AQA A-Level Chemistry: Thermodynamics (7405) – Revision Notes

These revision notes 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. All of it is A-level only. For full explanations and six worked examples, read the Thermodynamics study guide.

Course links: AQA A-level Chemistry hub | printable checklist | Thermodynamics practice questions | Energetics revision notes | free diagnostics

3.1.8.1 Definitions to learn word for word

Every term refers to 1 mole. Write the equation with state symbols whenever you define one.

Term	Definition	Sign
Enthalpy of formation, ΔfH⦵	1 mole of a compound formed from its elements in their standard states	Usually –
Enthalpy of atomisation, ΔatH⦵	1 mole of gaseous atoms formed from the element in its standard state	+
Bond enthalpy	1 mole of a covalent bond broken, all species gaseous	+
First ionisation energy	1 mole of electrons removed from 1 mole of gaseous atoms to form 1 mole of gaseous 1+ ions	+
Second ionisation energy	1 mole of electrons removed from 1 mole of gaseous 1+ ions to form gaseous 2+ ions	+
First electron affinity	Each atom in 1 mole of gaseous atoms gains one electron to form 1 mole of gaseous 1– ions	– (for Cl, O)
Second electron affinity	Each ion in 1 mole of gaseous 1– ions gains one electron to form gaseous 2– ions	+ (for O)
Lattice enthalpy of formation	1 mole of solid ionic compound formed from its gaseous ions	–
Lattice enthalpy of dissociation	1 mole of solid ionic compound separated into its gaseous ions	+
Enthalpy of hydration, ΔhydH⦵	1 mole of gaseous ions dissolved in water to form 1 mole of aqueous ions	–
Enthalpy of solution, ΔsolH⦵	1 mole of ionic solid dissolved in enough water that the ions no longer interact	+ or –

Equations to be able to write:

Atomisation of chlorine:      ½Cl₂(g) → Cl(g)
Second IE of calcium:         Ca⁺(g) → Ca²⁺(g) + e⁻
Second EA of oxygen:          O⁻(g) + e⁻ → O²⁻(g)
Lattice formation of MgO:     Mg²⁺(g) + O²⁻(g) → MgO(s)
Hydration of a chloride ion:  Cl⁻(g) → Cl⁻(aq)

Formula table

Use	Relationship	Notes
Born–Haber cycle	ΔfH = ΣΔatH + ΣIE + ΣEA + lattice enthalpy of formation	Multiply each step by how many times it happens
Atomisation of a gaseous diatomic element	ΔatH = ½ × bond enthalpy	Not for Br₂(l) or I₂(s)
Enthalpy of solution	ΔsolH = lattice enthalpy of dissociation + ΣΔhydH	Count each ion
Entropy change	ΔS = ΣS(products) – ΣS(reactants)	J K⁻¹ mol⁻¹; elements are not zero
Gibbs free-energy change	ΔG = ΔH – TΔS	T in K; ΔS in kJ K⁻¹ mol⁻¹
Feasibility	ΔG ≤ 0	“Zero or negative”
Temperature at which a reaction becomes feasible	T = ΔH / ΔS	Set ΔG = 0
Graph of ΔG against T	gradient = –ΔS; intercept = ΔH	Straight line

Born–Haber cycles: method in steps

1. Write the formation equation, e.g. Ca(s) + Cl₂(g) → CaCl₂(s).
2. Atomise the metal: ΔatH(metal) × number of metal atoms.
3. Atomise the non-metal: for Cl₂ forming 2 atoms, use the whole bond enthalpy; for 1 atom, half.
4. Ionise the metal: add every ionisation energy needed (first and second for a 2+ ion).
5. Add electron affinities: first EA × number of anions; add the second EA for O²⁻.
6. Close the cycle with the lattice enthalpy of formation.
7. Hess’s law: route via gaseous ions = direct route (ΔfH). Solve for the one unknown.
8. Check the sign makes sense: lattice formation is negative; ionisation energies are positive.

Small reminder. Na₂O needs 2 × ΔatH(Na) and 2 × first IE(Na), but only one oxygen atom: ½ × O=O bond enthalpy, then first EA(O) and second EA(O).

Perfect ionic model: what to say

• The model assumes perfectly spherical ions with evenly spread charge and purely electrostatic attraction.
• If the Born–Haber (experimental) value is close to the model value, the compound is close to perfectly ionic.
• If the Born–Haber value is more exothermic, there is covalent character: the cation polarises the anion’s electron cloud, giving extra attraction the model does not include.
• Polarisation is greater with a small, highly charged cation and a large anion.

Enthalpy of solution: method in steps

1. Write ΔsolH = lattice enthalpy of dissociation + hydration enthalpies of all ions.
2. If you are given lattice enthalpy of formation, change its sign.
3. Multiply each ΔhydH by the number of that ion in the formula.
4. Rearrange for the unknown if ΔsolH is given.

Hydration enthalpies are exothermic because water’s δ– oxygen is attracted to cations and its δ+ hydrogens to anions.

3.1.8.2 Entropy and Gibbs free energy

Why ΔH is not enough: some endothermic changes happen readily (melting ice above 273 K, dissolving ammonium nitrate), so feasibility also depends on entropy, a measure of disorder.

ΔS is positive when: solid → liquid → gas; the number of moles of gas increases. ΔS is negative when: gases combine into fewer moles of gas, or a gas or liquid forms a solid.

ΔG: method in steps

1. Calculate ΔS from absolute entropies, multiplying by coefficients.
2. Convert ΔS to kJ K⁻¹ mol⁻¹ (divide by 1000).
3. Convert T to kelvin.
4. Substitute into ΔG = ΔH – TΔS.
5. Feasible if ΔG is zero or negative.
6. For the temperature at which it becomes feasible, set ΔG = 0: T = ΔH/ΔS.
7. Decide “above” or “below” using the signs table.

ΔH	ΔS	Feasible
–	+	At all temperatures
+	–	At no temperature
–	–	Below T = ΔH/ΔS
+	+	Above T = ΔH/ΔS

Graph of ΔG against T

ΔG = ΔH – TΔS has the form y = c + mx. The gradient is –ΔS and the intercept at T = 0 is ΔH. A line sloping down means ΔS is positive. The line crosses ΔG = 0 at the temperature where the reaction becomes feasible.

ΔS of vaporisation using a kettle

At the boiling point, liquid and vapour are in equilibrium, so ΔG = 0 and ΔS = ΔH/T. Energy supplied = power × time; moles vaporised = mass lost ÷ 18.0. Heat lost to the surroundings makes ΔH, and so ΔS, too high.

Must-know distinctions

• Lattice formation vs dissociation: same size, opposite sign.
• Atomisation vs bond enthalpy: per mole of atoms vs per mole of bonds.
• First vs second electron affinity of oxygen: exothermic vs endothermic.
• Enthalpy of hydration vs enthalpy of solution: gaseous ions → aqueous ions vs solid → aqueous ions.
• Feasible vs fast: ΔG ≤ 0 says a reaction can happen, not how quickly.
• S vs ΔfH⦵ for elements: S of an element is not zero; ΔfH⦵ of an element in its standard state is.

Quick self-test

1. Write an equation for the lattice enthalpy of dissociation of CaF₂.
2. The F–F bond enthalpy is +158 kJ mol⁻¹. State ΔatH⦵(F).
3. Explain why ΔatH⦵(Br) is not half the Br–Br bond enthalpy.
4. State the sign of the second electron affinity of oxygen and explain it.
5. Lattice enthalpy of dissociation of LiF = +1037 kJ mol⁻¹; ΔhydH(Li⁺) = –519; ΔhydH(F⁻) = –506. Calculate ΔsolH(LiF).
6. ΔsolH(NaI) = –8 kJ mol⁻¹; lattice enthalpy of dissociation of NaI = +705; ΔhydH(Na⁺) = –406. Calculate ΔhydH(I⁻).
7. State the sign of ΔS for 2H₂(g) + O₂(g) → 2H₂O(l), with a reason.
8. For 2SO₂(g) + O₂(g) → 2SO₃(g), ΔH = –198 kJ mol⁻¹. S (J K⁻¹ mol⁻¹): SO₂ 248, O₂ 205, SO₃ 257. Calculate ΔS and the temperature below which the reaction is feasible.
9. ΔH = –50 kJ mol⁻¹ and ΔS = –100 J K⁻¹ mol⁻¹. Is the reaction feasible at 400 K?
10. A ΔG against T line has gradient –0.120 kJ K⁻¹ mol⁻¹. State ΔS.
11. A Born–Haber lattice enthalpy is much more exothermic than the perfect ionic model value. What does this show?

Answers

1. CaF₂(s) → Ca²⁺(g) + 2F⁻(g)
2. +79 kJ mol⁻¹ (half of 158).
3. Bromine’s standard state is liquid, so atomisation also includes vaporising Br₂(l).
4. Positive (endothermic): the electron is added to a negative ion, which repels it.
5. ΔsolH = 1037 – 519 – 506 = +12 kJ mol⁻¹
6. –8 = 705 – 406 + ΔhydH(I⁻), so ΔhydH(I⁻) = –307 kJ mol⁻¹
7. Negative: 3 moles of gas become a liquid, so disorder decreases.
8. ΔS = 2(257) – [2(248) + 205] = –187 J K⁻¹ mol⁻¹; T = 198/0.187 = 1059 K; feasible below this.
9. ΔG = –50 – 400(–0.100) = –10 kJ mol⁻¹, so yes, it is feasible.
10. ΔS = +120 J K⁻¹ mol⁻¹ (gradient = –ΔS).
11. The compound has covalent character: the cation polarises the anion.

Where marks are usually lost

• Defining lattice enthalpy without saying “gaseous ions” or “1 mole of solid”.
• Leaving out state symbols in definition equations, or writing Cl₂(g) → 2Cl(g) for atomisation.
• Using one ionisation energy for a 2+ ion.
• Forgetting the second electron affinity for O²⁻, or making it negative.
• Using lattice enthalpy of formation in an enthalpy-of-solution cycle without reversing its sign.
• Counting a hydration enthalpy once when the formula has two of that ion.
• Leaving ΔS in J K⁻¹ mol⁻¹ when ΔH is in kJ mol⁻¹.
• Using °C instead of K.
• Saying “feasible above” when ΔH and ΔS are both negative.
• Missing units on a final ΔS (J K⁻¹ mol⁻¹) or ΔG (kJ mol⁻¹).

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).