Consider an ionic compound, mx2, composed of generic metal m and generic, gaseous halogen x.
the enthalpy of formation of mx2 is δhf° = –799 kj/mol.
the enthalpy of sublimation of m is δhsub = 157 kj/mol.
the first and second ionization energies of m are ie1 = 785 kj/mol and ie2 = 1451 kj/mol. the electron affinity of x is δhea = –349 kj/mol.
the bond energy of x2 is be = 157 kj/mol. determine the lattice energy of mx2.

Lattice energy U(MX2) to calculate

Use Born-Haber(-Fajans) Cycle

http://en.wikipedia.org/wiki/Born%E2%80%...

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

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"