Draw a labelled Born-Haber cycle for potassium bromide, naming each step.

Calculate the value for the lattice energy of aluminium oxide, given that $∆{{\mathrm{H}}_{\mathrm{f}}}^{\circ }\left[{\mathrm{Al}}_2{\mathrm{O}}_3\right]=-1676 \mathrm{kJ} {\mathrm{mol}}^{-1}$. Remember that there are two moles of ${\mathrm{Al}}^{3+}$ ions and three moles of ${\mathrm{O}}^{2-}$ ions in one mole of ${\mathrm{Al}}_2{\mathrm{O}}_3$.

Given, $\mathrm{Al}\left(\mathrm{s}\right)\underset{+326 \mathrm{kJ} {\mathrm{mol}}^{-1}}{\overset{ ∆{{\mathrm{H}}_{\mathrm{at}}}^{\circ } }{\to }}\mathrm{Al}\left(\mathrm{g}\right)\underset{+577 \mathrm{kJ} {\mathrm{mol}}^{-1}}{\overset{ {\mathrm{IE}}_1 }{\to }}{\mathrm{Al}}^{+}\left(\mathrm{g}\right)\underset{+1820 \mathrm{kJ} {\mathrm{mol}}^{-1}}{\overset{ {\mathrm{IE}}_2}{\to }}{\mathrm{Al}}^{2+}\left(\mathrm{g}\right)\underset{+2740 \mathrm{kJ} {\mathrm{mol}}^{-1}}{\overset{{\mathrm{IE}}_3 }{\to }}{\mathrm{Al}}^{3+}\left(\mathrm{g}\right)$

           $\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\underset{+249 \mathrm{kJ} {\mathrm{mol}}^{-1}}{\overset{ ∆{{\mathrm{H}}_{\mathrm{at}}}^{\circ } }{\to }}\mathrm{O}\left(\mathrm{g}\right)\underset{-141 \mathrm{kJ} {\mathrm{mol}}^{-1}}{\overset{ {\mathrm{EA}}_1 }{\to }}{\mathrm{O}}^{-}\left(\mathrm{g}\right)\underset{+798 \mathrm{kJ} {\mathrm{mol}}^{-1}}{\overset{{\mathrm{EA}}_2 }{\to }}{\mathrm{O}}^{2-}\left(\mathrm{g}\right)$