AS Chemistry: Hess Law and Enthalpy Cycles — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — 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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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 100 kPa, 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].

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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.