JEE Main 2021ChemistryChemical EquilibriumChemical Equilibrium Law Of Mass Action And Equilibrium ConstantmediumNumerical

JEE Main 2021Chemical Equilibrium Question with Solution

From: JEE Main 2021 (Online) 20th July Morning Shift

Question

2SO2(g) + O2(g) 2SO3(g)

In an equilibrium mixture, the partial pressures are

PSO3 = 43 kPa; PO2 = 530 Pa and PSO2 = 45 kPa. The equilibrium constant KP = ___________ 102. (Nearest integer)

Enter your answer

Show full solutionCorrect answer: 172
Correct answer
172

Step-by-step explanation

On reaction, 2SO2(g) + O2(g) 2SO3(g)

Given values are : pSO3 = 45kPa, pSO2 = 530 Pa = 0.53 kPa

pSO2 = 43 kPa

Now,

On putting given values, we get









Hence, the equilibrium constant, Kp = 172.

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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.