Consider the following: $\operatorname{Li}(s)+\frac{1}{2} \mathrm{I}_{2}(g) \rightarrow \operatorname{LiI}(s) \Delta H=$$-292 \mathrm{kJ} .$ LiI(s) has a lattice energy of $-753 \mathrm{kJ} / \mathrm{mol} .$ The ionization energy of $\operatorname{Li}(g)$ is $520 . \mathrm{kJ} / \mathrm{mol}$, the bond energy of $\mathrm{I}_{2}(g)$ is $151 \mathrm{kJ} / \mathrm{mol},$ and the electron affinity of $\mathrm{I}(g)$ is $-295 \mathrm{kJ} / \mathrm{mol} .$ Use these data to determine the heat of sublimation of Li(s).