Class 11 Chemistry Notes Chapter 7 (Equilibrium) – Chemistry Part-I Book

Alright students, let's begin our detailed discussion on Chapter 7, 'Equilibrium'. This is a crucial chapter, not just for your Class 11 understanding, but also forms the basis for many concepts tested in various government examinations. Pay close attention to the definitions, principles, and formulas.

Chapter 7: Equilibrium - Detailed Notes for Government Exam Preparation

1. Introduction to Equilibrium

• Reversible Reactions: Reactions that can proceed in both forward and backward directions under given conditions. Represented by a double arrow (⇌).
  • Example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
• Irreversible Reactions: Reactions that proceed essentially to completion in only one direction. Represented by a single arrow (→).
  • Example: AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)
• Equilibrium: A state in a reversible process where the rates of the forward and reverse processes become equal. At equilibrium, the observable properties of the system (like concentration, pressure, temperature, colour) become constant over time.
• Dynamic Nature: Equilibrium is dynamic, meaning both forward and reverse reactions continue to occur, but at equal rates, resulting in no net change in the composition of the reaction mixture.

2. Equilibrium in Physical Processes

• Solid-Liquid Equilibrium: Rate of melting = Rate of freezing. Occurs at the melting point (at constant pressure).
  • Example: Ice ⇌ Water at 273 K (0 °C) and 1 atm.
• Liquid-Vapour Equilibrium: Rate of vaporization = Rate of condensation. Occurs in a closed container at a constant temperature. The pressure exerted by the vapour at equilibrium is called the vapour pressure. Occurs at the boiling point when external pressure equals vapour pressure.
  • Example: H₂O(l) ⇌ H₂O(g) at 373 K (100 °C) and 1 atm.
• Solid-Vapour Equilibrium: Rate of sublimation = Rate of deposition. Occurs in a closed container.
  • Example: I₂(s) ⇌ I₂(g)
• Solid-Solution Equilibrium: Rate of dissolution = Rate of precipitation. Occurs when a solution is saturated.
  • Example: Sugar(s) ⇌ Sugar(solution) in a saturated solution.
• Gas-Solution Equilibrium: Rate of gas dissolving = Rate of gas escaping. Governed by Henry's Law (The mass/mole fraction of a gas dissolved in a liquid is directly proportional to the partial pressure of the gas above the liquid).
  • Example: CO₂(g) ⇌ CO₂(aq) in soda water.

3. Equilibrium in Chemical Processes - Law of Chemical Equilibrium

• Characteristics of Chemical Equilibrium:
  • Achieved only in closed systems.
  • Dynamic in nature.
  • Concentrations of reactants and products become constant.
  • Can be approached from either direction.
  • A catalyst speeds up attainment but does not change the equilibrium position.
• Law of Mass Action (Guldberg & Waage): The rate of a chemical reaction is directly proportional to the product of the active masses (molar concentrations) of the reacting substances, raised to the power of their stoichiometric coefficients in the balanced chemical equation.
• Equilibrium Constant (K): For a general reversible reaction:
  aA + bB ⇌ cC + dD
  • Kc (in terms of Molar Concentration):
    Kc = ([C]^c [D]^d) / ([A]^a [B]^b)
    Units of Kc depend on Δn = (c+d) - (a+b). If Δn = 0, Kc is unitless.
  • Kp (in terms of Partial Pressure - for gaseous reactions):
    Kp = (P_C^c * P_D^d) / (P_A^a * P_B^b)
    Where P represents the partial pressure of the respective gas at equilibrium.
    Units of Kp depend on Δn. If Δn = 0, Kp is unitless.
• Relationship between Kp and Kc:
  Kp = Kc (RT)^Δn
  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)
  Δn = (Sum of stoichiometric coefficients of gaseous products) - (Sum of stoichiometric coefficients of gaseous reactants)
• Homogeneous Equilibrium: All reactants and products are in the same phase (e.g., all gases or all aqueous).
  • Example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
• Heterogeneous Equilibrium: Reactants and products are in different phases.
  • Example: 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] and Kp = P_CO₂

4. Applications of Equilibrium Constant

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

5. Factors Affecting Equilibrium - Le Chatelier's Principle

• Statement: If a change of condition (like concentration, pressure, or 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 concentration: Equilibrium shifts forward.
  • Increase product concentration: Equilibrium shifts backward.
  • Decrease reactant concentration: Equilibrium shifts backward.
  • Decrease product concentration: Equilibrium shifts forward.
• Effect of Pressure Change (Only for gaseous reactions where Δn ≠ 0):
  • Increase pressure (by decreasing volume): Equilibrium shifts in the direction with fewer moles of gas (smaller Δn).
  • Decrease pressure (by increasing volume): Equilibrium shifts in the direction with more moles of gas (larger Δn).
  • If Δn = 0, pressure change has no effect on the equilibrium position.
• Effect of Inert Gas Addition:
  • At Constant Volume: No effect on equilibrium (partial pressures/concentrations of reactants/products remain unchanged).
  • At Constant Pressure: Volume increases. Equilibrium shifts in the direction with more moles of gas (similar to decreasing pressure).
• Effect of Temperature Change:
  • Exothermic Reaction (ΔH = -ve): Heat is a product. Increasing temperature shifts equilibrium backward. Decreasing temperature shifts equilibrium forward. K decreases with increasing T.
  • Endothermic Reaction (ΔH = +ve): Heat is a reactant. Increasing temperature shifts equilibrium forward. Decreasing temperature shifts equilibrium backward. K increases with increasing T.
  • The value of the equilibrium constant (K) changes only with temperature.
• Effect of Catalyst:
  • A catalyst increases the rates of both forward and reverse reactions equally.
  • It helps attain equilibrium faster but does not change the equilibrium constant (K) or the equilibrium position.

6. Ionic Equilibrium in Solution

• Electrolytes: Substances that conduct electricity in molten state or aqueous solution.
  • Strong Electrolytes: Ionize completely (e.g., HCl, NaOH, NaCl).
  • Weak Electrolytes: Ionize partially (e.g., CH₃COOH, NH₄OH, H₂O).
• Acids, Bases, and Salts:
  • Arrhenius Concept: Acids give H⁺ ions in water; Bases give OH⁻ ions in water.
  • Bronsted-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).
  • Lewis Concept: Acids are electron pair acceptors; Bases are electron pair donors (e.g., Acid: BF₃, AlCl₃, H⁺; Base: NH₃, H₂O, OH⁻).
• Ionization of Acids and Bases:
  • Ionization Constant of Weak Acid (Ka): For HA ⇌ H⁺ + A⁻, Ka = [H⁺][A⁻] / [HA]
  • Ionization Constant of Weak Base (Kb): For BOH ⇌ B⁺ + OH⁻, Kb = [B⁺][OH⁻] / [BOH] or for B + H₂O ⇌ BH⁺ + OH⁻, Kb = [BH⁺][OH⁻] / [B]
  • pKa and pKb: pKa = -log(Ka), pKb = -log(Kb). Stronger acid → Larger Ka → Smaller pKa. Stronger base → Larger Kb → Smaller pKb.
• Ionic Product of Water (Kw): H₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)
  • Kw = [H₃O⁺][OH⁻] = 1.0 x 10⁻¹⁴ at 298 K (25 °C).
  • Kw increases with temperature (ionization of water is endothermic).
  • pKw = -log(Kw) = 14 at 298 K.
  • pKa + pKb = pKw = 14 (for a conjugate acid-base pair at 298 K).
• pH Scale: pH = -log[H₃O⁺] or -log[H⁺]
  • pOH = -log[OH⁻]
  • pH + pOH = pKw = 14 (at 298 K)
  • pH < 7: Acidic; pH > 7: Basic; pH = 7: Neutral (at 298 K).
• Ionization of Weak Acids/Bases:
  • For weak acid HA: [H⁺] ≈ √(Ka * C) ; Degree of ionization (α) = √(Ka / C) = [H⁺]/C
  • For weak base BOH: [OH⁻] ≈ √(Kb * C) ; Degree of ionization (α) = √(Kb / C) = [OH⁻]/C
  • (These approximations hold when α is small, usually < 5-10%)
• 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.
• Hydrolysis of Salts: Reaction of cation or anion (or both) of a salt with water to produce acidity or alkalinity.
  • Salt of Strong Acid + Strong Base (e.g., NaCl): No hydrolysis. Solution is neutral (pH = 7).
  • Salt of Weak Acid + Strong Base (e.g., CH₃COONa): Anionic hydrolysis. Solution is basic (pH > 7).
    pH = 7 + ½ (pKa + log C) ; Hydrolysis constant Kh = Kw / Ka
  • Salt of Strong Acid + Weak Base (e.g., NH₄Cl): Cationic hydrolysis. Solution is acidic (pH < 7).
    pH = 7 - ½ (pKb + log C) ; Hydrolysis constant Kh = Kw / Kb
  • Salt of Weak Acid + Weak Base (e.g., CH₃COONH₄): Both ions hydrolyze. Solution can be acidic, basic, or neutral depending on Ka and Kb.
    pH = 7 + ½ (pKa - pKb) ; Hydrolysis constant Kh = Kw / (Ka * Kb)
    (Note: pH is independent of concentration for this type).
• Buffer Solutions: Solutions that resist changes in pH upon addition of small amounts of acid or base, or upon dilution.
  • Acidic Buffer: Weak acid + its salt with a strong base (e.g., CH₃COOH + CH₃COONa).
    Henderson-Hasselbalch Equation: pH = pKa + log ([Salt]/[Acid])
  • Basic Buffer: Weak base + its salt with a strong acid (e.g., NH₄OH + NH₄Cl).
    Henderson-Hasselbalch Equation: pOH = pKb + log ([Salt]/[Base])
  • Buffer Action: Explained by the common ion effect and neutralization of added H⁺ or OH⁻ ions.
• Solubility Equilibria: For sparingly soluble salts (e.g., AgCl, BaSO₄).
  • Solubility Product (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) ⇌ xA^(y+)(aq) + yB^(x-)(aq)
    Ksp = [A^(y+)]x [B^(x-)]y
  • Solubility (s): Molar concentration of the salt dissolved in a saturated solution.
    Relationship between Ksp and s depends on stoichiometry:
    • AgCl (1:1): Ksp = s² => s = √Ksp
    • CaF₂ (1:2): Ksp = (s)(2s)² = 4s³ => s = ³√(Ksp/4)
    • Al(OH)₃ (1:3): Ksp = (s)(3s)³ = 27s⁴ => s = ⁴√(Ksp/27)
  • Ionic Product (Qsp): Calculated like Ksp but with initial or non-equilibrium ion concentrations.
    • If Qsp < Ksp: Solution is unsaturated, more salt can dissolve. No precipitation.
    • If Qsp > Ksp: Solution is supersaturated. Precipitation occurs until Qsp = Ksp.
    • If Qsp = Ksp: Solution is saturated. Equilibrium exists.
  • Common Ion Effect on Solubility: Solubility of a sparingly soluble salt decreases in the presence of a common ion.

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. Which of the following conditions will favour the forward reaction in the equilibrium: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g); ΔH = -198 kJ?
   a) Increasing temperature and decreasing pressure
   b) Decreasing temperature and increasing pressure
   c) Increasing temperature and increasing pressure
   d) Decreasing temperature and decreasing pressure

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

4. The conjugate base of H₂PO₄⁻ is:
   a) H₃PO₄
   b) HPO₄²⁻
   c) PO₄³⁻
   d) H₃O⁺

5. Which of the following is a Lewis acid?
   a) NH₃
   b) H₂O
   c) BF₃
   d) OH⁻

6. The pH of a 0.01 M NaOH solution is:
   a) 2
   b) 12
   c) 1
   d) 13

7. A buffer solution is prepared by mixing equal concentrations of a weak acid HA and its salt NaA. If the pKa of the acid is 4.8, the pH of the buffer is:
   a) 4.8
   b) 7.0
   c) 9.2
   d) 2.4

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

9. The hydrolysis of sodium acetate (CH₃COONa) in water produces:
   a) An acidic solution
   b) A basic solution
   c) A neutral solution
   d) A buffer solution

10. What happens to the solubility of AgCl in an aqueous solution when NaCl is added?
    a) Increases
    b) Decreases
    c) Remains unchanged
    d) First increases then decreases

Answer Key for MCQs:

1. b) Kp = Kc (RT)⁻² (Δn = 2 - (1+3) = -2)
2. b) Decreasing temperature (exothermic) and increasing pressure (fewer moles of gas on product side).
3. a) Amount of solid to decrease (Melting is usually endothermic, adding heat shifts equilibrium towards liquid).
4. b) HPO₄²⁻ (Loses one H⁺).
5. c) BF₃ (Boron has an incomplete octet, can accept an electron pair).
6. b) 12 ([OH⁻] = 0.01 M = 10⁻² M; pOH = -log(10⁻²) = 2; pH = 14 - pOH = 14 - 2 = 12).
7. a) 4.8 (pH = pKa + log([Salt]/[Acid]); Since [Salt] = [Acid], log(1) = 0, so pH = pKa).
8. a) √(1.8 x 10⁻¹⁰) (For AgCl, Ksp = s², so s = √Ksp).
9. b) A basic solution (Salt of weak acid CH₃COOH and strong base NaOH; anionic hydrolysis produces OH⁻ ions).
10. b) Decreases (Due to the common ion effect of Cl⁻ ions from NaCl).

Make sure you understand the underlying principles behind each point and MCQ answer. Revise the formulas and practice numerical problems, especially on Kp/Kc, pH calculations, buffers, and solubility products. Good luck with your preparation!