JEE Main 2023ChemistryChemical EquilibriumHardNumerical

JEE Main 2023 — Chemical Equilibrium Question with Solution

JEE Main 2023 (29 Jan Shift 2)

Question

At 298 K

N2g+3H2g⇌2NH3g,K1=4×105

N2g+O2g⇌2NOg,K2=1.6×1012

H2g+12O2g⇌H2O(g),K3=1.0×10-13

Based on above equilibria, the equilibrium constant of the reaction,

2NH3g+52O2( g)⇌2NOg+3H2Og
is _____ ×10-33   (Nearest integer)

Enter your answer

Show full solutionCorrect answer: 4
Correct answer
4

Step-by-step explanation

Given,
N2g3H2g⇌2NH3g,K1=4×105          …1

N2g+O2g⇌2NOg,K2=1.6×1012        …2

H2g+12O2g⇌H2Og,K3=1.0×10-13      …3
Using properties of equilibrium constant:

2+3×3-1

2NH3g+52O2g⇌2NOg+3H2Og

Keq=K2×K33K1=1.6×1012×10-1334×105

=1.64×10-32=4×10-33

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About this question

This is a previous-year question from JEE Main 2023, 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.