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Chemistry · Thermodynamics & Thermochemistry

JEE Main 2026 — 28 January, Evening Shift — Question 54

The plot of log⁡10 K\log _{10} \mathrm{~K} vs 1 T\frac{1}{\mathrm{~T}} gives a straight line. The intercept and slope respectively are (where K is equilibrium constant).

  1. Option A:

    2.303RΔH∘,2.303RΔS∘\frac{2.303 \mathrm{R}}{\Delta \mathrm{H}^{\circ}}, \frac{2.303 \mathrm{R}}{\Delta \mathrm{S}^{\circ}}

  2. Option B:

    ΔS∘2.303R,−ΔH∘2.303R\frac{\Delta \mathrm{S}^{\circ}}{2.303 \mathrm{R}},-\frac{\Delta \mathrm{H}^{\circ}}{2.303 \mathrm{R}}

    Correct
  3. Option C:

    −ΔS∘R2.303,ΔH∘R2.303-\frac{\Delta \mathrm{S}^{\circ} \mathrm{R}}{2.303}, \frac{\Delta \mathrm{H}^{\circ} \mathrm{R}}{2.303}

  4. Option D:

    −ΔH∘2.303R,ΔS∘2.303R-\frac{\Delta \mathrm{H}^{\circ}}{2.303 \mathrm{R}}, \frac{\Delta \mathrm{S}^{\circ}}{2.303 \mathrm{R}}

Answer: B

Step-by-step solution

log⁡10 K=−ΔHo2.303RT+ΔSo2.303R\log _{10} \mathrm{~K}=-\frac{\Delta \mathrm{H}^{\mathrm{o}}}{2.303 \mathrm{RT}}+\frac{\Delta \mathrm{S}^{\mathrm{o}}}{2.303 \mathrm{R}} y -intercept =ΔS∘2.303R=\frac{\Delta \mathrm{S}^{\circ}}{2.303 \mathrm{R}} Slope =−ΔHo2.303R=-\frac{\Delta \mathrm{H}^{\mathrm{o}}}{2.303 \mathrm{R}} Ans. (2) is correct.

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
28 January, Evening Shift
Subject
Chemistry
Chapter
Thermodynamics & Thermochemistry
Topic
Gibbs Free Energy - Relation with Equilibrium, Metallurgy and Electrochemistry
The plot of log 10 K vs frac 1 T gives a straight line. The intercept… | JEE Main 2026 PYQ with Solution · DhiX AI