Question

Consider an ionic compound, MX, composed of generic metal M and generic, gaseous halogen X.The enthalpy of formation of MX is ΔHf° = –423 kJ/mol. The enthalpy of sublimation of M is ΔHsub = 119 kJ/mol. The ionization energy of M is IE = 469 kJ/mol. The electron affinity of X is ΔHEA = –301 kJ/mol. (Refer to the hint). The bond energy of X2 is BE = 161 kJ/mol. Determine the lattice energy of MX.

Answer 1

Here are the given:
ΔHf° = –423 kJ/mol
ΔHsub = 119 kJ/mol
IE = 469 kJ/mol
ΔHEA = –301 kJ/mol
BE = 161 kJ/mol

The lattice energy of the compound is solved using the formula:
U = ΔHf° - ΔHsub - BE - IE - ΔHEA
U = -423 - 119 - 161 - 469 - (-301)
U = -871 kJ/mol

Therefore, the lattice energy is 871 kJ/mol (released).