\( 100 \mathrm{~cm}^{3} \) of \( 1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH} \) was mixed with \( 100 \mathrm{~cm}^{3} \) of \( 2 \mathrm{MCH}_{3} \mathrm{OH} \) to form an ester. The
change in the initial rate if each solution is diluted with equal volume of water would be

For the esterification reaction
\[
\begin{array}{l}
\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{CH}_{3} \mathrm{OH} \rightarrow \mathrm{CH}_{3} \mathrm{COOCH}_{3}+\mathrm{H}_{2} \mathrm{O} \\
\quad_{1 \mathrm{M}} \\
r_{1}=k\left[\mathrm{CH}_{3} \mathrm{COOH}\right]\left[\mathrm{CH}_{3} \mathrm{OH}\right] \\
=k[1][2] \\
r_{1}=2 k \rightarrow(1)
\end{array}
\]
when solution is diluted with equal volume of water, concentration is reduced to half
\[
\begin{array}{l}
r_{2}=k\left[\frac{1}{2}\right]\left[\frac{2}{2}\right] \\
r_{2}=\frac{k}{2} \rightarrow(2)
\end{array}
\]
Divide Eq. (1) by Eq. (2), we get
\[
\begin{array}{l}
\frac{r_{1}}{r_{2}}=\frac{2 k}{\frac{k}{2}}=4 \\
r_{2}=\frac{r_{1}}{4}=0.25 r_{1}
\end{array}
\]
Therefore, the rate of the reaction will reduce to \( 0.25 \) times of the initial rate.