JEE Main 2013ChemistryChemical KineticsEasyMCQ

JEE Main 2013 — Chemical Kinetics Question with Solution

JEE Main 2013 (07 Apr)

Question

The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of such a reaction will be:
​
R=8.314 JK-1mol-1 and log 2=0.301

Choose an option

Show full solutionCorrect option: C
Correct answer
C53.6 kJ mol -1

Step-by-step explanation

logk2k1=−Ea2.303R(1T2−1T1)

logk2k1=Ea2.303RT2-T1T1T2

log 2=Ea2.303×8.314×10300×310

Ea=0.301×2.303×8.314×300×31010=53603.9 J/mol

Ea=53.6 kJ mol-1.

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About this question

This is a previous-year question from JEE Main 2013, 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.