OCR A Level Chemistry A: Enthalpy and entropy (H432) – 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 section 5.2.2, Enthalpy and entropy, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. The questions test every learning outcome, 5.2.2(a) to (f), in Module 5 (Physical chemistry and transition elements) of this full A level. Questions 6, 10 and 11 are longer problems that combine outcomes.

Revise first with the Enthalpy and entropy study guide and the Enthalpy and entropy revision notes. See also the OCR chemistry hub and printable checklist.

All data are supplied in the questions: use these values. Units: S in J mol⁻¹ K⁻¹; ∆H and ∆G in kJ mol⁻¹.

Questions

1. Explain what is meant by the entropy of a system. [2]

2. The standard entropies of three substances are 72.1, 160.7 and 213.8 J mol⁻¹ K⁻¹. The substances are CO₂(g), NaCl(s) and C₂H₅OH(l). Match each substance to its value and explain your answer in terms of the particles. [3]

3. Predict the sign of ∆S for each reaction, or state that it cannot be predicted from the equation alone. Give a reason each time.

(a) 2Pb(NO₃)₂(s) → 2PbO(s) + 4NO₂(g) + O₂(g) [1] (b) 2Ca(s) + O₂(g) → 2CaO(s) [1] (c) CO(g) + NO₂(g) → CO₂(g) + NO(g) [1]

4. Burning magnesium continues to burn in carbon dioxide: 2Mg(s) + CO₂(g) → 2MgO(s) + C(s) ∆H = −809.7 kJ mol⁻¹ Entropies: Mg(s) 32.7, CO₂(g) 213.8, MgO(s) 26.9, C(s) 5.7.

(a) Calculate ∆S for the reaction. [2] (b) Use your answer to (a) to find ∆G at 298 K. Is the reaction feasible at this temperature? [2] (c) A clean strip of magnesium left in a jar of carbon dioxide at room temperature shows no visible change. Explain why this does not contradict your answer to (b). [2]

5. For H₂(g) + I₂(s) → 2HI(g), ∆S = +166.4 J mol⁻¹ K⁻¹. S(H₂(g)) = 130.7 and S(I₂(s)) = 116.1.

(a) Explain why ∆S is positive. [1] (b) Calculate the standard entropy of HI(g). [3]

6. Carbon dioxide can be turned into methane by reacting it with hydrogen: CO₂(g) + 4H₂(g) → CH₄(g) + 2H₂O(g)

Substance	CO₂(g)	H₂(g)	CH₄(g)	H₂O(g)
∆fH / kJ mol⁻¹	−393.5	0	−74.8	−241.8
S / J mol⁻¹ K⁻¹	213.8	130.7	186.3	188.8

(a) Calculate ∆H for the reaction. [2] (b) Calculate ∆S for the reaction. [2] (c) Explain the sign of ∆S in terms of the gas molecules. [1] (d) Calculate ∆G at 1100 K and state what it shows. [2] (e) Work out the range of temperatures at which this reaction is feasible. [2]

7. For 2CuO(s) + C(s) → 2Cu(s) + CO₂(g), ∆H = −78.9 kJ mol⁻¹. Entropies: CuO(s) 42.6, C(s) 5.7, Cu(s) 33.2, CO₂(g) 213.8.

(a) Calculate ∆S for the reaction. [2] (b) Use the signs of ∆H and ∆S to explain why this reaction is feasible at all temperatures. [2]

8. For the boiling of bromine, Br₂(l) → Br₂(g), ∆H = +30.9 kJ mol⁻¹. S(Br₂(l)) = 152.2 and S(Br₂(g)) = 245.5.

(a) Calculate ∆S for the change. [1] (b) Liquid and gas are in equilibrium at the boiling point, where ∆G = 0. Estimate the boiling point of bromine. [2] (c) Calculate ∆G at 298 K and explain what the sign tells you about bromine at this temperature. [1]

9. When a certain salt dissolves in water, the solution becomes colder. The salt dissolves fully at room temperature. Explain, in terms of ∆H, ∆S and temperature, how this process can be feasible. [3]

10. Blue copper(II) sulfate crystals lose their water on heating: CuSO₄·5H₂O(s) → CuSO₄(s) + 5H₂O(g)

Substance	CuSO₄·5H₂O(s)	CuSO₄(s)	H₂O(g)
∆fH / kJ mol⁻¹	−2279.7	−771.4	−241.8
S / J mol⁻¹ K⁻¹	300.4	109.2	188.8

(a) Predict the sign of ∆S without calculation, giving a reason. [1] (b) Calculate ∆H for the reaction. [2] (c) Calculate ∆S for the reaction. [2] (d) Find ∆G at 298 K. Is the dehydration feasible at room temperature? [2] (e) Find the lowest temperature at which dehydration becomes feasible. [2]

11. For a gas-phase reaction, ∆G = +21.0 kJ mol⁻¹ at 300 K and ∆G = −7.0 kJ mol⁻¹ at 500 K. Assume ∆H and ∆S do not change with temperature.

(a) A graph of ∆G against T is a straight line. Use the two points to calculate ∆S in J mol⁻¹ K⁻¹. [2] (b) Calculate ∆H. [2] (c) Give the temperatures at which the reaction is feasible. [1] (d) A student says: “At 600 K the reaction is feasible, so it will happen quickly.” Evaluate this statement. [2]

Answers

1. Entropy is a measure of the dispersal of energy in a system [1]; it is greater the more disordered the system is [1] Examiner insight: A bare “measure of disorder” is an incomplete answer; link the idea to how energy is spread out.

2. NaCl(s) 72.1, C₂H₅OH(l) 160.7, CO₂(g) 213.8 [1]; ions in a solid lattice only vibrate about fixed positions, so energy is least dispersed [1]; liquid molecules move past one another, while gas molecules move randomly through the whole volume, giving the most disorder [1] Examiner insight: Describe the motion and arrangement of particles in each state; listing “solid < liquid < gas” alone does not explain the order.

3. (a) Positive: no gas becomes 5 mol of gas [1] (b) Negative: 1 mol of gas is used up and only a solid forms [1] (c) Cannot be predicted from gas moles: 2 mol of gas on each side, so ∆S is small and data are needed [1] Examiner insight: Each sign needs its reason; a correct sign with no gas count or state argument may not be credited.

4. (a) ΣS(products) = 2(26.9) + 5.7 = 59.5 and ΣS(reactants) = 2(32.7) + 213.8 = 279.2 [1]; ∆S = −219.7 J mol⁻¹ K⁻¹ [1] (b) ∆G = −809.7 − 298 × (−0.2197) = −809.7 + 65.5 = −744.2 kJ mol⁻¹ [1]; negative, so the reaction is feasible [1] (c) The activation energy is very high [1]; so at room temperature the rate is negligible, and the magnesium must first be ignited before the reaction is seen [1] Examiner insight: In (c), name activation energy or rate explicitly; “it needs heat” alone does not explain why a feasible reaction is not seen.

5. (a) A solid becomes a gas, and gas moles rise from 1 to 2 [1] (b) 166.4 = 2S(HI) − (130.7 + 116.1) [1]; 2S(HI) = 166.4 + 246.8 = 413.2 [1]; S(HI) = 206.6 J mol⁻¹ K⁻¹ [1] Examiner insight: Write the full products-minus-reactants expression before rearranging, and remember to halve at the end because 2 mol of HI form.

6. (a) ∆H = [−74.8 + 2(−241.8)] − (−393.5) [1]; = −558.4 + 393.5 = −164.9 kJ mol⁻¹ [1] (b) ∆S = [186.3 + 2(188.8)] − [213.8 + 4(130.7)] = 563.9 − 736.6 [1]; = −172.7 J mol⁻¹ K⁻¹ [1] (c) 5 mol of gas become 3 mol, so the energy is less dispersed and ∆S is negative [1] (d) ∆G = −164.9 − 1100 × (−0.1727) = −164.9 + 190.0 = +25.1 kJ mol⁻¹ [1]; positive, so the reaction is not feasible at 1100 K [1] (e) T = −164.9 ÷ −0.1727 = 955 K [1]; ∆H and ∆S are both negative, so feasible below about 955 K [1] Examiner insight: In (e), a threshold temperature without “below” or “above” is an incomplete answer; justify the direction from the signs.

7. (a) ∆S = [2(33.2) + 213.8] − [2(42.6) + 5.7] = 280.2 − 90.9 [1]; = +189.3 J mol⁻¹ K⁻¹ [1] (b) ∆H is negative, and with ∆S positive the −T∆S term is negative at every temperature [1]; so ∆G = ∆H − T∆S is always negative and the reaction is always feasible [1] Examiner insight: Refer to ∆G = ∆H − T∆S in your explanation; quoting the signs alone without saying what happens to ∆G leaves out the key step.

8. (a) ∆S = 245.5 − 152.2 = +93.3 J mol⁻¹ K⁻¹ [1] (b) T = ∆H ÷ ∆S = 30.9 ÷ 0.0933 [1]; = 331 K [1] (c) ∆G = 30.9 − 298 × 0.0933 = +3.1 kJ mol⁻¹; positive, so boiling is not feasible at 298 K and bromine is a liquid at this temperature [1] Examiner insight: Convert 93.3 to 0.0933 kJ mol⁻¹ K⁻¹ before dividing; leaving it in J gives 0.331 K, which should prompt a check.

9. Dissolving is endothermic, so ∆H is positive and does not favour feasibility [1]; the ions become dispersed through the solution, so ∆S is positive [1]; at room temperature T∆S is larger than ∆H, so ∆G = ∆H − T∆S is negative and the process is feasible [1] Examiner insight: Give a clear comparison of T∆S with ∆H; stating “entropy increases” without linking it to the sign of ∆G gives an incomplete explanation.

10. (a) Positive: a solid forms 5 mol of gas [1] (b) ∆H = [−771.4 + 5(−241.8)] − (−2279.7) = −1980.4 + 2279.7 [1]; = +299.3 kJ mol⁻¹ [1] (c) ∆S = [109.2 + 5(188.8)] − 300.4 = 1053.2 − 300.4 [1]; = +752.8 J mol⁻¹ K⁻¹ [1] (d) ∆G = 299.3 − 298 × 0.7528 = 299.3 − 224.3 = +75.0 kJ mol⁻¹ [1]; positive, so not feasible at 298 K [1] (e) T = 299.3 ÷ 0.7528 [1]; = 398 K (feasible above this temperature) [1] Examiner insight: Multiply both the ∆fH and the S of water by 5; missing the balancing number in one sum but not the other is a frequent slip in multi-part data questions.

11. (a) Gradient = (−7.0 − 21.0) ÷ (500 − 300) = −0.140 kJ mol⁻¹ K⁻¹, and gradient = −∆S [1]; ∆S = +140 J mol⁻¹ K⁻¹ [1] (b) ∆H = ∆G + T∆S = 21.0 + 300 × 0.140 [1]; = +63.0 kJ mol⁻¹ [1] (c) Threshold T = 63.0 ÷ 0.140 = 450 K, so feasible above 450 K [1] (d) The reaction is feasible at 600 K, since ∆G is negative there [1]; but ∆G gives no information about rate, which depends on the activation energy, so the reaction may still be slow [1] Examiner insight: In (d), agree with the feasible part and then challenge the rate claim; an evaluation that only says “wrong” gains little.

Where marks are usually lost

• Using ∆S in J mol⁻¹ K⁻¹ directly with ∆H in kJ mol⁻¹ (questions 4, 6, 8 and 10).
• Leaving an element, such as H₂(g) in question 5, out of an entropy sum.
• Missing balancing numbers, such as the 5 in question 10.
• Giving a threshold temperature without saying “above” or “below” (questions 6 and 11).
• Explaining entropy as “chaos” with no reference to energy dispersal.
• Treating a negative ∆G as proof that a reaction is fast (questions 4 and 11).

Next steps

• Recap: Enthalpy and entropy revision notes
• Full explanations: Enthalpy and entropy study guide
• Hess cycles for ∆H: Enthalpy changes practice
• Check the whole course: all free 10-minute diagnostics
• More OCR chemistry topics: course hub
• Printable checklist: OCR A Level Chemistry checklist
• Book a free trial class: /trial/

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 5.2.2 Enthalpy and entropy, learning outcomes (a) to (f). Published by OCR.