NCERT Solutions for Class 11 Chemistry Chapter 7: Equilibrium
Chapter 7 of Class 11 Chemistry, Equilibrium, is a chapter that connects multiple ideas in chemistry — thermodynamics, kinetics, acid-base behaviour, and solubility. When a reversible chemical reaction reaches a state where the forward and reverse reactions occur at the same rate, the system is at equilibrium. But equilibrium is not static — it is a dynamic balance, and understanding how to quantify and shift it is at the heart of this chapter. At Myclass24, the NCERT solutions for Chapter 7 are written with a clear focus on the two major sections: chemical equilibrium and ionic equilibrium.
For chemical equilibrium, you will find solutions covering the equilibrium constant expression (Kc and Kp), the relationship between them, and Le Chatelier's principle — a beautifully intuitive rule that predicts how a system responds to stress. For ionic equilibrium, the solutions cover dissociation of acids and bases, pH calculations, buffer solutions, hydrolysis of salts, solubility product (Ksp), and common ion effect. These concepts appear regularly in CBSE board papers and in competitive entrance exams. The chapter has a healthy mix of conceptual questions and numerical problems, all of which are solved step by step in our solutions. Students from all parts of India — whether studying in Mysuru, Dehradun, Jabalpur, or Visakhapatnam — can benefit from the well-structured, mobile-friendly resources available at Myclass24.
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NCERT Solutions — Class 11 Chemistry Chapter 7: Equilibrium
Kc, Kp, Le Chatelier, pH, buffer and Ksp problems solved | Free PDF on Myclass24
Chapter 7 Complete Guide: Chemical & Ionic Equilibrium with Key Data
What is Equilibrium?
A reversible reaction reaches equilibrium when the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant (but not necessarily equal). This is dynamic equilibrium — both reactions are still occurring, just at equal rates. The equilibrium state is characterised by specific values of the equilibrium constant. One can check out all chapters of NCERT Solutions for Class 11 Chemistry and all subjects of NCERT Solutions for Class 11 from the Myclass24 page.
Equilibrium Constant (Kc and Kp)
For a general reaction: aA + bB ⇌ cC + dD, the equilibrium constant in terms of concentration is Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ. The subscript 'c' means concentrations. For gaseous reactions, Kp uses partial pressures. The relationship between them is: Kp = Kc(RT)^Δnᵍ, where Δnᵍ = moles of gaseous products − moles of gaseous reactants.
| Reaction Type | Δnᵍ | Kp vs Kc | Example |
|---|---|---|---|
| Equal moles of gas | 0 | Kp = Kc | H₂ + I₂ ⇌ 2HI |
| More moles in products | +ve | Kp > Kc | PCl₅ ⇌ PCl₃ + Cl₂ |
| Fewer moles in products | −ve | Kp < Kc | N₂ + 3H₂ ⇌ 2NH₃ |
Reaction Quotient (Q) and Direction of Reaction
The reaction quotient Q is calculated the same way as Kc, but using concentrations at any point (not at equilibrium). Comparing Q to K tells us which way the reaction will proceed. If Q < K, the reaction proceeds forward (more products form). If Q > K, the reaction proceeds in reverse. If Q = K, the system is at equilibrium.
Le Chatelier's Principle
Le Chatelier's Principle states that when a system at equilibrium is disturbed by a change in concentration, temperature, or pressure, it shifts to counteract that change and re-establish equilibrium. This is one of the most useful and intuitive rules in chemistry.
| Stress Applied | System's Response | Effect on K |
|---|---|---|
| Add reactant | Shifts forward (→) | No change |
| Remove product | Shifts forward (→) | No change |
| Increase pressure (gas) | Shifts to fewer moles of gas | No change |
| Increase temperature | Shifts in endothermic direction | Changes |
| Add inert gas (const. V) | No effect | No change |
Ionic Equilibrium — Acids, Bases, and pH
Acids donate protons (Brønsted-Lowry) or accept electron pairs (Lewis). Strong acids (HCl, H₂SO₄, HNO₃) ionise completely; weak acids (CH₃COOH, HF) ionise partially. The degree of ionisation depends on Ka (acid dissociation constant). The pH scale measures hydrogen ion concentration: pH = −log[H⁺]. At 25°C, pure water has pH = 7.
| Substance | Nature | Approximate pH |
|---|---|---|
| Gastric juice (HCl) | Strong acid | 1–2 |
| Lemon juice | Weak acid | 2–3 |
| Black coffee | Weakly acidic | 5 |
| Pure water | Neutral | 7 |
| Blood | Slightly basic | 7.4 |
| Baking soda solution | Basic | 8–9 |
| NaOH solution (0.1M) | Strong base | 13 |
Buffer Solutions
A buffer is a solution that resists significant changes in pH when small amounts of acid or base are added. An acidic buffer contains a weak acid and its conjugate base (e.g., CH₃COOH + CH₃COONa). A basic buffer contains a weak base and its conjugate acid (e.g., NH₃ + NH₄Cl). The Henderson-Hasselbalch equation gives buffer pH: pH = pKa + log([A⁻]/[HA]). Blood (pH ≈ 7.4) is maintained by the bicarbonate buffer system — a critical fact for NEET aspirants.
Solubility Product (Ksp) and Common Ion Effect
For a sparingly soluble salt like AgCl: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), Ksp = [Ag⁺][Cl⁻]. The Ksp value tells us the maximum extent of dissolution at a given temperature. The common ion effect states that adding an ion that is already present in equilibrium will suppress dissolution, reducing solubility. For example, adding NaCl to a saturated AgCl solution will cause more AgCl to precipitate.
- Kw (ionic product of water) at 25°C = 1 × 10⁻¹⁴ mol² L⁻²
- pOH + pH = 14 (at 25°C)
- If Ka × Kb = Kw, then pKa + pKb = 14
- Large K (>>1) = products strongly favoured; Small K (<<1) = reactants strongly favoured
- Haber process (N₂ + 3H₂ ⇌ 2NH₃): optimum conditions chosen using Le Chatelier's principle
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