Determine the lattice energy

Question

1. Consider an ionic compound, MX2, composed of generic metal M and generic, gaseous halogen X.The enthalpy of formation of MX2 is Δ𝐻∘f=−663 kJ/mol.
2. The enthalpy of sublimation of M is Δ𝐻sub=143 kJ/mol.
3. The first and second ionization energies of M are IE1=705 kJ/mol and IE2=1388 kJ/mol.
4. The electron affinity of X is Δ𝐻EA=−325 kJ/mol. (Refer to the hint.)
5. The bond energy of X2 is BE=237 kJ/mol.

Determine the lattice energy of MX2.

Answer 1

Born Haber Cycle

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

M(s) + X₂(g) ==> MX₂ .. ∆H_formation = -663 kJ/mol

M(s) ==> M(g) .. ∆H_sub = 143 kJ/mol

M(g) ==> M⁺(g) .. IE1 = 705 kJ/mol

M⁺(g) ==> M²⁺(g) .. IE2 = 1388 kJ/mol

X₂(g) ==> 2X(g) .. BE (bond energy) = 237 kJ/mol

2X(g) ==> 2X^- .. EA*2 = -325 * 2 = -650 kJ/mol^‡ (‡ times 2 since there are 2 electrons & is negative)

Mg²⁺(g) + 2X^-(g) ==> MgX₂ = Lattice Energy (LE)

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

-663 = 143 + 705 + 1388 + 237 -650 + LE

LE = -2486 kJ/mol