KCET · Chemistry · Chemical Kinetics
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
- A \(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\)
- B \(\mathrm{L} \mathrm{mol}^{-1} \mathrm{~s}^{-1}\)
- C \(\mathrm{s}^{-1}\)
- D \(\mathrm{mol}^{-2} \mathrm{~L}^2 \mathrm{~S}^{-1}\)
Answer & Solution
Correct Answer
(B) \(\mathrm{L} \mathrm{mol}^{-1} \mathrm{~s}^{-1}\)
Step-by-step Solution
Detailed explanation
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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