JEE Main 2015ChemistryThermodynamics (C)MediumMCQ

JEE Main 2015 — Thermodynamics (C) Question with Solution

JEE Main 2015 (04 Apr)

Question

The following reaction is performed at 298 K.

2NOg+O2g⇌2NO2g

The standard free energy of the formation of NOg is 86.6 kJ mol-1 at 298 K. What is the standard free energy of the formation of NO2g at 298 K? (KP=1.6×1012)

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Show full solutionCorrect option: A
Correct answer
A0.5[2×86,600-R298ln⁡(1.6×1012)]

Step-by-step explanation

Δ G reac ∘ = - 2.303 RT log K P

= - RT ln K P

=-R298ln 1.6×1012

ΔGreac∘=2ΔGfNO2∘-2ΔGfNO∘

⇒-R298ln 1.6×1012=2ΔGfNO2∘-2×86.6×103

⇒2ΔGfNO2∘=-R298ln1.6×1012+2×86,600

Δ G f NO 2 ∘ = 86,600 - R 2 9 8 2 ln 1.6 × 1 0 1 2

= 0.5 2 × 86,600 - R 2 9 8 ln 1.6 × 1 0 1 2

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About this question

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