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Q. Calculate the lattice energy of LiF(s), given that:
(i) $\Delta H _{\text {sub }}$ for Lithium is $155.2 \, kJ\, mol ^{-1}$
(ii) $\Delta H _{\text {diss }}$ of $1 / 2$ mole of $F _{2}( g )$ is $75.3\, kJ$
(iii) I.E of Lithium $=520\, kJ\, mol ^{-1}$
(iv) E.A. of Fluorine $=333\, kJ \, mol ^{-1}$
(v) $\Delta H _{ f }$ for $LiF ( s )=-594.1\, kJ \, mol ^{-1}$

Thermodynamics

Solution:

image

$\therefore \Delta_{f}=\Delta H _{sub}+\frac{1}{2} \Delta H _{diss}+ I.E. + E.A. +\Delta H _{I.E}$

$\therefore \Delta H _{I.E.}=-1011.6 kJ / mol$