If the enthalpy of sublimation of Li is 155 kJ mol –1, enthalpy of dissociation of F2 is 150 kJ mol –1, ionization enthalpy of Li is 520 kJ mol –1, electron gain enthalpy of F is –313 kJ mol –1, stand

Question

If the enthalpy of sublimation of Li is 155 kJ mol –1, enthalpy of dissociation of F2 is 150 kJ mol –1, ionization enthalpy of Li is 520 kJ mol –1, electron gain enthalpy of F is –313 kJ mol –1, standard enthalpy of formation of LiF is –594 kJ mol –1. The magnitude of lattice enthalpy of LiF is _________ kJ mol –1 (Nearest integer).

Prashant Kotian · Accepted Answer

Correct answer: .. Li(s) + ½ F2(g) 155 kJ –617 LiF(s) (150/2) Li(g) F(g) 520kJ –313kJ Li+(g) F–(g) L.E. –594 = 155 + 520 + 150 2 – 313 + (L.E.) L.E. = – 1031 kJ/mol

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