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NCERT Solutions for Class 11 Chemistry Chapter 6 Equilibrium

NCERT Solutions for Class 11 Chemistry Chapter 6 Equilibrium covers physical and chemical equilibrium, dynamic equilibrium, Law of Mass Action, equilibrium constants (Kc and Kp), Le Chatelier's Principle, ionic equilibrium, concepts of acids and bases, pH scale, ionization constants, common ion effect, buffer solutions, and solubility product (Ksp).

Overview of Topics & Subtopics (Click to Jump)

1. Equilibrium in Physical & Chemical Processes and Dynamic Nature

Short Note:

• Equilibrium State: A state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction, making the concentrations of reactants and products constant over time.

• Dynamic Nature of Equilibrium: Reactions do not stop at equilibrium; instead, both forward and backward reactions continue at equal speeds.

• Physical Equilibrium Examples:

  - Solid-Liquid: Ice ⇌ Water at 0°C.

  - Liquid-Vapor: Water ⇌ Water Vapor in a sealed container.

  - Solid-Vapor: Iodine (s) ⇌ Iodine (v).

2. Law of Mass Action & Equilibrium Constants (Kc and Kp)

Short Note:

• Law of Chemical Equilibrium: For a reversible reaction aA + bB ⇌ cC + dD, the equilibrium expression is:

  - In terms of concentration: Kc = [C]c[D]d / ([A]a[B]b).

  - In terms of partial pressure: Kp = (PCc · PDd) / (PAa · PBb).

• Relationship between Kp and Kc:

  - Kp = Kc (RT)Δng, where Δng = (moles of gaseous products) - (moles of gaseous reactants).

• Properties of K:

  - If a reaction is reversed, the new equilibrium constant is K' = 1/K.

  - If a reaction is multiplied by a factor n, the new equilibrium constant becomes K' = Kn.

3. Factors Affecting Equilibrium – Le Chatelier's Principle

Short Note:

• Le Chatelier's Principle: If a system at equilibrium is subjected to a change in concentration, pressure, temperature, or volume, the system adjusts itself to counteract the effect of the change.

• Effect of Changes:

  - Concentration: Adding reactant shifts equilibrium forward; adding product shifts it backward.

  - Pressure: Increasing pressure shifts equilibrium toward the side with fewer gaseous moles.

  - Temperature: Increasing temperature favors endothermic reactions (ΔH > 0); decreasing temperature favors exothermic reactions (ΔH < 0).

  - Catalyst: Increases the rates of both forward and reverse reactions equally; does not alter the equilibrium position or value of K.

4. Ionic Equilibrium: Acids, Bases, pH, Buffers & Solubility Product (Ksp)

Short Note:

• Acids and Bases Concepts:

  - Arrhenius: Acids give H+ ions; Bases give OH- ions in aqueous solution.

  - Brønsted-Lowry: Acids are proton (H+) donors; Bases are proton acceptors.

  - Lewis: Acids are electron-pair acceptors (e.g., BF3, AlCl3); Bases are electron-pair donors (e.g., NH3, H2O).

• pH Scale: pH = -log[H+] and pOH = -log[OH-]. At 25°C, pH + pOH = 14.

• Buffer Solutions: Solutions that resist changes in pH upon addition of small amounts of acid or base (e.g., CH3COOH + CH3COONa).

• Solubility Product Constant (Ksp): For a sparingly soluble salt AxBy(s) ⇌ xAy+(aq) + yBx-(aq), Ksp = [Ay+]x [Bx-]y.

Intext & Conceptual Questions

Q1: Write the expression for the equilibrium constant Kc for the reaction: N2(g) + 3H2(g) ⇌ 2NH3(g). Find the relationship between Kp and Kc for this reaction.
Answer:
1. Concentration Equilibrium Constant Expression (Kc):
Kc = [NH3]2 / ([N2] [H2]3)

2. Relationship between Kp and Kc:
Formula: Kp = Kc (RT)Δng
Calculate Δng (moles of gaseous products - moles of gaseous reactants):
Δng = 2 - (1 + 3) = 2 - 4 = -2.

Therefore, Kp = Kc (RT)-2 or Kp = Kc / (RT)2.

Textbook Chapter End Exercise Questions

Q1: What will be the conjugate base for the Brønsted acids HF, H2SO4, and HCO3-?
Answer:
A conjugate base is formed when a Brønsted acid donates a proton (H+):

• HF: HF - H+ = F- (Fluoride ion)
• H2SO4: H2SO4 - H+ = HSO4- (Hydrogen sulfate ion)
• HCO3-: HCO3- - H+ = CO32- (Carbonate ion)
Q2: The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10-3 M. What is its pH?
Answer:
Given:
[H+] = 3.8 × 10-3 M

Formula:
pH = -log[H+]
pH = -log(3.8 × 10-3)
pH = -[log(3.8) + log(10-3)]
Since log(3.8) ≈ 0.58 and log(10-3) = -3:
pH = -[0.58 - 3]
pH = -[-2.42] = 2.42.

The pH of the soft drink is 2.42 (Acidic).
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Class 11 › Chemistry › Equilibrium
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Topics in this chapter: Equilibrium in Physical & Chemical Processes and Dynamic NatureLaw of Mass Action & Equilibrium Constants (K c and K p )Factors Affecting Equilibrium – Le Chatelier's PrincipleIonic Equilibrium: Acids, Bases, pH, Buffers & Solubility Product (K sp )