Use the → Table of bond energies to recalculate the enthalpy of formation of $HCl(g)$.
$H-H(g)$ $+$ $Cl-Cl(g)$ $\longrightarrow$ $2 H-Cl (g)$ $\Delta H$ $= $ $E_{bond}(H-H)$ $ +$ $ E_{bond}(Cl-Cl)$ $-$ $ 2E_{bond}(H-Cl)$ $=$ $ -183\;kJ$ Enthalpy of formation (one mole !) of $HCl(g$): $\Delta H_f(HCl(g))$ $=$ $\frac{-183}{2}=-91.5kJ$ , an approximate value! (see the → Table of formation enthalpies!)