JEE Main 2018ChemistryThermodynamics (C)MediumMCQ

JEE Main 2018 — Thermodynamics (C) Question with Solution

JEE Main 2018 (16 Apr Online)

Question

At 320 K, a gas A2 is 20 % dissociated to A(g) . The standard Gibbs free energy change at 320 K and 1 atm in J mol–1 is approximately: (R = 8.314 JK–1 mol–1 ;ln2 = 0.693 ;ln3 = 1.098)

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Show full solutionCorrect option: D
Correct answer
D4763

Step-by-step explanation

Let assume initial moles is 1
               A2→                                  2A
t=0       1
t=t         1-0.2                   0.2×2

Partial pressure of A2=0.81.2×1 
Partial pressure of A=0.41.2×1 
kp=0.41.22×10.81.2×1 =16
∆G°= -RTlnk=-8.314×320 ln16
=8.314×320 0.693+1.098
⋍4763 J/mole

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About this question

This is a previous-year question from JEE Main 2018, covering the Thermodynamics (C) 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.