Chemistry · Class 11 · Chemical Thermodynamics

Chemical Thermodynamics for JEE

Everything you need for this topic on one page: the ideas, the formulas, the traps, and free lectures that explain it well.

Last updated: 26 October 2026

Direct answer

System and surroundings, first law, enthalpy, Hess's law, entropy, Gibbs free energy and spontaneity.

System and surroundings, first law, enthalpy, Hess's law, entropy, Gibbs free energy and spontaneity.

Key ideas

Chemical Thermodynamics: what you must understand

  • State functions: U, H, S, G (path-independent)
  • Hess's law: enthalpy changes add
  • ΔG < 0 means spontaneous at constant T and P
  • ΔG° = −RT ln K links thermodynamics and equilibrium
Formula sheet

Must-know formulas for Chemical Thermodynamics

  • ΔH = ΔU + Δn_g RT
  • ΔG = ΔH − TΔS
  • ΔS = q_rev / T
  • Isothermal reversible work W = −2.303 nRT log(V₂/V₁)
Print

One-page formula sheet

Printable Chemical Thermodynamics formula sheet: formulas, key ideas and traps on one page. Save it as a PDF for offline revision.

Avoid

Common mistakes students make

  • Sign convention for work (chemistry uses work done on the system)
  • Δn_g counting liquid or solid moles
  • Forgetting to convert kJ and J
For the exam

How this topic is asked in JEE

Appears frequently. Spontaneity with ΔH and ΔS signs is a favourite one-liner question.

Prerequisites: Mole Concept and Stoichiometry.

NCERT map

Where Chemical Thermodynamics is in NCERT

Chapter names follow the current rationalised NCERT textbooks. Read these sections before solving PYQs.

ClassBookChapterSections JEE asks from
Class 11Chemistry Part IIThermodynamicsSystem and Surroundings; Internal Energy as a State Function; Enthalpy; Enthalpies for Different Types of Reactions; Spontaneity; Gibbs Energy Change and Equilibrium
JEE Advanced level

A JEE Advanced-level example on Chemical Thermodynamics

Question. For the reaction 2SO₂(g) + O₂(g) → 2SO₃(g) at 298 K, given: ΔH°f(SO₂) = -297 kJ/mol, ΔH°f(SO₃) = -396 kJ/mol, S°(SO₂) = 248 J/mol·K, S°(O₂) = 205 J/mol·K, S°(SO₃) = 257 J/mol·K. Calculate (a) ΔH° for the reaction, (b) ΔS° for the reaction, (c) ΔG° at 298 K, and (d) the temperature at which the reaction reaches equilibrium.

Show solution
  1. ΔH°rxn = Σ ΔH°f(products) - Σ ΔH°f(reactants) = 2(-396) - [2(-297) + 0] = -792 + 594 = -198 kJ
  2. ΔS°rxn = Σ S°(products) - Σ S°(reactants) = 2(257) - [2(248) + 205] = 514 - 701 = -187 J/K
  3. ΔG° = ΔH° - TΔS° = -198 kJ - (298 K)(-0.187 kJ/K) = -198 + 55.7 = -142.3 kJ (reaction is spontaneous at 298 K)
  4. At equilibrium ΔG = 0, so T = ΔH°/ΔS° = (-198 kJ)/(-0.187 kJ/K) = 1059 K. Above this temperature the reaction becomes non-spontaneous.

Answer: (a) ΔH° = -198 kJ, (b) ΔS° = -187 J/K, (c) ΔG° = -142.3 kJ, (d) T = 1059 K

This example was solved twice independently and both solutions agree.

Free lectures

Best free lectures for Chemical Thermodynamics

Full-chapter free lectures picked from YouTube search results by views, from established JEE channels (mostly in Hinglish). Use the subtopic list to jump straight to the part you need (timestamps come from each video's own chapter list). Watch one, then solve PYQs the same day.

Practise

Where to practise for free

FAQs

Frequently asked questions

Is Chemical Thermodynamics important for JEE Main 2027?

Appears frequently. It is part of the official 'Chemical Thermodynamics' unit in the JEE Main 2027 syllabus.

What are the most important formulas in Chemical Thermodynamics?

ΔH = ΔU + Δn_g RT; ΔG = ΔH − TΔS; ΔS = q_rev / T.

How should I study Chemical Thermodynamics for JEE?

Understand the key ideas from NCERT or one full lecture, learn the formula sheet, then solve 25–30 past JEE questions on the topic. Spontaneity with ΔH and ΔS signs is a favourite one-liner question.

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