Use the bond enthalpies listed below to estimate the enthalpy change for the reaction
`H_(2)(g)+Br_(2)(g)rarr2HBr(g)`
Given:
`BE` of `H_(2), Br_(2)`, and `HBr` is `435, 192`, and `368 kJ mol^(-1)`, respectively.

Explanation

We have seen above
`Delta_(r)H^(Θ)=Sigma`bond `"enthalpies"_("reactants")-Sigma`bond `"enthalpies"_("products")`
`=[H_(H-H)^(Θ)+H_(Br-Br)^(Θ)]-2[H_(H-Br)^(Θ)]`
`=(435 kJ mol^(-1)+192 kJ mol^(-1))-(2xx368 kJ mol^(-1))`
`=627 mol^(-1)-736 kJ mol^(-1)=-109 kJ mol^(-1)`