JEE Main 2016ChemistryThermodynamics (C)MediumMCQ

JEE Main 2016 — Thermodynamics (C) Question with Solution

JEE Main 2016 (09 Apr Online)

Question

For the reaction,

Ag+Bg→Cg+Dg, ∆Ho and ∆So are, respectively, -29.8 kJ mol-1 and -0.100 kJ K-1 mol-1 at 298 K. The equilibrium constant for the reaction at 298 k is:

Choose an option

Show full solutionCorrect option: C
Correct answer
C1

Step-by-step explanation

∆G°= ∆H°-T∆S°

∆G°=-29.8 kJmol-1+0.1×298 kJmol-1=0

∆G= ∆G°+RTlnQ  at Equilibrium ∆G=0 & Q=k

∆G°=-RTln(k)

if ln(k)=0,  k=e0

∴K=1
 

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Thermodynamics (C) chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2016, 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.