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Chemical Kinetics

JEE Main level, numerical answer, medium

2 of 4

Question 2 of 4

A first order reaction has rate constant $k = 6.93 \times 10^{-3}\,\text{s}^{-1}$ . Its half life period $t_{1/2}$ is (in s):

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Answer: 100

t_{1/2} = \frac{0.693}{k} = \frac{0.693}{6.93 \times 10^{-3}} = 100\,\text{s}

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JEE Main 2025 Apr 2 Shift 1 question paper

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JEE Main Chemistry practice question on Chemical Kinetics, with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.

A first order reaction has rate constant k=6.93×10−3 s−1k = 6.93 \times 10^{-3}\,\text{s}^{-1}k=6.93×10−3s−1. Its half life period t1/2t_{1/2}t1/2​ is (in s):

JEE Main question paper PDF

More Chemistry practice

A first order reaction has rate constant $k = 6.93 \times 10^{-3}\,\text{s}^{-1}$ . Its half life period $t_{1/2}$ is (in s):