Calculate the enthalpy change for the reaction H_(2)(g) + Br_(2)rarr 2HBr (g) Give that the bond enthalpies ofH-H,Br-Br and H-Br are435, 192 and 364kJ mol^(-1) respectively.

Calculate the enthalpy change for the reaction H_(2)(g) + Br_(2)rarr 2

1.	Calculate the enthalpy change for the reaction H_(2)(g) + Br_(2)rarr 2HBr (g) Give that the bond enthalpies ofH-H,Br-Br and H-Br are435, 192 and 364kJ mol^(-1) respectively.
Answer» Solution :Energy absorbed for dissociation of 1 mole of H-H bonds =`435 KJ` Energy absorbed for dissociation OF1 mole of Br -Br bonds `= 192kJ` TOTAL energy absorbed `=435+ 192 = 627kJ` Energy released in the formationof1 mole of H- Br bonds= 364kJ `:.`Energy released in the formationof 2 moles of H - Br bonds `= 2 xx 364 kJ = 728 kJ` Energy released ` gt` Energy absorbed Hence, net result is the release of energy Energy released `= 728 kJ - 627 kJ = 101 kJ ` i.e., for the given reaction , `Delta_(r) H = - 101kJ` Alternatively, the problem may be solved by applying Hess's lawor by applying the following RELATION directly`Delta_(r) H = Sigma` B.E. ( Reactants ) `- Sigma ` B.E. ( Products) `=[B.E. (H_(2)) + ` B.E. ` (Br_(2))] - 2 B.E. ( HBr) = 435 + 192 - 2 xx 364 = - 101 kJ`