Q76 / 180 · AIIMS 2012 · Chemistry
For a reaction, $r=k\left(\mathrm{CH}_3 \mathrm{COCH}_3\right)^{3 / 2}$ then unit of rate of reaction and rate constant respectively is
- Correct answerA$\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}, \mathrm{~mol}^{-1 / 2} \mathrm{~L}^{1 / 2} \mathrm{~s}^{-1}$
- B$\mathrm{mol}^{-1} \mathrm{~L}^{-1} \mathrm{~s}^{-1}, \mathrm{~mol}^{-1 / 2} \mathrm{~L}^{-1 / 2} \mathrm{~s}^{-1}$
- C$\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}, \mathrm{~mol}^{+1 / 2} \mathrm{~L}^{1 / 2} \mathrm{~s}^{-1}$
- D$\mathrm{mol} \mathrm{L} \mathrm{s}, \mathrm{mol}^{+1 / 2} \mathrm{~L}^{1 / 2} \mathrm{~s}$
Solution
$\begin{aligned}
\text { : Rate } & =\frac{\text { concentration }}{\text { time }}=\frac{\mathrm{mol} \mathrm{L}^{-1}}{\mathrm{~s}} \\
= & \mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}
\end{aligned}$
$\begin{aligned}
: \text { Rate } & =\frac{\text { concentration }}{\text { time }}=\frac{\mathrm{mol} \mathrm{L}^{-1}}{\mathrm{~s}} \\
& =\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}
\end{aligned}$
The units of rate constant for the reaction of the $n^{\text {th }}$ order $=\left(\mathrm{mol} \mathrm{L}^{-1}\right)^{1-n} \mathrm{~s}^{-1}$ here, $=\frac{3}{2}$
$\therefore$ Unit of rate constant $=\left(\mathrm{mol} \mathrm{L}^{-1}\right)^{1-3 / 2} \mathrm{~s}^{-1}$
$=\left(\mathrm{mol} \mathrm{L}^{-1}\right)^{-1 / 2} \mathrm{~s}^{-1}$
$=\mathrm{mol}^{-1 / 2} \mathrm{~L}^{1 / 2} \mathrm{~s}^{-1}$
Hence, the correct answer is option: 'A'.