\(100\, \ cm^3\) of \(1\, \ M\) \(CH_3COOH\) and \(100\, \ cm^3\) of \(2\, \ M\) \(CH_3OH\) were mixed to form an ester which follow \(1^{st}\) order kinetic with respect to both \(CH_3COOH\) and \(CH_3OH\). The change in their initial rate, if each solution is mixed with equal number of volume of water would be:

0.25 times

The correct answer is option 2. 0.25 times.

To determine the change in the initial rate when each solution is mixed with an equal volume of water, let's first analyze the reaction.

The reaction is between acetic acid (\(CH_3COOH\)) and methanol (\(CH_3OH\)) to form an ester. The reaction follows first-order kinetics with respect to both reactants.

The rate equation for a first-order reaction is:

\[ \text{Rate} = k[A] \]

Where:

\( k \) is the rate constant.

\( [A] \) is the concentration of the reactant.

Since the reaction is first-order with respect to both reactants, the rate depends on the concentration of both \(CH_3COOH\) and \(CH_3OH\), and the rate equation can be written as:

\[ \text{Rate} = k[CH_3COOH][CH_3OH] \]

When equal volumes of water are added to each solution, the concentration of each reactant is halved. Therefore, the new rate equation becomes:

\[ \text{Rate} = k\left(\frac{1}{2}[CH_3COOH]\right)\left(\frac{1}{2}[CH_3OH]\right) = \frac{1}{4}k[CH_3COOH][CH_3OH] \]

So, the change in the initial rate is \( \frac{1}{4} \) times the original rate.

Therefore, the correct answer is: (2) 0.25 times.