Bond dissociation energies of $$H_{2}, Cl_{2}$$ and $$HCl_{(g)}$$ are $$104, 58$$ and $$103$$ kcal mol$$^{-1}$$ respectively. Calculate the enthalpy of formation of HCl gas.
Correct option is A. $$- 22 kcal$$
$$ \dfrac{1}{2} H_{2} + \dfrac{1}{2}Cl_{2} \rightarrow HCl $$
$$\Delta H =\dfrac12 B.E._{(H-H)} + \dfrac12 B.E._{(Cl-Cl)} -B.E._{(HCl)}$$
$$ \Delta H = \dfrac{1}{2} \times 104 + \dfrac{1}{2} \times 58 - 103 = - 22 \ kcal $$