JEE Main 2024ChemistryChemical KineticsMediumNumerical

JEE Main 2024 — Chemical Kinetics Question with Solution

JEE Main 2024 (30 Jan Shift 1)

Question

The rate of first order reaction is 0.04 mol L-1 s-1 at 10 minutes and 0.03 mol L-1 s-1 at 20 minutes after initiation. Half life of the reaction is ______ minutes. (Given log2=0.3010,log3=0.4771)

Round off your answer to the nearest integer.

Enter your answer

Show full solutionCorrect answer: 24
Correct answer
24

Step-by-step explanation

 Rate r =kA0.04=kA0100.03=kA20

⇒A010A20=0.040.03=43

From t=2.303klog A10A2010= 2.303klog 4310= 2.303k(0.6020-0.4771)k= 2.30310×0.1249=2.876×10-2 min-1

t1/2=0.693k=0.6932.876×10-2=0.24×102 min=24 min.

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About this question

This is a previous-year question from JEE Main 2024, covering the Chemical Kinetics chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.