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Equilibrium

NCERT Class 11 Chemistry Chapter 6

Equilibrium for Class 11 Chemistry, JEE and NEET: equilibrium constants, Le Chatelier’s principle, acids and bases, pH, buffers and solubility product.

14 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 6 “Equilibrium”, reprint 2026-27.

Open the official NCERT chapter (PDF, 5.7 MB, ncert.nic.in)

Checked on 19 Oct 2026.

Learn in this order

  1. Equilibrium in physical processes

    Solid–liquid, liquid–vapour and solid–vapour equilibria, vapour pressure, Henry’s law and solved examples.

    NCERT Section 6.1 15 min read 4 solved examples 8 practice questions

  2. Dynamic equilibrium in chemical processes

    Reversible reactions, equal forward and reverse rates, reaching equilibrium from both sides, the D test.

    NCERT Section 6.2 13 min read 4 solved examples 8 practice questions

  3. Law of chemical equilibrium and equilibrium constant

    Law of mass action, writing Kc for any reaction, reversed and multiplied equations, solved examples.

    NCERT Section 6.3 12 min read 4 solved examples 8 practice questions

  4. Homogeneous equilibria: Kp and Kc

    Homogeneous equilibria, Kp = Kc(RT)Δn derived, Δn, units of K, and NCERT Problems 6.3–6.5 solved.

    NCERT Section 6.4 12 min read 4 solved examples 8 practice questions

  5. Heterogeneous equilibria

    Why pure solids and liquids are left out of K, CaCO ⇌ CaO + CO, and NCERT Problem 6.6 solved.

    NCERT Section 6.5 11 min read 4 solved examples 8 practice questions

  6. Applications of equilibrium constants

    Extent of reaction from K, direction from Q and equilibrium concentrations, step by step.

    NCERT Section 6.6 13 min read 4 solved examples 8 practice questions

  7. Equilibrium constant, reaction quotient and Gibbs energy

    ΔG = ΔG° + RT ln Q, ΔG° = −RT ln K, the direction of reaction and NCERT Problems 6.10–6.11.

    NCERT Section 6.7 12 min read 4 solved examples 8 practice questions

  8. Factors affecting equilibria: Le Chatelier’s principle

    Effects of concentration, pressure, inert gas, temperature and catalyst on equilibrium, with examples.

    NCERT Section 6.8 17 min read 4 solved examples 8 practice questions

  9. Ionic equilibrium in solution

    Electrolytes, strong and weak electrolytes, ionisation in water, the dielectric constant and hydration.

    NCERT Section 6.9 11 min read 4 solved examples 8 practice questions

  10. Acids, bases and salts: Arrhenius, Brönsted and Lewis

    Conjugate acid–base pairs, amphiprotic species and NCERT problems solved.

    NCERT Section 6.10 12 min read 4 solved examples 8 practice questions

  11. Ionization of acids and bases: pH, Ka and Kb

    Polybasic acids, common ion effect, hydrolysis of salts and 10 NCERT problems solved.

    NCERT Section 6.11 23 min read 10 solved examples 8 practice questions

  12. Experiment: acid-base titration with an indicator

    Standard solutions, the burette, phenolphthalein or methyl orange, end point and a1M1V1 = a2M2V2, solved examples.

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

  13. Buffer solutions and the Henderson equation

    How buffers resist pH change, the Henderson–Hasselbalch equation, and worked acetate and ammonia buffers.

    NCERT Section 6.12 11 min read 4 solved examples 8 practice questions

  14. Solubility product and the common ion effect

    Ksp and molar solubility, precipitation from Qsp, the common ion effect and pH, NCERT problems solved.

    NCERT Section 6.13 15 min read 4 solved examples 8 practice questions

Key formulas in this chapter

Equilibrium in physical processes

  • Rate of evaporation = rate of condensation (liquid–vapour equilibrium)
  • Henry’s law: dissolved mass ∝ gas partial pressure (fixed solvent amount and temperature; dilute regime)

Dynamic equilibrium in chemical processes

  • At equilibrium: rate of forward reaction = rate of reverse reaction
  • Concentrations constant (not equal) at equilibrium

Law of chemical equilibrium and equilibrium constant

  • Kc = Cc Dd/(Aa Bb) for aA + bB ⇌ cC + dD, where A, B, C, D here stand for the equilibrium concentrations [A], [B], [C], [D] in mol/L
  • Reverse reaction: Kc(reverse) = 1/Kc
  • Equation multiplied by n: Kc^n; halved: √Kc
  • Equations added: K = K1 × K2

Homogeneous equilibria

  • p = cRT = [gas]RT, R = 0.0831 bar L mol K
  • Kp = Kc(RT)Δn
  • Δn = (moles of gaseous products) − (moles of gaseous reactants)

Heterogeneous equilibria

  • CaCO3(s) ⇌ CaO(s) + CO2(g): Kp = p(CO2)
  • Ni(s) + 4CO(g) ⇌ Ni(CO)4(g): Kc = c(Ni(CO)4)/c(CO)4, where c(X) = [X] in mol/L
  • Present pure bulk solids and liquids: activity ≈ 1, omitted from K

Applications of equilibrium constants

  • Qc = Cc Dd/(Aa Bb) for aA + bB ⇌ cC + dD, where A, B, C, D stand for the concentrations at any moment
  • Q < K: forward; Q > K: backward; Q = K: equilibrium
  • K alone does not determine percentage conversion; combine it with mass balance

Equilibrium constant, reaction quotient and Gibbs energy

  • ΔrG = ΔrG⊖ + RT ln Q
  • ΔrG⊖ = −RT ln K; K = exp(−ΔrG⊖/RT)
  • ΔrG = RT ln(Q/K)

Factors affecting equilibria

  • Halving the volume multiplies every gas concentration by 2, so Qc = Kc × 2Δn_g
  • Exothermic (ΔH < 0): K falls as T rises; endothermic (ΔH > 0): K rises as T rises
  • ΔrG⊖ = ΔrH⊖ − TΔrS⊖ = −RT ln K

Ionic equilibrium in solution

  • Continuum electrostatic model: force in medium = force in vacuum/relative permittivity (water ≈ 80)
  • Degree of ionisation = (moles ionised)/(moles dissolved)

Acids, bases and salts

  • Arrhenius: HX(aq) → H(aq) + X(aq); MOH(aq) → M(aq) + OH(aq)
  • Brönsted–Lowry: acid ⇌ conjugate base + H
  • Lewis: BF +:NH → BF:NH

Ionization of acids and bases

  • Ionic product of water: [H][OH] = 1.0 × 10 at 298 K
  • pH = −log [H]; pH + pOH = 14 at 298 K
  • Ka = cα2/(1 − α); α ≈ √(Ka/c) when α is small
  • Ka × Kb = Kw; pKa + pKb = 14 at 298 K
  • At 298 K, pH ≈ 7 + (pKa − pKb)/2 for a simple weak-acid/weak-base salt when water ionisation is negligible

Experiment: acid-base titration with an indicator

  • a1M1V1 = a2M2V2 (at equivalence; approximate at the indicator end point)
  • Strength (g/L) = molarity × molar mass
  • Dilution: M1V1 = M2V2

Buffer solutions

  • pH = pKa + log(salt/acid) (acidic buffer)
  • pOH = pKb + log(salt/base) (basic buffer)
  • Basic buffer: pH = pKa + log(base/salt), with pKa = 14 − pKb at 298 K

Solubility equilibria of sparingly soluble salts

  • Ksp = xx yy Sx+y for MxXy, with no added ions or side reactions
  • S = (Ksp/(xx yy))1/(x+y)
  • Qsp > Ksp: precipitation favoured; Qsp < Ksp: dissolution possible if solid is present

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