Thermodynamics Free Energy Equilibrium Question JEE Main 2026

Medium 27 Mar 2026
Nishant Kumar Gupta
Nishant Kumar Gupta
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Thermodynamics Free Energy Equilibrium Question JEE Main 2026

Quick Summary

Question Type: Thermodynamics and equilibrium constant calculation
Chapter: Thermodynamics – Gibbs Free Energy and Equilibrium
Difficulty: ⭐⭐⭐ Medium
Time to Solve: 3-6 minutes
Key Formula: ΔG° = ΔH° – TΔS° = -2.303RT log K
Correct Answer: 70
Why: First calculate ΔG° from the given equilibrium constant, then use ΔG° = ΔH° – TΔS° for the reaction A₂(g) + B₂(g) ⇌ 2AB(g). Solving the resulting equation gives x = 70 kJ mol⁻¹.

The Question

JEE Main 2026 (21 January Morning Shift) – Thermodynamics

Use the following data:

Substance ΔfH°(500 K) / kJ mol-1 S°(500 K) / J K-1 mol-1
AB(g) 32 222
A2(g) 6 146
B2(g) x 280

One mole each of A2(g) and B2(g) are taken in a 1 L closed flask and allowed to establish the equilibrium at 500 K.

A2(g) + B2(g) ⇌ 2AB(g)

The value of x (in kJ mol-1) is ________. (Nearest integer)

Given: log K = 2.2, R = 8.3 J K-1 mol-1

Quick Answer

Correct Answer: 70

Reasoning: For the reaction:

A2(g) + B2(g) ⇌ 2AB(g)

  • Use ΔG° = -2.303RT log K
  • Use ΔG° = ΔH° – TΔS°
  • Find ΔH° and ΔS° of reaction from formation enthalpy and entropy data
  • Then solve for x

Final calculation gives:

  • x = 70 kJ mol-1

Video Solution

If you want the full explanation in a clear step-by-step teaching format, watch the video solution below:

Watch Full Video Solution on YouTube

Understanding the Concept

Why This Is a Gibbs Free Energy Problem

This question connects equilibrium constant with thermodynamics. At constant temperature, the standard Gibbs free energy change of a reaction is related to equilibrium constant by:

ΔG° = -2.303RT log K

At the same time, Gibbs free energy is also related to enthalpy and entropy:

ΔG° = ΔH° – TΔS°

So this question becomes a bridge between equilibrium and thermodynamics.

The Key Principle

  1. Find ΔG° from the given K value
  2. Calculate ΔH° of reaction from formation enthalpies
  3. Calculate ΔS° of reaction from standard entropies
  4. Use ΔG° = ΔH° – TΔS° to solve for x

Detailed Step-by-Step Solution

Step 1: Calculate ΔG° from Equilibrium Constant

  • Given: log K = 2.2
  • Use: ΔG° = -2.303RT log K
  • ΔG° = -2.303 × 8.3 × 500 × 2.2
  • ΔG° ≈ -21020 J mol-1
  • ΔG° ≈ -21.0 kJ mol-1

Step 2: Calculate ΔS° of Reaction

Handwritten solution explaining thermodynamics free energy and equilibrium relation for JEE Main 2026 question


Download Handwritten Solution

For the reaction:

A2(g) + B2(g) ⇌ 2AB(g)

  • ΔS° = 2S°(AB) – [S°(A2) + S°(B2)]
  • ΔS° = 2(222) – (146 + 280)
  • ΔS° = 444 – 426
  • ΔS° = 18 J K-1 mol-1

Step 3: Calculate ΔH° of Reaction

  • ΔH° = 2ΔfH°(AB) – [ΔfH°(A2) + ΔfH°(B2)]
  • ΔH° = 2(32) – (6 + x)
  • ΔH° = 64 – 6 – x
  • ΔH° = 58 – x kJ mol-1

Step 4: Apply ΔG° = ΔH° – TΔS°

  • ΔG° = -21.0 kJ mol-1
  • ΔH° = 58 – x
  • TΔS° = 500 × 18 J = 9000 J = 9 kJ

So:

  • -21 = (58 – x) – 9
  • -21 = 49 – x
  • x = 70

Step 5: Final Answer

  • The value of x = 70 kJ mol-1

Final Answer

x = 70 ✓

The nearest integer value of x is 70 kJ mol-1.

Essential Formulas for This Topic

Primary Equations

  1. Gibbs Free Energy and Equilibrium:
    • ΔG° = -2.303RT log K
    • Connects equilibrium constant with spontaneity
    • Valid at the given temperature
  2. Thermodynamic Relation:
    • ΔG° = ΔH° – TΔS°
    • Links free energy with enthalpy and entropy
    • Useful for solving unknown thermodynamic quantities
  3. Reaction Property Formulae:
    • ΔH° = ΣΔfH°(products) – ΣΔfH°(reactants)
    • ΔS° = ΣS°(products) – ΣS°(reactants)
    • Apply stoichiometric coefficients carefully

Important Constants

  • Given: log K = 2.2
  • Given: R = 8.3 J K-1 mol-1
  • Given temperature: 500 K
  • Remember to convert J to kJ when needed

Common Mistakes to Avoid

❌ Mistake 1: Wrong Sign in ΔG° = -2.303RT log K

Wrong Thinking: “Since log K is positive, ΔG° must also be positive.”

Correct Approach: The formula has a negative sign. So positive log K gives negative ΔG°.

❌ Mistake 2: Forgetting Stoichiometric Coefficient 2 for AB

Wrong Approach: Taking ΔH° or ΔS° using only one mole of AB

Correct Approach: The reaction forms 2AB, so multiply both formation enthalpy and entropy of AB by 2.

❌ Mistake 3: Unit Conversion Error in TΔS°

Common Error:

  • Using ΔS° in J and ΔH° in kJ directly
  • Not converting 9000 J to 9 kJ

Correct Approach: Keep all terms in the same unit before substituting in ΔG° = ΔH° – TΔS°.

❌ Mistake 4: Using Reaction Enthalpy Formula Incorrectly

Wrong Thinking: “Add x instead of subtracting it.”

Correct Understanding:

  • ΔH° = products – reactants
  • So ΔH° = 2(32) – (6 + x)
  • That gives 58 – x
  • Then solve with the free energy equation

Key Concept Summary

What You Must Remember

  1. Equilibrium constant gives ΔG°: Use ΔG° = -2.303RT log K
  2. Thermodynamic quantities are linked: Use ΔG° = ΔH° – TΔS°
  3. Formation data must be used carefully: Products minus reactants
  4. Always maintain unit consistency: Convert J into kJ when needed
  5. Stoichiometric coefficients matter: 2AB means multiply by 2

The Golden Rule for Thermodynamics + Equilibrium Questions

“Find ΔG° from K first, then connect it to ΔH° and ΔS° using the Gibbs equation.”

Frequently Asked Questions

Q1: Why is ΔG° negative here?

A: Because log K is positive and the relation ΔG° = -2.303RT log K has a negative sign.

Q2: Why do we multiply AB values by 2?

A: Because the balanced equation forms 2 moles of AB.

Q3: How is ΔS° calculated here?

A: ΔS° = 2(222) – (146 + 280) = 18 J K-1 mol-1.

Q4: Why is TΔS° equal to 9 kJ?

A: TΔS° = 500 × 18 = 9000 J = 9 kJ.

Q5: What is the final value of x?

A: The final value of x is 70 kJ mol-1.

Prerequisites to Solve This Question

Before attempting this problem, you should understand:

  1. Gibbs free energy: Relation with spontaneity and equilibrium
  2. Equilibrium constant: Meaning of K and log K
  3. Reaction enthalpy and entropy: Products minus reactants method
  4. Unit conversion: J and kJ consistency
  5. Balanced chemical equations: Correct use of coefficients

After Solving This, You Can:

✅ Relate equilibrium constant to Gibbs free energy
✅ Use thermodynamic data to solve unknown quantities
✅ Calculate reaction entropy and enthalpy correctly
✅ Handle mixed thermodynamics-equilibrium questions confidently
✅ Avoid unit conversion mistakes in numerical problems
✅ Solve JEE Main Thermodynamics PYQs faster

Study Tips for This Topic

For JEE Main:

  1. Memorize both Gibbs relations: They are often used together
  2. Check units carefully: This is where most mistakes happen
  3. Write reaction property formulae clearly: Products minus reactants
  4. Practice mixed-concept questions: JEE often combines equilibrium with thermodynamics

Common JEE Variants:

  • Find ΔG° from equilibrium constant
  • Find unknown ΔH° or ΔS°
  • Use formation enthalpy data in equilibrium problems
  • Thermodynamics and equilibrium combined numericals

Difficulty Rating & Exam Frequency

Difficulty Level: ⭐⭐⭐ (3/5) – Medium
JEE Main Frequency: Medium – Mixed thermodynamics-equilibrium questions appear regularly
JEE Advanced Frequency: High – Such concept-linking questions are very common
Topic Importance: Very High – Gibbs free energy and equilibrium are core thermodynamic ideas

Written by Nishant Kumar
Chemistry Educator with 10+ Years of Experience Teaching JEE Aspirants
Founder – PadhoLikhoJEE

Last Updated: March 2026
Question Source: JEE Main 2026 PYQ
Topic: Thermodynamics – Gibbs Free Energy and Equilibrium

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