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Equilibrium – Class 11 Chemistry notes, important questions and MCQs

NCERT Class 11 Chemistry, chapter 6. Notes, key terms, important questions with answers and MCQs.

Chapter overview

In reversible reactions, forward and backward rates become equal at equilibrium. Equilibrium constants, Le Chatelier's principle and ionic equilibrium (acids, bases, pH, buffers, solubility) are central ideas.

Chemical equilibrium

Kc = [products]^coeff/[reactants]^coeff; Kp = Kc(RT)^Δn.
Q < K: reaction goes forward; Q > K: backward.
Le Chatelier: a system shifts to oppose a change in concentration, pressure or temperature; catalysts do not shift equilibrium.

Ionic equilibrium

pH = −log[H⁺]; at 25 °C Kw = [H⁺][OH⁻] = 10⁻¹⁴, pH + pOH = 14.
Buffer: weak acid + its salt; pH = pKa + log([salt]/[acid]).
Solubility product Ksp; precipitation occurs when ionic product > Ksp; common ion effect lowers solubility.

Key terms

Le Chatelier's principle: a system at equilibrium shifts to counteract an imposed change
Buffer solution: solution that resists change in pH
Ionic product of water: Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 298 K

Formulas

Kp = Kc(RT)^Δn
pH = pKa + log([salt]/[acid])
pH + pOH = 14

Important questions with answers

Write the expression of Kc for N₂(g) + 3H₂(g) ⇌ 2NH₃(g).2 marks

Kc = [NH₃]²/([N₂][H₂]³).

Explain the effect of adding an inert gas at constant volume and at constant pressure on an equilibrium.3 marks

At constant volume, partial pressures of reactants and products do not change, so the equilibrium does not shift. At constant pressure, the volume increases and partial pressures fall, so the equilibrium shifts towards the side with more moles of gas.

Calculate the pH of 0.1 M acetic acid (Ka = 1.8 × 10⁻⁵) and its degree of ionisation. If 0.1 M sodium acetate is added, what will be the new pH, and why does it change?5 marks

[H⁺] = √(Ka × C) = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ M; pH = 3 − log 1.34 ≈ 2.87. Degree of ionisation α = √(Ka/C) = √(1.8 × 10⁻⁴) ≈ 0.0134 (1.34%). With 0.1 M sodium acetate: pH = pKa + log([salt]/[acid]) = 4.74 + log 1 = 4.74. The pH rises because acetate ions (common ion) suppress the ionisation of acetic acid and the mixture becomes a buffer.

Quick revision

Catalyst does not change K.
Kw = 10⁻¹⁴ at 25 °C.
Common ion effect lowers solubility.

Exam tips

Increasing pressure favours the side with fewer gas moles.
For weak acids, use [H⁺] = √(Ka × C) when dissociation is small, then pH = −log[H⁺].

Practice MCQs

pH of pure water at 25 °C is
1. 0
2. 7
3. 14
4. 1
Show answer
Answer: (b) 7. Neutral.
A catalyst affects equilibrium by
1. shifting it forward
2. changing K
3. helping reach it faster
4. shifting it backward
Show answer
Answer: (c) helping reach it faster. Speeds both directions.
pH of 0.001 M HCl is
1. 1
2. 2
3. 3
4. 11
Show answer
Answer: (c) 3. −log 10⁻³.
For N₂ + 3H₂ ⇌ 2NH₃, increasing pressure
1. stops the reaction
2. favours N₂ and H₂
3. has no effect
4. favours NH₃
Show answer
Answer: (d) favours NH₃. Fewer gas moles.
pH of 0.01 M NaOH is
1. 2
2. 12
3. 10
4. 14
Show answer
Answer: (b) 12. pOH = 2.
For a reaction with Δn = 0, Kp equals
1. Kc/RT
2. Kc RT
3. Kc
4. zero
Show answer
Answer: (c) Kc. (RT)⁰ = 1.
Buffer with equal concentrations of acetic acid and sodium acetate (pKa = 4.76) has pH
1. 7
2. 9.24
3. 4.76
4. 2.38
Show answer
Answer: (c) 4.76. log 1 = 0.
Solubility of AgCl (Ksp = 1.8 × 10⁻¹⁰) in water is about
1. 1.34 × 10⁻¹⁰ mol/L
2. 1.8 × 10⁻¹⁰ mol/L
3. 1.34 × 10⁻⁵ mol/L
4. 3.6 × 10⁻⁵ mol/L
Show answer
Answer: (c) 1.34 × 10⁻⁵ mol/L. s = √Ksp.

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Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.