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The rate of the reaction, $\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5+\mathrm{NaOH} \longrightarrow \mathrm{CH}_3 \mathrm{COONa}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$ is given by the equation, rate $=k\left[\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5\right][\mathrm{NaOH}]$. If concentration is expressed in $\mathrm{mol}^{-1}$, the unit of $k$ is
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The correct answer is:
$\mathrm{L} \mathrm{mol}^{-1} \mathrm{~s}^{-1}$
Given that, rate $=k\left[\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5\right][\mathrm{NaOH}]$ That means the reaction is a second order
$$
\begin{aligned}
\therefore \quad \frac{d x}{d t} & =k[A]^2 \Rightarrow \frac{\text { conc }}{\text { time }}=k[\text { conc. }]^2 \\
\frac{\mathrm{mol} \mathrm{L}^{-1}}{\mathrm{~s}} & =k \mathrm{~mol} \mathrm{~L}^{-1} \times \mathrm{mol} \mathrm{L^{-1 }} \\
k & =\mathrm{L} \mathrm{mol}^{-1} \mathrm{~s}^{-1}
\end{aligned}
$$
$$
\begin{aligned}
\therefore \quad \frac{d x}{d t} & =k[A]^2 \Rightarrow \frac{\text { conc }}{\text { time }}=k[\text { conc. }]^2 \\
\frac{\mathrm{mol} \mathrm{L}^{-1}}{\mathrm{~s}} & =k \mathrm{~mol} \mathrm{~L}^{-1} \times \mathrm{mol} \mathrm{L^{-1 }} \\
k & =\mathrm{L} \mathrm{mol}^{-1} \mathrm{~s}^{-1}
\end{aligned}
$$
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