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JEE Main 2022 — Chemical Equilibrium Question with Solution

JEE Main 2022 (26 Jul Shift 1)

Question

At 298 K, the equilibrium constant is 2×1015 for the reaction:

Cus+2Ag+aq⇌Cu2+aq+2Ags

The equilibrium constant for the reaction

12Cu2+aq+Ags⇌12Cus+Ag+aq is x×10-8. The value of x is ______. (Round off the answer to the nearest integer)

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Show full solutionCorrect answer: 2
Correct answer
2

Step-by-step explanation

Cus+2Ag+aq⇌Cu2+aq+2Ags; Keq=2×1015

12Cu2+aq+Ags⇌12Cus+Ag+aq

The equilibrium reaction is reversed and multiplied with 12,

Hence, Keq'=1Keq=12×1015=x×10-8

⇒120×1107=x×10-8

⇒120×10-7=x×10-8

1020=x

⇒x=102=5=2.236

≃2.24

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About this question

This is a previous-year question from JEE Main 2022, 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.