JEE Main 2022ChemistryChemical EquilibriumHardMCQ

JEE Main 2022 — Chemical Equilibrium Question with Solution

JEE Main 2022 (29 Jun Shift 2)

Question

4.0 moles of argon and 5.0 moles of PCl5 are introduced into an evacuated flask of 100 litre capacity at 610 K. The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be 6.0 atm. The Kp for the reaction is [Given : R=0.082 L atm K-1 mol-1 ]

Choose an option

Show full solutionCorrect option: A
Correct answer
A2.25

Step-by-step explanation

PCl5=5 mole

Ar=4 mole

PTotal=9×0.082×610100=4.5 atm

PPCl5=5×4.59=2.5; PAr=4×4.59=2

PCl5⇌PCl3+Cl22.5-P P     P

Ptotal=2.5-P+P+P+PAr=6

P=1.5

Kp=1.5×1.51=2.25

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Chemical Equilibrium 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 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.