Given that $\Delta H_f(\mathrm{H})=218 \mathrm{~kJ} / \mathrm{mol}$, express the $\mathrm{H}-\mathrm{H}$…
Given that $\Delta H_f(\mathrm{H})=218 \mathrm{~kJ} / \mathrm{mol}$, express the $\mathrm{H}-\mathrm{H}$ bond energy in $\mathrm{kcal} / \mathrm{mol}$.
52.15
911
104
52153
Solution
Given : $\Delta H_f(\mathrm{H})=218 \mathrm{~kJ} / \mathrm{mol}$
ie, $\quad \frac{1}{2} \mathrm{H}_2 \longrightarrow \mathrm{H} ; \quad \Delta H=218 \mathrm{~kJ} / \mathrm{mol}$
$\begin{aligned} & \text { or } \quad \mathrm{H}_2 \longrightarrow 2 \mathrm{H} ; \quad \Delta H=436 \mathrm{~kJ} / \mathrm{mol} \\ & =\frac{436}{4.18}=104.3 \mathrm{kcal} / \mathrm{mol} \\ & \end{aligned}$
Thus, $104.3 \mathrm{kcal} / \mathrm{mol}$ energy is absorbed for breaking one mole of $\mathrm{H}-\mathrm{H}$ bonds. Hence, $\mathrm{H}-\mathrm{H}$ bond energy is $104.3 \mathrm{kcal} / \mathrm{mol}$.