JEE Main 2015ChemistryChemical EquilibriumMediumMCQ

JEE Main 2015 — Chemical Equilibrium Question with Solution

JEE Main 2015 (04 Apr)

Question

The standard Gibbs energy change at 300 K for the reaction 2A⇌B+C is 2494.2 J. At a given time, the composition of the reaction mixture is A=12, B=2 and C=12 . The reaction proceeds in the: [R=8.314 J/K-mol ,  e=2.718]  {Given antilog (-0.44)=0.36}

Choose an option

Show full solutionCorrect option: C
Correct answer
CReverse direction because Q>KC

Step-by-step explanation

Δ G ∘ = - RT ln K C

2494.2 = - 8.314 × 300 ln K C

2494.2 = - 8.314 × 300 × 2.303 log K C

- 2494.2 2.303 × 300 × 8.314 = - 0.44 = log K C

log KC=-0.44

K C = 0.36

Q=[B][C][A]2=2×12122=4

Q>KC, reaction will shift in reverse direction.

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About this question

This is a previous-year question from JEE Main 2015, 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.