The energy required for the following process is $1.96\times {10}^4\mathrm{kJ} {\mathrm{mol}}^{-1}$ . ${\mathrm{Li}}_{\left(\mathrm{g}\right)}\to 3{\mathrm{e}}^{-}+\mathrm{L}{{\mathrm{i}}^{3+}}_{\left(\mathrm{g}\right)}$ . If the first ionization energy of lithium is $520 \mathrm{kJ} {\mathrm{mol}}^{-1}$ , what is the second ionization energy?

$\begin{array}{r}\mathrm{Li}\left(\mathrm{g}\right)\to {\mathrm{Li}}^{3+}\left(\mathrm{g}\right)+\mathrm{e};\mathrm{\Delta H}=1.96\times {10}^4{\mathrm{kJmol}}^{-1}\dots .\text{ (i) }\\ \mathrm{Li}\left(\mathrm{g}\right)\to {\mathrm{Li}}^{+}\left(\mathrm{g}\right)+\mathrm{e};{\mathrm{IE}}_1=520\mathrm{kJ}{\mathrm{mol}}^{-1}\dots .\text{ (ii) }\\ {\mathrm{Li}}^{+}\left(\mathrm{g}\right)\to {\mathrm{Li}}^{2+}\left(\mathrm{g}\right)+\mathrm{e};{\mathrm{IE}}_2={\mathrm{a} \mathrm{kJmol}}^{-1}\dots .\text{ (iii) }\\ {\mathrm{Li}}^{2+}\left(\mathrm{g}\right)\to {\mathrm{Li}}^{3+}\left(\mathrm{g}\right)+\mathrm{e};{\mathrm{IE}}_3=\mathrm{b} \mathrm{kJ}{\mathrm{mol}}^{-1}\dots \dots \text{ (iv) }\end{array}$

$\begin{array}{r}\mathrm{b}={\mathrm{E}}_1\text{ for }{\mathrm{Li}}^{2+}\times {\mathrm{N}}_{\mathrm{A}}={\mathrm{E}}_1\mathrm{H}\times {\mathrm{Z}}^2\times {\mathrm{N}}_{\mathrm{A}} =2.18\times {10}^{-18}\times 3^2\times 6.023\times {10}^{23}{\mathrm{Jmol}}^{-1}\\ \\ \therefore \text{ eqs (i), (ii), (iii), (iv) }\\ 1.96\times {10}^4=520+\mathrm{a}+11.81\times {10}^3\\ \mathrm{a}=7270\mathrm{kJ}{\mathrm{mol}}^{-1}\end{array}$