JEE Main 2024ChemistryThermodynamics (C)EasyNumerical

JEE Main 2024Thermodynamics (C) Question with Solution

JEE Main 2024 (01 Feb Shift 2)

Question

For a certain reaction at 300 K, K=10, then ΔG° for the same reaction is – ________ ×10-1 kJ mol-1.  (Given R=8.314JK-1 mol-1)

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Show full solutionCorrect answer: 57
Correct answer
57

Step-by-step explanation

The standard Gibbs free energy of the formation of a compound is basically the change of Gibbs free energy that is followed by the formation of one  mole of that substance from its component element available at their standard states or the most stable form of the element which is at 25 °C and 100 kPa. Its symbol is ΔfG˚.

ΔG°=-RTℓnK

=-8.314×300ln(10)

=5744.14 J/mole

=57.44×10-1 kJ/mole

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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.