Bond enthalpy of bromine is `194 kJ mol^(-)`. If enthalpy of vapourisation of `Br_2` is `+30 kJ mol^(-)`, 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, `1/2Br_(2)(l) + e^(-) overset(aq)(rarr) Br^(-)(aq)`. 

Explanation

`(1)/(2)Br_(2(l))to(1)/(2)Br_(2_((g))),DeltaH=+15kJmol^(-1)`
`(1)/(2)Br_(2(g))toBr_((g)),DeltaH=+97kJmol^(-1)`
`Br_((g))toBr_((g))^(-),DeltaH=-325kJmol^(-1)`
`Br_((g))^(-)+aqtoBr_((aq))^(-),DeltaH=-339kJmol^(-1)`
Adding these equations we get ,
`(1)/(2)Br_(2(l))+e^(-)overset(aq)toBr_((aq))^(-),DeltaH=-552kJmol^(-1)`
The change in the enthalpy of the reaction is `-552kJmol^(-1)`.