Explain why the standard enthalpy of atomisation of chlorine is exactly half the bond enthalpy of the Cl–Cl bond, and explain how these two values are used in a Born–Haber cycle to determine the lattice enthalpy of potassium chloride.

The standard enthalpy of atomisation of chlorine is the enthalpy change for

½ Cl₂(g) → 2 Cl(g)

so it is the energy required to produce one mole of gaseous Cl atoms. The bond enthalpy of the Cl–Cl bond is the energy needed to break one mole of Cl–Cl bonds, giving two moles of Cl(g). Because the atomisation step produces only one mole of atoms, its value is exactly one‑half of the bond enthalpy.

In a Born–Haber cycle for KCl the steps are:
1. ½ Cl₂(g) → 2 Cl(g) (ΔH = ½ ΔH°atom,Cl)
2. ½ K₂(g) → K(g)     (ΔH = ½ ΔH°atom,K)
3. K(g) → K⁺(g) + e⁻   (ΔH = ΔH°IE,K)
4. Cl(g) + e⁻ → Cl⁻(g)  (ΔH = ΔH°EA,Cl)
5. K⁺(g) + Cl⁻(g) → KCl(s) (ΔH = ΔH°latt)

Adding the enthalpies of steps 1–4 and using Hess’s law gives:
ΔH°latt = –[½ ΔH°atom,Cl + ½ ΔH°atom,K + ΔH°IE,K + ΔH°EA,Cl – ΔH°f,KCl]

Thus the atomisation enthalpy of chlorine (½ ΔH°atom,Cl) is a key component of the cycle and, together with the other elemental enthalpies, allows the lattice enthalpy of KCl to be calculated.

The lattice enthalpy is the exothermic enthalpy change for forming one mole of KCl(s) from its gaseous ions and its magnitude reflects the strength of the ionic lattice.