JEE Main 2019 — Chemical Equilibrium Question with Solution
From: JEE Main 2019 (Online) 10th January Evening Slot
Question
5.1 g NH4SH is introduced in 3.0 L evacuated flask at 327ºC. 30% of the solid NH4SH decomposed to NH3 and H2S as gases . The Kp of the reaction at 327oC is (R = 0.082 L atm mol–1 K–1, Molar mass of S = 32 g mol–1 molar mass of N = 14 g mol–1)
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Show full solutionCorrect option: D
Correct answer
D0.242 atm2
Step-by-step explanation
NH4SH(s) NH3(g) + H2S(g)
so number of moles at equilibrium
Now use PV = nRT at equilibrium
Ptotal 3 lit = (.03 + .03) .082 600
Ptotal = .984 atm
At equilibrium
PNH3 = PH2S = = .492
So kp = PNH3 . PH2S = (.492) (.492)
kp = .242 atm2
so number of moles at equilibrium
Now use PV = nRT at equilibrium
Ptotal 3 lit = (.03 + .03) .082 600
Ptotal = .984 atm
At equilibrium
PNH3 = PH2S = = .492
So kp = PNH3 . PH2S = (.492) (.492)
kp = .242 atm2
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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.