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Thermodynamics

NCERT Class 11 Chemistry Chapter 5

Thermodynamics for Class 11 Chemistry, JEE and NEET: first law, enthalpy, calorimetry, Hess’s law, bond and lattice enthalpies, entropy and Gibbs energy.

9 topics Updated 8 Nov 2026

For: Class 11, JEE and NEET, unless a topic says otherwise

In the syllabus: JEE Class 11 syllabus, NEET Class 11 syllabus

With your textbook

NCERT Chemistry Part I (Class 11), Chapter 5 “Thermodynamics”, reprint 2026-27.

Open the official NCERT chapter (PDF, 5.1 MB, ncert.nic.in)

Checked on 19 Oct 2026.

Learn in this order

  1. Thermodynamic terms: system, state and internal energy

    Thermodynamic terms: system and surroundings, open, closed and isolated systems, state functions, internal energy, first law, for Class 11, JEE and NEET.

    NCERT Section 5.1 14 min read 4 solved examples 8 practice questions

  2. Work, enthalpy and heat capacity

    H = U + pV, ΔH = ΔU + ΔngRT, pressure–volume work, free expansion and heat capacity, with solved examples.

    NCERT Section 5.2 17 min read 5 solved examples 8 practice questions

  3. Calorimetry: measuring ΔU and ΔH

    How a bomb calorimeter gives ΔU and a coffee-cup calorimeter ΔH, and how to convert between them.

    NCERT Section 5.3 11 min read 4 solved examples 8 practice questions

  4. Reaction enthalpy, enthalpy of formation and Hess’s law

    Standard states, reaction enthalpy, phase changes and thermochemical equations.

    NCERT Section 5.4 19 min read 4 solved examples 8 practice questions

  5. Enthalpies of combustion, atomisation, bond, lattice and solution

    Bond enthalpy, lattice enthalpy and the Born–Haber cycle, with combustion, atomisation, solution and dilution enthalpies, for Class 11, JEE and NEET.

    NCERT Section 5.5 18 min read 4 solved examples 8 practice questions

  6. Enthalpy of neutralization

    Why strong acids and bases give about −57 kJ mol, weak acids, the calorimeter method and solved examples.

    JEE and NEET only 14 min read 4 solved examples 8 practice questions

  7. Experiment: enthalpy of solution of copper sulphate

    The calorimeter method, why the crystals cool water, the Hess cycle and solved examples.

    JEE and NEET only 14 min read 4 solved examples 8 practice questions

  8. Spontaneity, entropy and Gibbs energy

    Why a fall in enthalpy is not enough, entropy, Gibbs free energy ΔG = ΔH − TΔS and the third law.

    NCERT Section 5.6 17 min read 4 solved examples 8 practice questions

  9. Gibbs energy change and the equilibrium constant

    Why G is a minimum at equilibrium, ΔG° = −RT ln K and the 5.71 kJ rule.

    NCERT Section 5.7 12 min read 4 solved examples 8 practice questions

Key formulas in this chapter

Thermodynamic terms

  • ΔU = q + w (first law, IUPAC signs)
  • Adiabatic: q = 0, so ΔU = w
  • Isolated system: q = 0, w = 0, ΔU = 0

Work, enthalpy and heat capacity

  • w = −pex ΔV (constant external pressure)
  • wrev = −2.303 nRT log(Vf/Vi) (isothermal, ideal gas)
  • H = U + pV; ΔH = qp; ΔU = qV
  • ΔH = ΔU + Δng RT
  • q = C ΔT = m c ΔT; Cp − CV = R (per mole)

Calorimetry

  • q(calorimeter) = C ΔT; q(reaction) = −C ΔT
  • Bomb calorimeter: qV = ΔU
  • Constant-pressure calorimeter: qp = ΔH
  • ΔH = ΔU + Δng RT

Reaction enthalpy and Hess’s law

  • ΔrH⊖ = Σ a ΔfH⊖(products) − Σ b ΔfH⊖(reactants)
  • ΔfH⊖ = 0 for an element in its reference state
  • ΔsubH = ΔfusH + ΔvapH (same temperature)
  • Hess’s law: ΔH(overall) = ΔH1 + ΔH2 + ΔH3 + …

Combustion, bond and lattice enthalpies

  • ΔrH = Σ bond enthalpies (reactants) − Σ bond enthalpies (products), gases only
  • Mean bond enthalpy = ΔaH ÷ number of equivalent bonds
  • Born–Haber (NaCl): ΔlatticeH = −ΔfH + ΔsubH + ½ΔbondH + ΔiH + ΔegH
  • ΔsolH = ΔlatticeH + ΔhydH

Enthalpy of neutralization

  • H(aq) + OH(aq) → HO(l); ΔH ≈ −57 kJ mol (strong acid, strong base)
  • W = 4.184[Vh(th − tm) − Vc(tm − tc)]/(tm − tc)
  • q = (V × 4.184 + W) × Δt
  • ΔneutH = −q/n(water formed)
  • ΔionH = ΔneutH(weak) − ΔneutH(strong)

Experiment: enthalpy of solution of copper sulphate

  • q = (msoln × 4.184 + W) × Δt
  • ΔsolH = −q × M/msalt
  • W = 4.184[Vh(th − tm) − Vc(tm − tc)]/(tm − tc)
  • ΔH(hydrate formation) = ΔsolH(anhydrous) − ΔsolH(hydrate)

Spontaneity, entropy and Gibbs energy

  • ΔS = qrev/T (constant T)
  • ΔStotal = ΔSsystem + ΔSsurroundings > 0 (spontaneous)
  • ΔSsurr = −ΔHsys/T (constant p; only pV work; reservoir T)
  • G = H − TS; ΔG = ΔH − TΔS (constant T)
  • ΔG < 0 spontaneous; ΔG = 0 at equilibrium
  • Crossover: T = ΔH/ΔS (ΔG changes sign)
  • ΔrS⊖ = ΣS⊖(products) − ΣS⊖(reactants)

Gibbs energy and equilibrium

  • ΔrG = 0 at reaction equilibrium (constant T, p)
  • ΔrG⊖ = −RT ln K = −2.303 RT log K
  • ΔrG⊖ = ΔrH⊖ − TΔrS⊖
  • At 298 K: 2.303 RT = 5.71 kJ mol

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