One litre of 2 M acetic acid and one litre of 3 M ethyl alcohol are mixed to form ester according to the given equation: `CH_(3)COOH+C_(2)H_(5)OH hArr CH_(3)COOC_(2)H_(5)+H_(2)O` If each solution is diluted by adding equal volume (1 litre) of water by how many times the initial forward rate has changed?

Options

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

Answer

To solve the problem, we need to analyze the reaction and how the concentrations of the reactants change when we dilute the solutions.

### Step-by-Step Solution:

1. **Write the Reaction Equation:**
The reaction given is:
\[ \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \]

2. **Determine Initial Concentrations:**
- For acetic acid (CH₃COOH):
- Volume = 1 L, Molarity = 2 M
- Moles of CH₃COOH = 2 moles (since 1 L × 2 M = 2 moles)
- For ethyl alcohol (C₂H₅OH):
- Volume = 1 L, Molarity = 3 M
- Moles of C₂H₅OH = 3 moles (since 1 L × 3 M = 3 moles)

3. **Calculate Initial Forward Rate (RF1):**
The forward rate of the reaction can be expressed as:
\[ R_{F1} = k_F \cdot [\text{CH}_3\text{COOH}] \cdot [\text{C}_2\text{H}_5\text{OH}] \]
- Initial concentrations:
- \([\text{CH}_3\text{COOH}] = 2 \, \text{M}\)
- \([\text{C}_2\text{H}_5\text{OH}] = 3 \, \text{M}\)
- Therefore:
\[ R_{F1} = k_F \cdot 2 \cdot 3 = 6k_F \]

4. **Dilute the Solutions:**
After adding 1 L of water to each solution, the total volume becomes 2 L for each component:
- New concentration of CH₃COOH:
\[ [\text{CH}_3\text{COOH}] = \frac{2 \, \text{moles}}{2 \, \text{L}} = 1 \, \text{M} \]
- New concentration of C₂H₅OH:
\[ [\text{C}_2\text{H}_5\text{OH}] = \frac{3 \, \text{moles}}{2 \, \text{L}} = 1.5 \, \text{M} \]

5. **Calculate New Forward Rate (RF2):**
The new forward rate after dilution can be expressed as:
\[ R_{F2} = k_F \cdot [\text{CH}_3\text{COOH}] \cdot [\text{C}_2\text{H}_5\text{OH}] \]
- New concentrations:
- \([\text{CH}_3\text{COOH}] = 1 \, \text{M}\)
- \([\text{C}_2\text{H}_5\text{OH}] = 1.5 \, \text{M}\)
- Therefore:
\[ R_{F2} = k_F \cdot 1 \cdot 1.5 = 1.5k_F \]

6. **Calculate the Change in Rate:**
To find how many times the initial forward rate has changed, we take the ratio of the new rate to the initial rate:
\[ \frac{R_{F2}}{R_{F1}} = \frac{1.5k_F}{6k_F} = \frac{1.5}{6} = \frac{1}{4} \]

7. **Conclusion:**
The forward rate has changed by a factor of \(0.25\) times the initial forward rate.

### Final Answer:
The forward rate has changed by **0.25 times** the initial forward rate.