JEE PREP | APRIL 2026
JEE MAIN 2027 — CHAPTER-WISE PRACTICE & STRATEGY | CHEMISTRY
• Subject: Chemistry — Physical Chemistry (Class XI, Chapter 6)
• Expected questions: 2–4 per year in JEE Main
• Difficulty: Moderate (concepts) to High (numerical problems)
• High overlap with NEET — efficient dual-prep chapter
Thermodynamics is a cornerstone of Physical Chemistry in JEE Main. Appearing in virtually every session, it tests concepts from the laws of thermodynamics, enthalpy and Hess’s Law calculations, entropy and spontaneity via Gibbs free energy. NCERT Class 11 Chemistry Chapter 6 is the primary source, but JEE also demands numerical problem-solving fluency beyond what NCERT provides. This comprehensive guide covers all key concepts with formulas, data tables, and 10 JEE-pattern MCQs.
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1. Basic Concepts — System, Surroundings, and Processes
• Open system: exchanges both matter and energy with surroundings (e.g., open beaker)
• Closed system: exchanges only energy (e.g., sealed piston-cylinder)
• Isolated system: exchanges neither matter nor energy (e.g., thermos flask)
Types of Processes:
• Isothermal: constant temperature (ΔT = 0, ΔU = 0 for ideal gas)
• Adiabatic: no heat exchange (q = 0)
• Isobaric: constant pressure (ΔH = qp)
• Isochoric: constant volume (ΔU = qv)
• Reversible: carried out in infinitesimally small steps (maximum work)
2. Laws of Thermodynamics and Key Equations
• First Law: ΔU = q + w (q = heat added to system; w = work done ON system)
• Work by gas (expansion): w = −PextΔV (at constant external pressure)
• Reversible isothermal work: w = −nRT ln(V₂/V₁) = −2.303 nRT log(V₂/V₁)
• Enthalpy: H = U + PV; at constant P: ΔH = ΔU + PΔV = ΔU + ΔngRT
• Relationship: ΔH = ΔU + ΔngasRT (Δngas = moles of gaseous products − reactants)
Entropy and Gibbs Energy:
• Entropy change: ΔS = qrev/T (for reversible process)
• Gibbs Free Energy: ΔG = ΔH − TΔS
• At equilibrium: ΔG = 0; ΔG° = −RT ln K = −2.303 RT log K
• Spontaneity: ΔG < 0 (spontaneous); ΔG > 0 (non-spontaneous); ΔG = 0 (equilibrium)
ΔH negative + ΔS positive = ΔG always negative = Always Spontaneous
ΔH positive + ΔS negative = ΔG always positive = Never Spontaneous
ΔH negative + ΔS negative = ΔG negative at LOW temperature = Spontaneous when cold
ΔH positive + ΔS positive = ΔG negative at HIGH temperature = Spontaneous when hot
| ΔH | ΔS | ΔG = ΔH − TΔS | Spontaneity |
|---|---|---|---|
| Negative (−) | Positive (+) | Always negative | Always spontaneous |
| Positive (+) | Negative (−) | Always positive | Never spontaneous |
| Negative (−) | Negative (−) | Negative at low T | Spontaneous at low T only |
| Positive (+) | Positive (+) | Negative at high T | Spontaneous at high T only |
3. Hess’s Law, Bond Enthalpies, and Key Applications
• Hess’s Law: ΔH is independent of path — add/subtract reactions like algebraic equations
• Standard enthalpy of formation of elements in standard state = 0 (by definition)
• ΔH°rxn = Σ ΔHf°(products) − Σ ΔHf°(reactants)
• Bond energy method: ΔH = Σ (bond energies broken) − Σ (bond energies formed)
• Combustion enthalpy is always negative (exothermic)
• Resonance energy: actual ΔH(combustion) − calculated ΔH(combustion from bond energies)
• Third Law: Entropy of perfect crystal at 0 K = 0
• Kirchhoff’s equation: ΔH at T₂ = ΔH at T₁ + ΔCp(T₂ − T₁)
4. JEE Main 2027 — 10 MCQ Practice Questions
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5. Frequently Asked Questions (FAQ)
Q1: What is the difference between ΔH and ΔU in JEE Thermodynamics?
ΔU (internal energy change) accounts for all forms of energy. ΔH (enthalpy change) = ΔU + PΔV. At constant pressure, ΔH = qp (heat exchanged). The relationship ΔH = ΔU + ΔngasRT (where Δngas = moles of gaseous products − reactants) is essential for numerical problems. For reactions with no gaseous species or Δngas = 0, ΔH = ΔU.
Q2: What is the spontaneity criterion and how does temperature affect it?
A process is spontaneous when ΔG < 0. Since ΔG = ΔH − TΔS, temperature determines which factor dominates. When ΔH and ΔS have opposite signs, the reaction is either always or never spontaneous. When they have the same sign, there is a crossover temperature T = ΔH/ΔS at which ΔG = 0.
Q3: How do you apply Hess’s Law in numerical problems?
Write the target equation. Manipulate given equations (multiply by factors, reverse direction, and reverse sign of ΔH) to sum to the target. Add all ΔH values algebraically. The result is ΔH of the target reaction — regardless of path.
Q4: What is the standard state for thermodynamic measurements?
Standard state is: pure substance in its most stable form at 1 bar pressure and a specified temperature (usually 298 K). Standard enthalpy of formation of elements in their standard state is defined as zero. All ΔH° values are reported relative to these reference states.
Q5: What is the significance of ΔG° = −RT ln K?
This equation connects thermodynamics to chemical equilibrium. If ΔG° is large and negative, K is very large (reaction nearly complete). If ΔG° is large and positive, K is very small (reaction barely proceeds). At equilibrium ΔG = 0 (not ΔG°). This distinction is critical for JEE questions.
Last Updated: April 2026