JEE Main 2022ChemistryChemical EquilibriumHardNumerical

JEE Main 2022 — Chemical Equilibrium Question with Solution

JEE Main 2022 (25 Jun Shift 1)

Question

The standard free energy change ΔG° for 50% dissociation of N2O4 into NO2 at 27 °C and 1 atm pressure is -x J mol-1. The value of x is -..... J. (Nearest Integer)

[Given : R=8.31 J K-1 mol-1, log1.33=0.1239  ln10=2.3]

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Show full solutionCorrect answer: 710
Correct answer
710

Step-by-step explanation

N2O4g⇌2NO2g

1-α           2α

Kp=4α2p1-α2=4×0.52×11-0.52=10.75

Kp=43

ΔGo=-2.303 RT log Kp

=-2.3×8.31×300 log 43=-2.3×8.31×300 ×0.1239

=-710 J

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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.