JEE Main 2020ChemistryChemical KineticsMediumMCQ

JEE Main 2020 — Chemical Kinetics Question with Solution

JEE Main 2020 (08 Jan Shift 1)

Question

The rate of a certain biochemical reaction at physiological temperature T occurs 106 times faster with enzyme than without. The change in the activation energy upon adding enzyme is:

Choose an option

Show full solutionCorrect option: A
Correct answer
A-62.303RT

Step-by-step explanation

K=Ae-E/RT......(1)

106 K=Ae−EcRT…….(2)
 
On dividing eq, 2 by eq, 1

106=eE−Ec/RT
 
on taking logarithm both sides

6ln10=E−EcRT
 
2.303×6=E−EcRT
  E−Ec=2.303×(6RT)

ΔEa=Ec−E=−2.303×6RT

 

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About this question

This is a previous-year question from JEE Main 2020, 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.