If the enthalpy of sublimation of Li is $155\mathrm{~kJ~mol}^{-1}$, enthalpy of dissociation of $\mathrm{F}_{2}$ is $150\mathrm{~kJ~mol}^{-1}$, ionization enthalpy of Li is $520\mathrm{~kJ~mol}^{-1}$, electron gain enthalpy of F is $-313\mathrm{~kJ~mol}^{-1}$, standard enthalpy of formation of LiF is $-594\mathrm{~kJ~mol}^{-1}$. The magnitude of lattice enthalpy of LiF is _____ $\mathrm{kJ~mol}^{-1}$ (Nearest integer).

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Chemical Thermodynamics

If the enthalpy of sublimation of Li is

155 kJ mol^{- 1}

, enthalpy of dissociation of

F_{2}

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

The Born-Haber cycle gives

Δ H_{f}^{\circ} = Δ H_{sub} (Li) + IE (Li) + \frac{1}{2} Δ H_{diss} (F_{2}) + Δ H_{eg} (F) + U

. Substituting the data,

- 594 = 155 + 520 + \frac{1}{2} (150) + (- 313) + U = 437 + U

, so

U = - 594 - 437 = - 1031 kJ mo l^{- 1}

. The magnitude of the lattice enthalpy of

LiF

1031 kJ mo l^{- 1}

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Integer answer