Use bond energies to estimate the enthalpy of formation of HBr(g).
BE(H-H) = 436 kJ/mol
BE(Br-Br) = 192 kJ/mol
BE(H-Br) = 366 kJ/mol
Enthalpy of reaction is the difference in energy absorbed in breaking bonds in reactants and energy released when new bonds are formed in products.
Since two moles of HBr are formed in the reaction, for 1 mole 52 kJ/mol energy is required. Since energy is released during bond formation the value would be negative.
ΔHo= 436 + 192 - 2× 366 = -104 KJ/mol
Hence, option (B) is correct.