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 (Kc) for the hydrolysis of ester, we can follow these steps:

### Step 1: Identify the given values
The rate constants for the forward and backward reactions are provided:
- Rate constant for the forward reaction (kf) = \(1.1 \times 10^{-2} \, \text{min}^{-1}\)
- Rate constant for the backward reaction (kb) = \(1.5 \times 10^{-3} \, \text{min}^{-1}\)

### Step 2: Write the formula for the equilibrium constant
The equilibrium constant (Kc) is defined as the ratio of the rate constant of the forward reaction to the rate constant of the backward reaction:
\[ K_c = \frac{k_f}{k_b} \]

### Step 3: Substitute the values into the formula
Now we can substitute the given values into the formula:
\[ K_c = \frac{1.1 \times 10^{-2}}{1.5 \times 10^{-3}} \]

### Step 4: Perform the calculation
To calculate Kc, we can simplify the fraction:
\[ K_c = \frac{1.1}{1.5} \times \frac{10^{-2}}{10^{-3}} = \frac{1.1}{1.5} \times 10^{1} \]
Calculating \(\frac{1.1}{1.5}\):
\[ \frac{1.1}{1.5} \approx 0.7333 \]
Thus,
\[ K_c = 0.7333 \times 10^{1} = 7.333 \]

### Step 5: Round the result
Rounding the result to two decimal places, we get:
\[ K_c \approx 7.33 \]

### Final Answer
The equilibrium constant (Kc) for the hydrolysis of ester is approximately **7.33**.

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