Class 11 Notes

Class 11 Chemistry Notes Chapter 7 (Chapter 7) – Examplar Problems (English) Book

Philoid

Philoid

— 10 min read

Examplar Problems (English)
Alright class, let's dive into Chapter 7: Equilibrium. This is a crucial chapter, not just for your Class 11 understanding, but also because concepts from here frequently appear in various government competitive exams. Pay close attention to the principles and problem-solving techniques.

Chapter 7: Equilibrium - Detailed Notes for Competitive Exams

1. Introduction to Equilibrium

  • Equilibrium: A state in a process where observable properties like concentration, pressure, and temperature do not undergo any net change over time. It's a dynamic state where opposing processes occur at equal rates.
  • Types:
    • Physical Equilibrium: Equilibrium involving physical changes (phase transformations).
    • Chemical Equilibrium: Equilibrium involving chemical reactions.

2. Equilibrium in Physical Processes

  • Solid-Liquid Equilibrium: Rate of melting = Rate of freezing (e.g., Ice ⇌ Water at 273 K, 1 atm). At equilibrium, mass of solid and liquid remains constant.
  • Liquid-Vapour Equilibrium: Rate of evaporation = Rate of condensation (e.g., Water ⇌ Water Vapour in a closed container at a constant temperature). Pressure exerted by the vapour at equilibrium is called vapour pressure.
  • Solid-Vapour Equilibrium: Rate of sublimation = Rate of deposition (e.g., Solid Iodine ⇌ Iodine Vapour in a closed vessel).
  • Solid-Solution Equilibrium: Rate of dissolution = Rate of precipitation (e.g., Sugar (solid) ⇌ Sugar (in solution) in a saturated solution).

3. Equilibrium in Chemical Processes – Reversible Reactions

  • Reversible Reactions: Reactions that can proceed in both forward and backward directions under given conditions. Represented by '⇌'.
    • Example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
  • Chemical Equilibrium: The state in a reversible chemical reaction where the rate of the forward reaction equals the rate of the backward reaction. At equilibrium, the concentrations of reactants and products become constant (not necessarily equal).
  • Characteristics of Chemical Equilibrium:
    • Dynamic in nature.
    • Achieved only in closed systems.
    • Concentrations of reactants and products are constant.
    • Can be approached from either direction.
    • A catalyst speeds up attainment but does not alter the equilibrium position.

4. Law of Chemical Equilibrium and Equilibrium Constant

  • Law of Mass Action (Guldberg & Waage): The rate of a chemical reaction is directly proportional to the product of the molar concentrations (or active masses) of the reactants, each raised to the power equal to its stoichiometric coefficient in the balanced chemical equation.
  • For a general reversible reaction: aA + bB ⇌ cC + dD
    • Rate of forward reaction ∝ [A]ᵃ[B]ᵇ = k<0xE1><0xB5><0xA1>[A]ᵃ[B]ᵇ
    • Rate of backward reaction ∝ [C]ᶜ[D]ᵈ = k<0xE1><0xB5><0xAB>[C]ᶜ[D]ᵈ
  • At equilibrium, Rate<0xE1><0xB5><0xA1> = Rate<0xE1><0xB5><0xAB>
    • k<0xE1><0xB5><0xA1>[A]ᵃ[B]ᵇ = k<0xE1><0xB5><0xAB>[C]ᶜ[D]ᵈ
  • Equilibrium Constant (Kc): The ratio of the product of the equilibrium concentrations of products to the product of the equilibrium concentrations of reactants, with each concentration term raised to the power of its stoichiometric coefficient.
    • Kc = (k<0xE1><0xB5><0xA1>/k<0xE1><0xB5><0xAB>) = ([C]ᶜ[D]ᵈ) / ([A]ᵃ[B]ᵇ) (Concentrations in mol L⁻¹)
  • Equilibrium Constant (Kp): For gaseous reactions, equilibrium constant expressed in terms of partial pressures.
    • Kp = (P<0xE1><0xB5><0x9C>ᶜ * P<0xE1><0xB5><0x9D>ᵈ) / (P<0xE1><0xB5><0x9A>ᵃ * P<0xE1><0xB5><0x9B>ᵇ) (Partial pressures in bar or atm)
  • Relationship between Kp and Kc:
    • Kp = Kc (RT)Δng
    • Where:
      • R = Universal Gas Constant (0.0831 L bar K⁻¹ mol⁻¹ or 0.0821 L atm K⁻¹ mol⁻¹)
      • T = Absolute Temperature (in Kelvin)
      • Δng = (Sum of stoichiometric coefficients of gaseous products) - (Sum of stoichiometric coefficients of gaseous reactants)
    • If Δng = 0, Kp = Kc.
    • If Δng > 0, Kp > Kc.
    • If Δng < 0, Kp < Kc.

5. Homogeneous and Heterogeneous Equilibria

  • Homogeneous Equilibrium: All reactants and products are in the same phase (e.g., N₂(g) + 3H₂(g) ⇌ 2NH₃(g)).
  • Heterogeneous Equilibrium: Reactants and products are in different phases (e.g., CaCO₃(s) ⇌ CaO(s) + CO₂(g)).
    • Important Note: Concentrations (or partial pressures) of pure solids and pure liquids are considered constant (taken as 1) and do not appear in the equilibrium constant expression.
    • For CaCO₃(s) ⇌ CaO(s) + CO₂(g): Kc = [CO₂(g)] and Kp = P<0xE1><0xB5><0x9C><0xE1><0xB5><0x8C>₂

6. Applications of Equilibrium Constant

  • Predicting the Extent of Reaction:
    • High Kc (> 10³) : Reaction proceeds nearly to completion (products dominate).
    • Low Kc (< 10⁻³) : Reaction hardly proceeds (reactants dominate).
    • Intermediate Kc (10⁻³ to 10³) : Appreciable concentrations of both reactants and products exist.
  • Predicting the Direction of Reaction: Using the Reaction Quotient (Qc or Qp). Calculated using non-equilibrium concentrations/pressures in the same way as Kc/Kp.
    • If Qc < Kc (or Qp < Kp): Reaction proceeds in the forward direction.
    • If Qc > Kc (or Qp > Kp): Reaction proceeds in the backward direction.
    • If Qc = Kc (or Qp = Kp): The reaction is at equilibrium.
  • Calculating Equilibrium Concentrations: Using the ICE (Initial, Change, Equilibrium) table method.

7. Factors Affecting Equilibria – Le Chatelier’s Principle

  • Statement: If a change of condition (concentration, pressure, temperature) is applied to a system in equilibrium, the system will shift in a direction that tends to counteract the effect of the change.
  • Effect of Concentration Change:
    • Increase reactant conc.: Shifts forward.
    • Increase product conc.: Shifts backward.
    • Decrease reactant conc.: Shifts backward.
    • Decrease product conc.: Shifts forward.
  • Effect of Pressure Change (for gaseous reactions with Δng ≠ 0):
    • Increase pressure (by decreasing volume): Shifts towards the side with fewer moles of gas.
    • Decrease pressure (by increasing volume): Shifts towards the side with more moles of gas.
    • If Δng = 0, pressure change has no effect.
  • Effect of Inert Gas Addition:
    • At Constant Volume: No effect on equilibrium (partial pressures/concentrations don't change).
    • At Constant Pressure: Volume increases, partial pressures decrease. Shifts towards the side with more moles of gas.
  • Effect of Temperature Change:
    • Increase Temperature: Favours the endothermic reaction (ΔH > 0).
    • Decrease Temperature: Favours the exothermic reaction (ΔH < 0).
    • Note: Temperature change alters the value of the equilibrium constant (Kc/Kp). For endothermic reactions, K increases with T. For exothermic reactions, K decreases with T.
  • Effect of Catalyst:
    • Increases the rate of both forward and backward reactions equally.
    • Helps attain equilibrium faster.
    • Does not change the equilibrium constant or the equilibrium position.

8. Ionic Equilibrium in Solution

  • Electrolytes: Substances that conduct electricity in molten state or aqueous solution (e.g., acids, bases, salts).
    • Strong Electrolytes: Ionize almost completely (e.g., HCl, NaOH, NaCl).
    • Weak Electrolytes: Ionize partially (e.g., CH₃COOH, NH₄OH).
  • Acids, Bases and Salts:
    • Arrhenius Concept: Acids give H⁺ ions in water; Bases give OH⁻ ions in water.
    • Brønsted-Lowry Concept: Acids are proton (H⁺) donors; Bases are proton acceptors.
      • Conjugate Acid-Base Pair: An acid and a base that differ only by a proton (e.g., HCl/Cl⁻, NH₄⁺/NH₃, H₂O/OH⁻, H₃O⁺/H₂O).
      • Amphoteric/Amphiprotic: Substances that can act as both acid and base (e.g., H₂O, HCO₃⁻).
    • Lewis Concept: Acids are electron pair acceptors; Bases are electron pair donors (e.g., Acid: BF₃, AlCl₃, H⁺; Base: NH₃, H₂O, OH⁻).

9. Ionization of Acids and Bases

  • Ionization Constant of Weak Acid (Ka): For HA ⇌ H⁺ + A⁻
    • Ka = [H⁺][A⁻] / [HA]
    • Stronger acid → Larger Ka → Smaller pKa (pKa = -log Ka)
  • Ionization Constant of Weak Base (Kb): For BOH ⇌ B⁺ + OH⁻ or B + H₂O ⇌ BH⁺ + OH⁻
    • Kb = [B⁺][OH⁻] / [BOH] or Kb = [BH⁺][OH⁻] / [B]
    • Stronger base → Larger Kb → Smaller pKb (pKb = -log Kb)
  • Relationship between Ka and Kb for a Conjugate Pair: Ka × Kb = Kw

10. Ionization of Water and its Ionic Product (Kw)

  • Water undergoes auto-ionization: H₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)
  • Ionic Product of Water (Kw): Kw = [H₃O⁺][OH⁻]
    • At 298 K (25 °C), Kw = 1.0 × 10⁻¹⁴.
    • Kw increases with temperature (as ionization is endothermic).
    • In pure water at 298 K, [H₃O⁺] = [OH⁻] = 1.0 × 10⁻⁷ M.

11. pH Scale

  • A logarithmic scale to express the acidity or basicity of a solution.
  • pH = -log[H₃O⁺] or pH = -log[H⁺]
  • pOH = -log[OH⁻]
  • Relationship: pH + pOH = pKw (= 14 at 298 K)
    • Acidic solution: [H⁺] > 10⁻⁷ M, pH < 7
    • Neutral solution: [H⁺] = 10⁻⁷ M, pH = 7
    • Basic solution: [H⁺] < 10⁻⁷ M, pH > 7

12. Common Ion Effect

  • The suppression of the ionization of a weak electrolyte by the addition of a strong electrolyte containing a common ion.
  • Example: Ionization of CH₃COOH decreases upon adding CH₃COONa. Ionization of NH₄OH decreases upon adding NH₄Cl.
  • Application: Precipitation of salts, buffer solutions.

13. Hydrolysis of Salts

  • The reaction of the cation or anion (or both) of a salt with water to produce an acidic or basic solution.
  • Salt of Strong Acid & Strong Base (e.g., NaCl): No hydrolysis, pH = 7 (Neutral).
  • Salt of Strong Acid & Weak Base (e.g., NH₄Cl): Cationic hydrolysis, pH < 7 (Acidic). NH₄⁺ + H₂O ⇌ NH₄OH + H⁺
  • Salt of Weak Acid & Strong Base (e.g., CH₃COONa): Anionic hydrolysis, pH > 7 (Basic). CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻
  • Salt of Weak Acid & Weak Base (e.g., CH₃COONH₄): Both ions hydrolyse. pH depends on relative strengths (Ka and Kb).
    • If Ka = Kb, pH ≈ 7.
    • If Ka > Kb, pH < 7.
    • If Kb > Ka, pH > 7.
  • Hydrolysis Constant (Kh): Related to Kw, Ka, Kb.
    • For NH₄Cl: Kh = Kw / Kb
    • For CH₃COONa: Kh = Kw / Ka
    • For CH₃COONH₄: Kh = Kw / (Ka × Kb)

14. Buffer Solutions

  • Solutions that resist changes in pH upon addition of small amounts of acid or base, or upon dilution.
  • Types:
    • Acidic Buffer: Weak acid + its salt with a strong base (e.g., CH₃COOH + CH₃COONa). pH < 7.
    • Basic Buffer: Weak base + its salt with a strong acid (e.g., NH₄OH + NH₄Cl). pH > 7.
  • Mechanism: The added H⁺ or OH⁻ ions are consumed by the buffer components.
  • Henderson-Hasselbalch Equation:
    • For Acidic Buffer: pH = pKa + log ([Salt]/[Acid])
    • For Basic Buffer: pOH = pKb + log ([Salt]/[Base])
  • Buffer Capacity: Ability to resist pH change. Maximum when [Salt] = [Acid] or [Salt] = [Base], i.e., pH = pKa or pOH = pKb.

15. Solubility Equilibria of Sparingly Soluble Salts

  • Applies to salts that dissolve only slightly in water (e.g., AgCl, BaSO₄, PbI₂).
  • Equilibrium exists between the undissolved solid salt and its ions in a saturated solution.
    • Example: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
  • Solubility Product Constant (Ksp): The product of the molar concentrations of the constituent ions in a saturated solution, each raised to the power of its stoichiometric coefficient in the equilibrium equation.
    • For AxBy(s) ⇌ xAʸ⁺(aq) + yBˣ⁻(aq)
    • Ksp = [Aʸ⁺]ˣ [Bˣ⁻]ʸ
  • Solubility (s): Molar concentration of the salt dissolved in a saturated solution (mol L⁻¹).
  • Relationship between Ksp and s:
    • For AgCl (1:1 salt): Ksp = [Ag⁺][Cl⁻] = s * s = s² => s = √Ksp
    • For CaF₂ (1:2 salt): Ksp = [Ca²⁺][F⁻]² = s * (2s)² = 4s³ => s = ³√(Ksp/4)
    • For Al(OH)₃ (1:3 salt): Ksp = [Al³⁺][OH⁻]³ = s * (3s)³ = 27s⁴ => s = ⁴√(Ksp/27)
  • Common Ion Effect on Solubility: Solubility of a sparingly soluble salt decreases in the presence of a common ion.
  • Condition for Precipitation: If Ionic Product (Qsp) > Ksp, precipitation occurs until Qsp = Ksp.
    • Qsp is calculated using initial concentrations before equilibrium is reached.
    • If Qsp < Ksp: Solution is unsaturated, more salt can dissolve.
    • If Qsp = Ksp: Solution is saturated, equilibrium exists.

Multiple Choice Questions (MCQs)

  1. For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the relationship between Kp and Kc is:
    a) Kp = Kc
    b) Kp = Kc (RT)⁻²
    c) Kp = Kc (RT)²
    d) Kp = Kc (RT)⁻¹

  2. In which of the following equilibria will Kp be equal to Kc?
    a) H₂(g) + I₂(g) ⇌ 2HI(g)
    b) PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
    c) 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
    d) N₂O₄(g) ⇌ 2NO₂(g)

  3. According to Le Chatelier's principle, adding heat to a solid-liquid equilibrium will cause the:
    a) Amount of solid to decrease
    b) Amount of liquid to decrease
    c) Temperature to rise
    d) Temperature to fall

  4. Which of the following is a Lewis acid but not a Brønsted-Lowry acid?
    a) HCl
    b) H₂SO₄
    c) BF₃
    d) NH₄⁺

  5. The pH of a 0.01 M NaOH solution at 298 K is:
    a) 2
    b) 12
    c) 7
    d) 1

  6. The conjugate base of H₂PO₄⁻ is:
    a) H₃PO₄
    b) HPO₄²⁻
    c) PO₄³⁻
    d) H₂O

  7. The solubility product (Ksp) of AgCl is 1.8 × 10⁻¹⁰ at 298 K. Its molar solubility (in mol L⁻¹) is:
    a) 1.8 × 10⁻¹⁰
    b) √(1.8 × 10⁻¹⁰)
    c) 1.8 × 10⁻⁵
    d) (1.8 × 10⁻¹⁰)²

  8. For the reaction CaCO₃(s) ⇌ CaO(s) + CO₂(g), the equilibrium constant Kp is given by:
    a) P<0xE1><0xB5><0x9C><0xE1><0xB5><0x8C>₂
    b) [CaO][CO₂] / [CaCO₃]
    c) P<0xE1><0xB5><0x9C><0xE1><0xB5><0x8C>₂ / P<0xE1><0xB5><0x9C><0xE1><0xB5><0x8C><0xE1><0xB5><0x8C>₃
    d) P<0xE1><0xB5><0x9C><0xE1><0xB5><0x8C>₂ * P<0xE1><0xB5><0x9C><0xE1><0xB5><0x8C><0xE1><0xB5><0x8C>₃ / P<0xE1><0xB5><0x9C><0xE1><0xB5><0x8C><0xE1><0xB5><0x8C>₃

  9. Which of the following solutions will act as a buffer?
    a) HCl + NaCl
    b) NaOH + NaCl
    c) CH₃COOH + CH₃COONa
    d) H₂SO₄ + Na₂SO₄

  10. What happens to the pH of pure water when the temperature increases? (Given: Autoionization of water is endothermic)
    a) pH increases
    b) pH decreases
    c) pH remains 7
    d) Cannot be predicted

Answers to MCQs:

  1. b) Kp = Kc (RT)⁻² (Δng = 2 - (1+3) = -2)
  2. a) H₂(g) + I₂(g) ⇌ 2HI(g) (Δng = 2 - (1+1) = 0)
  3. a) Amount of solid to decrease (Melting is usually endothermic, adding heat shifts equilibrium towards liquid phase)
  4. c) BF₃ (Accepts electron pair, but has no proton to donate)
  5. b) 12 ([OH⁻] = 0.01 M = 10⁻² M => pOH = 2 => pH = 14 - 2 = 12)
  6. b) HPO₄²⁻ (Loses one H⁺)
  7. b) √(1.8 × 10⁻¹⁰) (For AgCl, Ksp = s², so s = √Ksp)
  8. a) P<0xE1><0xB5><0x9C><0xE1><0xB5><0x8C>₂ (Pure solids are not included in the Kp expression)
  9. c) CH₃COOH + CH₃COONa (Weak acid and its salt with a strong base)
  10. b) pH decreases (Endothermic process, increasing T increases Kw. Since Kw = [H⁺][OH⁻], [H⁺] increases, thus pH decreases below 7. Water remains neutral as [H⁺] still equals [OH⁻], but the neutral pH value itself changes.)

Make sure you understand the reasoning behind each answer. Go through these notes thoroughly, practice numerical problems, especially those involving Kp/Kc relationships, pH calculations, buffers, and solubility products. Good luck with your preparation!

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