The reaction
`CH_(3)COOC_(2)H_(5) + NaOH to CH_(3)COONa + C_(2)H_(5)OH` is allowed to take place with initial concentration of 0.2 mole/lit of each reactant. If the reaction mixture is diluted with water so that the initial concentration of each reactant becomes 0.1 mole/lit. The rate of the reaction will be

Options

A. 1/8 th of the original rate
B. 1/4th of the original rate
C. 1/2 th of the original rate
D. same as the original rate

Explanation

To solve the problem, we need to analyze the reaction and how the rate changes with the concentration of the reactants. The reaction given is:

\[ \text{CH}_3\text{COOC}_2\text{H}_5 + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{C}_2\text{H}_5\text{OH} \]

### Step-by-Step Solution:

1. **Identify the Initial Concentrations:**
The initial concentration of both reactants (CH₃COOC₂H₅ and NaOH) is given as 0.2 moles/liter.

2. **Determine the Effect of Dilution:**
The reaction mixture is diluted, and the new concentration of each reactant becomes 0.1 moles/liter. This means that the concentration has been halved.

3. **Write the Rate Law:**
The rate of the reaction can be expressed using the rate law:
\[ \text{Rate} = k [\text{CH}_3\text{COOC}_2\text{H}_5][\text{NaOH}] \]
where \( k \) is the rate constant.

4. **Substitute Initial Concentrations into the Rate Law:**
For the initial concentration of 0.2 moles/liter:
\[ \text{Rate}_{\text{initial}} = k (0.2)(0.2) = k (0.04) \]

5. **Substitute New Concentrations into the Rate Law:**
After dilution, the concentration of each reactant is 0.1 moles/liter:
\[ \text{Rate}_{\text{new}} = k (0.1)(0.1) = k (0.01) \]

6. **Compare the New Rate with the Initial Rate:**
To find the relationship between the new rate and the initial rate:
\[ \frac{\text{Rate}_{\text{new}}}{\text{Rate}_{\text{initial}}} = \frac{k (0.01)}{k (0.04)} = \frac{0.01}{0.04} = \frac{1}{4} \]
This means the new rate is \( \frac{1}{4} \) of the initial rate.

7. **Conclusion:**
Therefore, the rate of the reaction after dilution is \( \frac{1}{4} \) times the initial rate.

### Final Answer:
The rate of the reaction after dilution is \( \frac{1}{4} \) times the original rate.

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