Consider the following: Li(s) \(+\frac{1}{2} \mathrm{I}_{2}(g) \rightarrow\) Liil(s) \(\Delta H=\) \(-292 \mathrm{kJ} . \mathrm{LiI}(s)\) has a lattice energy of \(-753 \mathrm{kJ} / \mathrm{mol} .\) The ionization energy of 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).

In the Born-Haber cycle, the formation of a solid ionic compound is represented as a combination of several steps. For LiI(s), the Born-Haber cycle includes the following steps: 1. Sublimation of solid lithium (Li(s) to Li(g)) - This step represents the unknown heat of sublimation that we want to determine. 2. Ionization of gaseous lithium (Li(g) to Li+(g) + e-) - This step involves the ionization energy of Li. 3. Bond dissociation of diatomic iodine (I2(g) to 2 I(g)) - This step includes the bond energy of I2. 4. Electron affinity of iodine (I(g) + e- to I-(g)) - This step represents the electron affinity of I. 5. Formation of the solid lithium iodide (Li+(g) + I-(g) to LiI(s)) - This step is associated with the lattice energy of LiI.

According to the Hess's law, the overall energy change for the formation of LiI(s), which is the sum of the energy changes for all the steps mentioned above, should be equal to the enthalpy change for the given equation: Heat of sublimation (Li(s)) + Ionization energy (Li(g)) + Bond energy (I2(g)) + Electron affinity (I(g)) + Lattice energy (LiI(s)) = ∆H

Now plug in the given energy values into the equation: Heat of sublimation (Li(s)) + 520 kJ/mol + 151 kJ/mol + (-295 kJ/mol) + (-753 kJ/mol) = -292 kJ/mol Heat of sublimation (Li(s)) = -292 kJ/mol - 520 kJ/mol - 151 kJ/mol + 295 kJ/mol + 753 kJ/mol Heat of sublimation (Li(s)) = 85 kJ/mol So, the heat of sublimation for Li(s) is 85 kJ/mol.