OxfordAQA A-Level Chemistry: Thermodynamics (9620) – 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.

These questions cover section 3.1.8 Thermodynamics of the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: 3.1.8.1 Born–Haber cycles and 3.1.8.2 Gibbs free-energy change, ΔG, and entropy change, ΔS. The whole section is International A2 content, so these questions are International A-level only. All data are given in the questions and are for practice only.

Links: Thermodynamics study guide | Thermodynamics revision notes | OxfordAQA A-level Chemistry hub | printable checklist | free diagnostics

Enthalpy values are in kJ mol⁻¹ and absolute entropies in J K⁻¹ mol⁻¹ throughout.

Questions

1. This question is about the terms used in Born–Haber cycles.

(a) Define the first ionisation energy of an element. [2] (b) Define the lattice enthalpy of formation of an ionic compound. [2] (c) Write an equation, with state symbols, for the first electron affinity of sulfur. [1]

2. Use these data for lithium oxide, Li₂O. ΔfH⦵(Li₂O) –598; ΔatH⦵(Li) +159; first IE(Li) +520; O=O bond enthalpy +498; first EA(O) –141; second EA(O) +798.

(a) Calculate the lattice enthalpy of formation of Li₂O. [4] (b) Explain why only half the O=O bond enthalpy is used in the cycle. [1]

3. For rubidium bromide: ΔfH⦵(RbBr) –395; ΔatH⦵(Rb) +81; first IE(Rb) +403; ΔatH⦵(Br) +112; lattice enthalpy of formation of RbBr –660. Calculate the first electron affinity of bromine. [3]

4. The table gives lattice enthalpies of formation.

Compound	Born–Haber cycle	Perfect ionic model
KF	–821	–817
CuCl	–996	–921

(a) State two assumptions of the perfect ionic model. [2] (b) Which compound has the greater degree of covalent bonding? Justify your choice from the data. [3] (c) Predict whether LiF or LiI shows the larger difference between the two values. Explain your answer. [2]

5. This question is about strontium bromide.

(a) The lattice enthalpy of dissociation of SrBr₂ is +2075. ΔhydH(Sr²⁺) = –1445 and ΔhydH(Br⁻) = –335. Calculate the enthalpy of solution of SrBr₂. [3] (b) ΔhydH(F⁻) = –506. Explain why the hydration enthalpy of Br⁻ is less exothermic than that of F⁻. [2]

6. For potassium iodide, ΔsolH = +20, ΔhydH(K⁺) = –322 and ΔhydH(I⁻) = –293. Calculate the lattice enthalpy of formation of KI. [3]

7. Entropy and feasibility.

(a) State and explain the sign of ΔS for each change. (i) I₂(s) → I₂(g) (ii) 2H₂O₂(l) → 2H₂O(l) + O₂(g) (iii) 2Mg(s) + O₂(g) → 2MgO(s) [3] (b) Explain why the sign of ΔH alone cannot tell you whether a change is feasible. Use an example in your answer. [2]

8. Mercury(II) oxide decomposes on heating: 2HgO(s) → 2Hg(l) + O₂(g), ΔH⦵ = +182. Absolute entropies: HgO(s) 70; Hg(l) 76; O₂(g) 205.

(a) Calculate ΔS for the reaction. [2] (b) Calculate ΔG at 298 K. [2] (c) Calculate the lowest temperature at which the decomposition is feasible. [2]

9. Methanol can form from carbon monoxide and hydrogen: CO(g) + 2H₂(g) → CH₃OH(g), ΔH⦵ = –91. Absolute entropies: CO(g) 198; H₂(g) 131; CH₃OH(g) 240.

(a) Calculate ΔS for the reaction. [2] (b) Calculate ΔG at 298 K and state whether the reaction is feasible. [2] (c) Explain how ΔG changes as the temperature is raised. [2] (d) Calculate the temperature above which the reaction is no longer feasible. [2]

10. Tin is extracted from tin(IV) oxide by heating with carbon. Reaction A: SnO₂(s) + 2C(s) → Sn(s) + 2CO(g) Reaction B: SnO₂(s) + C(s) → Sn(s) + CO₂(g) ΔfH⦵: SnO₂(s) –581; CO(g) –111; CO₂(g) –394. Absolute entropies: SnO₂(s) 49; C(s) 6; Sn(s) 52; CO(g) 198; CO₂(g) 214.

(a) Calculate ΔH for reaction A. [2] (b) Calculate ΔS for reaction A. [2] (c) Calculate the lowest temperature at which reaction A is feasible. [2] (d) Calculate the lowest temperature at which reaction B is feasible, and state which reaction becomes feasible first as the mixture is heated. [3] (e) Suggest why a reaction with a negative ΔG may still not be observed. [1]

11. Potassium bromide dissolves in water and the solution cools.

(a) The lattice enthalpy of dissociation of KBr is +679. Use ΔhydH(K⁺) and ΔhydH(Br⁻) from questions 5 and 6 to calculate ΔsolH(KBr). [2] (b) Absolute entropies: KBr(s) 96; K⁺(aq) 103; Br⁻(aq) 82. Calculate ΔS for KBr(s) → K⁺(aq) + Br⁻(aq). [2] (c) Calculate ΔG at 298 K. [2] (d) Explain what your answers show about why KBr dissolves. [1]

Answers

1. (a) The enthalpy change to remove 1 mole of electrons from 1 mole of gaseous atoms [1] to form 1 mole of gaseous 1+ ions [1]. (b) The enthalpy change when 1 mole of an ionic solid is formed [1] from its gaseous ions [1]. (c) S(g) + e⁻ → S⁻(g) [1] Examiner insight: “gaseous” is part of both definitions; an answer that leaves out the state of the atoms or ions describes a different process and loses the point.

2. (a) Two Li atoms: 2 × 159 = 318 and 2 × 520 = 1040 [1] One O atom: ½ × 498 = 249; sum = 318 + 1040 + 249 – 141 + 798 = +2264 [1] Lattice enthalpy of formation = –598 – 2264 [1] = –2862 kJ mol⁻¹ [1] (b) Li₂O contains one O atom, and breaking one O=O bond gives two atoms [1]. Examiner insight: show each multiplier next to its term; a final answer with no visible working gives no way to recover credit if one step was doubled wrongly.

3. –395 = 81 + 403 + 112 + EA – 660 [1] EA = –395 – 81 – 403 – 112 + 660 [1] = –331 kJ mol⁻¹ [1] Examiner insight: bromine’s atomisation value is given directly because Br₂ is a liquid; halving a bond enthalpy here would be wrong.

4. (a) The ions are perfect spheres with evenly spread charge [1]. The only attraction is electrostatic, with no covalent bonding [1]. (b) CuCl [1]. Its Born–Haber value is 75 kJ mol⁻¹ more exothermic than the model, compared with 4 kJ mol⁻¹ for KF [1]. Cu⁺ polarises the Cl⁻ electron cloud, adding attraction the model ignores [1]. (c) LiI [1]. I⁻ is much larger than F⁻, so its electron cloud is more easily distorted (polarised) by the small Li⁺ ion [1]. Examiner insight: say that the experimental value is more exothermic; quoting the size of the gap without its direction does not show why the extra attraction exists.

5. (a) ΔsolH = lattice dissociation + ΔhydH(Sr²⁺) + 2 × ΔhydH(Br⁻) [1] = 2075 – 1445 – 670 [1] = –40 kJ mol⁻¹ [1] (b) Br⁻ is larger than F⁻ [1], so its charge is less concentrated and it attracts the δ+ hydrogen atoms of water less strongly [1]. Examiner insight: in (b), water’s δ+ hydrogens face an anion; describing δ– oxygen attracted to Br⁻ reverses the interaction.

6. ΔsolH = lattice dissociation + ΔhydH(K⁺) + ΔhydH(I⁻), so lattice dissociation = 20 + 322 + 293 = +635 [1] Lattice enthalpy of formation is the reverse [1] = –635 kJ mol⁻¹ [1] Examiner insight: the question asks for formation, so +635 alone answers a different question; check which definition is wanted before giving the sign.

7. (a) (i) Positive: a solid becomes a gas, so disorder increases [1]. (ii) Positive: a gas forms from liquids [1]. (iii) Negative: a gas is used up to form a solid [1]. (b) Some endothermic changes are feasible, so a positive ΔH does not rule a change out [1]. For example, ice melts above 273 K because the entropy increase outweighs the positive ΔH [1]. Examiner insight: a sign of ΔS with no reason is incomplete; refer to states or moles of gas each time.

8. (a) ΔS = [2(76) + 205] – 2(70) [1] = +217 J K⁻¹ mol⁻¹ [1] (b) ΔG = 182 – 298 × 0.217 [1] = +117 kJ mol⁻¹ [1] (c) T = ΔH/ΔS = 182 / 0.217 [1] = 839 K [1] Examiner insight: both ΔH and ΔS are positive here, so the reaction is feasible at 839 K and above; say “lowest” or “above” to show you know the direction.

9. (a) ΔS = 240 – [198 + 2(131)] [1] = –220 J K⁻¹ mol⁻¹ [1] (b) ΔG = –91 – 298 × (–0.220) = –25 kJ mol⁻¹ [1]; negative, so feasible [1]. (c) ΔS is negative, so –TΔS is positive [1]. As T rises, this term grows and ΔG becomes less negative, then positive [1]. (d) T = –91 / –0.220 [1] = 414 K [1] Examiner insight: subtracting a negative TΔS is where sign errors creep in; write ΔG = –91 + 0.220T before substituting.

10. (a) ΔH = 2(–111) – (–581) [1] = +359 kJ mol⁻¹ [1] (b) ΔS = [52 + 2(198)] – [49 + 2(6)] [1] = +387 J K⁻¹ mol⁻¹ [1] (c) T = 359 / 0.387 [1] = 928 K [1] (d) Reaction B: ΔH = –394 + 581 = +187 and ΔS = (52 + 214) – (49 + 6) = +211 [1] T = 187 / 0.211 = 886 K [1] Reaction B becomes feasible first (at the lower temperature) [1]. (e) The activation energy may be so high that the rate is too slow to observe [1]. Examiner insight: elements in their standard states have ΔfH⦵ = 0 but non-zero absolute entropies; setting S(C) or S(Sn) to zero changes ΔS.

11. (a) ΔsolH = 679 – 322 – 335 [1] = +22 kJ mol⁻¹ [1] (b) ΔS = (103 + 82) – 96 [1] = +89 J K⁻¹ mol⁻¹ [1] (c) ΔG = 22 – 298 × 0.089 [1] = –4.5 kJ mol⁻¹ [1] (d) Dissolving is endothermic but ΔG is negative, so the entropy increase makes it feasible [1]. Examiner insight: (d) needs a link between the two results; stating only that ΔG is negative does not explain why an endothermic change happens.

Where marks are usually lost

• Leaving “gaseous” out of ionisation energy, electron affinity or lattice enthalpy definitions.
• Doubling only some steps when a formula contains two of a particle (Li₂O, SrBr₂).
• Halving the bond enthalpy of an element whose standard state is not a gas.
• Giving a lattice enthalpy of dissociation when formation was asked for, or the reverse.
• Explaining covalent character without saying the experimental value is more exothermic.
• Counting a hydration enthalpy once for an ion that appears twice.
• Mixing J and kJ in ΔG = ΔH – TΔS, or using °C.
• Stating a temperature without saying whether the reaction is feasible above or below it.
• Setting the absolute entropy of an element to zero.

Next steps

• Thermodynamics revision notes
• Thermodynamics study guide
• Energetics study guide
• OxfordAQA A-level Chemistry hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: section 3.1.8 Thermodynamics (International A2).