For a hypothetical process , ${{\mathrm{X}}^{-}}_{\left(\mathrm{g}\right)}\to {\mathrm{X}}_{\left(\mathrm{g}\right)}\to {{\mathrm{X}}^{+}}_{\left(\mathrm{g}\right)} ; ∆\mathrm{H} =+900 \mathrm{KJ}$,  If Ionization potential of  ${\mathrm{X}}^{-}\left(\mathrm{g}\right)$ is 150KJ/mole. Then the Electron gain enthalpy of  ${\mathrm{X}}^{+}\left(\mathrm{g}\right)$ will be 

 

$$\begin{gathered} {{\mathrm{X}}^{-}}_{\left(\mathrm{g}\right)}\underset{\left(\mathrm{I}\right)}{\overset{{∆}_{\mathrm{IE}}\mathrm{H}=+150\mathrm{KJ}}{\to }}{\mathrm{X}}_{\left(\mathrm{g}\right)}\underset{\left(\mathrm{II}\right)}{\overset{∆\mathrm{H}=?}{\to }}{{\mathrm{X}}^{+}}_{\left(\mathrm{g}\right)} ; \\ \left(\mathrm{II}\right) \mathrm{is} \mathrm{IP} \mathrm{of} {\mathrm{X}}_{\left(\mathrm{g}\right)} \mathrm{which} \mathrm{is} \mathrm{endothermic} \\ ∆\mathrm{H} = ∆{\mathrm{H}}_1+∆{\mathrm{H}}_{2 } \\ 900\mathrm{KJ} = 150\mathrm{KJ} + ∆{\mathrm{H}}_{2 } \\ ∆{\mathrm{H}}_{2 }= +750\mathrm{KJ}/\mathrm{mole} \end{gathered}$$

Magnitude of  ${∆}_{\mathrm{eg}}\mathrm{H}$of  ${{\mathrm{X}}^{+}}_{\left(\mathrm{g}\right)}$ will be the same but will have opposite sign

$\therefore {{\mathrm{X}}^{+}}_{\left(\mathrm{g}\right)} +{\mathrm{e}}^{-}\to {\mathrm{X}}_{\left(\mathrm{g}\right)} ; {∆}_{\mathrm{eg}}\mathrm{H} = -750\mathrm{KJ}$