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 minutes
B. 23.03 minutes
C. 46.06 minutes
D. 460.6 minutes

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 is 6.93 minutes.

### Step-by-Step Solution:

**Step 1: Calculate the rate constant (k) using the half-life formula.**
For a first-order reaction, the half-life (t_half) is related to the rate constant (k) by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
Given that \( t_{1/2} = 6.93 \) minutes, we can rearrange the formula to find k:
\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{6.93} \approx 0.1 \, \text{min}^{-1} \]

**Step 2: Set up the equation to find the time (t) for 99% completion.**
For a first-order reaction, the relationship between time and concentration can be expressed as:
\[ t = \frac{2.303}{k} \log\left(\frac{[A_0]}{[A_0] - x}\right) \]
Where:
- \([A_0]\) is the initial concentration,
- \(x\) is the amount reacted.

Since 99% of the reaction is completed, if we assume the initial concentration \([A_0] = 100\), then:
\[ x = 99 \quad \text{(amount reacted)} \]
Thus, the remaining concentration \([A_0] - x = 100 - 99 = 1\).

**Step 3: Substitute the values into the equation.**
Now substituting into the equation:
\[ t = \frac{2.303}{0.1} \log\left(\frac{100}{1}\right) \]
Calculating the logarithm:
\[ \log(100) = 2 \quad \text{(since } 100 = 10^2\text{)} \]
Now substituting this back into the equation:
\[ t = \frac{2.303}{0.1} \times 2 = 23.03 \times 2 = 46.06 \, \text{minutes} \]

**Step 4: Conclusion.**
The time required for the completion of 99% of the chemical reaction is approximately **46.06 minutes**.