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

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

1.	Given that bond energies of H - H and Cl -Cl are 430 "kJ/mol"^(-1) and 240 "kJ/mol"^(-1) respectively and DeltaH_(f) for HCI is -90 "kJ/mol"^(-1), bond enthalpy of HCI is
Answer» 380 KJ/mol 425 kJ/mol 245 kJ/mol 290 kJ/mol Solution :`(1)/(2) H_(2) + (1)/(2) CI_(2) to HCI` `DeltaH_(HCI) = sum B.E` of reactant `- sumB.E` of products `-90= (1)/(2) xx 430 + (1)/(2) + (1)/(2) xx 240 - B.E.` of `HCI` `therefore B.E`. of HCI `= 215 + 210 + 90` `= 425 "kJ mol"^(-1)`