Chapter Introduction
Equilibrium is one of the most important and frequently tested chapters in CBSE Class 11 Chemistry. It bridges the concepts of thermodynamics with real chemical behaviour. The chapter covers two major areas — chemical equilibrium (reversible reactions, equilibrium constants, Le Chatelier's principle) and ionic equilibrium (acids, bases, pH, buffers, solubility product).
Mastering this chapter is essential because equilibrium concepts appear repeatedly in Class 12 (chemical kinetics, electrochemistry) and in competitive exams like NEET and JEE.
Key Definitions
| Term | Definition |
|---|---|
| Chemical equilibrium | The state of a reversible reaction where the rate of forward reaction equals the rate of backward reaction, and concentrations of reactants and products remain constant |
| Equilibrium constant (Kc) | Ratio of product concentrations to reactant concentrations at equilibrium, each raised to their stoichiometric coefficients |
| Equilibrium constant (Kp) | Equilibrium constant expressed in terms of partial pressures of gaseous species |
| Reaction quotient (Q) | Same expression as Kc but using initial (non-equilibrium) concentrations; Q = Kc at equilibrium |
| Le Chatelier's principle | If a system at equilibrium is disturbed, it shifts to counteract the disturbance |
| Strong electrolyte | Completely dissociates in solution (e.g., HCl, NaOH, NaCl) |
| Weak electrolyte | Partially dissociates in solution (e.g., CH₃COOH, NH₄OH) |
| pH | Negative logarithm of H⁺ concentration: pH = −log[H⁺] |
| Buffer | A solution that resists change in pH upon addition of small amounts of acid or base |
| Solubility product (Ksp) | Product of ion concentrations in a saturated solution of a sparingly soluble salt, each raised to their stoichiometric coefficients |
| Common ion effect | Suppression of dissociation of a weak electrolyte by adding a strong electrolyte sharing a common ion |
Chemical Equilibrium
Characteristics of Chemical Equilibrium
1. Dynamic in nature — forward and backward reactions continue at equal rates. 2. Attainable only from either direction — same equilibrium state is reached whether you start with reactants or products. 3. Constant macroscopic properties — concentration, pressure, colour, density remain constant at equilibrium. 4. Catalyst does not shift equilibrium — it speeds up attainment of equilibrium by affecting both rates equally.
Law of Chemical Equilibrium (Guldberg and Waage)
For the reaction: aA + bB ⇌ cC + dD
Kc = [C]c [D]d / [A]a [B]b
- Square brackets denote molar concentrations at equilibrium (mol/L).
- Pure solids and pure liquids have unit activity (omit from expression).
- Kc is dimensionless when activities are used (standard state = 1 M reference).
Kp and the Kp–Kc Relationship
Kp = Kc(RT)^Δn
Where:
- Δn = (moles of gaseous products) − (moles of gaseous reactants)
- R = 0.0831 L·bar/(mol·K)
- T = temperature in Kelvin
| Δn | Relationship | Example |
|---|---|---|
| 0 | Kp = Kc | H₂(g) + I₂(g) ⇌ 2HI(g) |
| > 0 | Kp > Kc | N₂O₄(g) ⇌ 2NO₂(g); Δn = 1 |
| < 0 | Kp < Kc | 2SO₂(g) + O₂(g) ⇌ 2SO₃(g); Δn = −1 |
Reaction Quotient (Q) and Predicting Direction
| Condition | Prediction |
|---|---|
| Q < Kc | Reaction proceeds in the forward direction (toward products) |
| Q = Kc | System is at equilibrium |
| Q > Kc | Reaction proceeds in the backward direction (toward reactants) |
Le Chatelier's Principle
Effect of Changes on Equilibrium
For the reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g); ΔH = −92.4 kJ/mol
| Change | Effect on Equilibrium | Reason |
|---|---|---|
| Increase [N₂] or [H₂] | Shifts right (toward NH₃) | Consumes added reactant |
| Increase [NH₃] | Shifts left (toward N₂ + H₂) | Consumes added product |
| Decrease volume / increase pressure | Shifts right (fewer moles: 4 → 2) | Reduces pressure |
| Increase volume / decrease pressure | Shifts left (more moles) | Increases pressure |
| Increase temperature | Shifts left (endothermic direction) | Absorbs added heat |
| Decrease temperature | Shifts right (exothermic direction) | Replaces lost heat |
| Add catalyst | No shift | Catalyst only speeds attainment |
Applications
1. Haber process (NH₃ synthesis): High pressure (200 atm), moderate temperature (~700 K), Fe catalyst. Compromise needed — low temperature favours yield but slows reaction. 2. Contact process (H₂SO₄): 2SO₂ + O₂ ⇌ 2SO₃. Excess O₂, moderate temperature (720 K), V₂O₅ catalyst, pressure ~2 atm. 3. CO + H₂O ⇌ CO₂ + H₂: Used in water-gas shift reaction. Δn = 0, so pressure has no effect.
Ionic Equilibrium
Acid-Base Theories
| Theory | Acid | Base | Example |
|---|---|---|---|
| Arrhenius | Produces H⁺ in water | Produces OH⁻ in water | HCl → H⁺ + Cl⁻; NaOH → Na⁺ + OH⁻ |
| Bronsted-Lowry | Proton (H⁺) donor | Proton (H⁺) acceptor | NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ |
| Lewis | Electron pair acceptor | Electron pair donor | BF₃ + NH₃ → F₃B←NH₃ |
Conjugate acid-base pairs: Every Bronsted acid has a conjugate base (formed after losing H⁺) and every Bronsted base has a conjugate acid (formed after gaining H⁺).
Example: CH₃COOH ⇌ CH₃COO⁻ + H⁺ (CH₃COOH/CH₃COO⁻ is a conjugate pair)
The pH Scale
pH = −log[H⁺] ; pOH = −log[OH⁻] ; pH + pOH = 14 (at 25 °C)
| Solution | [H⁺] | pH | Nature |
|---|---|---|---|
| 0.1 M HCl | 10⁻¹ | 1 | Strongly acidic |
| 0.01 M HCl | 10⁻² | 2 | Acidic |
| Pure water | 10⁻⁷ | 7 | Neutral |
| 0.01 M NaOH | 10⁻¹² | 12 | Strongly basic |
Ion product of water: Kw = [H⁺][OH⁻] = 10⁻¹⁴ at 25 °C
Strong vs Weak Acids and Bases
| Property | Strong Acid/Base | Weak Acid/Base |
|---|---|---|
| Dissociation | Complete (α ≈ 1) | Partial (α < 1) |
| Examples | HCl, HNO₃, H₂SO₄; NaOH, KOH | CH₃COOH, HCN; NH₄OH |
| pH of 0.1 M | ~1 (acid) or ~13 (base) | Higher (acid) or lower (base) |
| Ka/Kb | Very large | Small (Ka for CH₃COOH ≈ 1.8 × 10⁻⁵) |
Ionization of Weak Acids and Bases
For a weak acid HA: HA ⇌ H⁺ + A⁻
Ka = [H⁺][A⁻] / [HA]
If α is the degree of dissociation and C is the initial concentration:
- [H⁺] = Cα
- Ka = Cα² / (1 − α) ≈ Cα² (when α << 1)
- α = √(Ka/C)
Ostwald's dilution law: α = √(Ka/C) — dissociation increases with dilution.
For a weak base BOH: BOH ⇌ B⁺ + OH⁻
- Kb = [B⁺][OH⁻] / [BOH] = Cα² / (1 − α)
Buffer Solutions
| Buffer Type | Composition | pH Formula |
|---|---|---|
| Acidic buffer | Weak acid + its salt with strong base | pH = pKa + log([Salt]/[Acid]) |
| Basic buffer | Weak base + its salt with strong acid | pOH = pKb + log([Salt]/[Base]); pH = 14 − pOH |
Examples:
- Acidic: CH₃COOH + CH₃COONa (pH ≈ 4.74)
- Basic: NH₄OH + NH₄Cl (pH ≈ 9.25)
Buffer capacity: The amount of acid or base that can be added without significant pH change. Maximum when [Salt] = [Acid] or [Salt] = [Base].
Solubility Product (Ksp)
For a sparingly soluble salt AₓBᵧ: AₓBᵧ(s) ⇌ xAⁿ⁺ + yBᵐ⁻
Ksp = [Aⁿ⁺]ˣ [Bᵐ⁻]ʸ
If s is the solubility in mol/L:
- For AB type (e.g., AgCl): Ksp = s² → s = √Ksp
- For AB₂ type (e.g., CaF₂): Ksp = 4s³ → s = ∛(Ksp/4)
- For A₂B type (e.g., Ag₂CrO₄): Ksp = 4s³ → s = ∛(Ksp/4)
- For A₃B type (e.g., Al(OH)₃): Ksp = 27s⁴ → s = ∜(Ksp/27)
Predicting precipitation: If ionic product (IP) > Ksp → precipitate forms; IP < Ksp → no precipitate; IP = Ksp → saturated solution.
Common Ion Effect on Solubility
Adding a common ion decreases solubility of a sparingly soluble salt.
Example: Effect of NaCl on solubility of AgCl
- AgCl ⇌ Ag⁺ + Cl⁻; Ksp = [Ag⁺][Cl⁻] = 1.8 × 10⁻¹⁰
- Adding NaCl increases [Cl⁻], so [Ag⁺] must decrease to keep Ksp constant.
- Solubility of AgCl decreases in NaCl solution compared to pure water.
Salt Hydrolysis
| Salt Type | Hydrolysis | pH | Example |
|---|---|---|---|
| Strong acid + Strong base | No hydrolysis | 7 (neutral) | NaCl |
| Weak acid + Strong base | Anionic hydrolysis | > 7 (basic) | CH₃COONa |
| Strong acid + Weak base | Cationic hydrolysis | < 7 (acidic) | NH₄Cl |
| Weak acid + Weak base | Both hydrolyse | Depends on Ka and Kb | CH₃COONH₄ |
Important Numerical Patterns
Pattern 1: Finding Kc from equilibrium concentrations
For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), if at equilibrium [SO₂] = 0.4, [O₂] = 0.2, [SO₃] = 1.6 mol/L:
- Kc = [SO₃]² / ([SO₂]² × [O₂]) = (1.6)² / ((0.4)² × 0.2) = 2.56 / 0.032 = 80
Pattern 2: Kp from Kc
For N₂O₄(g) ⇌ 2NO₂(g), Kc = 0.04 at 300 K:
- Δn = 2 − 1 = 1
- Kp = Kc(RT)^Δn = 0.04 × (0.0831 × 300)¹ = 0.04 × 24.93 = 0.997 ≈ 1.0
Pattern 3: pH calculations
For 0.001 M HCl (strong acid): pH = −log(10⁻³) = 3 For 0.01 M NaOH (strong base): pOH = −log(10⁻²) = 2; pH = 14 − 2 = 12
Pattern 4: Buffer pH (Henderson-Hasselbalch)
Acidic buffer with [CH₃COONa] = 0.2 M, [CH₃COOH] = 0.1 M, Ka = 1.8 × 10⁻⁵:
- pKa = −log(1.8 × 10⁻⁵) = 4.74
- pH = 4.74 + log(0.2/0.1) = 4.74 + 0.301 = 5.04
Pattern 5: Ksp and solubility
For AgCl, Ksp = 1.8 × 10⁻¹⁰:
- AgCl ⇌ Ag⁺ + Cl⁻; Ksp = s²
- s = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ mol/L
Common Mistakes
1. Including pure solids/liquids in Kc: Do NOT include concentrations of pure solids or pure liquids in the equilibrium expression. Only gases and aqueous species count.
2. Confusing Kc and Kp: Kc uses concentrations (mol/L), Kp uses partial pressures (atm). They are equal only when Δn = 0.
3. Wrong Δn calculation: Count only GASEOUS moles, not total moles. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), Δn = 1 − 0 = 1 (only gas counted).
4. Henderson-Hasselbalch misuse: The equation pH = pKa + log([Salt]/[Acid]) applies only to buffer solutions, not to arbitrary acid or base solutions.
5. pH above 14 or below 0: pH can go outside 0–14 for very concentrated strong acids or bases. A 10 M HCl solution has pH = −1.
6. Assuming all salts are neutral: Only strong acid + strong base salts are neutral. Weak acid salts are basic, weak base salts are acidic.
7. Forgetting temperature dependence: Kc and Kp change with temperature. Le Chatelier's principle predicts the direction: for endothermic reactions, K increases with T; for exothermic, K decreases with T.
Exam Tips
1. Equilibrium constant expressions (3 marks): Practice writing Kc and Kp expressions. Remember to exclude pure solids and liquids.
2. Le Chatelier's principle (5 marks): Be ready to explain the effect of concentration, pressure, volume, and temperature changes with a specific example (Haber process is the most asked).
3. pH calculations (2–3 marks): Calculate pH of strong acids, strong bases, weak acids (using Ka), and buffer solutions. Show the formula, substitution, and final answer.
4. Ksp problems (2–3 marks): Calculate solubility from Ksp and vice versa. Practice the common ion effect on solubility.
5. Hybrid question: A common question combines chemical and ionic equilibrium — for example, finding pH at equilibrium after a reaction. Practice such combined problems from NCERT exercises.
6. Salt hydrolysis (1 mark): Memorize the four types of salts and their pH nature. A quick reasoning question on whether a given salt solution is acidic, basic, or neutral is very common.
7. Definitions: Bronsted-Lowry acids/bases, conjugate pairs, buffer, Ksp, and common ion effect — these are frequently asked as 1-mark or 2-mark questions.
Quick Revision Notes
- Equilibrium: Forward rate = backward rate; concentrations constant but not equal.
- Kc: Product concentrations / reactant concentrations (each raised to stoichiometric coefficient). Exclude pure solids/liquids.
- Kp = Kc(RT)^Δn where Δn = gaseous products − gaseous reactants.
- Q vs Kc: Q < Kc → forward; Q > Kc → backward; Q = Kc → equilibrium.
- Le Chatelier: System counteracts the disturbance.
- pH = −log[H⁺]; pOH = −log[OH⁻]; pH + pOH = 14 at 25 °C.
- Ka for weak acids, Kb for weak bases; pKa + pKb = 14 at 25 °C (for conjugate pairs).
- Buffer pH = pKa + log([Salt]/[Acid]) (acidic); pOH = pKb + log([Salt]/[Base]) (basic).
- Ksp = product of ion concentrations in saturated solution, raised to stoichiometric coefficients.
- Common ion effect reduces solubility of a sparingly soluble salt.
- Salt hydrolysis: SA+SB → neutral; WA+SB → basic; SA+WB → acidic; WA+WB → depends on Ka vs Kb.
- Ostwald's dilution law: α = √(Ka/C) for weak acid; increases with dilution.
- Degree of dissociation (α): fraction of molecules dissociated; 0 < α < 1 for weak electrolytes.
NCERT Exercise Focus Areas
The following NCERT exercise question types are most likely to appear in exams. Prioritize these during revision:
| Question Type | Chapter Examples | Marks |
|---|---|---|
| Write Kc and Kp expressions | Q1, Q2 | 1–2 |
| Calculate Kc from equilibrium concentrations | Q3, Q4, Q5 | 3 |
| Kp and Kc relationship (Δn calculations) | Q6, Q7 | 2–3 |
| Le Chatelier's principle application | Q11, Q12, Q13 | 3–5 |
| pH calculation (strong/weak acids/bases) | Q18, Q19, Q20 | 2–3 |
| Buffer pH using Henderson-Hasselbalch | Q21, Q22 | 2–3 |
| Ksp and solubility calculation | Q26, Q27 | 2–3 |
| Common ion effect on solubility | Q28 | 2–3 |
Key Derivations and Formulae
1. Kp = Kc(RT)^Δn
For gaseous reaction: aA + bB ⇌ cC + dD
Kc = [C]c[D]d / [A]a[B]b (concentrations in mol/L)
Kp = (pC)c(pD)d / (pA)a(pB)b (partial pressures in atm)
Using ideal gas equation: p = (n/V)RT = CRT
Therefore Kp = (CRT)^Δn × Kc where Δn = (c + d) − (a + b) for gaseous species only.
2. Henderson-Hasselbalch equation (derivation)
For a weak acid HA: HA ⇌ H⁺ + A⁻
Ka = [H⁺][A⁻] / [HA]
Taking −log of both sides: −log Ka = −log[H⁺] − log([A⁻]/[HA])
pKa = pH − log([A⁻]/[HA])
For weak base BOH: BOH ⇌ B⁺ + OH⁻
pOH = pKb + log([Salt]/[Base]); pH = 14 − pOH