a(ii))2 marksBond energies
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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.
Q1Calorimetry (q = mcΔT), neutralisation & combustion · Core · Paper 2 style · 5 marks
A student in a Lahore college lab mixes 50.0 cm³ of 1.00 mol/dm³ hydrochloric acid with 50.0 cm³ of 1.00 mol/dm³ sodium hydroxide in a polystyrene cup. Both solutions start at 24.3 °C. The highest temperature reached is 31.0 °C.
(a)Calculate the energy released, in J, in this experiment. Assume the density of the mixture is 1.00 g/cm³ and its specific heat capacity is 4.18 J g⁻¹ K⁻¹.[2]
(b)Calculate the enthalpy change of neutralisation, in kJ/mol. Give your answer to 3 significant figures.[2]
(c)Suggest why the experiment is carried out in an expanded polystyrene cup with a lid.[1]
(c) Polystyrene is a good insulator and the lid reduces heat loss to the surroundings.
B1 reduces heat loss / good insulator
Q2Calorimetry (q = mcΔT), neutralisation & combustion · Core · Paper 1 style · 1 mark
(a)25.0 cm³ of 2.00 mol/dm³ KOH(aq) is mixed with 25.0 cm³ of 2.00 mol/dm³ HNO₃(aq). The enthalpy change of neutralisation is −57.0 kJ/mol. Assume the mixture has a density of 1.00 g/cm³ and a specific heat capacity of 4.18 J g⁻¹ K⁻¹. What is the temperature rise?[1]
Show answer
(a) B 13.6 °C
n(H₂O) = 0.0500 mol, q = 0.0500 × 57 000 = 2850 J; ΔT = 2850 ÷ (50.0 × 4.18) = 13.6 °C. A (6.8) uses 100 g as the mass; C uses 25 g of solution only.
Q3Calorimetry (q = mcΔT), neutralisation & combustion · Core · Paper 1 style · 1 mark
(a)0.0100 mol of a liquid fuel is burned completely. The energy released raises the temperature of 100 g of water by 30.0 K. The specific heat capacity of water is 4.18 J g⁻¹ K⁻¹ and there is no heat loss. What is the enthalpy change of combustion of the fuel?[1]
Show answer
(a) C −1250 kJ/mol
q = 100 × 4.18 × 30.0 = 12 540 J = 12.54 kJ; ΔH = −12.54 ÷ 0.0100 = −1254 kJ/mol ≈ −1250 kJ/mol. D has the wrong sign: combustion is exothermic. A forgets to divide by the moles.
Q4Calorimetry (q = mcΔT), neutralisation & combustion · Standard · Paper 2 style · 5 marks
A spirit burner containing ethanol is used to heat 200 g of water in a copper can. The temperature of the water rises from 21.0 °C to 45.5 °C. The burner loses 1.12 g in mass. The specific heat capacity of water is 4.18 J g⁻¹ K⁻¹.
(a)Calculate the energy transferred to the water, in kJ.[1]
(b)Calculate the enthalpy change of combustion of ethanol, in kJ/mol, from these results. Give your answer to 3 significant figures. (Mr of ethanol = 46.0)[2]
(c)The data-book value is −1367 kJ/mol. Give two reasons why the experimental value is less exothermic.[2]
(c) Heat lost to the surroundings (air, can); incomplete combustion of ethanol.
1 mark each for any two of: heat loss to surroundings / heating the can and air; incomplete combustion (soot seen); evaporation of ethanol from the wick; conditions not standard (water formed as vapour)
Q5Calorimetry (q = mcΔT), neutralisation & combustion · Standard · Paper 3 style · 7 marks
A student places 50.0 cm³ of 0.500 mol/dm³ copper(II) sulfate solution in an insulated cup and records the temperature every minute. At 3.5 minutes an excess of zinc powder is added and the mixture is stirred.Time / min: 0, 1, 2, 3, 4, 5, 6, 7, 8Temperature / °C: 22.0, 22.0, 22.0, 22.0, 40.5, 44.0, 43.4, 42.8, 42.2Assume the solution has density 1.00 g/cm³ and specific heat capacity 4.18 J g⁻¹ K⁻¹.
(a)The cooling after 5 minutes is linear. By extrapolating this cooling line back to 3.5 minutes, find the maximum temperature the mixture would have reached with no heat loss. Hence find ΔT.[2]
(b)Calculate the enthalpy change, in kJ/mol, for Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s). Give your answer to 3 significant figures.[3]
(c)Explain why the zinc is used as a powder and in excess.[2]
Show answer
(a) 44.9 °C; ΔT = 22.9 °C
M1 cooling rate 0.6 °C per minute; temperature at 3.5 min = 44.0 + 0.6 × 1.5 = 44.9 °C; A1 ΔT = 44.9 − 22.0 = 22.9 °C
(c) Powder reacts fast, so the reaction finishes before much heat is lost; excess ensures all the CuSO₄ reacts, so moles reacting are known.
B1 powder: large surface area, fast reaction, less heat loss during reaction; B1 excess: all copper(II) ions react / CuSO₄ is limiting
Q6Calorimetry (q = mcΔT), neutralisation & combustion · Challenge · Paper 2 style · 6 marks
40.0 cm³ of 1.50 mol/dm³ ethanoic acid, CH₃COOH, is mixed with 60.0 cm³ of 1.20 mol/dm³ sodium hydroxide. The temperature rises by 7.9 °C. Assume the solution has a density of 1.00 g/cm³ and a specific heat capacity of 4.18 J g⁻¹ K⁻¹.
(a)Show by calculation which reagent is in excess.[1]
(b)Calculate the enthalpy change of neutralisation of ethanoic acid, in kJ/mol, to 3 significant figures.[3]
(c)The value for hydrochloric acid with sodium hydroxide is about −57 kJ/mol. Explain why the value for ethanoic acid is less exothermic.[2]
Show answer
(a) NaOH is in excess (0.0720 mol NaOH v 0.0600 mol CH₃COOH).
B1 n(acid) = 0.0400 × 1.50 = 0.0600; n(NaOH) = 0.0600 × 1.20 = 0.0720, so NaOH in excess
(c) Ethanoic acid is a weak acid, only partly ionised; some energy is absorbed to ionise (break O–H bonds in) the acid as it reacts.
B1 weak acid / partially dissociated; B1 energy needed to complete dissociation (endothermic), reducing the overall energy released
Q7Calorimetry (q = mcΔT), neutralisation & combustion · Challenge · Paper 2 style · 5 marks
On a trip to Murree, a camping stove burns propane, C₃H₈. The standard enthalpy change of combustion of propane is −2220 kJ/mol. The stove heats 1.50 kg of water from 8 °C to 100 °C. The specific heat capacity of water is 4.18 J g⁻¹ K⁻¹.
(a)Calculate the energy needed to heat the water, in kJ.[1]
(b)Only 45% of the energy released by burning propane reaches the water. Calculate the mass of propane, in g, burned. Give your answer to 3 significant figures. (Mr of propane = 44.0)[3]
(c)Suggest one reason why less than half of the energy reaches the water.[1]
Show answer
(a) 577 kJ
B1 q = 1500 × 4.18 × 92 = 576840 J = 577 kJ
(b) 25.4 g
M1 energy released = 576.84 ÷ 0.45 = 1281.9 kJ; M1 n = 1281.9 ÷ 2220 = 0.5774 mol; A1 mass = 25.4 g
(c) Heat escapes to the surrounding cold air / heats the pan / wind carries hot gases away.
B1 any sensible heat-loss route
Q8Hess's law cycles · Core · Paper 2 style · 4 marks
Standard enthalpy changes of combustion: carbon (graphite) −394 kJ/mol; carbon monoxide −283 kJ/mol.
(a)Write an equation, with state symbols, for the standard enthalpy change of formation of carbon monoxide.[1]
(b)Explain why this enthalpy change cannot easily be measured directly.[1]
(c)Use the data to calculate the standard enthalpy change of formation of CO, in kJ/mol.[2]
Show answer
(a) C(s) + ½O₂(g) → CO(g)
B1 correct equation with state symbols, 1 mol CO
(b) Burning carbon in limited oxygen also forms some CO₂ (and leaves unburnt carbon), so pure CO is not formed.
Q11Hess's law cycles · Standard · Paper 1 style · 1 mark
(a)Iron is extracted in a blast furnace by the reaction Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g).Standard enthalpy changes of formation / kJ/mol: Fe₂O₃(s) −824; CO(g) −111; CO₂(g) −394.What is the standard enthalpy change of this reaction?[1]
Show answer
(a) A −25 kJ/mol
ΔH = ΣΔHf(products) − ΣΔHf(reactants) = 3(−394) − [(−824) + 3(−111)] = −1182 + 1157 = −25 kJ/mol. B reverses the subtraction; C uses only one CO and one CO₂ (−394 + 824 + 111); D adds the reactant values instead of subtracting them.
Q12Hess's law cycles · Standard · Paper 2 style · 4 marks
A cement works near Dera Ghazi Khan needs the enthalpy change for making quicklime. Two reactions with hydrochloric acid are measured:CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g) ΔH₁ = −15 kJ/molCaO(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) ΔH₂ = −193 kJ/mol
(a)Calculate ΔH for the thermal decomposition CaCO₃(s) → CaO(s) + CO₂(g), in kJ/mol.[2]
(b)Explain why ΔH for this decomposition is found indirectly rather than by heating calcium carbonate in a calorimeter.[2]
(b) The reaction needs a high temperature (about 800 °C or more), so it is impossible to measure the heat absorbed with a simple calorimeter; and it is hard to make decomposition complete.
B1 high temperature needed / continuous heating; B1 cannot measure energy taken in / decomposition incomplete
Q13Hess's law cycles · Challenge · Paper 2 style · 6 marks
Two experiments are done. In each, the solid is added to 50.0 g of water in an insulated cup (specific heat capacity 4.18 J g⁻¹ K⁻¹).Experiment 1: 3.99 g of anhydrous CuSO₄ — temperature rises by 7.6 K.Experiment 2: 6.24 g of CuSO₄·5H₂O — temperature falls by 1.1 K.
(a)Calculate ΔH for dissolving each solid, in kJ/mol. (Mr: CuSO₄ = 159.6, CuSO₄·5H₂O = 249.6)[3]
(b)Use a Hess cycle to calculate ΔH for CuSO₄(s) + 5H₂O(l) → CuSO₄·5H₂O(s), in kJ/mol, to 3 significant figures.[2]
(c)Explain why this enthalpy change cannot be found by direct measurement.[1]
(c) Water cannot be added in exactly the right amount; the solid would dissolve/partly hydrate, so the reaction cannot be made to go to completion cleanly.
B1 cannot control hydration / excess water dissolves solid / slow, incomplete reaction
Q14Hess's law cycles · Challenge · Paper 2 style · 5 marks
Standard enthalpy changes of combustion / kJ/mol: C(graphite) −394; H₂(g) −286; C₃H₈(g) −2219.Standard enthalpy changes of formation / kJ/mol: C₂H₄(g) +52; CH₄(g) −75.
(a)Calculate the standard enthalpy change of formation of propane, in kJ/mol.[2]
(b)Propane is cracked: C₃H₈(g) → C₂H₄(g) + CH₄(g). Using your answer to (a), calculate ΔH for this reaction, in kJ/mol.[2]
(c)Suggest why cracking needs a high temperature, using your answer to (b).[1]
(c) It is endothermic (and has a high activation energy, C–C bonds are strong), so heat must be supplied.
B1 endothermic / energy needed to break bonds
Q15Standard enthalpy changes (definitions) · Core · Paper 2 style · 4 marks
(a)Define standard enthalpy change of formation.[2]
(b)Write an equation, with state symbols, for the standard enthalpy change of formation of ethanol, C₂H₅OH(l).[2]
Show answer
(a) The enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions (298 K, 101 kPa).
B1 one mole of compound formed from its elements; B1 in standard states under standard conditions
(b) 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)
B1 correct formula and balancing for 1 mol; B1 state symbols
Q16Standard enthalpy changes (definitions) · Core · Paper 2 style · 3 marks
(a)State the standard conditions of temperature and pressure used for standard enthalpy changes.[1]
(b)Define standard enthalpy change of combustion.[2]
Show answer
(a) 298 K (25 °C) and 101 kPa (1 atm).
B1 both values
(b) The enthalpy change when one mole of a substance is burned completely in oxygen under standard conditions.
B1 one mole of substance; B1 complete combustion in excess oxygen, standard conditions
Q17Standard enthalpy changes (definitions) · Standard · Paper 1 style · 1 mark
(a)Which equation represents both the standard enthalpy change of formation of a compound and the standard enthalpy change of combustion of an element?[1]
Show answer
(a) B C(graphite) + O₂(g) → CO₂(g)
Graphite is carbon in its standard state; one mole burns completely to form one mole of CO₂, which is also formation of CO₂ from its elements. C forms water as a gas, not its standard state (liquid). D forms two moles. A starts from a compound.
Q18Standard enthalpy changes (definitions) · Standard · Paper 2 style · 4 marks
(a)Define standard enthalpy change of neutralisation.[2]
(b)For H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l), ΔH = −114 kJ per mole of sulfuric acid. State the standard enthalpy change of neutralisation and explain your answer.[2]
Show answer
(a) The enthalpy change when an acid and an alkali react to form one mole of water under standard conditions.
B1 one mole of water formed; B1 acid + alkali (base) under standard conditions
(b) −57 kJ/mol, because two moles of water form per mole of H₂SO₄.
B1 −57 kJ/mol; B1 2 mol water formed so divide by 2
Q19Standard enthalpy changes (definitions) · Standard · Paper 2 style · 4 marks
The standard enthalpy change of combustion of propan-1-ol is −2021 kJ/mol.
(a)Write an equation, with state symbols, that represents the standard enthalpy change of combustion of propan-1-ol, C₃H₇OH(l).[2]
(b)Calculate the energy released, in kJ, when 15.0 g of propan-1-ol burns completely. Give your answer to 3 significant figures. (Mr of propan-1-ol = 60.0)[2]
Show answer
(a) C₃H₇OH(l) + 4½O₂(g) → 3CO₂(g) + 4H₂O(l)
B1 correctly balanced for 1 mol propan-1-ol; B1 state symbols with H₂O(l)
Q20Standard enthalpy changes (definitions) · Challenge · Paper 2 style · 5 marks
(a)For each equation, state whether its ΔH is a standard enthalpy change of formation, a standard enthalpy change of combustion, both, or neither. Give a reason in each case.(i) 2Na(s) + Cl₂(g) → 2NaCl(s)(ii) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)(iii) C(graphite) + 2H₂(g) → CH₄(g)(iv) CO(g) + ½O₂(g) → CO₂(g)[4]
(b)Explain why the standard enthalpy change of formation of oxygen gas, O₂(g), is zero but that of ozone, O₃(g), is not.[1]
Show answer
(a) (i) neither: two moles of NaCl formed. (ii) neither: water should be liquid under standard conditions. (iii) formation only: one mole of CH₄ from elements in standard states. (iv) combustion of CO only: CO₂ is formed from a compound, not from elements.
1 mark for each correct decision with reason
(b) O₂(g) is oxygen's standard state, so no change occurs in forming it; O₃ is a different form made from O₂ with an energy change.
B1 O₂ is the element in its standard state; O₃ is not the standard state
Q21Bond energies · Core · Paper 2 style · 4 marks
Bond energies / kJ/mol: C=C 610; C–C 350; C–H 410; H–H 436. This reaction is used to harden vegetable oils in ghee factories.
(a)Calculate the enthalpy change for C₂H₄(g) + H₂(g) → C₂H₆(g), in kJ/mol.[3]
(b)State why a value found from bond energies may differ from the accurate value.[1]
(b) Bond energies are averages; the standard value forms liquid water, releasing extra energy when water vapour condenses.
B1 average bond energies; B1 water as liquid in standard ΔHc (condensation exothermic)
Q23Bond energies · Standard · Paper 1 style · 1 mark
(a)Bond energies / kJ/mol: C–H 410; Cl–Cl 242; C–Cl 340; H–Cl 431.What is the enthalpy change for CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g)?[1]
Show answer
(a) B −119 kJ/mol
Only one C–H and one Cl–Cl bond break: 410 + 242 = 652; one C–Cl and one H–Cl bond form: 340 + 431 = 771; ΔH = 652 − 771 = −119 kJ/mol. A has the sign reversed; C leaves out the H–Cl bond formed; D counts only the bonds made.
Q24Bond energies · Challenge · Paper 2 style · 5 marks
(b) N≡N is very strong (944 kJ/mol), giving a high activation energy; the catalyst gives an alternative route with lower Ea so the rate is acceptable.
B1 strong triple bond / high Ea; B1 catalyst lowers Ea via alternative pathway
Q25Reaction pathway diagrams & activation energy · Core · Paper 2 style · 4 marks
For a reaction, the activation energy is +75 kJ/mol and ΔH = −40 kJ/mol.
(a)Describe the reaction pathway diagram for this reaction. Say where the reactants and products are and how Ea and ΔH are shown.[3]
(b)Calculate the activation energy of the reverse reaction, in kJ/mol.[1]
Show answer
(a) Products lower than reactants; curve rises to a maximum (transition state) then falls; Ea is from reactants up to the peak; ΔH is the drop from reactants to products (downward arrow).
B1 products below reactants (exothermic); B1 Ea arrow from reactant level to top of curve; B1 ΔH arrow from reactants down to products
(b) 115 kJ/mol
B1 Ea(reverse) = 75 + 40 = 115 kJ/mol
Q26Reaction pathway diagrams & activation energy · Standard · Paper 2 style · 5 marks
An endothermic reaction has ΔH = +30 kJ/mol. Without a catalyst the forward activation energy is 120 kJ/mol; with a catalyst it is 70 kJ/mol.
(a)Calculate the activation energy of the reverse reaction with and without the catalyst, in kJ/mol.[2]
(b)State the value of ΔH with the catalyst and explain.[1]
(c)Explain how a catalyst increases the rate of a reaction, using the idea of a Boltzmann distribution.[2]
Show answer
(a) Without catalyst: 90 kJ/mol; with catalyst: 40 kJ/mol.
B1 120 − 30 = 90; B1 70 − 30 = 40
(b) +30 kJ/mol; the catalyst does not change the energies of reactants or products.
B1 unchanged with reason
(c) It provides an alternative pathway with lower Ea, so a larger fraction of molecules have energy ≥ Ea; more successful collisions per second.
B1 alternative route, lower Ea; B1 more particles with E ≥ Ea (greater area under the curve beyond Ea) so more effective collisions
Q27Reaction pathway diagrams & activation energy · Challenge · Paper 4 style · 5 marks
A two-step reaction is described by these energies, measured relative to the reactants (0 kJ/mol):first transition state +85; intermediate +20; second transition state +60; products −45 (all in kJ/mol).
(a)State the overall enthalpy change of the reaction, and whether it is exothermic or endothermic.[1]
(b)Identify the rate-determining step and give a reason.[2]
(c)Calculate the activation energy for the overall reverse reaction (products → reactants), in kJ/mol.[1]
(d)State how the diagram shows that the intermediate could, in principle, be detected.[1]
Show answer
(a) −45 kJ/mol; exothermic.
B1 both
(b) Step 1 (reactants → intermediate), because its activation energy (85 kJ/mol) is larger than that of step 2 (60 − 20 = 40 kJ/mol).
B1 step 1; B1 compares Ea of each step: 85 v 40
(c) 130 kJ/mol
B1 from −45 up to highest transition state +85: 85 − (−45) = 130
(d) It sits in an energy minimum (a dip) between the two peaks, so it has a finite lifetime.
B1 local minimum / trough
Q28Exothermic & endothermic reactions · Core · Paper 2 style · 4 marks
A sports first-aid cold pack used at a cricket ground contains solid ammonium nitrate and a pouch of water. When the pouch is broken, the pack becomes cold.
(a)State whether dissolving ammonium nitrate in water is exothermic or endothermic, and give the sign of ΔH.[1]
(b)Explain, in terms of bonds broken and formed, why some reactions are exothermic.[2]
(c)Give one everyday exothermic reaction.[1]
Show answer
(a) Endothermic; ΔH is positive.
B1 both
(b) Breaking bonds absorbs energy and making bonds releases energy; if more energy is released forming new bonds than is absorbed breaking bonds, the reaction is exothermic.
B1 bond breaking endothermic, bond making exothermic; B1 energy released > energy absorbed
(c) Burning natural gas in a stove (or respiration, neutralisation, hand warmers).
B1 any valid example
Q29Exothermic & endothermic reactions · Standard · Paper 2 style · 5 marks
(a)Photosynthesis and respiration are reverse processes. Explain which is endothermic and which is exothermic.[2]
(b)Methane burning is very exothermic, yet a gas stove needs a spark to light. Explain why.[2]
(c)State what happens to the temperature of the surroundings during an endothermic reaction.[1]
Show answer
(a) Photosynthesis is endothermic: it absorbs light energy to make glucose; respiration is exothermic: it releases energy from glucose.
B1 photosynthesis endothermic with reason; B1 respiration exothermic
(b) The reaction has an activation energy; the spark supplies energy so some molecules can overcome Ea; once started the energy released keeps the reaction going.
B1 activation energy needed / bonds must break first; B1 heat released sustains reaction
(c) It falls.
B1
Q30Enthalpy calculations from experimental data · Challenge · Paper 5 style · 7 marks
To find ΔHf of magnesium oxide, two reactions with excess dilute hydrochloric acid (100 cm³ each time) are carried out in an insulated cup. Assume the solution has density 1.00 g/cm³ and specific heat capacity 4.18 J g⁻¹ K⁻¹.Experiment 1: 0.240 g Mg ribbon, temperature rise 10.9 °C.Experiment 2: 0.400 g MgO powder, temperature rise 3.5 °C.
(a)Calculate ΔH₁ for Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g) and ΔH₂ for MgO(s) + 2HCl(aq) → MgCl₂(aq) + H₂O(l), in kJ/mol, to 3 significant figures. (Ar: Mg = 24.3; Mr: MgO = 40.3)[3]
(b)Use a Hess cycle and ΔHf(H₂O(l)) = −286 kJ/mol to calculate the standard enthalpy change of formation of MgO, in kJ/mol.[2]
(c)Each thermometer reading has an uncertainty of ±0.05 °C. Calculate the percentage uncertainty in ΔT for experiment 2.[1]
(d)Suggest one change that would reduce the percentage uncertainty in experiment 2 and explain why.[1]
