Using Born-Haber cycle:
ΔHf(LiF)=ΔHsub(Li)+21ΔHdiss(F2)+ΔHie(Li)+ΔHeg(F)−Ulattice
Substituting values: −594=155+21(150)+520+(−313)−Ulattice
−594=155+75+520−313−Ulattice
−594=437−Ulattice
Ulattice=437+594=1031 kJ/mol
JEE Main 2026 — Chemistry Physical Chemistry
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 ____ kJmol−1. (Nearest Integer)
Held on 22 Jan 2026 · Verified 6 Jul 2026.
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