Question

The bond disssociation energies of gases $H_2,{Cl}_2$ and $HCl$ are $104,58$ and $103$ $kcal{mol}^{-1}$ respectively. Calculate the enthalpy of formation of $HCl\left(g\right)$.

Answer 1

The required equation is
$\frac{1}{2}{H}_2\left(g\right)+\frac{1}{2}{Cl}_2\left(g\right)\to HCl\left(g\right);\Delta H=?$
$\Delta H=[\frac{1}{2}\Delta H_{H-H}+\frac{1}{2}\Delta H_{Cl-Cl}]-\left[\Delta H_{H-Cl}\right]$
$=\frac{1}{2}\times 104+\frac{1}{2}\times 58-103=-22kcal{mol}^{-1}$