Q2 / 110 · VITEEE 2024 · Chemistry
The bond dissociation energies for $\mathrm{Cl}_2, \mathrm{I}_2$ and ICl are $242.3,151.0$ and $211.3 \mathrm{~kJ} / \mathrm{mol}$ respectively. The enthalpy of sublimation of iodine is $62.8 \mathrm{~kJ} / \mathrm{mol}$. What is the standard enthalpy of formation of $\mathrm{ICl}(\mathrm{g})$ nearly equal to?
- A$-211.3 \mathrm{~kJ} / \mathrm{mol}$
- B$-14.6 \mathrm{~kJ} / \mathrm{mol}$
- Correct answerC$16.8 \mathrm{~kJ} / \mathrm{mol}$
- D$33.5 \mathrm{~kJ} / \mathrm{mol}$
Solution
The given reaction is as follows,
$\frac{1}{2} \mathrm{I}_2(g)+\frac{1}{2} \mathrm{Cl}_2(g) \longrightarrow \mathrm{ICl}(g)$
We are given,
(i) $\mathrm{Cl}_2(g) \longrightarrow 2 \mathrm{Cl}(\mathrm{g}), \Delta \mathrm{H}=242.3 \mathrm{~kJ}$
(ii) $\mathrm{I}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{I}(\mathrm{g}), \Delta H=151 \mathrm{~kJ}$
(iii) $\mathrm{ICl}(\mathrm{g}) \longrightarrow \mathrm{I}(\mathrm{g})+\mathrm{Cl}(\mathrm{g}), \Delta H=211.3 \mathrm{~kJ}$
(iv) $\mathrm{I}_2(g) \longrightarrow \mathrm{I}_2(g), \Delta H=62.8 \mathrm{~kJ}$
$\frac{1}{2}$ (ii) $+\frac{1}{2}$ (i) - (iii) $+\frac{1}{2}$
(iv) gives the required equation.
$\begin{aligned} \therefore \Delta H & =\frac{1}{2}(151)+\frac{1}{2}(242.3)-211.3+\frac{1}{2}(62.8) \\ \Delta H & =16.8 \mathrm{~kJ}\end{aligned}$
Hence, the correct answer is option: 'C'.