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)) + Br_(2(g)) to 2HBr_((g)). Given that Bond energy of H_(2), Br_(2) and HBr is 435 "kJ mol"^(-1), 192 "kJ mol"^(-1) and 368 "kJ mol"^(-1) respectively.

The net enthalpy change of a reaction is the amount of energy required

1.	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)) + Br_(2(g)) to 2HBr_((g)). Given that Bond energy of H_(2), Br_(2) and HBr is 435 "kJ mol"^(-1), 192 "kJ mol"^(-1) and 368 "kJ mol"^(-1) respectively.
Answer» Solution :bond energy of `H_(2) = 435 "kJ mol"^(-1)` bond energy of `Br_(2) = 192 "kJ mol"^(-1)` bond energy of HBr `= 368 "kJ mol"^(-1)` For the reaction `H_(2(g)) + Br_(2(g)) to 2HBr_((g))` `Delta_(r) H^( THETA ) = sum BE ("Reactants") - sumBE ("products")` `= BE (H_(2) ) + BE(Br_2) - 2BE (HBr)` `= 435 + 192 - (2XX 368) "kJ mol"^(-1)` `= -109 "kJ mol"^(-1)`