The half-life period of a first order chemical reaction is 6.93 minutes. The time required for the completion of 99% of the chemical reaction will be

Options

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

Answer

To solve the problem of finding the time required for the completion of 99% of a first-order chemical reaction given a half-life of 6.93 minutes, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Half-Life**:
The half-life (t₁/₂) of a first-order reaction is the time required for the concentration of a reactant to decrease to half of its initial concentration.

2. **Use the Half-Life Formula for First-Order Reactions**:
For a first-order reaction, the relationship between the half-life and the rate constant (k) is given by:
\[ k = \frac{0.693}{t_{1/2}} \]
Here, \(t_{1/2} = 6.93 \text{ minutes}\).

3. **Calculate the Rate Constant (k)**:
Substitute the half-life into the formula:
\[ k = \frac{0.693}{6.93} \approx 0.1 \text{ min}^{-1} \]

4. **Determine the Remaining Concentration After 99% Completion**:
If 99% of the reaction is completed, then 1% of the reactant remains. If we assume the initial concentration is 100 (for simplicity), the remaining concentration after 99% completion is:
\[ [A] = 100 - 99 = 1 \]

5. **Use the First-Order Kinetics Equation**:
The first-order kinetics equation is:
\[ \ln\left(\frac{[A]_0}{[A]}\right) = kt \]
Where:
- \([A]_0\) is the initial concentration (100),
- \([A]\) is the concentration after time \(t\) (1),
- \(k\) is the rate constant,
- \(t\) is the time we want to find.

6. **Substitute the Values into the Equation**:
\[ \ln\left(\frac{100}{1}\right) = kt \]
This simplifies to:
\[ \ln(100) = kt \]

7. **Calculate \(\ln(100)\)**:
\[ \ln(100) \approx 4.605 \]

8. **Substitute k into the Equation**:
Now substituting \(k\) into the equation:
\[ 4.605 = (0.1)t \]

9. **Solve for t**:
\[ t = \frac{4.605}{0.1} = 46.05 \text{ minutes} \]

### Final Answer:
The time required for the completion of 99% of the chemical reaction is approximately **46.05 minutes**.