If the bond energies of  $H-H,Br-BrandH-Br$ are 433, 192 and 364 kJ mol^-1  respectively the $\Delta H^0$ for the reaction  $H_2\left(g\right)+B{r}_2\left(g\right)\to 2HBr\left(g\right)$ is 

$H_2\left(g\right)+B{r}_2\left(g\right)\to 2HBr\left(g\right)$  Enthalpy change for this reaction will equal to the difference in the energy consumed for breaking bonds of reactions and energy released when bonds of products are made
Enthalpy=(Sum of bond energies broken) - (sum of bond energies formed). 
 $$\begin{gathered} \Delta H^0={\left(BE\right)}_{reaction}-{\left(BE\right)}_{product} \\ =(433+192)-(2\times 364) \\ =625-728=-103kJ \end{gathered}$$