Search any question & find its solution
Question:
Answered & Verified by Expert
\( 1 \mathrm{~L} \) of \( 2 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH} \) is mixed with \( 1 \mathrm{~L} \) of \( 3 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \) to form an ester. The rate of the
reaction with respect to the initial rate when each solution is diluted with an equal volume of
water will be
Options:
reaction with respect to the initial rate when each solution is diluted with an equal volume of
water will be
Solution:
1390 Upvotes
Verified Answer
The correct answer is:
\( 0.25 \) times
(C)
Esterification is a second order reaction
$\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \rightarrow \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H} 5+\mathrm{H}_{2} \mathrm{O}$
$\mathrm{r}=\mathrm{k}\left[\mathrm{CH}_{3} \mathrm{COOH}\right]\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]$
Order $=2$
When equal volume of two solutions are mixed, concentration of the solutions reduces to half the initial value. Hence,
rate of reaction gets reduced to $\frac{1}{4}$ initial rate.
Esterification is a second order reaction
$\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \rightarrow \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H} 5+\mathrm{H}_{2} \mathrm{O}$
$\mathrm{r}=\mathrm{k}\left[\mathrm{CH}_{3} \mathrm{COOH}\right]\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]$
Order $=2$
When equal volume of two solutions are mixed, concentration of the solutions reduces to half the initial value. Hence,
rate of reaction gets reduced to $\frac{1}{4}$ initial rate.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.