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) rarr 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) rarr 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 :Given that 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) RARR 2HBr (g)` `Delta_(r) H^(Θ) = SigmaBE ("Reactants") - SigmaBE ("Products")` `= BE (H_(2)) + BE (Br_(2)) - 2BE (HBr)` `= 435 + 192 - (2 XX 368) kJ mol^(-1)` `= - 109 kJ mol^(-1)`