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. (Keep your answer in magnitude and in KJ/mol)

Explanation

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)`