The rate constant of a first order reaction is `60s^(-1)`. How much time it will take to reduce `75%` of its original concentration ?

Explanation

To solve the problem, we need to determine how much time it takes for a first-order reaction to reduce its original concentration by 75%.

### Step-by-Step Solution:

1. **Understand the Reaction Order**:
The reaction is first-order, which means the rate of reaction is directly proportional to the concentration of the reactant.

2. **Identify the Given Data**:
- Rate constant (k) = 60 s⁻¹
- Reduction in concentration = 75%
- Therefore, the remaining concentration after 75% reduction = 100% - 75% = 25%

3. **Use the First-Order Kinetics Formula**:
The integrated rate law for a first-order reaction is given by:
\[ t = \frac{2.303}{k} \log \left( \frac{[A_0]}{[A_t]} \right) \]
Where:
- \( [A_0] \) = initial concentration
- \( [A_t] \) = concentration at time t
- \( k \) = rate constant
- \( t \) = time

4. **Substitute the Values**:
- Let’s assume the initial concentration \( [A_0] = 100 \) (in arbitrary units).
- The concentration at time t \( [A_t] = 25 \) (after 75% reduction).
- Substitute these values into the formula:
\[ t = \frac{2.303}{60} \log \left( \frac{100}{25} \right) \]

5. **Calculate the Logarithm**:
\[ \frac{100}{25} = 4 \]
Therefore:
\[ \log(4) \approx 0.602 \]

6. **Calculate Time (t)**:
Now substituting the values back into the equation:
\[ t = \frac{2.303}{60} \times 0.602 \]
\[ t \approx \frac{1.384}{60} \]
\[ t \approx 0.02307 \text{ seconds} \]

### Final Answer:
The time it will take to reduce 75% of the original concentration is approximately **0.0231 seconds**.