Question

The half-life period of a first order reaction at 298K is 20 minutes. The time (in min.) required for 99.9% completion of the reaction at the same temperature, is

• A. 100
• B. 200
• C. 150
• D. 250
• E. 300

Answer 1

The Correct Option is B

For a first-order reaction, the relationship between the fraction of the reaction completed, the time, and the half-life is given by the formula: \[ \ln \left( \frac{[A]_0}{[A]} \right) = kt \] Where:
\( [A]_0 \) is the initial concentration,
\( [A] \) is the concentration after time \( t \),
\( k \) is the rate constant,
\( t \) is the time.
We are given that the half-life \( t_{1/2} \) is 20 minutes, and for a first-order reaction, the half-life is related to the rate constant \( k \) by the equation: \[ t_{1/2} = \frac{0.693}{k} \] Thus, the rate constant \( k \) can be calculated as: \[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{20} = 0.03465 \, \text{min}^{-1} \] To find the time required for 99.9% completion, we need the reaction to go from \( [A]_0 \) to 0.1% of \( [A]_0 \). This means \( [A] = 0.001[A]_0 \). Substituting into the first-order rate equation: \[ \ln \left( \frac{[A]_0}{0.001[A]_0} \right) = k \times t \] Simplifying: \[ \ln (1000) = k \times t \] \[ 6.907 = 0.03465 \times t \] \[ t = \frac{6.907}{0.03465} \approx 200 \, \text{minutes} \]

The correct option is (B) : \(200\)

Answer 2

For a first-order reaction, the relationship between the time required for a given percentage completion and the half-life is given by the formula:
\(t = \frac{2.303}{k} \log \left( \frac{[A]_{\text{initial}}}{[A]_{\text{final}}} \right)\)
Where: - \( t \) is the time required for the reaction to reach the desired completion, - \( k \) is the rate constant, - \([A]_{\text{initial}}\) is the initial concentration, - \([A]_{\text{final}}\) is the final concentration.

The half-life period of a first-order reaction is related to the rate constant \( k \) by: 
\( t_{\frac{1}{2}} = \frac{0.693}{k} \)
Given that the half-life \( t_{\frac{1}{2}} = 20 \) minutes, we can solve for \( k \):
\(k = \frac{0.693}{20} = 0.03465 \text{ min}^{-1}\)
Now, for 99.9% completion, the final concentration is 0.1% of the initial concentration. So, we can substitute this into the equation:
\(t = \frac{2.303}{0.03465} \log \left( \frac{1}{0.001} \right) = \frac{2.303}{0.03465} \times 3 = 200 \text{ minutes}\)

Therefore, the time required for 99.9% completion is 200 minutes.