For which of the following represents the expresison for theee/fourth life reaction?

Options

A. `k/2.303 log4//3`
B. `2.303/k log 3//4`
C. `2.303/k log 4`
D. `2.303/k log 3`

Explanation

To solve the question regarding the expression for the three-fourth life of a reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Half-Life and Three-Fourth Life:**
- The half-life of a reaction is the time required for the concentration of a reactant to decrease to half of its initial concentration.
- The three-fourth life (or 75% life) is the time required for the concentration of a reactant to decrease to one-fourth of its initial concentration.

2. **Identify the Reaction Order:**
- The question implies that we are dealing with a first-order reaction. For a first-order reaction, the rate constant \( k \) is related to the concentration of the reactants.

3. **Use the First-Order Reaction Formula:**
- The formula for the time \( t \) for a first-order reaction is given by:
\[ t = \frac{2.303}{k} \log \left( \frac{[A_0]}{[A]} \right) \]
where \( [A_0] \) is the initial concentration and \( [A] \) is the concentration at time \( t \).

4. **Determine the Concentration After Three-Fourths of the Reaction:**
- If 75% of the reactant has reacted, then 25% remains. Therefore, if the initial concentration is \( [A_0] \), the final concentration \( [A] \) after 75% of the reaction will be:
\[ [A] = [A_0] - 0.75[A_0] = 0.25[A_0] = \frac{[A_0]}{4} \]

5. **Substitute the Values into the Formula:**
- Now substituting \( [A_0] \) and \( [A] \) into the formula:
\[ t_{3/4} = \frac{2.303}{k} \log \left( \frac{[A_0]}{\frac{[A_0]}{4}} \right) \]
- This simplifies to:
\[ t_{3/4} = \frac{2.303}{k} \log \left( 4 \right) \]

6. **Final Expression:**
- Therefore, the expression for the three-fourth life of a first-order reaction is:
\[ t_{3/4} = \frac{2.303}{k} \log(4) \]
- This corresponds to option number C in the question.

### Final Answer:
The expression for the three-fourth life of the reaction is:
\[ t_{3/4} = \frac{2.303}{k} \log(4) \]
This is option C.