Question

Suggest and explain an indirect method to calculate lattice enthalpy of sodium chloride crystal.

Answer 1

The formation of NaCl can be considered in five steps. The sum of the enthalpy changes of these steps is equal to the enthalpy change for the overall reaction from which the lattice enthalpy of NaCl is calculated.

Let us calculate the lattice energy of sodium chloride using the Bom-Haber cycle

∆H_f = heat of formation of sodium chloride = – 411.3 kJ mol^–1

∆H₁ = heat of sublimation of Na_(g) = 108.7 kJ mol^–1

∆H₂ = ionisation energy of Na_(g) = 495.0 kJ mol^–1

∆H₃ = dissociation energy of Cl_2(g) = 244 kJ mol^–1

∆H₄ = Electron affinity of Cl_(S) = – 349 kJ mol^–1

U = lattice energy of NaCl

∆H_f = ∆H₁ + ∆H₂ + `1/2` ∆H₃ + ∆H₄ + U

∴ U = `(∆"H"_"f") – (∆"H"_1 + ∆"H"_2 + 1/2 ∆"H"_3 + ∆"H"_4)`

U = (– 411.3) – (108.7 + 495.0 + 122 – 349)

U = (– 411.3) – (376.7)

∴ U = – 788 kJ mol^–1