Q30 / 180 · JIPMER 2018 · Chemistry
Given that, for the reaction,
$\frac{1}{2} \mathrm{H}_2+\frac{1}{2} \mathrm{Cl}_2 \longrightarrow \mathrm{HCl}$
$\begin{aligned} & \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{HCl})=-93 \mathrm{~kJ} / \mathrm{mol} \\ & \text { B.E }\left(\mathrm{H}_2\right)=434 \mathrm{~kJ} / \mathrm{mol} \\ & \text { B.E }\left(\mathrm{Cl}_2\right)=242 \mathrm{~kJ} / \mathrm{mol}\end{aligned}$
The bond dissociation energy of HCl is
- A$232 \mathrm{~kJ} / \mathrm{mol}$
- B$331 \mathrm{~kJ} / \mathrm{mol}$
- Correct answerC$431 \mathrm{~kJ} / \mathrm{mol}$
- D$530 \mathrm{~kJ} / \mathrm{mol}$
Solution
Given,
$\frac{1}{2} \mathrm{H}_2+\frac{1}{2} \mathrm{Cl}_2 \longrightarrow \mathrm{HCl}, \Delta \mathrm{H}_{\mathrm{f}}=-93 \mathrm{~kJ} / \mathrm{mol}$
$\because$ In this reaction, one mole of HCl is formed
$\therefore \Delta \mathrm{H}_{\mathrm{f}}$ of the reaction will remain same, i.e. $-93 \mathrm{~kJ} / \mathrm{mol}$
Let the bond dissociation energy of $\mathrm{H}-\mathrm{Cl}$ bond be x
For the given reaction,
$\begin{aligned} \Delta \mathrm{H}_{\mathrm{f}} & =\Sigma \mathrm{B} \cdot \mathrm{E}_{\text {(reactants) }}-\Sigma \mathrm{B} \cdot \mathrm{E}_{\text {(products) }} \\ -93 & =\frac{1}{2}(434)+\frac{1}{2}(242)-\mathrm{x} \\ -93 & =+338-\mathrm{x} \\ \mathrm{X} & =431 \mathrm{~kJ} / \mathrm{mol}\end{aligned}$
Hence, the correct answer is option: 'C'.