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 a first-order reaction where the reactant is reduced 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 the Initial and Final Concentrations**:
- Let \( A_0 \) be the initial concentration of the reactant.
- Since the reactant is reduced to \( \frac{1}{4} \) of its initial value, we have:
\[ A_t = \frac{A_0}{4} \]

2. **Substitute the Values into the Rate Constant Formula**:
- The time \( t \) is given as 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 Logarithmic Expression**:
- The expression inside the logarithm simplifies to:
\[ \frac{A_0}{\frac{A_0}{4}} = 4 \]
- Therefore, we have:
\[ k = \frac{1}{10} \ln(4) \]

4. **Calculate \( \ln(4) \)**:
- The natural logarithm of 4 can be calculated as:
\[ \ln(4) \approx 1.386 \]

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

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