Born-Haber Cycles, Entropy and Free Energy

A-Level Chemistry · Energetics and Thermodynamics

Lattice enthalpy and Born-Haber cycles

Lattice enthalpy measures the strength of ionic bonding. Lattice formation enthalpy is the energy released when one mole of an ionic solid forms from its gaseous ions (exothermic); lattice dissociation is the reverse.

A Born-Haber cycle is an enthalpy cycle (an application of Hess's law) used to find the lattice enthalpy indirectly, because it can't be measured directly. It links the enthalpy of formation to a series of steps:

Atomisation (of the metal and non-metal) — form gaseous atoms.
Ionisation energy (metal) — form positive gaseous ions.
Electron affinity (non-metal) — form negative gaseous ions.
Lattice formation enthalpy — gaseous ions → solid lattice.

Applying Hess's law around the cycle lets you solve for the unknown term.

Factors affecting lattice enthalpy

Lattice enthalpy is more exothermic (stronger ionic bonding) when:

ions are smaller (charges are closer together);
ions have a higher charge.

So MgO (2+ and 2−, small ions) has a much larger lattice enthalpy than NaCl (1+, 1−).

Perfect ionic model vs reality

Comparing the experimental (Born-Haber) lattice enthalpy with the theoretical (perfect ionic model) value shows how ionic a compound really is — a big difference indicates covalent character (polarisation of the anion by the cation).

Entropy (ΔS)

Entropy is a measure of disorder. Systems tend towards greater disorder.

Entropy increases: solid → liquid → gas; more gas moles produced; dissolving a solid.
Units: J K⁻¹ mol⁻¹.

ΔS = Σ S(products) − Σ S(reactants)

Gibbs free energy (ΔG)

Whether a reaction is feasible (spontaneous) depends on both enthalpy and entropy:

ΔG = ΔH − TΔS

(T in kelvin). A reaction is feasible when ΔG ≤ 0.

Exothermic (ΔH negative) + increasing entropy (ΔS positive) → always feasible.
Temperature can make an unfavourable reaction feasible (e.g. endothermic reactions with positive ΔS become feasible at high T).

Worked example

A reaction has ΔH = +30 kJ mol⁻¹ and ΔS = +100 J K⁻¹ mol⁻¹. Find the minimum temperature for feasibility.

Feasible when ΔG ≤ 0 → ΔH = TΔS → T = ΔH ÷ ΔS = 30000 ÷ 100 = 300 K. ✓ (Convert ΔH to J.)

Common mistakes

Forgetting to convert ΔH to J (or ΔS to kJ) so the units match in ΔG = ΔH − TΔS.
Mixing up lattice formation (exothermic) and dissociation (endothermic).
Forgetting temperature is in kelvin.

Exam tips

Draw Born-Haber cycles carefully with correct arrow directions and apply Hess's law.
Relate lattice enthalpy to ionic size and charge.
Use ΔG = ΔH − TΔS for feasibility; find the temperature where ΔG = 0.

Key facts to remember

Born-Haber cycle finds lattice enthalpy via atomisation, ionisation, electron affinity and formation (Hess's law).
Lattice enthalpy more exothermic with smaller, higher-charged ions.
Entropy (ΔS) = disorder; ΔG = ΔH − TΔS, feasible when ΔG ≤ 0 (watch units and kelvin).