For the first order reaction, the taken to reduce the initial concentration to a factor of `1/4` is 10 minutes if the reduction in concentration is carried out to a factor of `1/16`, then time required will be

Options

A. 10 minutes
B. 20 minutes
C. 40 minutes
D. 60 minutes

Answer

To solve the problem, we need to understand the concept of half-lives in the context of first-order reactions.

### Step-by-Step Solution:

1. **Understanding the Problem:**
We are given that for a first-order reaction, the time taken to reduce the initial concentration to a factor of \( \frac{1}{4} \) is 10 minutes. We need to find the time required to reduce the concentration to a factor of \( \frac{1}{16} \).

2. **Relating Concentration Reductions to Half-Lives:**
- Reducing the concentration to \( \frac{1}{4} \) means that the concentration has gone through 2 half-lives. This is because:
\[ \frac{A_0}{2} \to \frac{A_0}{4} \quad \text{(2 half-lives)} \]
- Since it takes 10 minutes to reduce the concentration to \( \frac{1}{4} \), this means:
\[ \text{Time for 2 half-lives} = 10 \text{ minutes} \]

3. **Calculating the Time for 1 Half-Life:**
- From the above, we can find the time for one half-life (\( t_{1/2} \)):
\[ t_{1/2} = \frac{10 \text{ minutes}}{2} = 5 \text{ minutes} \]

4. **Finding Time for Reduction to \( \frac{1}{16} \):**
- Reducing the concentration to \( \frac{1}{16} \) means that the concentration has gone through 4 half-lives:
\[ \frac{A_0}{2} \to \frac{A_0}{4} \to \frac{A_0}{8} \to \frac{A_0}{16} \quad \text{(4 half-lives)} \]
- Therefore, the time required for 4 half-lives is:
\[ \text{Time for 4 half-lives} = 4 \times t_{1/2} = 4 \times 5 \text{ minutes} = 20 \text{ minutes} \]

5. **Final Calculation:**
- Thus, the time required to reduce the concentration to a factor of \( \frac{1}{16} \) is:
\[ \text{Time} = 4 \times 5 \text{ minutes} = 20 \text{ minutes} \]

### Conclusion:
The time required to reduce the concentration to a factor of \( \frac{1}{16} \) is **20 minutes**.