Determine lattice energy

Question

Consider an ionic compound, Mx2 , composed of generic metal M  and generic, gaseous halogen X.

The enthalpy of formation MX2 is delta H°f= -991 kj/mol

The enthalpy of sublimation of M delta H sub= 127 kj/mol

The first and second ionization energies of M are IE₁= 711 kj/mol and IE₂= 1404 kj/mol

The electron configuration of X is delta H_EA= -305 kj/mol

The bond energy of X₂ is BE= 225 kj/mol

Determine the lattice energy MX₂ in kj/mol

Answer 1

Use the Born Haber cycle:

∆Hf = heat of formation

∆Hsub = heat of sublimation

IE = ionization energy

BE = bond energy (dissociation energy)

EA = electron affinity

LE = lattice energy

∆Hf = ∆Hsub + IE1 + IE2 + BE + EA + LE

Plugging in the values we have...

-991 = 127 + 711 +1404 + 225 -305 + LE

-991 = 2162 + LE

Solving for LE (lattice energy), we have...

LE = -3153 kJ/mol