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. `4.33`
B. `5.33`
C. `6.33`
D. `7.33`

Explanation

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

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Rate constant for the forward reaction (k_f) = \(1.1 \times 10^{-2}\) per minute
- Rate constant for the backward reaction (k_b) = \(1.5 \times 10^{-3}\) per minute

2. **Write the Formula for Equilibrium Constant**:
The equilibrium constant (K) is defined as the ratio of the rate constant of the forward reaction to the rate constant of the backward reaction:
\[ K = \frac{k_f}{k_b} \]

3. **Substitute the Given Values into the Formula**:
\[ K = \frac{1.1 \times 10^{-2}}{1.5 \times 10^{-3}} \]

4. **Perform the Division**:
To simplify the division, we can express it as:
\[ K = \frac{1.1}{1.5} \times \frac{10^{-2}}{10^{-3}} = \frac{1.1}{1.5} \times 10^{1} \]
Now calculate \( \frac{1.1}{1.5} \):
\[ \frac{1.1}{1.5} \approx 0.7333 \]
Therefore:
\[ K \approx 0.7333 \times 10^{1} = 7.333 \]

5. **Final Result**:
The equilibrium constant \( K \) is approximately \( 7.33 \).