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JEE Main 202622 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2026 · 22 January · Evening Shift · Chemistry · Thermodynamics (C)

If the enthalpy of sublimation of Li is 155 ~kJ ~mol^ -1 , enthalpy...

If the enthalpy of sublimation of Li is 155 kJ mol−1155 \mathrm{~kJ} \mathrm{~mol}^{-1}, enthalpy of dissociation of F2\mathrm{F}_{2} is 150 kJ mol−1150 \mathrm{~kJ} \mathrm{~mol}^{-1}, ionization enthalpy of Li is 520 kJ mol−1520 \mathrm{~kJ} \mathrm{~mol}^{-1}, electron gain enthalpy of F is −313 kJ mol−1-313 \mathrm{~kJ} \mathrm{~mol}^{-1}, standard enthalpy of formation of LiF is −594 kJ mol−1-594 \mathrm{~kJ} \mathrm{~mol}^{-1}. The magnitude of lattice enthalpy of LiF is ____\_\_\_\_ kJmol−1\mathrm{kJ} \mathrm{mol}^{-1}. (Nearest Integer)

Correct answer

0

Step-by-step solution

Using Born-Haber cycle:
ΔHf(LiF)=ΔHsub(Li)+12ΔHdiss(F2)+ΔHie(Li)+ΔHeg(F)−Ulattice\Delta H_f(LiF) = \Delta H_{sub}(Li) + \frac{1}{2}\Delta H_{diss}(F_2) + \Delta H_{ie}(Li) + \Delta H_{eg}(F) - U_{lattice}

Substituting values: −594=155+12(150)+520+(−313)−Ulattice-594 = 155 + \frac{1}{2}(150) + 520 + (-313) - U_{lattice}

−594=155+75+520−313−Ulattice-594 = 155 + 75 + 520 - 313 - U_{lattice}

−594=437−Ulattice-594 = 437 - U_{lattice}

Ulattice=437+594=1031U_{lattice} = 437 + 594 = 1031 kJ/mol
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