JEE Main 2024ChemistryChemical EquilibriumMediumNumerical

JEE Main 2024 — Chemical Equilibrium Question with Solution

JEE Main 2024 (29 Jan Shift 1)

Question

For the reaction N2O4( g)⇌2NO2( g), Kp=0.492 atm at 300 K.Kc for the reaction at same temperature is ______×10-2. (Given :R=0.082 L atm mol-1 K-1)

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Show full solutionCorrect answer: 2
Correct answer
2

Step-by-step explanation

The given reaction is,

 N2O4( g)⇌2NO2( g)

Kp=0.492 atm

T=300 K

KP=KC·(RT)Δng

Δng= number of gaseous molecules of products – number of gaseous molecules of reactants.

Δng=1

By putting the values in the above equation,

⇒Kc=KpRT=0.4920.082×300=2×10-2

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About this question

This is a previous-year question from JEE Main 2024, 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.