Calculate the enthalpy change for the reaction H2(g) + Br2(g) → 2HBr(g)
Given that the bond enthalpies of H–H, Br–Br and H–Br are 435, 192 and 364 kJ mol–1 respectively.
Energy absorbed for dissociation of 1 mole of H–H bon = 435 kJ
Energy absorbed of dissociation of 1 mole of Br–Br bond = 192 kJ
Total energy absorbed = 435 + 192 = 627 kJ
Energy released in the formation of 1 mole of H–Br bonds = 364 kJ
∴ Energy released in the formation of 2 moles of H–Br bonds = 2 × 364 kJ = 728 kJ.
Energy released > Energy absorbed
Hence, net result is the release of energy.
Energy released = 728 kJ –627 kJ = 101 kJ
i.e., for the given reaction, ∆rH = – 101 kJ