The rate equation for the reaction
`CH_3COOC_2H_5+H_2O to^([H^+]) CH_3COOH+C_2H_5OH` is: `(dx)/(dt)=k[CH_3 COOC_2H_5][H_2O]^0`
If the concentration of ester is doubled the rate becomes

Options

A. Half
B. Twice
C. No change
D. Thrice

Answer

To solve the problem, we need to analyze the given rate equation and determine how the rate of the reaction changes when the concentration of the ester is doubled.

### Step-by-Step Solution:

1. **Identify the Rate Equation**:
The rate equation for the reaction is given as:
\[ \frac{dx}{dt} = k [CH_3COOC_2H_5][H_2O]^0 \]
Here, \( [H_2O]^0 \) indicates that the reaction rate does not depend on the concentration of water.

2. **Determine the Order of the Reaction**:
From the rate equation, we can see that:
- The order with respect to the ester \( [CH_3COOC_2H_5] \) is 1 (since it is raised to the power of 1).
- The order with respect to water \( [H_2O] \) is 0 (since it is raised to the power of 0).
- Therefore, the overall order of the reaction is \( 1 + 0 = 1 \).

3. **Define the Initial Rate**:
Let the initial concentration of the ester \( [CH_3COOC_2H_5] \) be \( C \). The initial rate \( R \) can be expressed as:
\[ R = k [C] \]

4. **Change in Concentration**:
If the concentration of the ester is doubled, the new concentration becomes \( 2C \).

5. **Calculate the New Rate**:
The new rate \( R' \) when the concentration of the ester is doubled is:
\[ R' = k [2C] = 2k [C] \]
Thus, we can express the new rate as:
\[ R' = 2R \]

6. **Conclusion**:
Therefore, when the concentration of the ester is doubled, the rate of the reaction also doubles.

### Final Answer:
The rate becomes twice the initial rate.

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