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

`1/2Br_(2(l)) to 1/2Br_(2(g)) , Delta H = + 15 kJ mol^(-1)`
`1/2 Br_(2(g)) to Br_((g)) , Delta H = + 97 kJ mol^(-1)`
`Br_((g)) to Br_((g))^(-) , Delta H = -325 kJ mol^(-1)`
`Br_((g))^(-) + aq to Br_((aq))^(-) , Delta H = -339 kJ mol^(-1)`
Adding these equations we get,
`1/2 Br_(2(l)) + e^(-) overset(aq)(rarr) Br_((aq))^(-) , Delta H = -552 kJ mol^(-1)`
The change in the enthalpy of the reaction is `-552 kJ mol^(-1)`.