Lattice Energy and Born–Haber Cycles

Lattice energy ( $\Delta H_\text{latt}$ ) is the enthalpy change when 1 mol of an ionic solid forms from its gaseous ions (exothermic). It can't be measured directly, so we use a Born–Haber cycle (Hess's law): $\Delta_fH=\Delta_\text{at}H(\text{metal})+\text{IE}+\Delta_\text{at}H(\text{non-metal})+\text{EA}+\Delta H_\text{latt}$ . Lattice energy is more exothermic for smaller ions and higher charges (stronger attraction). Differences between experimental and theoretical (perfect ionic) values show polarisation / covalent character. For dissolving: $\Delta_\text{sol}H=-\Delta H_\text{latt}+\sum\Delta_\text{hyd}H$ .

Cambridge 9701 Paper 4Edexcel YCH11 WCH14AQA 7405 Paper 1≈ 50 minRevision 1 · updated 2026-10-30

Section 1 of 14

Learning objectives

Define the enthalpy terms in a Born–Haber cycle.
Calculate lattice energy or another unknown from a cycle.
Explain the effect of ionic charge and radius.
Use hydration enthalpies to find enthalpy of solution.

Pause and think

Why is the lattice energy of MgO about four times larger than that of NaF, even though the ions are similar in size?

Section 2 of 14

Core idea

Start hereMeasuring the unmeasurable

Nobody can take a mole of gaseous Na⁺ and Cl⁻ ions and let them snap together in a calorimeter. Yet we know the energy released to within a few kJ. Hess's law makes it possible.

The big idea

A Born–Haber cycle builds an ionic solid two ways: directly from its elements, or step by step through gaseous atoms and ions. Both routes have the same total enthalpy change.

1. The cycle for NaCl

Born–Haber cycle for sodium chloride

Drawing diagram…

Endothermic steps go up; exothermic steps go down.

Finding the lattice energy

$\Delta_fH=\Delta_\text{at}H_\mathrm{Na}+\text{IE}+\Delta_\text{at}H_\mathrm{Cl}+\text{EA}+\Delta H_\text{latt}$
Hess's law: both routes from elements to solid are equal.
$-411=107+496+122-349+\Delta H_\text{latt}$
Substitute the data.
$\Delta H_\text{latt}=-411-376=-787\ \text{kJ mol}^{-1}$
Rearrange.

Worked exampleEnthalpy of solution

Lattice energy of NaCl $=-787$ ; hydration enthalpies Na⁺ $-406$ , Cl⁻ $-377$ kJ mol⁻¹. Find $\Delta_\text{sol}H$ .

Goal: $\Delta_\text{sol}H$ .
Strategy: Break the lattice (reverse lattice energy), then hydrate the ions.

$\text{break lattice}=+787$
Reverse of lattice formation.
$\Delta_\text{sol}H=+787-406-377=+4\ \text{kJ mol}^{-1}$
Add hydration enthalpies.

Answer+4 kJ mol⁻¹ (slightly endothermic).

✓ CheckSalt feels very slightly cold when dissolving ✓.

Key takeawayCharge and size Lattice energy is more exothermic for higher charges and smaller ions: MgO ≫ NaCl > KBr.

Summary

Lattice energy from gaseous ions, exothermic.
Born–Haber cycle = Hess's law.
Charge ↑, radius ↓ → larger lattice energy.
Solution = −lattice + hydration.

Lattice energy: Enthalpy change when 1 mol of ionic solid forms from its gaseous ions.
Electron affinity: Enthalpy change when 1 mol of gaseous atoms each gain one electron.
Hydration enthalpy: Enthalpy change when 1 mol of gaseous ions dissolves to form aqueous ions.

HistoryBorn and Haber, 1919. Max Born and Fritz Haber independently built the cycle in 1919. Haber is better known for ammonia synthesis; Born later won a Nobel Prize for quantum mechanics.

Fun factWhy MgO lines furnaces. Magnesium oxide melts at about 2850 °C because of its huge lattice energy, so it's used to line furnaces and kilns.

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