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

For the reaction `H_(2)(g)+Cl_(2)(g) to 2HCl(g)`
For reactants
Bond energy of 1 mole of H-H bond `=104` kcal
Bond energy of 1 mole of Cl-Cl bond `=58` kcal
For products :
Energy of formation of 2 moles of H-Cl bond `=-2xx103` kcal
Thus `DeltaH` of the above reaction `=104+58-206=-44` kcal
Since, for
`H_(2)+Cl_(2) to 2HCl` , `DeltaH=-44kcal`
then for, `(1)/(2)H_(2)+(1)/(2)Cl_(2) to HCl` , `DeltaH=-22kcal`