← Back to subjects
⭐ 0
AQA A-level Chemistry (7405) · Thermodynamics
Mini-Lesson

Thermodynamics

This mini-lesson covers AQA 3.1.8 Thermodynamics: Born–Haber cycles and lattice enthalpy, enthalpies of solution and hydration, why the perfect ionic model fails, entropy ΔS, and Gibbs free energy ΔG = ΔH − TΔS — the true test of feasibility.

Born–Haber & lattice enthalpy entropy ΔS free energy ΔG a reaction is feasible when ΔG is negative — not just when it is exothermic

Work through each screen, answer the questions as you go (some are wordy, most are calculations) and collect ⭐ stars. Everything here is A-level standard — the maths is done properly, not skipped. Press Start when you're ready.

Thermodynamics · lattice enthalpy

Lattice enthalpy and the Born–Haber cycle

The lattice enthalpy of formation is the enthalpy change when 1 mol of a solid ionic lattice is formed from its gaseous ions. It is strongly exothermic. (The lattice enthalpy of dissociation is the exact reverse — same number, opposite sign. Read the question carefully.)

Na⁺(g) + Cl⁻(g) → NaCl(s) ΔHLE = −787 kJ mol⁻¹

Lattice enthalpy gets more exothermic when:

  • the ionic charges are larger (this matters most — the lattice enthalpy of MgO, about −3791 kJ mol⁻¹, is nearly five times that of NaCl, −787 kJ mol⁻¹), and
  • the ionic radii are smaller, so the centres of charge are closer together.

Lattice enthalpy cannot be measured directly, so we find it with a Born–Haber cycle — Hess’s law applied to an ionic solid:

ΔHf = ΔHat(metal) + IE + ΔHat(non-metal) + EA + ΔHLErearrange to find whichever term is unknown

Sign traps: ionisation energies are always positive. The first electron affinity is negative (exothermic) but the second is positive — you are forcing an electron onto an already negative ion.

Calculate

Your turn

1For NaCl: ΔHf = −411, ΔHat(Na) = +107, IE₁(Na) = +496, ΔHat(Cl) = +122, EA(Cl) = −349 (all kJ mol⁻¹). Calculate the lattice enthalpy of formation.
kJ mol⁻¹
Hint: ΔH_LE = ΔHf − (ΔHat(Na) + IE + ΔHat(Cl) + EA) = −411 − (107 + 496 + 122 − 349) = −411 − 376.
Quick check

Quick check

?Why is the lattice enthalpy of MgO far more exothermic than that of NaCl?
Thermodynamics · the ionic model

Perfect ionic model, polarisation and enthalpies of solution

A theoretical lattice enthalpy can be calculated by assuming the perfect ionic model: the ions are perfect spheres with their charge evenly distributed over the surface, and the bonding is 100% ionic.

The experimental (Born–Haber) value is usually more exothermic than the theoretical one. The difference reveals covalent character: a small, highly charged cation polarises a large anion, distorting its electron cloud so that electron density builds up between the ions. The bigger the discrepancy, the more covalent character — largest for compounds like AgI, and negligible for compounds of large 1+ cations with small anions.

Two more enthalpies dissolve the picture:

  • Enthalpy of hydration — 1 mol of gaseous ions dissolve in water to give an infinitely dilute solution. Always exothermic (ion–dipole attractions form).
  • Enthalpy of solution — 1 mol of solid dissolves to give an infinitely dilute solution.
ΔHsol = −ΔHLE(formation) + Σ ΔHhydbreak the lattice apart (endothermic), then hydrate the gaseous ions (exothermic)
Quick check

Quick check

?The experimental lattice enthalpy of AgI is much more exothermic than the value from the perfect ionic model. What does this show?
Thermodynamics · entropy

Entropy, ΔS

Entropy (S) measures the number of ways the particles and their energy can be arranged — loosely, the disorder of the system. Its units are J K⁻¹ mol⁻¹ (note: joules, not kilojoules — a constant source of lost marks).

  • solid < liquid < gas — gases have by far the highest entropy.
  • More moles of gas on the product side → ΔS positive.
  • Dissolving a solid usually increases entropy; a perfect crystal at 0 K has S = 0.
ΔS = Σ S(products) − Σ S(reactants)multiply each S° by its balancing number first
Worked example — N₂ + 3H₂ → 2NH₃

S°: N₂ = 191.6, H₂ = 130.6, NH₃ = 192.3 J K⁻¹ mol⁻¹

ΔS = 2(192.3) − [191.6 + 3(130.6)] = 384.6 − (191.6 + 391.8) = 384.6 − 583.4 = −198.8 J K⁻¹ mol⁻¹

Negative, as expected: 4 moles of gas become 2.

Calculate

Your turn

2For N₂ + 3H₂ → 2NH₃, S° values (J K⁻¹ mol⁻¹) are N₂ = 191.6, H₂ = 130.6, NH₃ = 192.3. Calculate ΔS for the reaction.
J K⁻¹ mol⁻¹
Hint: ΔS = 2(192.3) − [191.6 + 3(130.6)] = 384.6 − 583.4.
Sort it

Sort each change by the sign of its entropy change

Tap a change, then tap the sign of ΔS. Think about the change in the number of moles of gas.

🟩 ΔS positive

🟪 ΔS negative

🟦 ΔS about zero

Thermodynamics · free energy

Gibbs free energy: ΔG = ΔH − TΔS

Being exothermic is not the true test of feasibility. The correct test is the Gibbs free energy change:

ΔG = ΔH − TΔSa reaction is feasible when ΔG ≤ 0 · ΔH in kJ mol⁻¹ · ΔS must be converted from J to kJ (÷1000) · T in K

The four cases:

  • ΔH negative, ΔS positive → always feasible, at any temperature.
  • ΔH positive, ΔS negative → never feasible.
  • ΔH negative, ΔS negative → feasible only at low T (e.g. the Haber process).
  • ΔH positive, ΔS positive → feasible only at high T (e.g. the thermal decomposition of CaCO₃).

At the changeover, ΔG = 0, so:

T = ΔH / ΔSthe temperature at which the reaction just becomes (or stops being) feasible

The same caveat as E°cell: a negative ΔG means the reaction is thermodynamically feasible, but says nothing about the rate. Diamond → graphite has a negative ΔG at 298 K, yet diamonds do not visibly turn to graphite: the activation energy is enormous.

Calculate

Your turn

3For the Haber process, ΔH = −92.2 kJ mol⁻¹ and ΔS = −198.8 J K⁻¹ mol⁻¹. Calculate ΔG at 298 K.
kJ mol⁻¹
Hint: Convert ΔS to kJ: −0.1988 kJ K⁻¹ mol⁻¹. ΔG = ΔH − TΔS = −92.2 − (298 × −0.1988) = −92.2 + 59.2.
Calculate

Your turn

4Using the same data, calculate the temperature at which the Haber reaction stops being feasible (ΔG = 0).
K
Hint: At ΔG = 0, T = ΔH ÷ ΔS = (−92 200 J mol⁻¹) ÷ (−198.8 J K⁻¹ mol⁻¹).
Match it

Match the Born–Haber term to its definition

Tap a term on the left, then its definition on the right.

Born–Haber term
Definition (per mole)
Quick check

Quick check

?A reaction has ΔH = +178 kJ mol⁻¹ and ΔS = +161 J K⁻¹ mol⁻¹. Which statement is correct?
Quick check

Quick check

?ΔG for the conversion of diamond to graphite at 298 K is negative. Why do diamonds not turn into graphite?
Thermodynamics · exam traps

Units, signs and the entropy question

  • ΔS is in J K⁻¹ mol⁻¹; ΔH is in kJ mol⁻¹. Before using ΔG = ΔH − TΔS you must divide ΔS by 1000. This single error wrecks more thermodynamics answers than anything else.
  • Lattice enthalpy of FORMATION is negative; of DISSOCIATION is positive. Read which the question wants.
  • The first electron affinity is exothermic; the second is endothermic (you are forcing an electron onto an already negative ion).
  • ΔHat of a diatomic element makes ONE mole of gaseous atoms, so it is half the bond dissociation enthalpy: ½Cl₂(g) → Cl(g).
ΔHsol = −ΔHLE(formation) + Σ ΔHhydbreak the lattice up (endothermic), then hydrate the ions (exothermic) — and remember Σ means every ion, so 2 × the hydration enthalpy for MgCl₂
Quick check

Quick check

?Which combination makes a reaction feasible at all temperatures?
Calculate

Your turn

5For NaCl: lattice enthalpy of dissociation = +787, ΔHhyd(Na⁺) = −406, ΔHhyd(Cl⁻) = −363 kJ mol⁻¹. Calculate the enthalpy of solution.
kJ mol⁻¹
Hint: ΔH_sol = +787 + (−406) + (−363) = 787 − 769.
Recap

The big ideas to know

Lattice enthalpy: more exothermic for higher charge and smaller ionic radius — MgO ≫ NaCl

Born–Haber: ΔHf = ΔHat + IE + ΔHat(X) + EA + ΔHLE — rearrange for the unknown

Perfect ionic model: assumes perfect spheres with charge evenly spread; the experimental value is more exothermic because of covalent character (polarisation)

Solution: ΔHsol = −ΔHLE(formation) + Σ ΔHhyd

Entropy: ΔS = ΣS(products) − ΣS(reactants), in J K⁻¹ mol⁻¹ (note the J, not kJ)

Free energy: ΔG = ΔH − TΔS; feasible when ΔG ≤ 0; at the changeover T = ΔH / ΔS

That is the whole of AQA 3.1.8. Press Finish to see your score.

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Thermodynamics for AQA A-level Chemistry (7405). 🎉

Your stars: 0 / 0

Next: test yourself in the Evaluate stage Confidence Quiz, then lock it in with Verify.

📣 Smashed it? Share your score

Challenge a mate to beat your stars, or show a parent how you got on.

→ Back to all subjects