Given that bond energies of $\mathrm{H}-\mathrm{H}$ and $\mathrm{Cl}-\mathrm{Cl}$ are $430 \mathrm{~kJ}…
Given that bond energies of $\mathrm{H}-\mathrm{H}$ and $\mathrm{Cl}-\mathrm{Cl}$ are $430 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $240 \mathrm{~kJ}$ $\mathrm{mol}^{-1}$ respectively and $\Delta \mathrm{H}_{\mathrm{f}}$ for $\mathrm{HCl}$ is $-\mathrm{kJ}$ $\mathrm{mol}^{-1}$, bond enthalpy of $\mathrm{HCl}$ is:
$380 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$425 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$245 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$290 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Solution
$\frac{1}{2} \mathrm{H}_2+\frac{1}{2} \mathrm{Cl}_2 \longrightarrow \mathrm{HCl}$
$\Delta \mathrm{H}_{\mathrm{HCl}}=\Sigma$ B.E. of reactant $-\Sigma$ B.E. of products
$-90=\frac{1}{2} \times 430+\frac{1}{2} \times 240-$ B.E. of $\mathrm{HCl}$
$\therefore \quad$ B.E. of $\mathrm{HCl}=215+120+90$
$=425 \mathrm{~kJ} \mathrm{~mol}^{-1}$