Equilibrium
'All chemical reactions are reversible in principle. Equilibrium is the state where forward and reverse reactions balance.' — Chemical Equilibrium
1. Chapter Overview
EQUILIBRIUM is a state where FORWARD and REVERSE reaction rates are EQUAL, and concentrations of reactants and products remain CONSTANT. This chapter covers PHYSICAL EQUILIBRIUM (solid-liquid, liquid-vapour), CHEMICAL EQUILIBRIUM (Kc, Kp), LE CHATELIER'S PRINCIPLE (predicting how equilibrium shifts), ACID-BASE EQUILIBRIA (pH, Ka, Kb), BUFFER SOLUTIONS, and the SOLUBILITY PRODUCT principle.
2. Equilibrium in Physical Processes
Solid-Liquid Equilibrium
- At melting point: Rate of melting = Rate of freezing
- Dynamic equilibrium: molecules constantly exchange between phases
Liquid-Vapour Equilibrium
- Vapour pressure: Pressure of vapour in EQUILIBRIUM with liquid
- Increases with temperature
- Boiling point: T at which vapour pressure = external pressure
Factors Affecting Vapour Pressure
- Nature of liquid (intermolecular forces)
- Temperature (↑ T → ↑ VP)
3. Law of Chemical Equilibrium
For a General Reaction
- aA + bB ⇌ cC + dD
- Equilibrium Constant: K_c = [C]ᶜ[D]ᵈ/[A]ᵃ[B]ᵇ
- K_p = (P_C)ᶜ(P_D)ᵈ/(P_A)ᵃ(P_B)ᵇ
Relationship between Kp and Kc
- K_p = K_c(RT)^Δn (Δn = moles of gaseous products — moles of gaseous reactants)
Characteristics of Equilibrium Constant
- K is CONSTANT at a given temperature
- K depends on the FORM of the equation (reverse: K' = 1/K)
- K does NOT depend on initial concentrations or catalyst
Reaction Quotient (Q)
- Q has the SAME expression as K, but with INITIAL (non-equilibrium) concentrations
- If Q < K: Forward reaction proceeds
- If Q > K: Reverse reaction proceeds
- If Q = K: System at equilibrium
Worked Problem
Q: For H₂ + I₂ ⇌ 2HI, K_c = 54.8 at 425°C. If 0.5 M H₂ and 0.5 M I₂ are mixed, find [HI] at equilibrium. A: 54.8 = (2x)²/(0.5-x)(0.5-x) = 4x²/(0.5-x)². Taking square root: 7.4 = 2x/(0.5-x). x = 0.393. [HI] = 0.786 M.
4. Le Chatelier's Principle
- Statement: If a system at equilibrium is DISTURBED, the system SHIFTS in a direction that COUNTERACTS the disturbance
Effect of Concentration
- ↑ [reactant]: Equilibrium shifts FORWARD
- ↑ [product]: Equilibrium shifts BACKWARD
Effect of Pressure (for Gaseous Reactions)
- ↑ Pressure: Shift toward SIDE with FEWER gas molecules
- No change if Δn_g = 0
- Example: N₂ + 3H₂ ⇌ 2NH₃ (4 → 2 moles), ↑ P → forward shift (more NH₃)
Effect of Temperature
- ↑ T: Shift in ENDOTHERMIC direction
- ↑ T for exothermic forward reaction → shift BACKWARD
- ↑ T for endothermic forward reaction → shift FORWARD
Effect of Catalyst
- Speeds up BOTH forward and reverse reactions EQUALLY
- Does NOT change equilibrium position or K value
- Equilibrium is reached FASTER
Application: Haber Process for NH₃
- N₂ + 3H₂ ⇌ 2NH₃, ΔH = -92.4 kJ (exothermic)
- Optimal conditions: HIGH pressure (200 atm), MODERATE temperature (450°C), Fe catalyst
5. Acid-Base Equilibria
Arrhenius Concept
- Acid: Produces H⁺ in water
- Base: Produces OH⁻ in water
Bronsted-Lowry Concept
- Acid: PROTON (H⁺) donor
- Base: PROTON acceptor
- Conjugate acid-base pair: Differ by ONE H⁺
- Example: HCl (acid) + H₂O (base) ⇌ H₃O⁺ (conjugate acid) + Cl⁻ (conjugate base)
Lewis Concept
- Acid: Electron PAIR acceptor
- Base: Electron PAIR donor
Ionisation of Weak Acids/Bases
- K_a = [H⁺][A⁻]/[HA]; K_b = [B⁺][OH⁻]/[BOH]
- pK_a = -log K_a; pK_b = -log K_b
- For a conjugate pair: K_a × K_b = K_w
6. Ionisation of Water and pH
Self-Ionisation of Water
- H₂O + H₂O ⇌ H₃O⁺ + OH⁻
- K_w = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C
- In neutral water: [H⁺] = [OH⁻] = 10⁻⁷ M
pH Scale
- pH = -log[H⁺]; pOH = -log[OH⁻]
- pH + pOH = 14
- pH < 7: Acidic; pH = 7: Neutral; pH > 7: Basic
Worked Problem
Q: Find pH of 0.001 M HCl and 0.001 M NaOH. A: HCl (strong acid): [H⁺] = 0.001 M. pH = -log(10⁻³) = 3. NaOH (strong base): [OH⁻] = 0.001 M. pOH = 3. pH = 14-3 = 11.
7. Buffer Solutions
- Definition: Solutions that RESIST change in pH when small amounts of acid or base are added
- Types:
- Acidic buffer: Weak acid + its salt (CH₃COOH + CH₃COONa)
- Basic buffer: Weak base + its salt (NH₄OH + NH₄Cl)
Henderson-Hasselbalch Equation
- For acidic buffer: pH = pK_a + log([Salt]/[Acid])
- For basic buffer: pOH = pK_b + log([Salt]/[Base])
Buffer Capacity
- Depends on the ratio [Salt]/[Acid] and absolute concentrations
- Maximum when [Salt] = [Acid] → pH = pK_a
Worked Problem
Q: Find pH of a buffer containing 0.1 M CH₃COOH and 0.2 M CH₃COONa (pK_a = 4.74). A: pH = 4.74 + log(0.2/0.1) = 4.74 + log(2) = 4.74 + 0.30 = 5.04.
8. Solubility Product (K_sp)
- Definition: EQUILIBRIUM constant for a SPARINGLY SOLUBLE salt
- For AₓBᵧ ⇌ xA^y⁺ + yB^x⁻: K_sp = [A^y⁺]ˣ[B^x⁻]ʸ
Predicting Precipitation
- Ionic Product (IP): Q calculated from actual concentrations
- IP < K_sp: Unsaturated solution (NO precipitate)
- IP = K_sp: SATURATED solution
- IP > K_sp: SUPERSATURATED (PRECIPITATE forms)
Common Ion Effect
- Addition of a common ion DECREASES solubility of a salt
- Example: Adding NaCl to AgCl solution → [Cl⁻] ↑ → AgCl precipitates MORE
Worked Problem
Q: K_sp of AgCl is 1.6 × 10⁻¹⁰. Find solubility in water and in 0.1 M NaCl. A: In water: s = √K_sp = √(1.6×10⁻¹⁰) = 1.26 × 10⁻⁵ M. In 0.1 M NaCl: s = K_sp/[Cl⁻] = 1.6×10⁻¹⁰/0.1 = 1.6 × 10⁻⁹ M (much LOWER).
9. Common Mistakes
- Equilibrium ≠ equal concentrations: It means EQUAL RATES, not equal concentrations. Products and reactants can have VERY different concentrations at equilibrium
- Pure solids and liquids are OMITTED from K expression: Their concentration is constant
- K changes with TEMPERATURE, NOT with concentration/pressure: Only T changes K
- pH of a weak acid is NOT the same as its concentration: For weak acid, pH > -log[HA]
- Buffer capacity is MAXIMUM when pH = pK_a: Not at any random ratio
10. CBSE Exam Focus
- Equilibrium constant expression — Kc and Kp problems (3/5-mark)
- Le Chatelier's principle — applications (3/5-mark)
- pH calculations for strong/weak acids and bases (3-mark)
- Buffer solution — Henderson-Hasselbalch equation (5-mark)
- Solubility product — precipitation predictions (5-mark)
- Relationship between Kp and Kc
11. Key Formulas
- K_c = [C]ᶜ[D]ᵈ/[A]ᵃ[B]ᵇ
- K_p = K_c(RT)^Δn
- pH + pOH = 14
- pH = pK_a + log([Salt]/[Acid]) (buffer)
- K_sp = solubility product
- IP = ionic product (compare with K_sp)
12. Self-Test (5+ Q&A)
Q1: For the reaction 2SO₂ + O₂ ⇌ 2SO₃, K_p = 0.0041 at 1000 K. Express K_c. (R = 0.083 L·bar/mol·K) A: Δn = 2 — 3 = -1. K_p = K_c(RT)^Δn → K_c = K_p/(RT)^(-1) = K_p × RT = 0.0041 × 0.083 × 1000 = 0.34.
Q2: What is the pH of a 0.1 M CH₃COOH solution? (K_a = 1.8 × 10⁻⁵) A: [H⁺] = √(K_a × C) = √(1.8×10⁻⁵ × 0.1) = √(1.8×10⁻⁶) = 1.34 × 10⁻³. pH = -log(1.34×10⁻³) = 2.87.
Q3: Predict the effect of increasing temperature on N₂ + 3H₂ ⇌ 2NH₃ (ΔH = -92.4 kJ). A: ΔH negative → forward is exothermic. ↑ T favours ENDOTHERMIC direction (backward). LESS NH₃ formed.
Q4: K_sp of BaSO₄ is 1.5 × 10⁻⁹. Will a precipitate form when 0.01 M BaCl₂ and 0.01 M Na₂SO₄ are mixed? A: IP = [Ba²⁺][SO₄²⁻] = (0.01)(0.01) = 10⁻⁴. IP (10⁻⁴) >> K_sp (1.5×10⁻⁹) → PRECIPITATE forms.
Q5: A buffer has pH = 5.2. When 0.01 mol HCl is added to 1 L, pH changes to 5.1. Is this a good buffer? A: Change of 0.1 pH unit for 0.01 mol HCl is REASONABLE. Buffer is working effectively — a good buffer resists pH change within ±1 unit of pK_a.
13. Conclusion
Equilibrium is a DYNAMIC state that pervades chemistry. The equilibrium constant QUANTIFIES the extent of reaction. Le Chatelier's principle helps PREDICT how equilibria respond to disturbances — essential for INDUSTRIAL chemistry (Haber process, Contact process). Acid-base equilibria determine pH, which is CRITICAL for biological systems. Buffers are VITAL in living organisms (blood pH = 7.4). Solubility product explains precipitation. These concepts are UNIVERSALLY important in analytical, biological, and environmental chemistry.
