If the enthalpy of sublimation of Li is 155 kJ mol β1, enthalpy of dissociation of F2 is 150 kJ mol β JEE Mains Chemistry Past Papers Chemistry Question
Question
If the enthalpy of sublimation of Li is 155 kJ mol β1, enthalpy of dissociation of F2 is 150 kJ mol β1, ionization enthalpy of Li is 520 kJ mol β1, electron gain enthalpy of F is β313 kJ mol β1, standard enthalpy of formation of LiF is β594 kJ mol β1. The magnitude of lattice enthalpy of LiF is _________ kJ mol β1 (Nearest integer).
Answer: .
π‘ Solution & Explanation
Li(s) + Β½ F2(g) 155 kJ β617 LiF(s) (150/2) Li(g) F(g) 520kJ β313kJ Li+(g) Fβ(g) L.E. β594 = 155 + 520 + 150 2 β 313 + (L.E.) L.E. = β 1031 kJ/mol
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