consider an ionic compound, mx, composed of generic metal m and generic, gaseous halogen x.

Consider an ionic compound, MX2, composed of generic metal M and generic, gaseous halogen X?

Given:

The enthalpy of formation of MX2 is ΔHf° = –957 kJ/mol.

The enthalpy of sublimation of M is ΔHsub = 123 kJ/mol.

The first and second ionization energies of M are IE1 = 757 kJ/mol and IE2 = 1451 kJ/mol.

The electron affinity of X is ΔHEA = –313 kJ/mol.

The bond energy of X2 is BE = 153 kJ/mol.

Answer:

Lattice energy U(MX2) to calculate

Use Born-Haber(-Fajans) Cycle

for more info: wikipedia

M(s) + X2 → MX2(s) ΔHf°MX2 = -957 kJ mol^-1

M(s) →ΔHsub(M)→ M(g) ΔHsub = 123kJ mol^-1

M(g) →ΔHIE1(M)→ M^+ + e⁻ 757 kJ mol^-1

M^+→ΔHIE2(M)→ M^2+ + e⁻ 1451 kJ mol^-1

X2 →ΔHBE(X2)→ 2X +153 kJ mol^-1

X + e⁻→ΔHEA(X)→ X^- -313 kJ mol^-1

ΔHfo = U + ΔHsub(M) + IE1 + IE2 + ΔHdissX2 + 2EA1(X)

-957= U + 123 + 757 + 1451 + 153 – 626 = U + 1858

U = -2815 kJ mol^-1 The –ve sign is often dropped

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