Consider an ionic compound, MX, composed of generic metal M and generic, gaseous halogen X.

Question

1. The enthalpy of formation of MX is Δ𝐻∘f=−529 kJ/mol.
2. The enthalpy of sublimation of M is Δ𝐻sub=113 kJ/mol.
3. The ionization energy of M is IE=411 kJ/mol.
4. The electron affinity of X is Δ𝐻EA=−339 kJ/mol. (Refer to the hint.)
5. The bond energy of X2 is BE=171 kJ/mol.

Determine the lattice energy of MX.

Answer 1

This problem involves an understanding of the Born Haber cycle. There are a couple of ways to look at this cycle, but in essence it will show us that ...

1. ∆H_formation = ∆H_sub + IE + BE + EA + LE

So to find LE (lattice energy) we just solve the above equation for that quantity. But the way to look at this is via the Born Haber cycle

M(s) + 1/2 X₂(g) ==> MX(s) ... ∆H_formation = -529 kJ/mol

1. M(s) ==> M(g) ∆H_sub = 113 kJ/mol
2. M(g) ==> M⁺(g) IE = 411 kJ/mol
3. 1/2 X₂(g) ==> X(g) BE = 1/2 x 171 kJ/mol = 85.5 kJ
4. X(g) ==> X^-(g) EA = -339 kJ/mol
5. M⁺(g) + X^-(g) ==> MX(s) LE = ?

LE = ∆H_formation - ∆H_sub -IE - BE - EA

LE = -529 - 113 -411 - 85.5 + 339

LE = -799.5 kJ/mol