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 how many times the half-life (T_half) is required for 93.75% of a first-order reaction to occur.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- We are dealing with a first-order reaction.
- We need to find the time required for 93.75% of the reaction to occur in terms of the half-life (T_half).

2. **Calculating the Remaining Concentration**:
- If 93.75% of the reaction has occurred, then the remaining concentration (A - x) is:
\[ A - x = A - 0.9375A = 0.0625A \]
- If we assume the initial concentration (A) is 100, then:
\[ A - x = 100 - 93.75 = 6.25 \]

3. **Using the First-Order Kinetics Equation**:
- The time (t) for a first-order reaction can be calculated using the formula:
\[ t = \frac{2.303}{k} \log\left(\frac{A}{A - x}\right) \]
- Substituting the values:
\[ t = \frac{2.303}{k} \log\left(\frac{100}{6.25}\right) \]

4. **Calculating the Logarithm**:
- We can simplify \(\frac{100}{6.25}\):
\[ \frac{100}{6.25} = 16 \]
- Thus, we need to calculate:
\[ t = \frac{2.303}{k} \log(16) \]

5. **Using Logarithmic Properties**:
- We know that \(16 = 2^4\), so:
\[ \log(16) = \log(2^4) = 4 \log(2) \]
- Therefore, we can rewrite t as:
\[ t = \frac{2.303}{k} \cdot 4 \log(2) \]

6. **Relating to Half-Life**:
- The half-life (T_half) for a first-order reaction is given by:
\[ T_{half} = \frac{0.693}{k} \]
- Substituting this into the equation for t:
\[ t = 4 \cdot \frac{0.693}{k} \cdot \log(2) \]
- Since \( \log(2) \approx 0.301\), we can express t in terms of T_half:
\[ t = 4 \cdot T_{half} \]

7. **Conclusion**:
- Therefore, the time required for 93.75% of the reaction to occur is 4 times the half-life:
\[ x = 4 \]

### Final Answer:
The value of \(x\) is \(4\).

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