If the bond energies of $H-H$,$Br-Br$ and $H-Br$ are $433$,$192$ and $364 \ kJ \ mol^{-1}$ respectively,then $\Delta H^{\circ}$ for the reaction: $H_{2(g)} + Br_{2(g)} \rightarrow 2HBr_{(g)}$ is:
- A$-261 \ kJ$
- B$+103 \ kJ$
- C$+261 \ kJ$
- D$-103 \ kJ$
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