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ChemistryInteger+4 / -1Question 75 of 75

JEE Main 2026 · 6 Apr · Shift 1, Chemistry question 75

Consider the reaction X⇌Y\ce{X <=> Y} at 300 K. If ΔH∘\Delta H^\circ and K are 28.40 kJ mol−1^{-1} and 1.8×10−71.8\times10^{-7} at the same temperature, then the magnitude of ΔS∘\Delta S^\circ for the reaction in J K−1^{-1} mol−1^{-1} is ________. (Nearest integer)

(Given : R = 8.3 J K−1^{-1} mol−1^{-1}, ln 10 = 2.3, log 3 = 0.48, log 2 = 0.30)

Answer and worked solution

Official answer: 34

Free energy.
log⁡(1.8×10−7)=log⁡18−8=1.26−8=−6.74.\log(1.8\times10^{-7})=\log18-8=1.26-8=-6.74.
ΔG∘=−2.3RTlog⁡K=2.3×8.3×300×6.74≈38 600 J mol−1.\Delta G^\circ=-2.3RT\log K=2.3\times8.3\times300\times6.74\approx38\,600\text{ J mol}^{-1}.

Entropy.
ΔS∘=ΔH∘−ΔG∘T=28 400−38 600300≈−34 J K−1 mol−1.\Delta S^\circ=\frac{\Delta H^\circ-\Delta G^\circ}T=\frac{28\,400-38\,600}{300}\approx-34\text{ J K}^{-1}\text{ mol}^{-1}.

Magnitude: 34.

Answer: 34

Topics:Gibbs energy and spontaneityEquilibrium and Gibbs energy
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