Bond dissociation enthalpy of H_(2), CI_(2) and HCI are 434, 242 and 431 "kJ mol"^(-1) respectively. Enthalpy of formation of HCl is

Bond dissociation enthalpy of H_(2), CI_(2) and HCI are 434, 242 and 4

1.	Bond dissociation enthalpy of H_(2), CI_(2) and HCI are 434, 242 and 431 "kJ mol"^(-1) respectively. Enthalpy of formation of HCl is
Answer» `93 "kJ mol"^(-1)` `-245 "kJ mol"^(-1)` `-93 "kJ mol"^(-1)` `245 "kJ mol"^(-1)` Solution :The REACTION for formation of HO can be written as `H_(2) + CI_(2) to 2HCI` `H- H + CI- CI to 2(H- CI)` Substituting the given values, we GET enthalpy of formation of `2HCI = - (862-676) = - 186` kJ `therefore` Enthalpy of formation of `HCI = (-186)/(2) "kJ"= -93` kJ