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

Explanation

To solve the problem, we need to determine the expression for \( t_{1/4} \), which is the time taken for the concentration of a reactant to drop to \( \frac{3}{4} \) of its initial value in a first-order reaction.

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- Let the initial concentration of the reactant \( A \) be \( [A]_0 \).
- After time \( t_{1/4} \), the concentration of \( A \) drops to \( \frac{3}{4} [A]_0 \).
- This means that the amount of reactant that has reacted is \( [A]_0 - \frac{3}{4} [A]_0 = \frac{1}{4} [A]_0 \).

2. **Use the First-Order Reaction Rate Equation**:
- For a first-order reaction, the rate constant \( k \) is given by:
\[ k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]} \right) \]
- Here, \( [A] \) at time \( t_{1/4} \) is \( \frac{3}{4} [A]_0 \).

3. **Substituting the Values**:
- Substitute \( [A] = \frac{3}{4} [A]_0 \) into the equation:
\[ k = \frac{2.303}{t_{1/4}} \log \left( \frac{[A]_0}{\frac{3}{4} [A]_0} \right) \]
- The \( [A]_0 \) cancels out:
\[ k = \frac{2.303}{t_{1/4}} \log \left( \frac{1}{\frac{3}{4}} \right) \]

4. **Simplifying the Logarithm**:
- The logarithm can be simplified:
\[ \log \left( \frac{1}{\frac{3}{4}} \right) = \log \left( \frac{4}{3} \right) \]

5. **Rearranging the Equation**:
- Now, rearranging the equation gives:
\[ t_{1/4} = \frac{2.303 \log \left( \frac{4}{3} \right)}{k} \]

6. **Calculating the Logarithm**:
- Calculate \( \log \left( \frac{4}{3} \right) \):
\[ \log \left( \frac{4}{3} \right) \approx 0.1249 \]
- Therefore:
\[ t_{1/4} \approx \frac{2.303 \times 0.1249}{k} \approx \frac{0.288}{k} \approx \frac{0.29}{k} \]

### Final Result:
Thus, the expression for \( t_{1/4} \) is:
\[ t_{1/4} = \frac{0.29}{k} \]