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.

Question

Maya

Chemistry

17 September, 01:18

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.

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Danielle Santiago · 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).