Bond dissociation enthalpy of H₂, Cl₂ and HCl are 434 , 242 and 431 kJ mol^–1 respectively. Enthalpy of formation of HCl is:

The reaction for formation of HCl can be written as 

$\begin{array}{l}{\mathrm{H}}_2+{\mathrm{Cl}}_2\to 2\mathrm{HCI}\\ \mathrm{H}-\mathrm{H}+\mathrm{Cl}-\mathrm{Cl}\to 2(\mathrm{H}-\mathrm{Cl})\end{array}$

Substituting the given values, we get enthalpy of formation of $2\mathrm{HCl}=-(862-676)=-186\mathrm{kJ}.$

$\therefore$ Enthalpy of formation of $\mathrm{HCl}=\frac{-186}{2}\mathrm{kJ}=-93\mathrm{kJ}$