CUET · CHEMISTRY · PYQ PAPER 2023
\(t_{99\%}\) with respect to \(t_{90\%}\) for a first order reaction is :
- A four times of \(t_{50\%}\)
- B one and half time
- C It is same
- D Double
Answer & Solution
Correct Answer
(D) Double
Step-by-step Solution
Detailed explanation
\(t_{90\%} = \frac{2.303}{k} \log \frac{[A]_0}{0.10[A]_0} = \frac{2.303}{k} \times 1\) \(t_{99\%} = \frac{2.303}{k} \log \frac{[A]_0}{0.01[A]_0} = \frac{2.303}{k} \times 2\) \(\frac{t_{99\%}}{t_{90\%}} = \frac{2}{1} = 2\)
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