In the case of a first order reaction, the time required for `93.75%` of reaction to take place is `x` time required for half of the reaction. Find the value of `x`.

Explanation

To solve the problem, we need to determine the relationship between the time required for 93.75% completion of a first-order reaction and the half-life of that reaction.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- We are dealing with a first-order reaction. The time required for a certain percentage of the reaction to occur can be calculated using the first-order kinetics formula.

2. **Half-Life of a First-Order Reaction**:
- The half-life (\( t_{1/2} \)) for a first-order reaction is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
where \( k \) is the rate constant.

3. **Calculating Time for 93.75% Completion**:
- For 93.75% of the reaction to occur, the remaining concentration is:
\[ 100\% - 93.75\% = 6.25\% \]
- We can use the first-order kinetics equation:
\[ t = \frac{2.303}{k} \log\left(\frac{[A]_0}{[A]}\right) \]
where \( [A]_0 \) is the initial concentration and \( [A] \) is the remaining concentration.

4. **Substituting Values**:
- Let’s assume the initial concentration \( [A]_0 = 100\% \) and the remaining concentration \( [A] = 6.25\% \):
\[ t = \frac{2.303}{k} \log\left(\frac{100}{6.25}\right) \]

5. **Calculating the Logarithm**:
- The fraction \( \frac{100}{6.25} = 16 \).
- Therefore, we need to calculate \( \log(16) \):
\[ \log(16) = \log(2^4) = 4 \log(2) \]
- Using \( \log(2) \approx 0.301 \):
\[ \log(16) = 4 \times 0.301 = 1.204 \]

6. **Substituting Back into the Time Equation**:
- Now substituting back into the time equation:
\[ t = \frac{2.303}{k} \times 1.204 \]
- This simplifies to:
\[ t = \frac{2.303 \times 1.204}{k} = \frac{2.775}{k} \]

7. **Relating to Half-Life**:
- The half-life \( t_{1/2} = \frac{0.693}{k} \).
- We can express \( t \) in terms of \( t_{1/2} \):
\[ t = \frac{2.775}{k} = 4 \times \frac{0.693}{k} = 4 \times t_{1/2} \]

8. **Conclusion**:
- Thus, the time required for 93.75% completion of the reaction is \( 4 \) times the half-life of the reaction. Therefore, the value of \( x \) is:
\[ x = 4 \]