At standard conditions, if the change in the enthalpy for the following reaction is -109 kJ mol^-1. H_2( g ) + Br_2( g ) → 2 HBr_( g ) Given that…

At standard conditions, if the change in the enthalpy for the following reaction is $\displaystyle -109 \mathrm{~kJ} \mathrm{~mol}^{-1}$. $$\mathrm{H}_{2(\mathrm{~g})}+\mathrm{Br}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{HBr}_{(\mathrm{g})} $$ Given that bond energy of $\displaystyle \mathrm{H}_2$ and $\displaystyle \mathrm{Br}_2$ is $\displaystyle 435 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\displaystyle 192 \mathrm{~kJ} \mathrm{~mol}^{-1}$, respectively, what is the bond energy (in $\displaystyle \mathrm{kJ} \mathrm{mol}^{-1}$ ) of HBr ?