JEE Main 2024ChemistryChemical EquilibriumMediumNumerical

JEE Main 2024Chemical Equilibrium Question with Solution

JEE Main 2024 (29 Jan Shift 2)

Question

The following concentrations were observed at 500 K for the formation of NH3 from N2 and H2. At equilibrium : N2=2×10-2M,H2=3×10-2M and NH3=1.5×10-2M. Equilibrium constant for the reaction is ______.

Enter your answer

Show full solutionCorrect answer: 417
Correct answer
417

Step-by-step explanation

The equilibrium constant of concentration (denoted by Kc) of a chemical reaction at equilibrium can be defined as the ratio of the concentration of products to the concentration of the reactants, each raised to their respective stoichiometric coefficients.

The equilibrium reaction is 

N2g+3H2g2NH3g

Kc=NH32H23N2

  =32×10-223×10-23×2×10-2

   =32×10-422×33×10-6×2×10-2

 =14×2×3×104=1000024=416.6417

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 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.