JEE Main 2022 · ChemistryNumerical answerNumerical valueEasyCalculation

JEE Main 27 June 2022, Shift 1, Chemistry Q86

Question 86 of 90 in this shift, Chemistry question 26 of 30, Section B.

The rate constant for a first order reaction is given by the following equation :

\ln k = 33.24 - \frac{2.0\times10^4\ \mathrm{K}}{T}

The Activation energy for the reaction is given by __________ kJ mol

^{-1}

. (In Nearest integer) (Given : R = 8.3 J K

^{-1}

mol

^{-1}

)

Official answer

166

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.