JEE Main 2022ChemistryChemical KineticsMediumNumerical

JEE Main 2022 — Chemical Kinetics Question with Solution

JEE Main 2022 (29 Jun Shift 1)

Question

The activation energy of one of the reactions in a biochemical process is 532611 J mol-1. When the temperature falls from 310 K to 300 K, the change in rate constant observed is k300=x×10-3k310°. The value of x is
[Given: ln10=2.3 R=8.3 J K-1 mol-1]

Enter your answer

Show full solutionCorrect answer: 1
Correct answer
1

Step-by-step explanation

lnK2 K1=EaR1 T1-1 T2

lnK2 K1=5326118.3×10310×300

where K2 is at 310 K & K1 is at 300 K

lnK2 K1=6.9

=3×ln10

lnK2 K1=ln103

K2=K1×103

K1=K2×103

So 

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Chemical Kinetics chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2022, 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.