`t_(1//4)` can be taken as the time taken for concentration of reactant to drop to `.^(3)//_(4)` of its initial value. If the rate constant for a first order reaction is `K`, then `t_(1//4)` can be written as:

Options

A. `0.10//k`
B. `0.29//k`
C. `0.69//k`
D. `0.75//k`

Explanation

To solve the problem, we need to find the time \( t_{1/4} \) for a first-order reaction, where the concentration of the reactant drops to \( \frac{3}{4} \) of its initial value.

### Step-by-Step Solution:

1. **Understanding the First-Order Reaction:**
For a first-order reaction, the rate constant \( K \) is related to the concentration of the reactant over time using the formula:
\[ K = \frac{2.303}{t} \log\left(\frac{[A_0]}{[A]}\right) \]
where \( [A_0] \) is the initial concentration and \( [A] \) is the concentration at time \( t \).

2. **Setting Up the Concentration Values:**
Let \( [A_0] = A \) (initial concentration). According to the problem, at time \( t_{1/4} \), the concentration drops to \( \frac{3}{4} A \). Thus, we have:
\[ [A] = \frac{3}{4} A \]

3. **Substituting into the Rate Equation:**
Substitute \( [A_0] \) and \( [A] \) into the rate equation:
\[ K = \frac{2.303}{t_{1/4}} \log\left(\frac{A}{\frac{3}{4} A}\right) \]
This simplifies to:
\[ K = \frac{2.303}{t_{1/4}} \log\left(\frac{A}{\frac{3}{4} A}\right) = \frac{2.303}{t_{1/4}} \log\left(\frac{4}{3}\right) \]

4. **Rearranging to Solve for \( t_{1/4} \):**
Rearranging the equation to solve for \( t_{1/4} \):
\[ t_{1/4} = \frac{2.303 \log\left(\frac{4}{3}\right)}{K} \]

5. **Calculating \( \log\left(\frac{4}{3}\right) \):**
Using logarithmic properties:
\[ \log\left(\frac{4}{3}\right) = \log(4) - \log(3) \]
The approximate values are:
\[ \log(4) \approx 0.602 \quad \text{and} \quad \log(3) \approx 0.477 \]
Thus:
\[ \log\left(\frac{4}{3}\right) \approx 0.602 - 0.477 = 0.125 \]

6. **Final Expression for \( t_{1/4} \):**
Substituting this back into the equation for \( t_{1/4} \):
\[ t_{1/4} = \frac{2.303 \times 0.125}{K} \approx \frac{0.287875}{K} \approx \frac{0.29}{K} \]

### Final Answer:
Thus, the time \( t_{1/4} \) can be written as:
\[ t_{1/4} \approx \frac{0.29}{K} \]