The half life period of a first order chemical reaction is `6.93` minutes. The time required for the completion of `99%` of the reaction will be `(log2=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 determine the time required for 99% completion of a first-order reaction given its half-life period. Here’s a step-by-step solution:

### Step 1: Understand the relationship between half-life and rate constant (k)
For a first-order reaction, the half-life (t₁/₂) is related to the rate constant (k) by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]

### Step 2: Calculate the rate constant (k)
Given that the half-life of the reaction is 6.93 minutes, we can rearrange the formula to find k:
\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{6.93 \text{ minutes}} \]

Calculating this gives:
\[ k = 0.1 \text{ min}^{-1} \]

### Step 3: Use the first-order kinetics equation to find the time for 99% completion
The first-order kinetics equation is:
\[ t = \frac{2.303}{k} \log \left( \frac{[A]_0}{[A]} \right) \]
Where:
- \([A]_0\) is the initial concentration
- \([A]\) is the concentration at time t

For 99% completion, if we assume the initial concentration \([A]_0 = 100\), then:
\[ [A] = 100 - 99 = 1 \]

### Step 4: Substitute values into the equation
Now substituting the values into the equation:
\[ t = \frac{2.303}{0.1} \log \left( \frac{100}{1} \right) \]
\[ t = \frac{2.303}{0.1} \log(100) \]

Since \(\log(100) = 2\):
\[ t = \frac{2.303}{0.1} \times 2 \]
\[ t = 2.303 \times 20 \]
\[ t = 46.06 \text{ minutes} \]

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

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