Q6 / 180 · COMEDK 2021 · Chemistry
Determine the specific rate constant of a first-order reaction if its half-life period is 1402 s.
- A$4.94 \times 10^{-3} \mathrm{~s}^{-1}$
- Correct answerB$0.49 \times 10^{-3} \mathrm{~s}^{-1}$
- C$0.49 \times 10^{-4} \mathrm{~s}^{-1}$
- D$4.94 \times 10^{-5} \mathrm{~s}^{-1}$
Solution
For a first-order reaction: $k = \dfrac{0.693}{t_{1/2}}$
$k = \dfrac{0.693}{1402} \approx 4.943 \times 10^{-4} \ s^{-1}$
$k \approx 0.49 \times 10^{-3} \ s^{-1}$
$k = \dfrac{0.693}{1402} \approx 4.943 \times 10^{-4} \ s^{-1}$
$k \approx 0.49 \times 10^{-3} \ s^{-1}$
Hence, the correct answer is option: 'B'.