JEE Main 2017ChemistryChemical KineticsMediumMCQ

JEE Main 2017 — Chemical Kinetics Question with Solution

JEE Main 2017 (08 Apr Online)

Question

The rate of a reaction A doubles on increasing the temperature from 300 to 310 K. By how much, the temperature of reaction B should be increased from 300 K so that rate doubles if activation energy of the reaction B is twice to that of reaction A.

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Show full solutionCorrect option: A
Correct answer
A4.92 K

Step-by-step explanation

log 2=EaR1300-1310 .......(i)

log  2=2EaR 1300-1T ......(ii)

2EaR 1300-1T=EaR 1300-1310

1300+1310=2T

⇒     T=300×310610×2

=304.92

Hence, the temperature of reaction B should be increased from 300 K by 304.92−300=4.92 K.

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About this question

This is a previous-year question from JEE Main 2017, 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.