CUET · CHEMISTRY · PYQ PAPER 2023
The rate constant for the first order reaction is \(60\) \(s^{-1}\). The time taken to reduce the initial concentration of the reactant to its \(1/26^{th}\) value is:
[Given \(\log 26 = 1.4150\)]
- A \(7.1 \times 10^{-2}\) s
- B \(5.43 \times 10^{-2}\) s
- C \(7.6 \times 10^2\) s
- D \(2.9 \times 10^{-2}\) s
Answer & Solution
Correct Answer
(B) \(5.43 \times 10^{-2}\) s
Step-by-step Solution
Detailed explanation
\(t = \frac{2.303}{k} \log \frac{[A]_0}{[A]}\) \(t = \frac{2.303}{60 \, s^{-1}} \log 26\) \(t = \frac{2.303}{60} \times 1.4150\) \(t = \frac{3.262745}{60}\) \(t = 0.054379 \, s\) \(t \approx 5.43 \times 10^{-2} \, s\)
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