Given that the bond energies of H - H and CI - CI are 430 kJ"mol"^(-1) and 240 kJ"mol"^(-1) respectively and Delta_(f)H for HCI is -90 kJ"mol"^(-1), bond enthalpy of HCI is

Given that the bond energies of H - H and CI - CI are 430 kJ"mol"^(-1)

1.	Given that the bond energies of H - H and CI - CI are 430 kJ"mol"^(-1) and 240 kJ"mol"^(-1) respectively and Delta_(f)H for HCI is -90 kJ"mol"^(-1), bond enthalpy of HCI is
Answer» `254 kJ"mol"^(-1)` `290 kJ"mol"^(-1)` `380 kJ"mol"^(-1)` `425 kJ"mol"^(-1)` Solution :`1/2 H_(2) + 1/2 CI_(2) to HCI` `DeltaH = 1/2 B.E.(H-H) + 1/2B.E(CI - CI) - B.E.(H- ci)` `-90 = 1/2(430) + 1/2(240) - B.E (H - CI)` `therefore B.E.(H - CI) = 215 + 120+ 90 = 425 kJ"mol"^(-1)`