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Chemistry Question 54 – JEE-MAIN 2026

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Consider the following data for the reaction X2(g)+Y2(g)=2XY(g) at 600 K. The ΔrG∘ (in kJ mol−1) for the reaction is : CompoundΔfH600K∘ (kJ mol−1)S600K∘ (J mol−1 K−1)XY(g)42200X2(g)8140Y2(g)80250

To determine the Gibbs free energy change for a reaction, you need to calculate the enthalpy and entropy changes first.

● Video Walkthrough
Step 1: Calculate the standard enthalpy change of the reaction (ΔrH∘)✦ Active

For the reaction X2(g)+Y2(g)=2XY(g), the standard enthalpy change is calculated using the standard enthalpies of formation of products and reactants:

ΔrH∘=[2×ΔfHXY(g)∘]−[ΔfHX2(g)∘+ΔfHY2(g)∘] ΔrH∘=[2×42 kJ mol−1]−[8 kJ mol−1+80 kJ mol−1] ΔrH∘=84 kJ mol−1−88 kJ mol−1=−4 kJ mol−1
Step 2: Calculate the standard entropy change of the reaction (ΔrS∘)○ Expand

The standard entropy change is calculated using the standard molar entropies of products and reactants:

ΔrS∘=[2×SXY(g)∘]−[SX2(g)∘+SY2(g)∘] ΔrS∘=[2×200 J mol−1 K−1]−[140 J mol−1 K−1+250 J mol−1 K−1] ΔrS∘=400 J mol−1 K−1−390 J mol−1 K−1=10 J mol−1 K−1

Convert ΔrS∘ from J to kJ for consistency with ΔrH∘:

ΔrS∘=10 J mol−1 K−1=0.010 kJ mol−1 K−1
💡 Teacher's Secret Hint

Ensure units are consistent (kJ vs J) before the final calculation.

Step 3: Calculate the standard Gibbs free energy change (ΔrG∘)○ Expand

Using the Gibbs-Helmholtz equation at T=600 K:

ΔrG∘=ΔrH∘−TΔrS∘ ΔrG∘=−4 kJ mol−1−(600 K×0.010 kJ mol−1 K−1) ΔrG∘=−4 kJ mol−1−6 kJ mol−1=−10 kJ mol−1
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