Consider an ionic compound, MX3 , composed of generic metal M and generic gaseous halogen X. The enthalpy of formation of MX3 is Δ∘f=−917 kJ/mol. The enthalpy of sublimation of M is Δsub=131 kJ/mol. The first, second, and third ionization energies of M are IE1=697 kJ/mol, IE2=1561 kJ/mol, and IE3=2723 kJ/mol . The electron affinity of X is ΔEA=−305 kJ/mol . (Refer to the hint). The bond energy of X2 is BE=243 kJ/mol. Determine the lattice energy of MX3.

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pstnonsonjoku pstnonsonjoku · Answer 1 · 2021-11-12T20:07:35+01:00

Lattice energy is obtained from Hess law of constant heat summation. The lattice energy of MX3 is -6272 kJ/mol.

We have the following information from the question;

Heat of formation (ΔH∘f) = −917 kJ/mol

Heat of sublimation ( ΔHsub) = 131 kJ/mol

Sum of ionization energies of the meta (IE)= (IE1 + 1E2 + IE3) = 4981 kJ/mol

Electron affinity (EA) = −305 kJ/mol

Bond energy of X2 (BE) = 243 kJ/mol

Lattice energy (U) = ?

Using Hess law of constant heat summation;

ΔH∘f = ΔHsub + BE + IE + EA + U

Substituting values;

−917 kJ/mol = 131 kJ/mol + 243 kJ/mol + 4981 kJ/mol + U

U = (−917 kJ/mol) - [ 131 kJ/mol + 243 kJ/mol + 4981 kJ/mol]

U = -6272 kJ/mol

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