Question

Use the following data to estimate $\mathbf{∆}{\mathbf{H}}_{\mathbf{f}}^{\mathbf{°}}$ for potassium chloride.

$$\begin{gathered} \mathbf{K}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{Cl}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\oplus }\mathbf{KCl}\mathbf{\left(}\mathbf{s}\mathbf{\right)} \\ \mathbf{Lattice}\mathbf{}\mathbf{energy}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{-}\mathbf{690}\mathbf{}\mathbf{KJ}\mathbf{/}\mathbf{mol} \\ \mathbf{Ionization}\mathbf{}\mathbf{energy}\mathbf{}\mathbf{for}\mathbf{}\mathbf{K}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{419}\mathbf{}\mathbf{KJ}\mathbf{/}\mathbf{mol} \\ \mathbf{Electron}\mathbf{}\mathbf{affinity}\mathbf{}\mathbf{of}\mathbf{}\mathbf{Cl}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{-}\mathbf{349}\mathbf{}\mathbf{KJ}\mathbf{/}\mathbf{mol} \\ \mathbf{Bond}\mathbf{}\mathbf{energy}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{Cl}}_{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{239}\mathbf{KJ}\mathbf{/}\mathbf{mol} \\ \mathbf{Enthalpy}\mathbf{}\mathbf{of}\mathbf{}\mathbf{sublimation}\mathbf{}\mathbf{for}\mathbf{}\mathbf{K}\mathbf{}\mathbf{}\mathbf{90}\mathbf{.}\mathbf{KJ}\mathbf{/}\mathbf{mol} \end{gathered}$$

Short answer

$-410.5\mathrm{KJ}/\mathrm{mol}$

Step-by-step solution

Energy is a step function so we can break the secretion into steps and then add them up:

The first step will be the sublimation of K(s).

Sublimation is changing of the state of a substance from a solid to a gaseous state.

$\mathrm{K}\left(\mathrm{s}\right)\to \mathrm{K}\left(\mathrm{g}\right)$

Enthalpy of sublimation for K(g) $=90$ kJ/mol

Ionization of potassium atoms to form K+  ions in the gas phase:

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

Ionization energy $=419$kJ/mol

Dissociation of Cl2  molecules:

$\frac{1}{2}C{l}_2\left(g\right)\to \mathrm{Cl}\left(\mathrm{g}\right)$

The bond energy of $C{l}_2\mathit{=}\mathit{239}\mathit{}KJ\mathit{/}\mathrm{mol}$

But we are breaking the bonds in a half mole of chlorine.

So, energy will be $239/2=119.5\mathrm{KJ}/\mathrm{mol}$

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

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

Electron affinity of Cl $=-349\mathrm{KJ}/\mathrm{mol}$

Formation of solid KCl from gaseous K+ and Cl-  ions:

${\mathrm{K}}^{+}\left(\mathrm{g}\right)+{\mathrm{Cl}}^{-}\left(\mathrm{g}\right)\to \mathrm{KCl}\left(\mathrm{g}\right)$

The lattice energyof KCl $=-690\mathrm{KJ}/\mathrm{mol}$

Add up all the five processes to obtain overall energy change:

$\mathrm{K}\left(\mathrm{s}\right)+\frac{1}{2}{\mathrm{Cl}}_2\left(\mathrm{g}\right)\to \mathrm{KCl}\left(\mathrm{s}\right)$

Energy change $(∆{\mathrm{H}}_{\mathrm{f}}^{°})=90+419+119.5-349-690=-410.5\mathrm{KJ}/\mathrm{mol}$kJ/mol.