Bond enthalpy of bromine is 194 kJ mol^-1. If enthalpy of vapourisation of Br₂ is +30 kJ mol^-1, electron gain enthalpy of Br is -325 kJ mol^-1 and hydration enthalpy of bromide is -339 kJ mol^-1 , calculate the change in enthalpy for the reaction, $\frac{1}{2}{\mathrm{Br}}_2\left(\mathrm{l}\right) +{\mathrm{e}}^{-}\stackrel{\mathrm{aq}}{\to }{\mathrm{Br}}^{-}\left(\mathrm{aq}\right).$

$$\begin{gathered} \frac{1}{2}{\mathrm{Br}}_2\left(\mathrm{l}\right) \to \frac{1}{2}{\mathrm{Br}}_2\left(\mathrm{g}\right); ∆\mathrm{H} = +15 \mathrm{kJ} {\mathrm{mol}}^{-1} \\ \frac{1}{2}{\mathrm{Br}}_2\left(\mathrm{g}\right)\to \mathrm{Br}\left(\mathrm{g}\right) ; ∆\mathrm{H} = +97 \mathrm{kJ} {\mathrm{mol}}^{-1} \\ \mathrm{Br}\left(\mathrm{g}\right) \to {\mathrm{Br}}^{-}\left(\mathrm{g}\right) ; ∆\mathrm{H} = -325 \mathrm{kJ} {\mathrm{mol}}^{-1} \\ {\mathrm{Br}}^{-}\left(\mathrm{g}\right) +\mathrm{aq}\to {\mathrm{Br}}^{-}\left(\mathrm{aq}\right); ∆\mathrm{H} = -339 \mathrm{kJ} {\mathrm{mol}}^{-1} \end{gathered}$$
Adding these equations we get, $\frac{1}{2}{\mathrm{Br}}_2\left(\mathrm{l}\right) + {\mathrm{e}}^{-}\stackrel{\mathrm{aq}}{\to } {\mathrm{Br}}^{-}\left(\mathrm{aq}\right); ∆\mathrm{H} = -552 \mathrm{kJ} {\mathrm{mol}}^{-1}$