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Chemistry · Thermodynamics & Thermochemistry

JEE Main 2026 — 5 April, Evening Shift — Question 52

Consider the following data for the reaction X2( g)+Y2( g)⇌2XY(g)\mathrm{X}_{2}(\mathrm{~g})+\mathrm{Y}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{XY}(\mathrm{g}) at 600 K. The ΔrG⊖(inkJmol−1)600 \mathrm{~K}_{\text {. The } \Delta_{\mathrm{r}} \mathrm{G}^{\ominus}}\left(\mathrm{in} \mathrm{kJ} \mathrm{mol}^{-1}\right) for the reaction is:

CompoundΔfH600K∘ (kJ mol−1)S600K∘ (J mol−1K−1)XY(g)42200XX2(g)8140YX2(g)80250\begin{array}{c|c|c} \text{Compound} & \Delta_f H^\circ_{600\text{K}} \, (\text{kJ mol}^{-1}) & S^\circ_{600\text{K}} \, (\text{J mol}^{-1}\text{K}^{-1}) \\ \hline \ce{XY(g)} & 42 & 200 \\ \ce{X2(g)} & 8 & 140 \\ \ce{Y2(g)} & 80 & 250 \end{array}
  1. Option A:

    -21000

  2. Option B:

    -10

    Correct
  3. Option C:

    -1000

  4. Option D:

    -9.012

Answer: B

Step-by-step solution

ΔrG=ΔrH−TΔrS\Delta_{\mathrm{r}}G=\Delta_{\mathrm{r}}H-T\Delta_{\mathrm{r}}S

ΔrH=(2×42−80−8)=−4 kJ/molΔrS=(400−250−140)=+10 J/(K⋅mol)ΔrG=−4000−600(10)=−10,000 J/mol=−10 kJ/mol\begin{aligned} \Delta_{\mathrm{r}}H &= (2\times42-80-8)=-4\,\mathrm{kJ/mol}\\ \Delta_{\mathrm{r}}S &= (400-250-140)=+10\,\mathrm{J/(K\cdot mol)}\\ \Delta_{\mathrm{r}}G &= -4000-600(10)\\ &=-10{,}000\,\mathrm{J/mol}\\ &=-10\,\mathrm{kJ/mol} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
5 April, Evening Shift
Subject
Chemistry
Chapter
Thermodynamics & Thermochemistry
Topic
Gibbs Free Energy - Relation with Equilibrium, Metallurgy and Electrochemistry

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