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)rarr(1)/(2)Br_(2)(g),,Delta=+15kJmol^(-1)),((1)/(2)Br_(2)(g)rarrBr(g),,DeltaH=+97kJmol^(-1)),(Br(g)rarrBr^(-)(g),,DeltaH=-325kJmol^(-1)),(Br^(-)(g)+aq rarr Br^(-)(aq),,DeltaH=-339kJ mol^(-1)):}`
Adding these equations we get, `(1)/(2)Br_(2)(l)+e^(-)overset(aq)rarrBr^(-)(aq)),DeltaH=-552kJ mol^(-1)`