MCQ
For the reaction$PC{l_5}(g) \to PC{l_3}(g) + C{l_2}(g)$
- A$\Delta H = \Delta E$
- ✓$\Delta H > \Delta E$
- C$\Delta H < \Delta E$
- DNone of these
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$\frac{1}{2}C{l_2}(g)\xrightarrow{{\frac{1}{2}{\Delta _{diss}}{H^\Theta }}}Cl(g)\xrightarrow{{{\Delta _{eg}}{H^\Theta }}}$ $C{l^ - }(g)\xrightarrow{{{\Delta _{Hyd}}{H^\Theta }}}C{l^ - }(aq)$
(using the data,
${{\Delta _{diss}}H_{C{l_2}}^\Theta } = 240\,kJ\,mol^{-1}, {{\Delta _{eg}}H_{C{l}}^\Theta }= -349 \,kJ\,mol^{-1},$${{\Delta _{Hyd}}H_{C{l}}^\Theta }= -381 \,kJ\,mol^{-1}$ ) will be ............. $\mathrm{kJ\,mol}^{-1}$
$\begin{array}{|l|l|} \hline Product & Sulphide\,\,Solubility \\ \hline (I) & 1.74\,\,\times\,\,10^{-16} \\ \hline (II) & 1.2\,\,\times\,\,10^{-22} \\ \hline (III) & 8.2\,\,\times\,\,10^{-46} \\ \hline (IV) & 5.0\,\,\times\,\,10^{-34} \\ \hline \end{array}$