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Equilibrium – Class 11 Chemistry Chapter 6 Notes & NCERT PDF

Class 11ChemistryChapter 6NCERT book: Chemistry Part I - Class 11

Notes and a simple summary of Chapter 6 of the NCERT Class 11 Chemistry book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

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Chapter 6: Equilibrium

Introduction

Unit 6 of NCERT Chemistry Part I studies equilibrium: the state reached in a closed system when the rate of the forward process equals the rate of the reverse process. The measurable properties then stop changing, even though molecules keep reacting. This dynamic equilibrium appears in physical changes (ice ⇌ water, liquid ⇌ vapour, solutions) and in chemical reactions (the Haber process, dissociation of weak acids). The unit turns the idea into numbers through the equilibrium constant, uses Le Chatelier's principle to predict responses to change, and then applies everything to ionic equilibria: acids and bases, pH, buffers and solubility products.

Key Concepts

6.1 Equilibrium in Physical Processes

  • Solid–liquid (6.1.1): ice and water at 273 K and 1 atm. The rate of melting equals the rate of freezing, and the normal melting point is where the two coexist at 1 atm.
  • Liquid–vapour (6.1.2): in a closed box the rate of evaporation equals the rate of condensation, and the pressure is the equilibrium vapour pressure. The normal boiling point is where the vapour pressure equals 1.013 bar.
  • Solid–vapour (6.1.3): sublimation of iodine or camphor in a closed vessel.
  • Dissolution (6.1.4): in a saturated solution the solid dissolves and crystallises at equal rates. For gases, Henry's law says the mass of gas dissolved is proportional to its pressure over the solution (why soda water fizzes when opened).
  • General characteristics (6.1.5): equilibrium is possible only in a closed system at a given temperature, it is dynamic, and the measurable properties stay constant.

    • 6.2–6.3 Chemical Equilibrium and the Law of Chemical Equilibrium

  • For aA + bB ⇌ cC + dD, the law of mass action gives Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ, using equilibrium concentrations.
  • If the reaction is reversed, K becomes 1/K. If the equation is multiplied by n, K becomes Kⁿ.

    • 6.4 Homogeneous Equilibria

  • All species are in the same phase. For gases, Kp uses partial pressures.
  • Kp = Kc(RT)^Δn, where Δn = moles of gaseous products − moles of gaseous reactants, and R = 0.0831 bar L mol⁻¹ K⁻¹.

    • 6.5 Heterogeneous Equilibria

  • More than one phase is present. Pure solids and pure liquids are left out of K, because their "concentration" is constant.
  • Example: for CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kp = p(CO₂).

    • 6.6 Applications of Equilibrium Constants

  • Extent of reaction (6.6.1): if K > 10³ the products dominate; if K < 10⁻³ the reactants dominate; between these, both are appreciable.
  • Direction (6.6.2): compare the reaction quotient Q, which has the same expression as K but uses the current concentrations:

    • - Q < K: the net reaction goes forward.

    - Q > K: the net reaction goes backward.

    - Q = K: the system is at equilibrium.

  • Calculating equilibrium concentrations (6.6.3): use an ICE table (initial, change, equilibrium) and solve, often a quadratic.

    • 6.7 Relationship between K, Q and G

  • ΔG = ΔG⁰ + RT ln Q. At equilibrium ΔG = 0, so ΔG⁰ = −RT ln K.

    • 6.8 Factors Affecting Equilibria (Le Chatelier's Principle)

    A change in any of the factors that determine the equilibrium conditions causes the system to shift so as to reduce or counteract that change.
  • Concentration (6.8.1): adding a reactant, or removing a product, shifts the equilibrium forward.
  • Pressure (6.8.2): raising the pressure (lowering the volume) favours the side with fewer moles of gas. There is no effect if Δn_g = 0.
  • Inert gas at constant volume (6.8.3): no effect, because the partial pressures do not change.
  • Temperature (6.8.4): this is the only factor that changes K.

    • - For an exothermic reaction, raising T decreases K.

    - For an endothermic reaction, raising T increases K.

    - Example from the book: the brown NO₂ / colourless N₂O₄ system darkens on heating.

  • Catalyst (6.8.5): speeds up both forward and backward reactions equally. Equilibrium is reached sooner, but K is unchanged.

    • 6.9–6.10 Ionic Equilibrium; Acids, Bases and Salts

  • Strong electrolytes ionise almost completely; weak electrolytes ionise only partly.
  • Arrhenius: acids give H⁺ in water and bases give OH⁻.
  • Brönsted–Lowry: an acid is a proton donor and a base is a proton acceptor. Each acid has a conjugate base differing by one H⁺. The stronger the acid, the weaker its conjugate base.
  • Lewis: an acid is an electron-pair acceptor (BF₃, AlCl₃, H⁺) and a base is an electron-pair donor (NH₃, OH⁻).

    • 6.11 Ionisation of Acids and Bases

  • Ionic product of water: Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 298 K.
  • pH = −log[H₃O⁺] and pH + pOH = 14 at 298 K. Neutral pH is 7 at 298 K.
  • Weak acid: Ka = cα²/(1 − α). When α is small, α ≈ √(Ka/c) and [H⁺] ≈ √(Ka·c). Also pKa = −log Ka.
  • Weak base: Kb, pKb. For a conjugate pair, Ka × Kb = Kw, so pKa + pKb = 14.
  • Polybasic acids ionise in steps, with Ka₁ > Ka₂ > …
  • Acid strength depends on H–A bond strength and the polarity of the bond.
  • Common-ion effect (6.11.8): adding a common ion suppresses ionisation, e.g. acetate added to acetic acid.
  • Salt hydrolysis (6.11.9):

    • - Salt of a strong acid and a weak base gives an acidic solution.

    - Salt of a weak acid and a strong base gives a basic solution.

    - Salt of a weak acid and a weak base: pH = 7 + ½(pKa − pKb).

    6.12 Buffer Solutions

  • A buffer resists changes in pH on dilution or on adding small amounts of acid or base.
  • An acidic buffer is a weak acid plus its salt (CH₃COOH + CH₃COONa). A basic buffer is a weak base plus its salt (NH₄OH + NH₄Cl).
  • Henderson–Hasselbalch equation: pH = pKa + log([conjugate base]/[acid]). When the two concentrations are equal, pH = pKa.

    • 6.13 Solubility Equilibria of Sparingly Soluble Salts

  • For M_xX_y(s) ⇌ xM^(y+) + yX^(x−), Ksp = [M^(y+)]^x [X^(x−)]^y.
  • A precipitate forms when the ionic product Qsp > Ksp.
  • The common-ion effect lowers solubility.
  • Salts of weak acids (such as phosphates) become more soluble at low pH.

    • Worked Examples (from the book)

  • Problem 6.1. N₂ + 3H₂ ⇌ 2NH₃ at 500 K, with [N₂] = 1.5 × 10⁻², [H₂] = 3.0 × 10⁻² and [NH₃] = 1.2 × 10⁻² M.

    • Kc = (1.2 × 10⁻²)² / [(1.5 × 10⁻²)(3.0 × 10⁻²)³] ≈ 3.55 × 10².

  • Problem 6.7. For 2A ⇌ B + C, Kc = 2 × 10⁻³, and at a given time [A] = [B] = [C] = 3 × 10⁻⁴ M.

    • Qc = (3 × 10⁻⁴)²/(3 × 10⁻⁴)² = 1. Since Qc > Kc, the reaction proceeds in the reverse direction.

  • Problem 6.8. 13.8 g N₂O₄ (0.15 mol) in 1 L at 400 K gives an initial p = 0.15 × 0.083 × 400 = 4.98 bar.

    • - The total pressure at equilibrium is 9.15 bar, so x = 4.17 bar.

    - p(N₂O₄) = 0.81 bar and p(NO₂) = 8.34 bar.

    - Kp = 8.34²/0.81 ≈ 85.87, and Kc = Kp/(RT) ≈ 2.6.

  • Problem 6.9. 3.00 mol PCl₅ in 1 L at 380 K with Kc = 1.80.

    • - x²/(3 − x) = 1.80 gives x ≈ 1.59.

    - So [PCl₅] ≈ 1.41 M and [PCl₃] = [Cl₂] ≈ 1.59 M.

  • Problem 6.18. For 0.02 M HF with Ka = 3.2 × 10⁻⁴, solving the quadratic gives α ≈ 0.12.

    • [H₃O⁺] = [F⁻] = 2.4 × 10⁻³ M, so pH ≈ 2.62.

  • Problem 6.26. For A₂X₃ with Ksp = 1.1 × 10⁻²³: Ksp = (2S)²(3S)³ = 108S⁵, so S ≈ 1.0 × 10⁻⁵ mol L⁻¹.

    • Problem 6.28: Ni(OH)₂ in 0.10 M NaOH has S = Ksp/(0.10)² = 2.0 × 10⁻¹³ M, a clear example of the common-ion effect.

    Note on Problem 6.27: the conclusion is correct. Ni(OH)₂ (S ≈ 7.9 × 10⁻⁶ M) is more soluble than AgCN (S ≈ 7.8 × 10⁻⁹ M). The book prints the Ni(OH)₂ solubility as "0.58 × 10⁻⁴", but solving 4S³ = 2.0 × 10⁻¹⁵ actually gives about 7.9 × 10⁻⁶ M.

    Formula / Key-Facts Sheet

  • Kc = [products]^coeff / [reactants]^coeff; Kp = Kc(RT)^Δn
  • Q < K forward; Q > K backward; Q = K equilibrium
  • ΔG⁰ = −RT ln K; ΔG = ΔG⁰ + RT ln Q
  • Kw = 1.0 × 10⁻¹⁴ (298 K); pH = −log[H⁺]; pH + pOH = 14
  • Weak acid: α ≈ √(Ka/c), [H⁺] ≈ √(Ka·c); Ka·Kb = Kw
  • Weak acid + weak base salt: pH = 7 + ½(pKa − pKb)
  • Buffer: pH = pKa + log([A⁻]/[HA])
  • Ksp for AB₂: Ksp = 4S³; for A₂X₃: Ksp = 108S⁵

    • Exam Focus

  • Ice-table problems: write the balanced equation, make a table of initial / change / equilibrium values, substitute into K, and reject the physically impossible root. In Problem 6.4, the root 0.194 is rejected because it would give negative concentrations.
  • Le Chatelier questions: always name the factor, state which side is favoured, and give the reason (fewer gas moles, exothermic direction, and so on). Mention that only temperature changes K.
  • pH questions: watch out for very dilute strong acids. For 1.0 × 10⁻⁸ M HCl you must include the H⁺ from water, which gives pH ≈ 6.98, not 8.
  • Salt hydrolysis: ammonium acetate with pKa = 4.76 and pKb = 4.75 gives pH = 7 + ½(0.01) = 7.005.

    • Key Terms

  • Dynamic equilibrium (गतिक साम्य)
  • Equilibrium constant (साम्य स्थिरांक)
  • Reaction quotient (अभिक्रिया भागफल)
  • Homogeneous equilibrium (समांगी साम्य)
  • Heterogeneous equilibrium (विषमांगी साम्य)
  • Le Chatelier's principle (ले-शातेलिए सिद्धांत)
  • Conjugate acid–base pair (संयुग्मी अम्ल–क्षार युग्म)
  • Lewis acid (लुईस अम्ल)
  • Ionic product of water (जल का आयनिक गुणनफल)
  • pH scale (pH पैमाना)
  • Degree of ionisation (आयनन की मात्रा)
  • Common-ion effect (सम-आयन प्रभाव)
  • Hydrolysis (जल-अपघटन)
  • Buffer solution (उभय-प्रतिरोधी / बफ़र विलयन)
  • Solubility product (विलेयता गुणनफल)

    • Common Mistakes

  • Including pure solids or liquids in K. Leave them out.
  • Getting Δn in Kp = Kc(RT)^Δn wrong. Count only gaseous moles, products minus reactants.
  • Thinking a catalyst or added reactant changes K. Only temperature changes K. A catalyst changes only how fast equilibrium is reached.
  • Assuming inert gas added at constant volume shifts equilibrium. It does not.
  • Forgetting that pH + pOH = 14 only at 298 K.
  • Using α ≈ √(Ka/c) when α is not small. Use the full quadratic instead, as in the HF problem.

    • 💡 Key Learning Points

    • ✓Recognise dynamic equilibrium in physical processes (phase changes, dissolution, Henry's law) and in chemical reactions
    • ✓Write Kc and Kp expressions for homogeneous and heterogeneous equilibria and interconvert them using Kp = Kc(RT)^Δn
    • ✓Use the reaction quotient Q and ICE tables to predict the direction of a reaction and calculate equilibrium concentrations
    • ✓Apply Le Chatelier's principle to changes in concentration, pressure, inert gas, temperature and catalyst, knowing that only temperature changes K
    • ✓Solve ionic-equilibrium problems on pH, weak acids and bases (Ka, Kb), salt hydrolysis, buffers and solubility products (Ksp, common-ion effect)

    👨‍🏫 Teaching Tips

    • →Demonstrate dynamic equilibrium with two measuring cylinders and two different-width tubes transferring water back and forth until the levels stop changing
    • →Heat and cool a sealed tube of NO₂/N₂O₄ (or show the book photograph) so students see the colour shift with temperature
    • →Drill ICE tables as a fixed three-row routine before letting students solve any quadratic
    • →Build a 'what changes K?' checklist on the board: concentration no, pressure no, catalyst no, temperature yes
    • →Use universal indicator on household samples (lemon juice, soap solution, soda water) to connect pH values to real substances
    • →Show buffer action with a quick demo: add a few drops of acid to water and to an acetate buffer and compare the pH change
    • →Practise Ksp relations for AB, AB₂ and A₂X₃ salts as a table on the board (4S³, 108S⁵) and then remove the table for recall

    📋 Assessment Questions

    1. Can the student write correct Kc/Kp expressions, excluding pure solids and liquids?
    2. Can the student decide the direction of a reaction from Q versus K?
    3. Can the student predict and justify the effect of pressure, temperature and a catalyst using Le Chatelier's principle?
    4. Can the student calculate pH for strong acids, weak acids and weak bases, including very dilute solutions?
    5. Can the student relate Ksp to molar solubility and explain the common-ion effect?

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    Keep learning

    All chapters in Chemistry Part I - Class 11

    1. 1. Some Basic Concepts of Chemistry
    2. 2. Structure of Atom
    3. 3. Classification of Elements and Periodicity in Properties
    4. 4. Chemical Bonding and Molecular Structure
    5. 5. Thermodynamics
    6. 6. Equilibrium