Half life of a first order reaction is `69.3` minutes. Time required to complete `99.9%` of the reaction will be `:`

Options

A. 693 minutes
B. 999 minutes
C. `99.9` minutes
D. 691 minutes

Explanation

To find the time required to complete 99.9% of a first-order reaction, we can use the relationship between the rate constant \( k \), the half-life \( t_{1/2} \), and the fraction of the reaction completed.

### Step 1: Determine the rate constant \( k \)

The half-life \( t_{1/2} \) for a first-order reaction is given by:
\[ t_{1/2} = \frac{\ln 2}{k} \]

Given:
\[ t_{1/2} = 69.3 \text{ minutes} \]

We can solve for \( k \):
\[ k = \frac{\ln 2}{t_{1/2}} = \frac{\ln 2}{69.3} \]

### Step 2: Calculate \( k \)

Using the value of \( \ln 2 \approx 0.693 \):
\[ k = \frac{0.693}{69.3} \]
\[ k = 0.01 \text{ min}^{-1} \]

### Step 3: Use the first-order reaction formula

The formula for the concentration of a reactant at time \( t \) for a first-order reaction is:
\[ [A]_t = [A]_0 e^{-kt} \]

We need to find the time \( t \) when 99.9% of the reaction is complete. This means 0.1% of the reactant remains:
\[ [A]_t = 0.001 [A]_0 \]

### Step 4: Set up the equation

Substitute \( [A]_t = 0.001 [A]_0 \) into the first-order reaction formula:
\[ 0.001 [A]_0 = [A]_0 e^{-kt} \]

### Step 5: Simplify and solve for \( t \)

Divide both sides by \( [A]_0 \):
\[ 0.001 = e^{-kt} \]

Take the natural logarithm of both sides:
\[ \ln(0.001) = -kt \]

Since \( \ln(0.001) = \ln(10^{-3}) = -3 \ln(10) \) and \( \ln(10) \approx 2.303 \):
\[ \ln(0.001) = -3 \times 2.303 = -6.909 \]

So,
\[ -6.909 = -kt \]

Substitute \( k = 0.01 \text{ min}^{-1} \):
\[ -6.909 = -0.01 t \]

Solve for \( t \):
\[ t = \frac{6.909}{0.01} \]
\[ t = 690.9 \text{ minutes} \]

### Final Answer:
The time required to complete 99.9% of the reaction is approximately \( 691 \) minutes.