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JEE Main 27 June 2022, Shift 2, Chemistry Q88

Question 88 of 90 in this shift, Chemistry question 28 of 30, Section B.

It has been found that for a chemical reaction with rise in temperature by 9 K the rate constant gets doubled. Assuming a reaction to be occurring at 300 K, the value of activation energy is found to be __________

\mathrm{kJ\ mol^{-1}}

. [nearest integer] (Given ln10 = 2.3, R = 8.3 J

\mathrm{K^{-1}\ mol^{-1}}

, log 2 = 0.30)

Official answer

NTA final key.

Chapter: Chemical Kinetics
Topic: Arrhenius Equation
Idea tested: Activation Energy Calculation

Same idea in other shifts

Asked 7× in all

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Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.