Question

Consider the following: $\mathbf{Li}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{I}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{LiI}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{∆}\mathbf{H}\mathbf{=}\mathbf{-}\mathbf{292}\mathbf{}\mathbf{kJ}\mathbf{.}\mathbf{}\mathbf{Lil}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$ . LiI has a lattice energy of $\mathbf{-}\mathbf{753}\mathbf{}\mathbf{kJ}\mathbf{/}\mathbf{mol}$. The ionization energy of Li(g) is 520 kJ/mol , the bond energy of I₂ is 151 kJ/mol , and the electron affinity of I(g) is -295 KJ/mol . Use these data to determine the heat of sublimation of Li(s) .

Short answer

$142.5\mathrm{kJ}/\mathrm{mol}$

Step-by-step solution

The first step will be the sublimation of Li(s) to Li(g):

$$\begin{gathered} \mathrm{Li}\left(\mathrm{s}\right)\to \mathrm{Li}\left(\mathrm{g}\right) \\ ∆{\mathrm{H}}_{\mathrm{sub}}=? \end{gathered}$$

Next will be the ionization of gaseous Li:

$\mathrm{Li}\left(\mathrm{g}\right)\to {\mathrm{Li}}^{+}\left(\mathrm{g}\right)$

Ionization energy of$\mathrm{Li}\left(\mathrm{g}\right)=520\mathrm{kJ}/\mathrm{mol}$

The Iodine bond will break:

$\frac{1}{2}{I}_2\left(g\right)\to {\mathrm{I}}_2\left(\mathrm{g}\right)$

Bond energy of${\mathrm{I}}_2=151\mathrm{kJ}/\mathrm{mol}$

But we are breaking the bonds in a half mole of iodine, so bond energy will be $151/2=75.5$ kJ/mol

Formation of   I- from I atoms in the gas phase:

$\mathrm{I}\left(\mathrm{g}\right)+{\mathrm{e}}^{-}\to {\mathrm{I}}^{-}\left(\mathrm{g}\right)$

Electron affinity of$\mathrm{I}=-295\mathrm{kJ}/\mathrm{mol}$

Formation of solid LiI from gaseous ions:

${\mathrm{Li}}^{+}\left(\mathrm{g}\right)+{\mathrm{I}}^{-}\left(\mathrm{g}\right)\to \mathrm{LiI}\left(\mathrm{s}\right)$

two attractions Lattice energy of$\mathrm{LiI}\left(\mathrm{s}\right)=-735\mathrm{kJ}/\mathrm{mol}$

Add up all the values to get ∆Hfo:

$$\begin{gathered} \mathrm{Li}\left(\mathrm{s}\right)+\frac{1}{2}{\mathrm{I}}_2\left(\mathrm{g}\right)\to \mathrm{LiI}\left(\mathrm{s}\right) \\ ∆{\mathrm{H}}_{\mathrm{f}}^{°}=-292\mathrm{kJ}/\mathrm{mol} \\ ∆{\mathrm{H}}_{\mathrm{f}}^{°}=∆{\mathrm{H}}_{\mathrm{sub}}+520+75.5+(-295)+(-735) \\ ∆{\mathrm{H}}_{\mathrm{sub}}=142.5\mathrm{kJ}/\mathrm{mol} \end{gathered}$$

So, the enthalpy of sublimation is 142.5 kJ/mol .