The half-life period of a first-order chemical reaction is `6.93 min`. The time required for the completion of `99%` of the chemical reaction will be `(log 2 = 0.301)`

Options

A. 230.3 min
B. 23.03 min
C. 46.06 min
D. 460.6 min

Explanation

To solve the problem, we need to find the time required for the completion of 99% of a first-order chemical reaction, given that the half-life period (t_half) is 6.93 minutes.

### Step-by-Step Solution:

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

2. **Calculate the Rate Constant (k):**
Rearranging the formula to find \( k \):
\[ k = \frac{0.693}{t_{1/2}} \]
Substituting the given half-life:
\[ k = \frac{0.693}{6.93 \text{ min}} = 0.1 \text{ min}^{-1} \]

3. **Using the First-Order Reaction Time Formula:**
For a first-order reaction, the time (t) required for the reaction to go from an initial concentration \( A \) to a final concentration \( A - x \) is given by:
\[ t = \frac{1}{k} \ln \left( \frac{A}{A - x} \right) \]
Here, \( A \) is the initial concentration and \( x \) is the amount reacted.

4. **Setting Up the Concentrations:**
If 99% of the reaction is complete, then:
- Initial concentration \( A = 100 \) (arbitrary units)
- Final concentration \( A - x = 100 - 99 = 1 \)

5. **Substituting Values into the Time Formula:**
\[ t = \frac{1}{0.1} \ln \left( \frac{100}{1} \right) = 10 \ln(100) \]

6. **Calculating \( \ln(100) \):**
We can express \( 100 \) as \( 10^2 \):
\[ \ln(100) = \ln(10^2) = 2 \ln(10) \]
Given \( \ln(10) = 2.303 \):
\[ \ln(100) = 2 \times 2.303 = 4.606 \]

7. **Final Calculation:**
Now substituting back into the time formula:
\[ t = 10 \times 4.606 = 46.06 \text{ minutes} \]

### Conclusion:
The time required for the completion of 99% of the chemical reaction is **46.06 minutes**.