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Born–Haber cycle: lattice enthalpy of NaCl
A-Level (AQA) · unreviewed
Born–Haber cycle for NaCl with every step except the lattice enthalpy known.
Worked example
Construct a Born–Haber cycle for NaCl and find the lattice formation enthalpy. Data: ΔfH(NaCl) −410 kJ mol⁻¹; atomisation of Na 108 kJ mol⁻¹; IE₁(Na) 496 kJ mol⁻¹; atomisation of chlorine (per mole of Cl atoms) 122 kJ mol⁻¹; EA₁(Cl) −350 kJ mol⁻¹.
1. Weighted sum of 1·ΔfH(NaCl) − 1·ΔatH(Na) − 1·IE₁(Na) − 1·ΔatH(Cl) − 1·EA₁(Cl) → Lattice formation enthalpy
= −786.0 kJ mol⁻¹
Answer: ΔLH (formation) = −786 kJ mol⁻¹
Method
Going round the cycle: ΔfH = ΔatH(Na) + IE₁ + ΔatH(Cl) + EA₁ + ΔLH(formation). Rearranging, ΔLH = ΔfH − (ΔatH(Na) + IE₁ + ΔatH(Cl) + EA₁).
Common mistakes
- That is the lattice dissociation enthalpy. Lattice formation (gaseous ions → solid) is exothermic, so it is negative.
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