If the bond energies of H–H, Br–Br and H–Br are 433, 192 and 364 kJ mol$^{-1}$ respectively the $\Delta H^\circ$ for the reaction $H_2(g) + Br_2(g) \rightarrow 2HBr(g)$ is-

$\Delta H = \sum$ (bond energies of bonds broken in reactants) $- \sum$ (bond energies of bonds formed in products)
Bonds broken: one H–H and one Br–Br $= 433 + 192 = 625$ kJ
Bonds formed: two H–Br $= 2 \times 364 = 728$ kJ
$\Delta H^\circ = 625 - 728 = -103$ kJ
So the reaction is exothermic, with $\Delta H^\circ = -103$ kJ.