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.