JEE Main 2019ChemistryChemical EquilibriumHardMCQ

JEE Main 2019 — Chemical Equilibrium Question with Solution

JEE Main 2019 (10 Jan Shift 1)

Question

What are the values of KpKc for the following reactions at 300 K respectively?

(At 300 K, RT=24.62 dm2atm mol-1 )

N2 g+O2 g⇌2NO g

N2O4 g⇌2NO g

N2 g+3H2 g⇌2NH3 g

Choose an option

Show full solutionCorrect option: C
Correct answer
C1,24.62 dm3atm mol-11.65×10-3dm-6atm-2mol2

Step-by-step explanation

∵ KP=KCRTΔng

KPKC=RTΔng

(i)  N2+O2⇌2NO Δng=0∴KPKC=1

(ii)  N2O4⇌2Na2 Δng=1 ∴KPKC=RT1=24.621

(iii)  N2+3H2⇌2 NH3 Δng=-2  ∴KPKC=RT-2=24.62-2=1.659×10-3 ≃1.65× 10 −3

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About this question

This is a previous-year question from JEE Main 2019, 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.