The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus the amount of energy required to form all the bonds in the product molecules. What will be the enthalpy change for the following reaction: $H_{2(g)} + Br_{2(g)} \to 2HBr_{(g)}$? Given that the bond energy of $H_2$,$Br_2$,and $HBr$ is $435 \ kJ \ mol^{-1}$,$192 \ kJ \ mol^{-1}$,and $368 \ kJ \ mol^{-1}$ respectively.
- A$-109 \ kJ \ mol^{-1}$
- B$109 \ kJ \ mol^{-1}$
- C$259 \ kJ \ mol^{-1}$
- D$-259 \ kJ \ mol^{-1}$
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