JEE Main 2024ChemistryThermodynamics (C)HardNumerical

JEE Main 2024 — Thermodynamics (C) Question with Solution

JEE Main 2024 (27 Jan Shift 2)

Question

For a certain thermochemical reaction M→N at T=400 K,ΔHo=77.2 kJ mol-1, ΔSo=122JK-1, log equilibrium constant (logK) is - _____×10-1.

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Show full solutionCorrect answer: 37
Correct answer
37

Step-by-step explanation

Given,

 T=400 K,ΔHo=77.2 kJ mol-1,ΔSo=122JK-1

ΔGo=ΔHo-TΔSo

=77.2×103-400×122=28400 J

ΔG°=-2.303 R T log K

Where K is the equilibrium constant.

By putting the values in above equation,

⇒28400=-2.303×8.314×400×log K

⇒log K=-3.708=-37.08×10-1

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About this question

This is a previous-year question from JEE Main 2024, 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.