AQA A-Level Chemistry: Thermodynamics (7405)

This study guide covers 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. It teaches both subsections: 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 marked “A-level only”, so it is not part of the AS.

Use it with the Thermodynamics revision notes and the Thermodynamics practice questions. The AQA A-level Chemistry hub lists every topic, and the printable checklist lets you tick off each outcome. To find gaps quickly, try the free 10-minute diagnostics.

It builds on Energetics (enthalpy of formation, Hess’s law, bond enthalpy), Atomic structure (ionisation energy) and Bonding.

What this topic covers

Spec section	What you must be able to do	Status
3.1.8.1 Born–Haber cycles	Define lattice enthalpy (dissociation or formation) and the five other enthalpy terms; construct cycles to calculate a lattice enthalpy or any other single term; compare with perfect ionic model values as evidence for covalent character	A-level only
3.1.8.1 Born–Haber cycles	Define enthalpy of hydration; calculate enthalpies of solution and related changes from lattice and hydration enthalpies	A-level only
3.1.8.2 ΔG and ΔS	Explain why ΔH alone does not decide feasibility; calculate ΔS from absolute entropies; use ΔG = ΔH – TΔS for how ΔG varies with T and the temperature at which a reaction becomes feasible; find ΔH and ΔS from a ΔG against T graph	A-level only

The specification lists 3.1.8 under A-level Paper 1. A-level Paper 3 can test any content.

3.1.8.1 The enthalpy changes in a Born–Haber cycle

A Born–Haber cycle is a Hess’s law cycle for forming an ionic solid. Learn each definition with its equation.

Term	Definition	Example equation
Enthalpy of formation, ΔfH⦵	Enthalpy change when 1 mole of a compound is formed from its elements in their standard states	K(s) + ½Cl₂(g) → KCl(s)
Enthalpy of atomisation, ΔatH⦵	Enthalpy change when 1 mole of gaseous atoms is formed from the element in its standard state	K(s) → K(g)
Bond enthalpy	Enthalpy change to break 1 mole of a covalent bond, with reactants and products gaseous	Cl₂(g) → 2Cl(g)
First ionisation energy	Enthalpy change when 1 mole of electrons is removed from 1 mole of gaseous atoms to form 1 mole of gaseous 1+ ions	K(g) → K⁺(g) + e⁻
Second ionisation energy	Enthalpy change when 1 mole of electrons is removed from 1 mole of gaseous 1+ ions to form gaseous 2+ ions	Mg⁺(g) → Mg²⁺(g) + e⁻
First electron affinity	Enthalpy change when each atom in 1 mole of gaseous atoms gains one electron to form 1 mole of gaseous 1– ions	Cl(g) + e⁻ → Cl⁻(g)
Second electron affinity	Enthalpy change when each ion in 1 mole of gaseous 1– ions gains one electron to form gaseous 2– ions	O⁻(g) + e⁻ → O²⁻(g)
Lattice enthalpy of formation	Enthalpy change when 1 mole of a solid ionic compound is formed from its gaseous ions	K⁺(g) + Cl⁻(g) → KCl(s)
Lattice enthalpy of dissociation	Enthalpy change when 1 mole of a solid ionic compound is separated into its gaseous ions	KCl(s) → K⁺(g) + Cl⁻(g)

Points that cost marks:

• Lattice enthalpy has two definitions. Formation is negative; dissociation is positive, with the same size. Always say which one you mean.
• Atomisation of a diatomic gas is half the bond enthalpy. Cl₂(g) → 2Cl(g) makes 2 moles of atoms, so ΔatH⦵(Cl) = ½ × bond enthalpy. This only works when the element’s standard state is the gaseous molecule. Bromine is a liquid, so ΔatH⦵(Br) also includes vaporisation.
• Ionisation energies are always endothermic.
• First electron affinities are exothermic for non-metals such as Cl and O: the incoming electron is attracted by the nucleus. The second electron affinity of oxygen is endothermic: an electron is being added to a negative ion, which repels it.

Constructing a Born–Haber cycle

Draw the cycle as an energy level diagram: each species on its own horizontal level, endothermic steps as up arrows and exothermic steps as down arrows. For KCl, the indirect route is:

K(s) + ½Cl₂(g)                     start: elements in standard states
  → K(g) + ½Cl₂(g)       ΔatH(K)          up (endothermic)
  → K(g) + Cl(g)         ΔatH(Cl)         up
  → K⁺(g) + e⁻ + Cl(g)   first IE(K)      up
  → K⁺(g) + Cl⁻(g)       first EA(Cl)     down (exothermic)
  → KCl(s)               lattice enthalpy of formation, down
Direct route: K(s) + ½Cl₂(g) → KCl(s)     ΔfH, down

The route from the elements to the solid can go two ways: directly (ΔfH⦵), or through gaseous atoms and gaseous ions. Hess’s law makes the two routes equal:

ΔfH⦵ = ΔatH(metal) + ΔatH(non-metal) + IE(s) + EA(s) + lattice enthalpy of formation

Worked example 1: lattice enthalpy of potassium chloride

Data (kJ mol⁻¹): ΔfH⦵(KCl) –437; ΔatH⦵(K) +89; first IE(K) +419; Cl–Cl bond enthalpy +242; first EA(Cl) –349.

ΔatH(Cl) = ½ × 242 = +121
Sum of steps to gaseous ions = 89 + 419 + 121 + (–349) = +280
Lattice enthalpy of formation = ΔfH – (sum of steps)
                              = –437 – 280 = –717 kJ mol⁻¹

The lattice enthalpy of dissociation of KCl is therefore +717 kJ mol⁻¹.

Worked example 2: finding a missing term (electron affinity)

A cycle can be rearranged for any single unknown. For MgCl₂, every chlorine step happens twice.

Data (kJ mol⁻¹): ΔfH⦵(MgCl₂) –641; ΔatH⦵(Mg) +148; first IE(Mg) +738; second IE(Mg) +1451; Cl–Cl bond enthalpy +242; lattice enthalpy of formation of MgCl₂ –2524.

Two Cl atoms: 2 × ΔatH(Cl) = 242 (one whole Cl–Cl bond)
ΔfH = 148 + 738 + 1451 + 242 + 2EA + (–2524)
–641 = 2579 + 2EA – 2524
2EA = –641 – 2579 + 2524 = –696
EA(Cl) = –348 kJ mol⁻¹

Both ionisation energies appear because Mg must become Mg²⁺, and two EAs because two Cl⁻ ions form.

Comparing with the perfect ionic model

A lattice enthalpy can also be calculated from a model. The perfect ionic model assumes:

• the ions are perfect spheres with their charge spread evenly;
• the only attraction is electrostatic between oppositely charged ions (no covalent character).

The Born–Haber value comes from experimental data, so it reflects the real bonding.

• Values close together: the compound is close to perfectly ionic.
• Born–Haber value more exothermic than the model value: there is extra attraction the model leaves out. This is evidence for covalent character.

Covalent character arises when a small or highly charged cation polarises a large anion, pulling its electron cloud towards the cation.

Compound (illustrative data)	Born–Haber (kJ mol⁻¹)	Perfect ionic model (kJ mol⁻¹)	Conclusion
Compound A	–760	–755	Nearly perfectly ionic
Compound B	–890	–820	Significant covalent character

Enthalpies of solution and hydration

Enthalpy of hydration, ΔhydH⦵: the enthalpy change when 1 mole of gaseous ions dissolves in water to form 1 mole of aqueous ions. Example: K⁺(g) → K⁺(aq). It is always exothermic, because water molecules are attracted to the ion: the δ– oxygen towards cations and the δ+ hydrogens towards anions.

Enthalpy of solution, ΔsolH⦵: the enthalpy change when 1 mole of an ionic solid dissolves in enough water that the ions no longer interact. Example: KCl(s) → K⁺(aq) + Cl⁻(aq).

The cycle goes from the solid to gaseous ions (lattice dissociation, endothermic) and then from gaseous ions to aqueous ions (hydration, exothermic):

ΔsolH = lattice enthalpy of dissociation + Σ ΔhydH(ions)

Worked example 3: enthalpy of solution of potassium chloride

Data (kJ mol⁻¹): lattice dissociation of KCl +717 (worked example 1); ΔhydH(K⁺) –340; ΔhydH(Cl⁻) –360.

ΔsolH = +717 + (–340) + (–360) = +17 kJ mol⁻¹

The positive value means the solution cools as KCl dissolves. For MgCl₂, include ΔhydH(Cl⁻) twice. If you are given lattice enthalpy of formation, reverse its sign first.

3.1.8.2 Entropy and feasibility

Why ΔH is not enough

Some endothermic changes are feasible: ice melts at room temperature, and ammonium nitrate dissolves in water as the solution gets colder. ΔH alone cannot explain this. The missing factor is entropy.

Entropy, S, is a measure of disorder: the number of ways particles and their energy can be arranged. Units: J K⁻¹ mol⁻¹. Every substance has a positive absolute entropy, including elements.

Entropy increases when:

• a solid melts or a liquid boils (solid < liquid < gas);
• a reaction produces more moles of gas than it uses.

Calculating ΔS from absolute entropies

ΔS = ΣS(products) – ΣS(reactants)

Multiply each value by its coefficient in the equation. Do not set elements to zero; that rule is for ΔfH⦵, not for S.

The Gibbs free-energy equation

The balance between enthalpy and entropy is given by

ΔG = ΔH – TΔS        (derivation not required)

with T in kelvin. A reaction is feasible when ΔG is zero or negative. ΔH is usually in kJ mol⁻¹ and ΔS in J K⁻¹ mol⁻¹, so divide ΔS by 1000 before substituting.

At the temperature where ΔG = 0, the reaction is just feasible:

T = ΔH / ΔS        (ΔS in kJ K⁻¹ mol⁻¹)

How ΔG varies with temperature depends on the signs:

ΔH	ΔS	Effect of temperature	Feasible
–	+	ΔG negative at every T	Always
+	–	ΔG positive at every T	Never
–	–	ΔG becomes less negative as T rises	Below T = ΔH/ΔS
+	+	ΔG becomes more negative as T rises	Above T = ΔH/ΔS

“Feasible” means thermodynamically possible. A feasible reaction can still be very slow if its activation energy is high.

Worked example 4: thermal decomposition of calcium carbonate

CaCO₃(s) → CaO(s) + CO₂(g), ΔH⦵ = +178 kJ mol⁻¹. Absolute entropies (J K⁻¹ mol⁻¹): CaCO₃(s) 93; CaO(s) 40; CO₂(g) 214.

ΔS = (40 + 214) – 93 = +161 J K⁻¹ mol⁻¹ = +0.161 kJ K⁻¹ mol⁻¹
ΔG at 298 K = 178 – (298 × 0.161) = +130 kJ mol⁻¹   → not feasible
T at which feasible = 178 / 0.161 = 1106 K

ΔS is positive because a gas forms from a solid. The reaction is feasible at 1106 K and above.

Worked example 5: ΔH and ΔS from a graph of ΔG against T

ΔG = ΔH – TΔS has the form y = c + mx, so a straight-line graph of ΔG against T has:

• gradient = –ΔS
• intercept on the ΔG axis (at T = 0) = ΔH

A line passes through (300 K, +40 kJ mol⁻¹) and (500 K, +10 kJ mol⁻¹).

gradient = (10 – 40) / (500 – 300) = –0.150 kJ K⁻¹ mol⁻¹
ΔS = +0.150 kJ K⁻¹ mol⁻¹ = +150 J K⁻¹ mol⁻¹
ΔH = ΔG + TΔS = 40 + (300 × 0.150) = +85 kJ mol⁻¹
Feasible from T = 85 / 0.150 = 567 K, where the line crosses ΔG = 0

Worked example 6: ΔS of vaporisation using a kettle

The specification suggests finding ΔS for vaporising water with a kettle. At the boiling point, liquid and gas are in equilibrium, so ΔG = 0 and ΔS = ΔH / T.

A 2.00 kW kettle keeps water boiling at 373 K for 60 s, and the mass falls by 50.0 g.

Energy supplied = 2000 W × 60 s = 120 000 J = 120 kJ
n(H₂O) = 50.0 / 18.0 = 2.78 mol
ΔvapH = 120 / 2.78 = 43.2 kJ mol⁻¹
ΔS = 43 200 / 373 = +116 J K⁻¹ mol⁻¹

Some energy heats the surroundings rather than vaporising water, so this method overestimates ΔvapH and ΔS.

Common errors

• Forgetting the second ionisation energy for a 2+ ion, or the second electron affinity for O²⁻.
• Including only one hydration enthalpy for an ion that appears twice in the formula.
• Setting the absolute entropy of an element to zero.
• Mixing ΔS in J K⁻¹ mol⁻¹ with ΔH in kJ mol⁻¹, or using T in °C.
• Taking the graph gradient as +ΔS: the gradient is –ΔS.

Next steps

Revise from the revision notes, test yourself with the practice questions, and read the exam preparation guide.

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).