For given first order reaction, the reactant reduced to 1/4th its initial value in 10 min. The rate constant of the reaction is

Options

A. `"0.1386 min"^(-1)`
B. `"0.0693 min"^(-1)`
C. `"0.1386 mol L"^(-1)"min"^(-1)`
D. `"0.0693 mol L"^(-1)"min"^(-1)`

Answer

To find the rate constant \( k \) for the given first-order reaction where the reactant reduces to \( \frac{1}{4} \) of its initial value in 10 minutes, we can use the first-order rate equation:

\[ k = \frac{1}{t} \ln \left( \frac{A_0}{A_t} \right) \]

### Step-by-Step Solution:

1. **Identify Initial and Final Concentrations**:
- Let \( A_0 \) be the initial concentration of the reactant.
- After 10 minutes, the concentration reduces to \( A_t = \frac{A_0}{4} \).

2. **Substitute Values into the Rate Equation**:
- We know that \( t = 10 \) minutes.
- Substitute \( A_0 \) and \( A_t \) into the equation:
\[ k = \frac{1}{10} \ln \left( \frac{A_0}{\frac{A_0}{4}} \right) \]

3. **Simplify the Equation**:
- The fraction simplifies to:
\[ \frac{A_0}{\frac{A_0}{4}} = 4 \]
- Therefore, we have:
\[ k = \frac{1}{10} \ln(4) \]

4. **Calculate \( \ln(4) \)**:
- We know that \( \ln(4) = \ln(2^2) = 2 \ln(2) \).
- The value of \( \ln(2) \) is approximately \( 0.693 \).
- Thus, \( \ln(4) \approx 2 \times 0.693 = 1.386 \).

5. **Final Calculation of \( k \)**:
- Substitute \( \ln(4) \) back into the equation for \( k \):
\[ k = \frac{1}{10} \times 1.386 = 0.1386 \text{ per minute} \]

### Conclusion:
The rate constant \( k \) for the reaction is approximately \( 0.1386 \text{ min}^{-1} \).