Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below :  

$\frac{1}{2}\mathrm{C}{\mathrm{l}}_{2_{\left(\mathrm{g}\right)}}\stackrel{\frac{1}{2}{∆}_{\mathrm{diss}}\mathrm{H}\odot }{\to }{\mathrm{Cl}}_{\left(\mathrm{g}\right)}\stackrel{{∆}_{\mathrm{eg}}\mathrm{H}\odot }{\to }{\mathrm{Cl}}^{-1}\stackrel{{∆}_{\mathrm{hyd}}\mathrm{H}\odot }{\to }{{\mathrm{Cl}}^{-1}}_{2_{\left(\mathrm{g}\right)}}$ The energy involved in the conversion of $\frac{1}{2}\mathrm{C}{1}_{2_{\left(\mathrm{g}\right)}}$ to $\mathrm{C}{1^1}_{\left(\mathrm{aq}\right)}$ (using the data, ${∆}_{\mathrm{diss}}{\mathrm{H}}_{{\mathrm{Cl}}_2}^{\odot }=240{\mathrm{KJmol}}^{-1}; {∆}_{\mathrm{egg}}{\mathrm{H}}_{\mathrm{Cl}}^{\odot }=-349{\mathrm{KJmol}}^{-1}$ and ${∆}_{\mathrm{hyd}}{\mathrm{H}}_{{\mathrm{Cl}}^{-}}=-381{\mathrm{KJmol}}^{-1}),$ will be - ____ KJ mol^–1. 

Oxidising power of chlorine in aqueous solution can be estimated by calculating the overall enthalpy change of reaction.
The reaction takes place with following steps:

$$\begin{gathered} \frac{1}{2}{\mathrm{Cl}}_2 \stackrel{ }{\to } \mathrm{Cl} ; {\mathrm{\Delta H}}_1 \\ {\mathrm{Cl}}_{\left(\mathrm{g}\right)} \stackrel{ }{\to } \mathrm{C}{{\mathrm{l}}^{-}}_{\left(\mathrm{g}\right)} ; {\mathrm{\Delta H}}_2 \\ \mathrm{C}{{\mathrm{l}}^{-}}_{\left(\mathrm{g}\right)} \stackrel{ }{\to } {\mathrm{Cl}}_{\left(\mathrm{aq}\right)}^{-} ; {\mathrm{\Delta H}}_3 \end{gathered}$$

ΔH will be the sum of the energy involved in these
steps.
$$\begin{gathered} \mathrm{Hence}, \mathrm{\Delta H} = {\mathrm{\Delta H}}_1 +{\mathrm{\Delta H}}_2 +{\mathrm{\Delta H}}_3 \\ \mathrm{\Delta H}1 = 240/2 = 120\mathrm{kJ}/\mathrm{mol}, \\ \mathrm{\Delta H}2 = -349\mathrm{kJ}/\mathrm{mol}, \\ {\mathrm{\Delta H}}_3 = -381\mathrm{kJ}/\mathrm{mol} \\ \mathrm{Hence}, \mathrm{\Delta H} = 120-349-381 = -610\mathrm{kJ}/\mathrm{mol} \end{gathered}$$