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Question: Answered & Verified by Expert
Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below:
$$
\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \stackrel{\frac{1}{2} \Delta_{\mathrm{ass}} \mathrm{H}^{\ominus}}{\longrightarrow} \mathrm{Cl}(\mathrm{g}) \stackrel{\Delta_{\mathrm{eg} H^{\ominus}}}{\longrightarrow} \mathrm{Cl}^{-}(\mathrm{g}) \stackrel{\Delta_{\mathrm{m} \mathrm{on}^{H^{\ominus}}}^{\longrightarrow}}{\longrightarrow} \mathrm{Cl}^{-}(\mathrm{aq}) .
$$
The energy involved in the conversion of $\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g})$ to $\mathrm{Cl}^{-}(\mathrm{g})$ (using the data, $\left.\Delta_{\text {diss }} H_{\mathrm{Cl}_2}^{\ominus}=240 \mathrm{kJmol}^{-1}, \Delta_{\mathrm{eg}} H_{\mathrm{cl}}^{\ominus}=-349 \mathrm{kJmol}^{-1}, \Delta_{\mathrm{hyd}} \mathrm{H}_{\mathrm{Cl}}^{\ominus}=-381 \mathrm{kJmol}^{-1}\right)$ will be
ChemistryThermodynamics (C)JEE MainJEE Main 2008
Options:
  • A
    $+152 \mathrm{kJmol}^{-1}$
  • B
    $-610 \mathrm{kJmol}^{-1}$
  • C
    $-850 \mathrm{kJmol}^{-1}$
  • D
    $+120 \mathrm{kJmol}^{-1}$
Solution:
2068 Upvotes Verified Answer
The correct answer is:
$-610 \mathrm{kJmol}^{-1}$
$$
\begin{aligned}
& \text { For the process } \frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \longrightarrow \mathrm{Cl}_{\mathrm{aq}}^{-} \\
& \Delta \mathrm{H}=\frac{1}{2} \Delta \mathrm{H}_{\text {diss }} \text { of } \mathrm{Cl}_2+\Delta_{\mathrm{eg}} \mathrm{Cl}+\Delta_{\mathrm{hyd}} \mathrm{Cl}^{-} \\
& =+\frac{240}{2}-349-381 \\
& =-610 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$$

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