Q128 / 200 · MHT CET - Medical 2026 (21 April Shift 2) · Chemistry
The rate constant for a first order reaction is $6.93 \times 10^{-2}$ per minutes. What is the time required for the completion of $99.9\%$ of the reaction?
- A$693$ minutes
- B$20$ minutes
- C$50$ minutes
- Correct answerD$99.7$ minutes
Solution
For a first order reaction, the time required is given by:
$t = \dfrac{2.303}{k} \log \left( \dfrac{a}{a-x} \right)$
Given $k = 6.93 \times 10^{-2} \text{ min}^{-1}$ and for $99.9\%$ completion, $x = 0.999a$.
Remaining amount $a - x = a - 0.999a = 0.001a$
$t = \dfrac{2.303}{6.93 \times 10^{-2}} \log \left( \dfrac{a}{0.001a} \right)$
$t = \dfrac{2.303}{6.93 \times 10^{-2}} \log (10^3)$
$t = \dfrac{2.303 \times 3}{6.93 \times 10^{-2}}$
$t = \dfrac{6.909}{0.0693} \approx 99.7 \text{ minutes}$
Answer: $99.7$ minutes
$t = \dfrac{2.303}{k} \log \left( \dfrac{a}{a-x} \right)$
Given $k = 6.93 \times 10^{-2} \text{ min}^{-1}$ and for $99.9\%$ completion, $x = 0.999a$.
Remaining amount $a - x = a - 0.999a = 0.001a$
$t = \dfrac{2.303}{6.93 \times 10^{-2}} \log \left( \dfrac{a}{0.001a} \right)$
$t = \dfrac{2.303}{6.93 \times 10^{-2}} \log (10^3)$
$t = \dfrac{2.303 \times 3}{6.93 \times 10^{-2}}$
$t = \dfrac{6.909}{0.0693} \approx 99.7 \text{ minutes}$
Answer: $99.7$ minutes
Hence, the correct answer is option: 'D'.