OxfordAQA A-Level Chemistry: Thermodynamics (9620) – Revision Notes

These revision notes 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. All of it is International A2 content, so it is International A-level only. For full explanations and worked examples, read the Thermodynamics study guide.

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

3.1.8.1 The terms you must define

The specification lists five enthalpy terms used as data in a Born–Haber cycle, plus lattice enthalpy and enthalpy of hydration. Every definition is per mole and comes with an equation.

Term	Key words in the definition	Equation pattern	Sign
Enthalpy of formation	1 mol of compound from its elements in standard states	Sr(s) + Br₂(l) → SrBr₂(s)	Usually –
Enthalpy of atomisation	1 mol of gaseous atoms from the element in its standard state	½Br₂(l) → Br(g)	+
Bond enthalpy	1 mol of a covalent bond broken; all species gaseous	Br₂(g) → 2Br(g)	+
First ionisation energy	1 mol of electrons removed from 1 mol of gaseous atoms	Sr(g) → Sr⁺(g) + e⁻	+
Second ionisation energy	1 mol of electrons removed from 1 mol of gaseous 1+ ions	Sr⁺(g) → Sr²⁺(g) + e⁻	+
First electron affinity	Each atom in 1 mol of gaseous atoms gains one electron	Br(g) + e⁻ → Br⁻(g)	–
Lattice enthalpy of dissociation	1 mol of ionic solid → its gaseous ions	SrBr₂(s) → Sr²⁺(g) + 2Br⁻(g)	+
Lattice enthalpy of formation	Gaseous ions → 1 mol of ionic solid	Sr²⁺(g) + 2Br⁻(g) → SrBr₂(s)	–
Enthalpy of hydration	1 mol of gaseous ions → 1 mol of aqueous ions	Sr²⁺(g) → Sr²⁺(aq)	–

Remember: lattice enthalpy can be defined either way. The two have the same size and opposite signs, so always say which one you mean.

Formula table

Use	Relationship	Watch for
Born–Haber cycle	ΔfH = ΣΔatH + ΣIE + ΣEA + lattice enthalpy of formation	Multiply each step by how often it occurs
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 every ion in the formula
Entropy change	ΔS = ΣS(products) – ΣS(reactants)	Units J K⁻¹ mol⁻¹
Gibbs free-energy change	ΔG = ΔH – TΔS	T in K; ΔS ÷ 1000 to match kJ
Feasibility	ΔG ≤ 0	“Zero or negative”
Temperature at which a reaction becomes feasible	T = ΔH / ΔS	From ΔG = 0

The specification states that the derivation of ΔG = ΔH – TΔS is not required.

Born–Haber cycle: method in steps

1. Write the formation equation for the solid, with state symbols.
2. Atomise the metal (× number of metal atoms).
3. Atomise the non-metal. For a gaseous diatomic element, two atoms need one whole bond enthalpy.
4. Ionise the metal: one ionisation energy per electron lost.
5. Add an electron affinity for each electron gained (× number of anions).
6. Close the cycle with the lattice enthalpy of formation.
7. Set the two routes equal (Hess’s law) and solve for the single unknown.
8. Sense-check: atomisations and ionisations positive; lattice formation negative.

Small reminder (Al₂O₃ counting): 2 × ΔatH(Al); 2 × (first + second + third IE of Al); 3 × ΔatH(O), which is 1½ O=O bonds; 3 × (first + second EA of O).

Perfect ionic model: what to write

• The model: perfectly spherical ions, charge evenly spread, purely electrostatic attraction.
• Born–Haber value close to the model value: almost purely ionic bonding.
• Born–Haber value more exothermic than the model value: extra attraction, so covalent character.
• Cause: the cation polarises the anion, distorting its electron cloud towards the cation.
• Most polarisation: small, highly charged cation with a large anion (for example, iodides more than fluorides of the same cation).

Enthalpy of solution: method in steps

1. ΔsolH = lattice enthalpy of dissociation + hydration enthalpies of all the ions.
2. Given a lattice enthalpy of formation? Reverse its sign first.
3. Multiply each ΔhydH by the number of those ions in the formula.
4. If ΔsolH is given, rearrange for the unknown.

Why hydration is exothermic: cations attract the δ– oxygen end of water molecules and anions attract the δ+ hydrogen end. Smaller and more highly charged ions attract water more strongly, so their ΔhydH is more exothermic.

3.1.8.2 Entropy and Gibbs free energy

Why ΔH is not sufficient: some endothermic changes happen on their own (melting of ice above 273 K, many salts dissolving). Feasibility also depends on entropy, the degree of disorder.

ΔS is positive for:

• physical changes towards more disorder: solid → liquid → gas;
• chemical changes that increase the moles of gas, or turn a solid or liquid into a gas.

ΔS is negative for the reverse of these.

ΔG: method in steps

1. Calculate ΔS from absolute entropies, using coefficients.
2. Divide ΔS by 1000 to get kJ K⁻¹ mol⁻¹.
3. Use T in kelvin.
4. ΔG = ΔH – TΔS. Feasible if ΔG ≤ 0.
5. For the temperature at which a reaction becomes feasible, set ΔG = 0: T = ΔH/ΔS.
6. Use the signs to say “above” or “below” that temperature.

How ΔG varies with temperature

ΔH	ΔS	Effect of raising T	Feasible
–	+	ΔG more negative	Always
+	–	ΔG more positive	Never
+	+	ΔG falls through zero	Above T = ΔH/ΔS
–	–	ΔG rises through zero	Below T = ΔH/ΔS

Small reminder: a reaction with ΔH = +30 kJ mol⁻¹ and ΔS = +80 J K⁻¹ mol⁻¹ has ΔG = 30 – 300 × 0.080 = +6 kJ mol⁻¹ at 300 K (not feasible) but ΔG = 30 – 400 × 0.080 = –2 kJ mol⁻¹ at 400 K (feasible).

Must-know distinctions

• Atomisation vs bond enthalpy: per mole of atoms formed vs per mole of bonds broken.
• Lattice formation vs dissociation: same size, opposite sign.
• Hydration vs solution: gaseous ions → aqueous ions vs solid → aqueous ions.
• First vs second electron affinity of oxygen: exothermic vs endothermic.
• Feasible vs fast: ΔG ≤ 0 shows a reaction is thermodynamically possible but says nothing about its rate. A high activation energy can make a feasible reaction too slow to observe.
• Absolute entropy vs ΔfH⦵ of an element: S of an element is never zero; ΔfH⦵ of an element in its standard state is zero.

Quick self-test

1. Write an equation for the lattice enthalpy of dissociation of BaF₂.
2. The H–H bond enthalpy is +436 kJ mol⁻¹. State ΔatH⦵ of hydrogen.
3. Write the equation for the enthalpy of atomisation of iodine.
4. How many first electron affinities of chlorine appear in the Born–Haber cycle for AlCl₃?
5. Lattice enthalpy of dissociation of CsCl = +676 kJ mol⁻¹; ΔhydH(Cs⁺) = –276; ΔhydH(Cl⁻) = –381. Calculate ΔsolH(CsCl).
6. ΔsolH(BaBr₂) = –25 kJ mol⁻¹; lattice enthalpy of dissociation = +1985; ΔhydH(Br⁻) = –335. Calculate ΔhydH(Ba²⁺).
7. A compound’s Born–Haber lattice enthalpy of formation is –735 kJ mol⁻¹ and its perfect ionic model value is –733 kJ mol⁻¹. What does this show?
8. State the sign of ΔS for 2KClO₃(s) → 2KCl(s) + 3O₂(g), with a reason.
9. For 2NO(g) + O₂(g) → 2NO₂(g), ΔH = –114 kJ mol⁻¹. Absolute entropies (J K⁻¹ mol⁻¹): NO 211, O₂ 205, NO₂ 240. Calculate ΔS and the temperature range over which the reaction is feasible.
10. State what happens to ΔG as temperature rises when ΔS is negative.

Answers

1. BaF₂(s) → Ba²⁺(g) + 2F⁻(g)
2. +218 kJ mol⁻¹ (half of 436, per mole of H atoms).
3. ½I₂(s) → I(g)
4. Three: three Cl⁻ ions form.
5. ΔsolH = 676 – 276 – 381 = +19 kJ mol⁻¹
6. –25 = 1985 + ΔhydH(Ba²⁺) + 2(–335), so ΔhydH(Ba²⁺) = –1340 kJ mol⁻¹
7. The values almost match, so the compound is very close to purely ionic (little covalent character).
8. Positive: a solid forms 3 moles of gas, so disorder increases.
9. ΔS = 2(240) – [2(211) + 205] = –147 J K⁻¹ mol⁻¹; T = 114/0.147 = 776 K; feasible below 776 K.
10. –TΔS is positive and grows, so ΔG becomes more positive (less negative).

Where marks are usually lost

• Defining lattice enthalpy without “1 mole” of solid or without “gaseous ions”.
• Writing Br₂(g) → 2Br(g) or I₂(s) → 2I(g) for an atomisation instead of forming 1 mole of atoms.
• Missing the second (or third) ionisation energy for a 2+ (or 3+) ion.
• Treating a second electron affinity as exothermic.
• Using lattice enthalpy of formation in a solution cycle without reversing the sign.
• Counting a hydration enthalpy once when the formula holds two of that ion.
• Saying a Born–Haber value that is less exothermic than the model shows covalent character.
• Substituting ΔS in J K⁻¹ mol⁻¹ with ΔH in kJ mol⁻¹, or using °C.
• Giving T = ΔH/ΔS with no statement of “above” or “below”.

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).