JEE Main 2021ChemistryChemical EquilibriumHardNumerical

JEE Main 2021 — Chemical Equilibrium Question with Solution

JEE Main 2021 (16 Mar Shift 1)

Question

For the reaction Ag⇌Bg at 495 K, ΔrG°=-9.478 kJ mol-1

If we start the reaction in a closed container at 495 K with 22 millimoles of A, the amount of B is the equilibrium mixture is ________ millimoles. (Round off to the Nearest Integer).

R=8.314 J mol-1 K-1;ℓn10=2.303

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Show full solutionCorrect answer: 20
Correct answer
20

Step-by-step explanation

ΔG°=-RTℓnKeq

Given ΔG°=-9.478 KJ/mole

T=495 K  R=8.314 J mol-1

So -9.478×103=-495×8.314×ℓnKeq

ℓnKeq=2.303

=ℓn10

So Keq=10

Now Ag⇌Bgt=0220t=022-xx

Keq=BC=x22-x=10

or x=20

So millimoles of B=20

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About this question

This is a previous-year question from JEE Main 2021, covering the Chemical Equilibrium 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.