The enthalpy change for the conversion of $\frac{1}{2} \mathrm{Cl}_{2}(\mathrm{~g})$ to $\mathrm{Cl}^{-}$(aq) is ($-$) ___________ $\mathrm{kJ} \mathrm{mol}^{-1}$ (Nearest integer)
Given : $$\Delta_{\mathrm{dis}} \mathrm{H}_{\mathrm{Cl}_{2(\mathrm{~g})}^{\theta}}^{\ominus}=240 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta_{\mathrm{eg}} \mathrm{H}_{\mathrm{Cl_{(g)}}}^{\ominus}=-350 \mathrm{~kJ} \mathrm{~mol}^{-1}$$,
${\mathrm{\Delta _{hyd}}H_{Cl_{(g)}^ - }^\Theta = - 380}$ $\mathrm{kJ~mol^{-1}}$
Answer (integer)
610
Solution
$$
\begin{aligned}
& \frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \longrightarrow \mathrm{Cl}_{\text {(aq) }}^{-} \quad \Delta \mathrm{H}=? \\\\
& \begin{aligned}
\Delta \mathrm{H} & =\frac{1}{2} \Delta_{\mathrm{diss}} \mathrm{H}_{\mathrm{Cl}_2}^{\circ}+\Delta_{\mathrm{eg}} \Delta \mathrm{H}_{\mathrm{Cl}(\mathrm{g})}^{\circ}+\Delta_{\mathrm{hyd}} \mathrm{H}_{\mathrm{Cl}_{(\mathrm{g})}^{-}}^{\circ} \\\\
& =\frac{1}{2} \times 240+(-350)+(-380) \\\\
& =-610 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
\end{aligned}
$$
About this question
Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law
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