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. **Write the Given Rate Equation**:
The rate equation for the reaction is given as:
\[ \frac{dx}{dt} = k [CH_3COOC_2H_5][H_2O]^0 \]
Here, \( k \) is the rate constant, \([CH_3COOC_2H_5]\) is the concentration of the ester, and \([H_2O]^0\) indicates that the concentration of water does not affect the rate (since any number raised to the power of 0 is 1).

2. **Understand the Impact of Doubling the Ester Concentration**:
If the concentration of the ester \([CH_3COOC_2H_5]\) is doubled, we can denote the new concentration as:
\[ [CH_3COOC_2H_5]_{new} = 2[CH_3COOC_2H_5] \]

3. **Substitute the New Concentration into the Rate Equation**:
The new rate of the reaction, denoted as \( R_2 \), can be expressed as:
\[ R_2 = k [CH_3COOC_2H_5]_{new} [H_2O]^0 \]
Substituting the new concentration:
\[ R_2 = k (2[CH_3COOC_2H_5]) [H_2O]^0 \]
Since \([H_2O]^0\) is still equal to 1, we can simplify this to:
\[ R_2 = 2k [CH_3COOC_2H_5] [H_2O]^0 \]

4. **Relate the New Rate to the Original Rate**:
The original rate, denoted as \( R_1 \), is:
\[ R_1 = k [CH_3COOC_2H_5] [H_2O]^0 \]
Therefore, we can express \( R_2 \) in terms of \( R_1 \):
\[ R_2 = 2R_1 \]

5. **Conclusion**:
When the concentration of the ester is doubled, the rate of the reaction becomes:
\[ R_2 = 2R_1 \]
This means the rate of the reaction increases to twice its original value.

### Final Answer:
The rate becomes **twice**.

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