AQA A-Level Chemistry: Thermodynamics (7405) – 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 AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 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 A-level only, so these questions are for the A-level, not the AS. All data you need is given in the questions; values are for practice and are not quoted from a data book.

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

All enthalpy values are in kJ mol⁻¹ and all absolute entropies are in J K⁻¹ mol⁻¹. Take Mr(H₂O) = 18.0.

Questions

1. Define each term.

(a) Enthalpy of atomisation [2] (b) Enthalpy of hydration [2]

2. Write an equation, with state symbols, for each of these enthalpy changes.

(a) The first electron affinity of oxygen [1] (b) The second ionisation energy of strontium [1] (c) The lattice enthalpy of formation of potassium oxide, K₂O [1]

3. Use these data for sodium oxide, Na₂O. ΔfH⦵(Na₂O) –414; ΔatH⦵(Na) +107; first IE(Na) +496; O=O bond enthalpy +498; first EA(O) –141; second EA(O) +798.

(a) Calculate the lattice enthalpy of formation of Na₂O. [4] (b) Explain why the second electron affinity of oxygen is endothermic. [2]

4. Use the data to calculate the second ionisation energy of calcium. ΔfH⦵(CaO) –635; ΔatH⦵(Ca) +178; first IE(Ca) +590; ½ × O=O bond enthalpy +249; first EA(O) –141; second EA(O) +798; lattice enthalpy of formation of CaO –3454. [3]

5. The table gives lattice enthalpies of formation for two ionic compounds, X and Y.

Compound	From a Born–Haber cycle	From the perfect ionic model
X	–688	–684
Y	–916	–834

(a) State two assumptions of the perfect ionic model. [2] (b) Deduce which compound has more covalent character. Explain your answer. [3]

6. This question is about dissolving Group 2 chlorides.

(a) The lattice enthalpy of dissociation of MgCl₂ is +2524. ΔhydH(Mg²⁺) = –1921 and ΔhydH(Cl⁻) = –364. Calculate the enthalpy of solution of MgCl₂. [3] (b) Explain why the hydration enthalpy of Mg²⁺ is more exothermic than that of Na⁺. [2] (c) ΔsolH(CaCl₂) = –81 and its lattice enthalpy of dissociation is +2258. Use ΔhydH(Cl⁻) from part (a) to calculate the hydration enthalpy of Ca²⁺. [2]

7. For sodium bromide, ΔsolH = –1, the lattice enthalpy of formation is –747 and ΔhydH(Na⁺) = –406. Calculate the enthalpy of hydration of the bromide ion. [3]

8. State whether ΔS is positive or negative for each change. Give a reason in each case.

(a) H₂O(l) → H₂O(g) [1] (b) N₂(g) + 3H₂(g) → 2NH₃(g) [1] (c) NH₄Cl(s) → NH₃(g) + HCl(g) [1]

9. Ethene reacts with hydrogen: C₂H₄(g) + H₂(g) → C₂H₆(g), ΔH⦵ = –137. Absolute entropies: C₂H₄(g) 220; H₂(g) 131; C₂H₆(g) 230.

(a) Calculate ΔS for the reaction. [2] (b) Calculate ΔG at 298 K. [2] (c) Calculate the temperature above which the reaction is no longer feasible. [2]

10. Zinc oxide can be reduced by carbon: ZnO(s) + C(s) → Zn(s) + CO(g), ΔH⦵ = +238. Absolute entropies: ZnO(s) 44; C(s) 6; Zn(s) 42; CO(g) 198.

(a) Explain, without calculation, why ΔS is positive. [1] (b) Calculate ΔS for the reaction. [2] (c) Calculate the lowest temperature at which the reaction is feasible. [3] (d) Calculate ΔG at 1500 K and state what it shows. [2]

11. A straight-line graph of ΔG (kJ mol⁻¹) against T (K) for a reaction passes through (400, –12) and (700, –36).

(a) Use the graph to find ΔS, in J K⁻¹ mol⁻¹. [2] (b) Find ΔH. [2] (c) State the temperature range over which the reaction is feasible. [1]

12. A student finds the entropy change for vaporising water. A 2.20 kW kettle keeps water boiling at 373 K. In 90 s, the mass of water falls by 82.0 g.

(a) Explain why ΔG = 0 for H₂O(l) → H₂O(g) at 373 K. [1] (b) Calculate ΔH for the vaporisation of water in kJ mol⁻¹. [3] (c) Calculate ΔS for the vaporisation of water. [1] (d) Explain whether your values from (b) and (c) are likely to be too high or too low. [2]

Answers

1. (a) The enthalpy change when 1 mole of gaseous atoms is formed [1] from the element in its standard state [1]. (b) The enthalpy change when 1 mole of gaseous ions [1] dissolves in water to form 1 mole of aqueous ions [1]. Examiner insight: “1 mole of gaseous atoms” is the key idea in (a); a definition based on breaking one mole of bonds describes bond enthalpy instead.

2. (a) O(g) + e⁻ → O⁻(g) [1] (b) Sr⁺(g) → Sr²⁺(g) + e⁻ [1] (c) 2K⁺(g) + O²⁻(g) → K₂O(s) [1] Examiner insight: the second ionisation energy starts from the 1+ ion, not the atom; writing Sr(g) → Sr²⁺(g) + 2e⁻ describes the sum of two ionisation energies.

3. (a) Steps to gaseous ions: 2 × 107 = 214 and 2 × 496 = 992 [1] ½ × 498 = 249 for one O atom; sum = 214 + 992 + 249 – 141 + 798 = +2112 [1] Lattice enthalpy of formation = ΔfH – sum = –414 – 2112 [1] = –2526 kJ mol⁻¹ [1] (b) The electron is being added to an O⁻ ion, which is already negative [1]. Energy is needed to overcome the repulsion between the ion and the incoming electron [1]. Examiner insight: doubling only one of the sodium steps is the usual slip; label each term with its multiplier so the working can be followed.

4. Hess’s law: –635 = 178 + 590 + IE₂ + 249 – 141 + 798 – 3454 [1] IE₂ = –635 – 178 – 590 – 249 + 141 – 798 + 3454 [1] = +1145 kJ mol⁻¹ [1] [3] Examiner insight: write the full Hess’s law equation before rearranging; a sign error made while moving terms is then easy to spot.

5. (a) Ions are perfect spheres with evenly spread charge [1]. Bonding is purely electrostatic, with no covalent character [1]. (b) Y has more covalent character [1]. Its Born–Haber value is 82 kJ mol⁻¹ more exothermic than the model value, compared with only 4 kJ mol⁻¹ for X [1]. The cation in Y polarises the anion, distorting its electron cloud and adding attraction the model ignores [1]. Examiner insight: the comparison needs the direction, not just the size of the gap: the experimental value is more exothermic, so extra bonding is present.

6. (a) ΔsolH = lattice dissociation + ΔhydH(Mg²⁺) + 2 × ΔhydH(Cl⁻) [1] = 2524 – 1921 – 728 [1] = –125 kJ mol⁻¹ [1] (b) Mg²⁺ has a higher charge and a smaller radius than Na⁺ [1], so it attracts the δ– oxygen atoms of water molecules more strongly [1]. (c) –81 = 2258 + ΔhydH(Ca²⁺) + 2(–364), so ΔhydH(Ca²⁺) = –81 – 2258 + 728 [1] = –1611 kJ mol⁻¹ [1] Examiner insight: using ΔhydH(Cl⁻) once instead of twice gives +239 in (a), with the wrong sign as well as the wrong size.

7. Lattice enthalpy of dissociation = +747 [1] –1 = 747 – 406 + ΔhydH(Br⁻) [1] ΔhydH(Br⁻) = –342 kJ mol⁻¹ [1] [3] Examiner insight: the data give lattice formation, so reverse the sign before you build the solution cycle.

8. (a) Positive: a liquid becomes a gas, so disorder increases [1]. (b) Negative: 4 moles of gas become 2 moles of gas [1]. (c) Positive: a solid forms 2 moles of gas [1]. Examiner insight: a sign with no reason is incomplete; refer to states or to moles of gas.

9. (a) ΔS = 230 – (220 + 131) [1] = –121 J K⁻¹ mol⁻¹ [1] (b) ΔG = –137 – 298 × (–0.121) [1] = –101 kJ mol⁻¹ [1] (c) T = ΔH/ΔS = –137 / –0.121 [1] = 1132 K [1] Examiner insight: with ΔH and ΔS both negative, the reaction is feasible below 1132 K; reversing “above” and “below” turns a correct number into a wrong conclusion.

10. (a) A gas (CO) forms from solids only [1]. (b) ΔS = (42 + 198) – (44 + 6) [1] = +190 J K⁻¹ mol⁻¹ [1] (c) At the limit ΔG = 0, so T = ΔH/ΔS [1] = 238 / 0.190 [1] = 1253 K [1] (d) ΔG = 238 – 1500 × 0.190 = –47 kJ mol⁻¹ [1]. ΔG is negative, so the reaction is feasible at 1500 K [1]. Examiner insight: dividing 238 by 190 gives 1.25 K; check that a “feasible above” temperature for a reduction like this is a high temperature in kelvin.

11. (a) Gradient = (–36 – (–12)) / (700 – 400) = –0.080 kJ K⁻¹ mol⁻¹ [1] ΔS = –gradient = +80 J K⁻¹ mol⁻¹ [1] (b) ΔH = ΔG + TΔS = –12 + 400 × 0.080 [1] = +20 kJ mol⁻¹ [1] (c) Feasible at 250 K and above (ΔG ≤ 0 when T ≥ 20/0.080) [1] Examiner insight: the gradient is –ΔS, so a line sloping downwards means ΔS is positive; quoting –80 J K⁻¹ mol⁻¹ is a common sign error.

12. (a) At the boiling point, liquid and vapour are in equilibrium [1]. (b) Energy = 2200 × 90 = 198 000 J = 198 kJ [1] n(H₂O) = 82.0 / 18.0 = 4.56 mol [1] ΔH = 198 / 4.56 = +43.5 kJ mol⁻¹ [1] (c) ΔS = ΔH/T = 43 500 / 373 = +117 J K⁻¹ mol⁻¹ [1] (d) Too high [1]: some of the energy heats the kettle and surroundings instead of vaporising water, so the energy per mole vaporised is overestimated [1]. Examiner insight: in (c) keep ΔH in J mol⁻¹ to get ΔS in J K⁻¹ mol⁻¹; dividing 43.5 by 373 gives 0.117, which needs kJ K⁻¹ mol⁻¹ as its unit.

Where marks are usually lost

• Defining atomisation per mole of bonds broken instead of per mole of gaseous atoms formed.
• Writing a second ionisation energy from the atom rather than from the 1+ ion.
• Doubling only some of the steps when the formula contains two of a particle (Na₂O, MgCl₂).
• Using lattice enthalpy of formation in a solution cycle without changing its sign.
• Giving the sign of ΔS with no reason.
• Mixing J and kJ in ΔG = ΔH – TΔS.
• Stating the temperature but not whether the reaction is feasible above or below it.
• Reading the gradient of a ΔG–T graph as +ΔS.
• Missing units on ΔS (J K⁻¹ mol⁻¹) and signs on enthalpy values.

Next steps

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

Official syllabus

AQA, AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: section 3.1.8 Thermodynamics (A-level only).