If the rate constant for a first order reaction is k, the time (t) required for the completion of 99% of the reaction is given by :

1. \(t = 0.693/k\)

2. \(t = 6.909/k\)

3. \(t = 4.606/k\)

4. \(t = 2.303/k\)

Doubtly Asked 30 Oct 2026 NEET UG 2019 Practice test neet Save WhatsApp Report

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Answer: 3. \(t = 4.606/k\)

For a first order reaction \(t = \dfrac{2.303}{k}\log\dfrac{[A]_0}{[A]}\).

When 99% is complete, \([A] = 0.01[A]_0\), so \(t = \dfrac{2.303}{k}\log 100 = \dfrac{2.303\times2}{k} = \dfrac{4.606}{k}\). Option (3).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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