Lattice energy and Born-Haber cycles

Chemistry A2 · Chemical energetics. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

Lattice energy is the enthalpy change when one mole of an ionic solid forms from its gaseous ions. It is always exothermic, because oppositely charged ions attract. It cannot be measured directly, so we find it with a Born-Haber cycle, which is Hess's law applied to an ionic solid.

The cycle links the enthalpy change of formation to all the steps that turn the elements into gaseous ions: atomisation of each element, ionisation of the metal and electron affinity of the non-metal. Lattice energy is the last step.

Small ions and highly charged ions attract more strongly, so the lattice energy is more exothermic.

Rules to remember

Lattice energy, ΔH_latt: gaseous ions → one mole of solid, e.g. Mg²⁺(g) + 2Cl⁻(g) → MgCl₂(s). Always negative.
Enthalpy change of atomisation, ΔH_at: one mole of gaseous atoms formed from the element in its standard state, e.g. ½Cl₂(g) → Cl(g). Always positive.
First electron affinity, EA₁: X(g) + e⁻ → X⁻(g), per mole. Usually exothermic.
Second electron affinity is endothermic: the electron is added to a negative ion, which repels it.
ΔH_latt = ΔH_f − (sum of all atomisation, ionisation and electron affinity steps)
Electron affinity becomes less exothermic down Groups 16 and 17 (more shells, more shielding, weaker attraction). Exceptions: O and F are less exothermic than S and Cl, because the atoms are so small that the added electron is repelled strongly by the electrons already in the outer shell.
Higher ionic charge or smaller ionic radius → more exothermic lattice energy.

Common mistake

Students forget to multiply by the number of atoms or ions in the formula. For CaCl₂ you need 2 × ΔH_at(Cl) and 2 × EA₁(Cl), and both ionisation energies of calcium.

Worked example

Find the lattice energy of NaCl. Data in kJ mol⁻¹: ΔH_f(NaCl) = −411; ΔH_at(Na) = +107; IE₁(Na) = +496; ΔH_at(Cl) = +122; EA₁(Cl) = −349.

Add the steps from elements to gaseous ions: 107 + 496 + 122 + (−349) = +376 kJ mol⁻¹.
Hess's law: ΔH_f = (sum of steps) + ΔH_latt, so ΔH_latt = ΔH_f − (sum of steps).
ΔH_latt = −411 − 376 = −787 kJ mol⁻¹.

Answer: ΔH_latt(NaCl) = −787 kJ mol⁻¹

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Watch: 2 | Chemical Energetics 2 | Electron Affinity & Lattice Energy

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