AS Chemistry: Hess Law and Enthalpy Cycles — 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 available free from your board.

Related: Hess’s Law and Enthalpy Cycles revision notes

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Section A

1. Define standard enthalpy of combustion, including the standard conditions it is measured under. [2]

2. State Hess’s law, and explain what it allows chemists to do that direct measurement often cannot. [2]

3. Explain why bond enthalpy calculations give only approximate values. [2]

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Section B

4. Use the enthalpies of formation below to calculate ΔH for:

CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)

delta-Hf: CH4 = -75 kJ mol-1,  CO2 = -394 kJ mol-1,  H2O(l) = -286 kJ mol-1

(a) State the value of ΔH_f for O₂(g) and explain why. [2]

(b) Calculate ΔH for the reaction. [3]

5. A student burns 0.740 g of ethanol (M_r = 46.0) and uses the heat released to warm 150 g of water. The temperature rises from 19.0 °C to 40.5 °C. (c of water = 4.18 J g⁻¹ K⁻¹)

(a) Calculate the heat energy transferred to the water. [2]

(b) Calculate the enthalpy of combustion of ethanol. [3]

(c) The data book value is −1367 kJ mol⁻¹. Calculate the percentage difference and give three reasons for it. [4]

(d) Suggest two specific improvements to the experiment, with reasons. [2]

6. Define standard enthalpy of neutralisation. [2]

7. Calculate the standard enthalpy of formation of methanol, CH₃OH(l), from the standard enthalpies of combustion below.

C(graphite) + 2H2(g) + 1/2 O2(g) -> CH3OH(l)

delta-Hc: C(graphite) = -394 kJ mol-1, H2(g) = -286 kJ mol-1, CH3OH(l) = -726 kJ mol-1

(a) State which Hess cycle expression applies here, and explain why. [2]

(b) Calculate ΔH_f for methanol. [3]

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8. Use the bond energies below to calculate the enthalpy change for the reaction C₂H₄ + H₂ → C₂H₆.

Bond energies in kJ mol⁻¹: C=C 610, C–C 350, C–H 410, H–H 436 [3]

Answers

1. The enthalpy change when one mole of a substance [1] is completely burned in oxygen under standard conditions [1], namely 298 K and 101 kPa (the values the 9701 syllabus assumes), with all substances in their standard states.

2. The total enthalpy change is independent of the route taken, provided the initial and final conditions are the same [1]. This lets chemists calculate an enthalpy change that cannot be measured directly (e.g. formation of methane) by building a Hess cycle from other, measurable enthalpy changes instead [1].

3. Bond enthalpies are mean values averaged over many different compounds [1], and they apply only to species in the gaseous state [1].

4. (a) Zero [1], because O₂ is an element in its standard state [1].

(b) ΔH = ΣΔH_f(products) − ΣΔH_f(reactants) [1] = [(−394) + 2(−286)] − [(−75) + 0] [1] = (−966) − (−75) = −891 kJ mol⁻¹ [1].

5. (a) q = mcΔT = 150 × 4.18 × 21.5 [1] = 13 480 J = 13.48 kJ [1]. m is the mass of water, not of ethanol.

(b) n(ethanol) = 0.740 ÷ 46.0 = 0.01609 mol [1]. ΔH = −13.48 ÷ 0.01609 [1] = −838 kJ mol⁻¹ [1]. The minus sign is required — the reaction is exothermic.

(c) Difference = (1367 − 838) ÷ 1367 × 100 = 38.7% [1]. Any three: heat lost to the surroundings [1]; incomplete combustion producing carbon monoxide or soot [1]; evaporation of ethanol from the wick [1]; heat absorbed by the apparatus; non-standard conditions.

(d) Any two with reasons: use a draught shield to reduce heat loss to the surroundings [1]; use a lid on the beaker to reduce heat loss by evaporation; reduce the distance between flame and container; use a copper calorimeter rather than glass because it conducts heat to the water more efficiently [1].

6. The enthalpy change when an acid and alkali react [1] to form one mole of water, under standard conditions [1].

7. (a) Combustion data: ΔH_r = ΣΔH_c(reactants) − ΣΔH_c(products) [1], because combustion enthalpies are given for the elements/starting materials rather than formation enthalpies for a product, so the cycle’s arrows point downward from reactants and products alike to the same combustion products [1]. (b) ΔH_f = [(−394) + 2(−286)] − (−726) [1] = (−966) − (−726) [1] = −240 kJ mol⁻¹ [1].

8. Bonds broken: C=C + H–H = 610 + 436 = 1046 kJ mol⁻¹ [1]. Bonds formed: C–C + 2 C–H = 350 + (2 × 410) = 1170 kJ mol⁻¹ [1]. ΔH = 1046 − 1170 = −124 kJ mol⁻¹ [1]. (Counting all bonds on both sides gives the same answer: 2686 − 2810.)

Examiner insight (June 2025): a common mistake was to forget the energy needed to break the H–H bond, or to break the wrong number of H–H bonds. List every bond broken and formed before calculating. Try the real question next: Cambridge International AS & A Level Chemistry 9701, June 2025, Paper 22, Question 3.

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Where marks are usually lost

• Using the mass of fuel instead of the mass of water in q = mcΔT.
• Omitting the minus sign in ΔH = −q/n.
• Subtracting the wrong way round for the type of data given.
• Not converting J to kJ.
• Giving vague improvements such as “be more careful” instead of specific ones with reasons.
• Using the formation-data expression (products − reactants) on a question that gives combustion data instead — the two cycles subtract in opposite directions, and mixing them up flips the sign of the final answer.
• Forgetting O₂ contributes nothing to either side of a combustion-data cycle, since it isn’t itself a fuel with a combustion enthalpy.
• Assigning the supplied combustion-enthalpy values to CO₂ and H₂O — these are the shared products that every combustion route in the cycle ends at, not the substances whose combustion enthalpies were measured. The supplied ΔH_c values belong to the combustible reactants (and any combustible product) named in the target equation; CO₂ and liquid water are simply the common endpoint the cycle uses to connect them.
• Omitting “one mole of water” from the neutralisation definition, or giving the definition of enthalpy of formation instead by mistake.

Questions 6 and 7 draw on the standard enthalpy of neutralisation definition and the combustion-data Hess cycle from the Hess’s Law and Enthalpy Cycles revision notes — the formation-data direction used in question 4 is the opposite cycle, so question 7 exercises the direction question 4 doesn’t reach.