Question

Consider an ionic compound, MX2MX2, composed of generic metal MM and generic, gaseous halogen XX.

-The enthalpy of formation of MX2MX2 is Δ?∘f=−691ΔHf∘=−691 kJ/mol.

-The enthalpy of sublimation of MM is Δ?sub=127ΔHsub=127 kJ/mol.

-The first and second ionization energies of MM are IE1=671IE1=671 kJ/mol and IE2=1448IE2=1448 kJ/mol.

-The electron affinity of XX is Δ?EA=−311ΔHEA=−311 kJ/mol. (Refer to the hint).

-The bond energy of X2X2 is BE=167BE=167 kJ/mol.

Determine the lattice energy of MX2MX2.

Answer 1

Given ionic compound, MX2 can be formed as M(s) + X2 → MX2

From given data

Enthalpy of formation for M(s) + X2 → MX2 ; ΔHf(MX2) = -929 kJ /mol

Enthalpy of sublimation for M(s) → M(g) ΔH(sub) = 109 kJ /mol

1st and 2nd ionization energies of M, M(g) →→ M2+ IE1+ IE (M) = 747+ 1449 = 2196 kJ/ mol

Bond energy of X, X2 → 2X ΔH(diss) = 217 kJ

The electron affinity of X, X + e- → X- EA (X) = -311 kJ/ mol (need 2X)

According to Hess's law (Born-Haber cycle) U is the lattice energy

ΔHf(MX2) = U + ΔH(sub)+ (IE1+ IE2)(M) + ΔH(diss) + 2*EA(X)

   -929   = U + 109 + 2196 + 217 + 2*(-311)

-929 = U + 1900

           U = -2829 kJ / mol

~2800 kj mol^-1 is a typical lattice energy for a MX2 salt.