Q27 / 150 · MHT CET 2026 (11 April Shift 1) · Chemistry
The rate constant for a first order reaction is $60 \text{ s}^{-1}$. How much time will it take to reduce the concentration of the reactant to $1/20^{\text{th}}$ of its initial value?
- A$0.0529$ s
- B$0.0852$ s
- Correct answerC$0.0499$ s
- D$0.0357$ s
Solution
For a first order reaction, the time required is given by the integrated rate equation:
$t = \dfrac{2.303}{k} \log \left( \dfrac{[R]_0}{[R]} \right)$
Given $k = 60 \text{ s}^{-1}$ and the final concentration $[R] = \dfrac{[R]_0}{20}$.
Substituting the values into the equation:
$t = \dfrac{2.303}{60} \log \left( \dfrac{[R]_0}{\dfrac{[R]_0}{20}} \right)$
$t = \dfrac{2.303}{60} \log(20)$
Using $\log(20) = \log(2 \times 10) = \log 2 + \log 10 = 0.3010 + 1 = 1.3010$:
$t = \dfrac{2.303 \times 1.3010}{60}$
$t = \dfrac{2.9962}{60}$
$t \approx 0.0499 \text{ s}$
Answer: $0.0499$ s
$t = \dfrac{2.303}{k} \log \left( \dfrac{[R]_0}{[R]} \right)$
Given $k = 60 \text{ s}^{-1}$ and the final concentration $[R] = \dfrac{[R]_0}{20}$.
Substituting the values into the equation:
$t = \dfrac{2.303}{60} \log \left( \dfrac{[R]_0}{\dfrac{[R]_0}{20}} \right)$
$t = \dfrac{2.303}{60} \log(20)$
Using $\log(20) = \log(2 \times 10) = \log 2 + \log 10 = 0.3010 + 1 = 1.3010$:
$t = \dfrac{2.303 \times 1.3010}{60}$
$t = \dfrac{2.9962}{60}$
$t \approx 0.0499 \text{ s}$
Answer: $0.0499$ s
Hence, the correct answer is option: 'C'.