Chemical Kinetics — Medium JEE chemistry MCQ
The half-life of a first-order reaction is $20$ minutes. The time required for $75\%$ completion of the reaction is:
- A. $30$ min
- B. $40$ min
- C. $60$ min
- D. $80$ min
Solution
For a first-order reaction:
$$t_{1/2} = \frac{0.693}{k}$$
$$k = \frac{0.693}{20} = 0.0347 \text{ min}^{-1}$$
For 75% completion, 25% remains.
$$t = \frac{2.303}{k}\log\frac{[A]_0}{[A]}$$
$$t = \frac{2.303}{0.0347}\log\frac{100}{25} = \frac{2.303}{0.0347}\log 4$$
$$t = \frac{2.303}{0.0347} \times 0.602 = \frac{1.386}{0.0347} = 40 \text{ min}$$
Alternatively: For 75% completion, we need 2 half-lives.
$t = 2 \times t_{1/2} = 2 \times 20 = 40$ min