• # question_answer 77)   The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus amount of energy required to form all the bonds in the product molecules. What will be the enthalpy change for the following reaction. ${{H}_{2}}(g)+B{{r}_{2}}(g)\to 2HBr(g)$ Given that bond energy of ${{H}_{2}},B{{r}_{2}}$ and $HBr$is $435\text{ }kJmo{{l}^{-1}},\text{ }192\text{ }kJ\text{ }mo{{l}^{-1}}\text{ }and\text{ }368\text{ }kJ\text{ }mo{{l}^{-1}}$respectively.

${{H}_{2}}(g)+B{{r}_{2}}(g)\to 2HBr(g)$ Heat of reaction ${{\Delta }_{r}}H=\Sigma$ Bond energy of reactants $-\Sigma$ Bond energy of products $=\text{ }\left[ B.E.\left( H-\text{ }H \right)\text{ }+\text{ }B.E.\left( Br-\text{ }Br \right) \right]-\text{ }\left[ 2B.E.\left( H-\text{ }Br \right) \right]$$=[435+192]-[2\times 368]$ $=-109\text{ }kJ\,mo{{l}^{-1}}$