Q51 / 150 · JEE Main 2006 · Chemistry
The enthalpy changes for the following processes are listed below:
$\begin{array}{ll}\mathrm{Cl}_2(\mathrm{~g})=2 \mathrm{Cl}(\mathrm{g}), & 242.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \mathrm{I}_2(\mathrm{~g})=2 \mathrm{I}(\mathrm{g}), & 151.0 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \mathrm{ICl}(\mathrm{g})=\mathrm{I}(\mathrm{g})+\mathrm{Cl}(\mathrm{g}), & 211.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \mathrm{I}_2(\mathrm{~s})=\mathrm{I}_2(\mathrm{~g}), & 62.76 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{array}$
Given that the standard states for iodine and chlorine are $\mathrm{I}_2(\mathrm{~s})$ and $\mathrm{Cl}_2(\mathrm{~g})$, the standard enthalpy of formation for $\mathrm{ICl}(\mathrm{g})$ is
- A$-14.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- B$-16.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- Correct answerC$+16.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- D$+244.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Solution
$$
\begin{aligned}
& \frac{1}{2} \mathrm{I}_2(\mathrm{~s})+\frac{1}{2} \mathrm{Cl}_2 \longrightarrow \mathrm{ICl}(\mathrm{g}) \\
& \Delta \mathrm{H}=\left[\frac{1}{2} \Delta \mathrm{H}_{\mathrm{b}_2(s) \rightarrow k_2(g)}+\frac{1}{2} \mu_{1-1}+\frac{1}{2} \mu_{\mathrm{cl}-\mathrm{cl}}\right]-\left[\mu_{1-a}\right] \\
& =\left(\frac{1}{2} \times 62.76+\frac{1}{2} \times 151.0+\frac{1}{2} \times 242.3\right)-(211.3) \\
& =228.03-211.3 \\
& \Delta \mathrm{H}=16.73 \\
&
\end{aligned}
$$
Hence, the correct answer is option: 'C'.