JEE Main 2020 — Chemical Equilibrium Question with Solution
From: JEE Main 2020 (Online) 6th September Evening Slot
Question
The value of KC is 64 at 800 K for the reaction
N2(g) + 3H2(g) ⇌ 2NH3(g)
The value of KC for the following reaction is :
NH3(g) ⇌ N2(g) + H2(g)
N2(g) + 3H2(g) ⇌ 2NH3(g)
The value of KC for the following reaction is :
NH3(g) ⇌ N2(g) + H2(g)
Choose an option
Show full solutionCorrect option: B
Correct answer
B
Step-by-step explanation
N2(g) + 3H2(g) ⇌ 2NH3(g) ; KC
2NH3(g) ⇌ N2(g) + 3H2(g) ;
Multiplying by , reaction becomes
NH3(g) ⇌ N2(g) + H2(g) ;
New KC = = =
2NH3(g) ⇌ N2(g) + 3H2(g) ;
Multiplying by , reaction becomes
NH3(g) ⇌ N2(g) + H2(g) ;
New KC = = =
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This is a previous-year question from JEE Main 2020, 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.