The bond dissociation energy of gaseous `H_(2),Cl_(2)` and `HCl` are `104,58` and `103 kcal mol^(-1)` respecitvely. Calculate the enthalpy of formation for `HCl` gas.

Explanation

The required equation is
`(1)/(2)H_(2)(g)+(1)/(2)Cl_(2)(g)toHCl(g), " "DeltaH=?`
`DeltaH=[(1)/(2)DeltaH_(H-H)+(1)/(2)DeltaH_(Cl-Cl)]-[DeltaH_(H-Cl)]`
`=(1)/(2)xx104+(1)/(2)xx58-103=-22` kcal `mol^(-1)`