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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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  1. 17 September, 01:47
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    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).
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