`100" cm"^(3)" of "1" M "CH_(3)COOH` was mixed with `100" cm"^(3)" of "2" M "CH_(3)OH` to form an ester. The change in the initial rate if each solution is diluted with equal volume of water would be

Options

A. 4 times
B. 0.25 times
C. 2 times
D. 0.5 times

Explanation

To solve the problem, we need to analyze how the dilution of the reactants affects the rate of the reaction. We start with the initial concentrations of acetic acid (CH₃COOH) and methanol (CH₃OH) and then see how the dilution changes these concentrations.

### Step-by-Step Solution:

1. **Initial Concentrations**:
- We have 100 cm³ of 1 M CH₃COOH and 100 cm³ of 2 M CH₃OH.
- The initial concentration of CH₃COOH (C₁) = 1 M.
- The initial concentration of CH₃OH (C₂) = 2 M.

2. **Volume After Mixing**:
- When we mix both solutions, the total volume becomes:
\[ V_{\text{total}} = 100 \, \text{cm}^3 + 100 \, \text{cm}^3 = 200 \, \text{cm}^3 \]

3. **Calculating Initial Molar Concentrations**:
- The number of moles of CH₃COOH (n₁) = Molarity × Volume = 1 M × 0.1 L = 0.1 moles.
- The number of moles of CH₃OH (n₂) = 2 M × 0.1 L = 0.2 moles.
- After mixing, the new concentrations are:
\[ C_{\text{CH}_3\text{COOH}} = \frac{n_1}{V_{\text{total}}} = \frac{0.1 \, \text{moles}}{0.2 \, \text{L}} = 0.5 \, \text{M} \]
\[ C_{\text{CH}_3\text{OH}} = \frac{n_2}{V_{\text{total}}} = \frac{0.2 \, \text{moles}}{0.2 \, \text{L}} = 1 \, \text{M} \]

4. **Rate of Reaction**:
- The rate of reaction (ROR) can be expressed as:
\[ \text{ROR}_1 = k \cdot [\text{CH}_3\text{COOH}] \cdot [\text{CH}_3\text{OH}] = k \cdot (0.5) \cdot (1) = 0.5k \]

5. **Dilution with Water**:
- If each solution is diluted with an equal volume of water (100 cm³), the new total volume becomes:
\[ V_{\text{new}} = 200 \, \text{cm}^3 + 100 \, \text{cm}^3 + 100 \, \text{cm}^3 = 400 \, \text{cm}^3 \]

6. **New Concentrations After Dilution**:
- The number of moles remains the same, but the volume has increased:
\[ C'_{\text{CH}_3\text{COOH}} = \frac{0.1 \, \text{moles}}{0.4 \, \text{L}} = 0.25 \, \text{M} \]
\[ C'_{\text{CH}_3\text{OH}} = \frac{0.2 \, \text{moles}}{0.4 \, \text{L}} = 0.5 \, \text{M} \]

7. **New Rate of Reaction After Dilution**:
- The new rate of reaction (ROR₂) is:
\[ \text{ROR}_2 = k \cdot [\text{CH}_3\text{COOH}]' \cdot [\text{CH}_3\text{OH}]' = k \cdot (0.25) \cdot (0.5) = 0.125k \]

8. **Comparing the Rates**:
- The ratio of the new rate to the initial rate is:
\[ \frac{\text{ROR}_2}{\text{ROR}_1} = \frac{0.125k}{0.5k} = \frac{0.125}{0.5} = \frac{1}{4} \]
- Therefore, the new rate is one-fourth of the initial rate.

### Final Answer:
The change in the initial rate if each solution is diluted with equal volume of water would be that the new rate is **1/4 of the original rate**.