If the enthalpy of sublimation of Li is 155 ~kJ ~mol ^ -1 , enthalpy of dissociation of F _ 2 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 lat

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If the enthalpy of sublimation of Li is 155 ~kJ ~mol ^ -1 , enthalpy of dissociation of F _ 2 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)

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Correct answer: A. A

Using Born-Haber cycle: H_f(LiF) = H_ sub (Li) + 1 2 H_ diss (F_2) + H_ ie (Li) + H_ eg (F) - U_ lattice Substituting values: -594 = 155 + 1 2 (150) + 520 + (-313) - U_ lattice -594 = 155 + 75 + 520 - 313 - U_ lattice -594 = 437 - U_ lattice U_ lattice = 437 + 594 = 1031 kJ/mol

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