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)
Answer
1031
Solution
Born–Haber cycle: Li(s) + 12F2(g) LiF(s)
f H = subH + 12 dissH + i H + egH + U (U = enthalpy of formation of the lattice from gaseous ions)
-594 = 155 + 12(150) + 520 + (-313) + U
-594 = 437 + U
U = -1031 kJ mol-1
Magnitude of lattice enthalpy = 1031 kJ mol-1
Answer: 1031