The kinetic data for the reaction $\mathrm{OCl}^{-}+\mathrm{I}^{-}…
$\begin{array}{cccc}
\frac{\left[\mathrm{OCl}^{-}ight]}{\mathrm{mol} \mathrm{dm}^{-3}} & \frac{\left[\mathrm{I}^{-}ight]}{\mathrm{mol} \mathrm{dm}^{-3}} & \frac{\left[\mathrm{OH}^{-}ight]}{\mathrm{mol} \mathrm{dm}^{-3}} & \frac{10^{-4} \times \mathrm{d}\left[\mathrm{IO}^{-}ight] / \mathrm{d} t}{\mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}} \\
\hline 0.0017 & 0.0017 & 1.0 & 1.75 \\
0.0034 & 0.0017 & 1.0 & 3.50 \\
0.0017 & 0.0034 & 1.0 & 3.50 \\
0.0017 & 0.0017 & 0.5 & 3.50
\end{array}$
The rate low for the formation of $\mathrm{OI}^{-}$is
- $r=k\left[\mathrm{OCl}^{-}ight]\left[\mathrm{I}^{-}ight]$
- $r=k\left[\mathrm{OCl}^{-}ight]^{2}$
- $r=k\left[\mathrm{I}^{-}ight]^{2}\left[\mathrm{OCl}^{-}ight]$
- $r=k\left[\mathrm{OCl}^{-}ight]\left[\mathrm{I}^{-}ight] /\left[\mathrm{OH}^{-}ight]$
Solution
On doubling the $\left[\mathrm{I}^{-}ight]$keeping the $\left[\mathrm{OCl}^{-}ight]$and $\left[\mathrm{OH}^{-}ight]$constant, the rate of formation of $\left[\mathrm{OI}^{-}ight]$is doubled, hence the order of reaction with respect to $\mathrm{I}^{-}$is one.
On changing the $\left[\mathrm{OH}^{-}ight]$to a half-value, the rate of formation of $\left[\mathrm{OI}^{-}ight]$is doubled. Hence, the order of reaction with respect to $\left[\mathrm{OH}^{-}ight]$is $-1$.
Hence the rate law is $\quad r=k\left[\mathrm{OCl}^{-}ight]\left[\mathrm{I}^{-}ight]\left[\mathrm{OH}^{-}ight]^{-1}=k\left[\mathrm{OCl}^{-}ight]\left[\mathrm{I}^{-}ight] /\left[\mathrm{OH}^{-}ight]$
Asked in: JEE-TOPICTESTS-CHEMISTRY