How do you determine the lattice energy of MX 2?

Question

The enthalpy of formation of MX 2 is Δ H ∘ f = − 973 kJ/mol. The enthalpy of sublimation of M is Δ H sub = 141 kJ/mol. The 1st and 2nd ionization energies of M are IE 1 = 705 kJ/mol and IE 2 = 1391 kJ/mol. The electron affinity of X is Δ H EA = − 311 kJ/mol. The bond energy of X 2 is BE = 199 kJ/mol.

Answer 1

Born-Haber Cycle and Hess' Law:

M(s) + X₂(g) ---------------------------> MX₂(s)

M(s) ==> M(g) ∆H_sub = 141 kJ/mol

M(g) ==> M⁺(g) IE₁ = 705 kJ/mol

M+(g) ==> M²⁺(g) IE₂ = 1391 kJ/mol

X₂(g) ==> 2X(g) BE = 199 kJ/mol

2X(g) + 2e- ==> 2X^-(g) EA = 2x-311 kJ/mol = -622 kJ/mol

M²⁺(g) + 2X^-(g) ==> MX₂(s) Lattice Energy = LE = ?

∆H_f = ∆H_sub + IE₁ + IE₂ + BE + 2EA + LE

-973 = 141 + 705 + 1391 + 199 -622 + LE

-973 = 1814 + LE

LE = -2787 kJ/mol