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This page provides complete, step-by-step NCERT solutions for CBSE Class 11 Chemistry Chapter 7 — Equilibrium. Every intext question (grouped by page number) and every chapter-end exercise question is answered with clear explanations, worked-out numerical solutions, and chemical reasoning. Use the section headings to jump to the question you need help with.
Key Concepts (Quick Recap)
- Dynamic Equilibrium: Rate of forward reaction = Rate of backward reaction; concentrations remain constant.
- Law of Chemical Equilibrium: Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ for aA + bB ⇌ cC + dD.
- Kp and Kc: Kp = Kc(RT)^Δn, where Δn = (moles of gaseous products) − (moles of gaseous reactants).
- Le Chatelier's Principle: A system at equilibrium shifts to counteract any imposed change.
- Reaction Quotient (Q): Q has the same expression as K but uses current concentrations. Q < K → forward shift; Q > K → backward shift; Q = K → equilibrium.
- Acid-Base Theories: Arrhenius (H⁺/OH⁻ in water), Bronsted-Lowry (proton donor/acceptor), Lewis (electron pair acceptor/donor).
- pH Scale: pH = −log[H⁺]; pOH = −log[OH⁻]; pH + pOH = 14 (at 25 °C).
- Buffer Solutions: Resist pH change. Acidic buffer = weak acid + its salt with strong base. Basic buffer = weak base + its salt with strong acid.
- Solubility Product (Ksp): Ksp = [cation]ˣ[anion]ʸ for a sparingly soluble salt AₓBᵧ.
- Common Ion Effect: Solubility of a sparingly soluble salt decreases in the presence of a common ion.
- Hydrolysis of Salts: Strong acid + strong base → neutral; weak acid + strong base → basic; strong acid + weak base → acidic; weak acid + weak base → depends on Ka and Kb.
Intext Questions
Page 168–169
Q1. When the reaction attains equilibrium, which of the following is true? (a) Forward reaction stops (b) Backward reaction stops (c) Both forward and backward reactions stop (d) Both forward and backward reactions continue at equal rates
Answer(d) At equilibrium, both forward and backward reactions continue at equal rates. This is dynamic equilibrium — the reactions do not stop, but the net change is zero.
Q2. In the equilibrium N₂(g) + 3H₂(g) ⇌ 2NH₃(g), if the pressure is increased, what happens?
AnswerIncreasing pressure shifts the equilibrium toward the side with fewer moles of gas. Reactant side: 1 + 3 = 4 moles. Product side: 2 moles. Equilibrium shifts toward NH₃ (products) to reduce pressure.
Q3. What is the effect of adding a catalyst on the equilibrium constant?
AnswerA catalyst has no effect on the equilibrium constant. It only speeds up both the forward and backward reactions equally, helping the system reach equilibrium faster without changing the position of equilibrium or the value of K.
Q4. The equilibrium constant for the reaction H₂(g) + I₂(g) ⇌ 2HI(g) is 50. If the concentrations of H₂ and I₂ are both 0.1 M, what is the concentration of HI at equilibrium?
Answer
Kc = [HI]² / ([H₂][I₂]) = 50
50 = [HI]² / (0.1 × 0.1) = [HI]² / 0.01
[HI]² = 50 × 0.01 = 0.5
[HI] = √0.5 = 0.707 M
Page 179–180
Q5. For the reaction N₂O₄(g) ⇌ 2NO₂(g), Kp = 0.14 at 298 K. If the initial pressure of N₂O₄ is 1 atm, calculate the equilibrium partial pressures.
Answer
Let degree of dissociation = α. Initially: N₂O₄ = 1 atm, NO₂ = 0.
At equilibrium: P(N₂O₄) = (1 − α)/(1 + α) × P(total); P(NO₂) = 2α/(1 + α) × P(total)
Kp = [P(NO₂)]² / P(N₂O₄) = 0.14
Using the approximation for small α:
Kp = 4α²/(1 − α²) × 1 ≈ 4α² (when α is small)
α² = 0.14/4 = 0.035; α = 0.187
P(N₂O₄) = (1 − 0.187)/(1 + 0.187) × 1 = 0.813/1.187 = 0.685 atm P(NO₂) = 2(0.187)/1.187 = 0.315 atm
Q6. State Le Chatelier's principle. What happens when temperature is increased for an exothermic reaction at equilibrium?
AnswerLe Chatelier's principle states that if a system at equilibrium is subjected to a change in concentration, pressure, or temperature, the system adjusts to counteract the change.
For an exothermic reaction (ΔH < 0), increasing temperature shifts equilibrium in the backward direction (endothermic direction) to absorb the added heat.
Q7. What is the reaction quotient Q? How is it related to K?
AnswerThe reaction quotient Q has the same form as the equilibrium constant K, but uses current (non-equilibrium) concentrations or pressures instead of equilibrium values.
- Q < K: Reaction proceeds in the forward direction.
- Q > K: Reaction proceeds in the backward direction.
- Q = K: The system is at equilibrium.
Q8. For the reaction CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g), the initial concentrations are [CO] = 0.1 M, [H₂O] = 0.1 M, [CO₂] = 0.0 M, [H₂] = 0.0 M and Kc = 4.24. Find the equilibrium concentrations.
Answer
Let x moles of CO react.
At equilibrium: [CO] = 0.1 − x; [H₂O] = 0.1 − x; [CO₂] = x; [H₂] = x
Kc = x² / (0.1 − x)² = 4.24
x / (0.1 − x) = √4.24 = 2.059
x = 2.059(0.1 − x) = 0.2059 − 2.059x
3.059x = 0.2059
x = 0.0673 M
[CO] = 0.1 − 0.0673 = 0.0327 M; [H₂O] = 0.0327 M; [CO₂] = 0.0673 M; [H₂] = 0.0673 M
Page 192–194
Q9. Calculate the pH of a 0.01 M HCl solution.
Answer
pH = −log[H⁺] = −log(0.01) = −log(10⁻²) = 2.0
Q10. Calculate the pH of a 0.05 M H₂SO₄ solution (assuming complete dissociation of both H⁺).
Answer
[H⁺] = 2 × 0.05 = 0.10 M
pH = −log(0.10) = −log(10⁻¹) = 1.0
Q11. Calculate the pH of a 0.001 M NaOH solution.
Answer
[OH⁻] = 0.001 M; pOH = −log(0.001) = 3.0
pH = 14 − 3 = 11.0
Q12. The pH of a solution is 6. Is the solution acidic or basic?
Answer
pH = 6 < 7 → The solution is acidic (at 25 °C).
Q13. What is a buffer solution? Give one example of an acidic buffer.
AnswerA buffer solution is one that resists changes in pH when small amounts of acid or base are added. An acidic buffer consists of a weak acid and its salt with a strong base, e.g., CH₃COOH + CH₃COONa.
Q14. Calculate the pH of a buffer containing 0.1 M CH₃COOH and 0.1 M CH₃COONa. (Ka = 1.8 × 10⁻⁵)
Answer
pH = pKa + log([salt]/[acid]) = −log(1.8 × 10⁻⁵) + log(0.1/0.1)
= 4.74 + 0 = 4.74
Chapter-End Exercise
Q1. A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased. What happens initially?
AnswerWhen the volume is suddenly increased at constant temperature, the vapour pressure decreases momentarily (pressure is inversely proportional to volume at constant temperature, Boyle's law). More liquid evaporates to restore equilibrium vapour pressure. Equilibrium is re-established at the same vapour pressure as before (since temperature is unchanged).
Q2. What is the effect of: (a) Adding more H₂ on the equilibrium: H₂(g) + I₂(g) ⇌ 2HI(g)? (b) Adding more HI on the same equilibrium?
Answer(a) Adding H₂ (reactant) shifts the equilibrium forward (toward HI) to consume the added H₂. (b) Adding HI (product) shifts the equilibrium backward (toward H₂ and I₂) to consume the added HI.
Q3. For the equilibrium: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92.4 kJ. What happens when: (a) Temperature is increased? (b) Pressure is increased? (c) NH₃ is removed?
Answer(a) The reaction is exothermic. Increasing temperature shifts equilibrium backward (toward N₂ + H₂) — the endothermic direction. (b) Increasing pressure shifts equilibrium toward fewer moles — forward (toward NH₃), since 4 moles become 2 moles. (c) Removing NH₃ (product) shifts equilibrium forward to produce more NH₃.
Q4. Write the equilibrium constant expression for: 2NO(g) + Cl₂(g) ⇌ 2NOCl(g)
Answer
Kc = [NOCl]² / ([NO]² × [Cl₂])
Q5. Write Kp expression for: N₂(g) + O₂(g) ⇌ 2NO(g)
Answer
Kp = (P_NO)² / (P_N₂ × P_O₂)
Since Δn = 2 − 1 − 1 = 0, Kp = Kc.
Q6. Derive the relation between Kp and Kc.
Answer
For aA(g) + bB(g) ⇌ cC(g) + dD(g):
Kp = Kc(RT)^Δn where Δn = (c + d) − (a + b)
Derivation: For ideal gas, Pi = (ni/V)RT = CiRT. Substituting partial pressures in Kp expression in terms of concentrations gives Kp = Kc(RT)^Δn.
Q7. The equilibrium constant Kc for A(g) + B(g) ⇌ C(g) + D(g) is 100. If the initial concentrations of A and B are both 1 M, what are the equilibrium concentrations?
Answer
Let x moles react per litre.
At equilibrium: [A] = 1 − x; [B] = 1 − x; [C] = x; [D] = x
Kc = x² / (1 − x)² = 100
x/(1 − x) = 10
x = 10 − 10x → 11x = 10 → x = 10/11 = 0.909
[A] = [B] = 1 − 0.909 = 0.091 M; [C] = [D] = 0.909 M
Q8. For the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), Kc = 1.8 × 10⁻³ at 500 K. If 1 mole of PCl₅ is placed in a 5 L vessel, calculate the equilibrium concentrations.
Answer
Initial: [PCl₅] = 1/5 = 0.2 M; [PCl₃] = 0; [Cl₂] = 0
Let x dissociate: [PCl₅] = 0.2 − x; [PCl₃] = x; [Cl₂] = x
Kc = x² / (0.2 − x) = 1.8 × 10⁻³
x² + 1.8 × 10⁻³x − 0.36 × 10⁻³ = 0
Using quadratic formula: x = 0.018 M (taking positive root)
[PCl₅] = 0.2 − 0.018 = 0.182 M; [PCl₃] = [Cl₂] = 0.018 M
Q9. Define Bronsted-Lowry acids and bases with examples.
AnswerA Bronsted-Lowry acid is a proton (H⁺) donor, and a Bronsted-Lowry base is a proton (H⁺) acceptor.
Example: In NH₃ + H₂O → NH₄⁺ + OH⁻, H₂O donates a proton (acid) and NH₃ accepts a proton (base). A conjugate acid-base pair differs by one proton: NH₄⁺/NH₃ and H₂O/OH⁻ are conjugate pairs.
Q10. What are Lewis acids and bases? Give examples.
AnswerA Lewis acid is an electron pair acceptor; a Lewis base is an electron pair donor.
Lewis acids: BF₃ (incomplete octet), AlCl₃, H⁺, Ag⁺ Lewis bases: NH₃ (lone pair), H₂O, OH⁻, F⁻
Q11. Calculate the pH of 0.002 M HNO₃.
Answer
pH = −log[H⁺] = −log(2 × 10⁻³) = 3 − log 2 = 3 − 0.301 = 2.699 ≈ 2.70
Q12. Calculate the pH of a solution having [H⁺] = 5.6 × 10⁻⁴ M.
Answer
pH = −log(5.6 × 10⁻⁴) = 4 − log 5.6 = 4 − 0.748 = 3.25
Q13. Calculate the pH of 0.1 M CH₃COOH. (Ka = 1.8 × 10⁻⁵)
Answer
[H⁺] = √(Ka × C) = √(1.8 × 10⁻⁵ × 0.1) = √(1.8 × 10⁻⁶) = 1.342 × 10⁻³ M
pH = −log(1.342 × 10⁻³) = 3 − log 1.342 = 3 − 0.128 = 2.87
Q14. The Ka of HF is 6.8 × 10⁻⁴. Calculate the pH of 0.05 M HF.
Answer
[H⁺] = √(Ka × C) = √(6.8 × 10⁻⁴ × 0.05) = √(3.4 × 10⁻⁵) = 5.83 × 10⁻³ M
pH = −log(5.83 × 10⁻³) = 3 − log 5.83 = 3 − 0.766 = 2.23
Q15. Calculate the pH of a buffer containing 0.2 M NH₄OH and 0.3 M NH₄Cl. (Kb = 1.8 × 10⁻⁵)
Answer
pOH = pKb + log([salt]/[base]) = −log(1.8 × 10⁻⁵) + log(0.3/0.2)
= 4.74 + log(1.5) = 4.74 + 0.176 = 4.916
pH = 14 − 4.916 = 9.08
Q16. What is the solubility product? Write Ksp expression for AgCl.
AnswerThe solubility product (Ksp) is the product of the molar concentrations of the ions of a sparingly soluble salt, each raised to the power of its stoichiometric coefficient, at equilibrium.
For AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq): Ksp = [Ag⁺][Cl⁻]
Q17. The Ksp of BaSO₄ is 1.1 × 10⁻¹⁰. Calculate its solubility in water.
Answer
BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq)
Let solubility = s mol/L; then [Ba²⁺] = [SO₄²⁻] = s
Ksp = s² = 1.1 × 10⁻¹⁰
s = √(1.1 × 10⁻¹⁰) = 1.05 × 10⁻⁵ mol/L
Q18. Calculate the solubility of BaSO₄ in 0.1 M BaCl₂ solution. (Ksp = 1.1 × 10⁻¹⁰)
Answer
[Ba²⁺] = 0.1 + s ≈ 0.1 (since s is very small) [SO₄²⁻] = s
Ksp = [Ba²⁺][SO₄²⁻] = 0.1 × s = 1.1 × 10⁻¹⁰
s = 1.1 × 10⁻¹⁰/0.1 = 1.1 × 10⁻⁹ mol/L
The solubility decreases dramatically due to the common ion (Ba²⁺) — this is the common ion effect.
Q19. Explain the common ion effect with an example.
AnswerThe common ion effect is the suppression of the dissociation of a weak electrolyte by the presence of a common ion from a strong electrolyte.
Example: Adding NaCl (strong electrolyte) to a saturated solution of AgCl decreases the solubility of AgCl because Cl⁻ is the common ion. The increased [Cl⁻] shifts the equilibrium AgCl(s) ⇌ Ag⁺ + Cl⁻ to the left, reducing [Ag⁺] and hence solubility.
Q20. Predict whether the following salts will be acidic, basic, or neutral in aqueous solution: (a) NaCl (b) NaCN (c) NH₄NO₃ (d) CH₃COONH₄
Answer(a) NaCl: Strong acid (HCl) + strong base (NaOH) → Neutral (b) NaCN: Weak acid (HCN) + strong base (NaOH) → Basic (CN⁻ hydrolyzes) (c) NH₄NO₃: Strong acid (HNO₃) + weak base (NH₄OH) → Acidic (NH₄⁺ hydrolyzes) (d) CH₃COONH₄: Weak acid (CH₃COOH) + weak base (NH₄OH) → Depends on Ka and Kb; since Ka(CH₃COOH) ≈ Kb(NH₄OH), the solution is approximately neutral.
Q21. Write the hydrolysis reaction and calculate the pH of 0.1 M CH₃COONa. (Ka of CH₃COOH = 1.8 × 10⁻⁵)
Answer
CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻
Kh = Kw/Ka = 10⁻¹⁴/1.8 × 10⁻⁵ = 5.56 × 10⁻¹⁰
[OH⁻] = √(Kh × C) = √(5.56 × 10⁻¹⁰ × 0.1) = √(5.56 × 10⁻¹¹) = 7.46 × 10⁻⁶ M
pOH = −log(7.46 × 10⁻⁶) = 6 − 0.873 = 5.13
pH = 14 − 5.13 = 8.87 (basic, as expected)
Q22. The Ksp of Ag₂CrO₄ is 1.1 × 10⁻¹². Calculate its solubility.
Answer
Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq)
Let solubility = s; then [Ag⁺] = 2s; [CrO₄²⁻] = s
Ksp = [Ag⁺]²[CrO₄²⁻] = (2s)²(s) = 4s³ = 1.1 × 10⁻¹²
s³ = 2.75 × 10⁻¹³
s = (2.75 × 10⁻¹³)^(1/3) = 6.5 × 10⁻⁵ mol/L
Q23. The equilibrium constant for the reaction H₂(g) + I₂(g) ⇌ 2HI(g) is 49 at 700 K. If the initial amounts are [H₂] = 0.5 M and [I₂] = 0.5 M, find the equilibrium concentrations.
Answer
Let x moles react per litre.
At equilibrium: [H₂] = 0.5 − x; [I₂] = 0.5 − x; [HI] = 2x
Kc = [HI]²/([H₂][I₂]) = 4x²/(0.5 − x)² = 49
2x/(0.5 − x) = 7
2x = 3.5 − 7x → 9x = 3.5 → x = 0.389
[H₂] = [I₂] = 0.5 − 0.389 = 0.111 M; [HI] = 2 × 0.389 = 0.778 M
Q24. Explain: In the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), adding an inert gas at constant volume has no effect, but at constant pressure it shifts the equilibrium.
AnswerAt constant volume, adding inert gas increases total pressure but does not change the partial pressures or concentrations of the reactants/products (since volume is constant). Q = K and equilibrium is unaffected.
At constant pressure, adding inert gas increases the total volume (to keep total pressure constant). This decreases partial pressures of all gases. Since Δn > 0 for this reaction, the decrease in pressure shifts equilibrium forward (toward more moles of gas).
Internal Links
- [Chapter 7 Guide — Equilibrium](/school/cbse/class/11/chemistry/cbse-equilibrium)
- [All Class 11 Chemistry Solutions](/school/cbse/class/11/chemistry/solutions)
- [Class 11 Chemistry Subject Guide](/school/cbse/class/11/chemistry)
