Chemistry · Thermodynamics & Thermochemistry

JEE Main 2026 — 22 January, Evening Shift — Question 63

If the enthalpy of sublimation of Li is 155 kJ mol−1155 \mathrm{~kJ} \mathrm{~mol}^{-1}, enthalpy of dissociation of F2F_{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 FF 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).

Answer: 1031

Numerical answer — enter this value.

Step-by-step solution

−594=155+520+1502−313+(-594=155+520+\frac{150}{2}-313+( L.E. )) L.E. =−1031 kJ/mol=-1031 \mathrm{~kJ} / \mathrm{mol}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Subject
Chemistry
Chapter
Thermodynamics & Thermochemistry
Topic
Thermochemistry and Enthalpy Changes
If the enthalpy of sublimation of Li is 155 kJ mol -1 , enthalpy of… | JEE Main 2026 PYQ with Solution · DhiX AI