Calculate the enthalpy change during the reaction `:`
`H_(2(g))+Br_(2(g))rarr 2HBr_((g))`
Given, `e_(H-H)=435kJ mol^(-1),e_(Br-Br)=192kJ mol^(-1)` and `e_(H-Br)=368kJ mol^(-1).`

Explanation

Using the bond energy form of Hess's law, i.e., ,we have
`Delta_(r) H^(@) = [Delta_(H - H) H^(@) + Delta_(Br - Br) H^(@)] - [2 Delta_(H - Br) H^(@)]`
`= (435 + 192) - (2 xx 368)`
`= (627) - (736) = - 109 kJ mol^(1)`
This type of calculation can be combined with `Delta_(f) H^(@)` data to measure other average bond enthalpies.