The rate constant for forward and backward reactions of hydrolysis of ester are `1.1xx10^(-2)` and `1.5xx10^(-3)` per minute respectively. Equilibrium constant for the reaction is

Options

A. 7.33
B. 0.733
C. 73.3
D. 733

Answer

To find the equilibrium constant for the hydrolysis of an ester, we can use the relationship between the rate constants of the forward and backward reactions. The equilibrium constant (K) is given by the formula:

\[ K = \frac{k_f}{k_r} \]

where:
- \(k_f\) is the rate constant for the forward reaction,
- \(k_r\) is the rate constant for the backward reaction.

### Step-by-Step Solution:

1. **Identify the rate constants**:
- Given \(k_f = 1.1 \times 10^{-2} \, \text{min}^{-1}\)
- Given \(k_r = 1.5 \times 10^{-3} \, \text{min}^{-1}\)

2. **Substitute the values into the equilibrium constant formula**:
\[ K = \frac{1.1 \times 10^{-2}}{1.5 \times 10^{-3}} \]

3. **Perform the division**:
- First, divide the coefficients:
\[ \frac{1.1}{1.5} \approx 0.7333 \]
- Next, divide the powers of ten:
\[ 10^{-2} \div 10^{-3} = 10^{(-2) - (-3)} = 10^{1} = 10 \]
- Combine the results:
\[ K \approx 0.7333 \times 10 = 7.333 \]

4. **Round the result**:
\[ K \approx 7.33 \]

### Final Answer:
The equilibrium constant for the reaction is approximately \(7.33\).

---