$\frac{{d\ln k}}{{dt}} = \frac{{\Delta {H^o}}}{{R{T^2}}}$
or $\ln {k_p} = - \frac{{\Delta {H^o}}}{{RT}} + I$.
Hence $(a)$ is correct $(b)$ is also correct as plot of $log\,(X)$ vs $time$ is linear.
$ (c) $ is wrong because $p \propto T$ at constant volume. $(d)$ is correct by Boyle’s law.
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$(a)$ tetraamminecobalt $(III)$ dim-hydroxidotetra amminecobalt $(III)$ ion
$(b)$ di $-\mu -$ hydroxidobis(tetraammine)dicobalt $(III)$
$(c)$ Bis $-(\mu -$ hydroxidotetraamminecobalt $(III))$
$(d)$ di $-\mu-$ hydroxidooctaamminedicobalt $(III)$
correct option is
$\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - C - CH = C{H_2}\,\,\,\,\,} \\
{\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,\,\,\,C{H_{3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}}
\end{array}\xrightarrow[{(ii)\,NaB{H_4}}]{{(i)\,Hg{{(OAc)}_2};{H_2}O}}$
$Cl_2 + 2Y^-\to 2Cl^-+ Y_2$
The incorrect match regarding above reaction is
