Equilibrium — Class 11 Chemistry NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Chemistry chapter "Equilibrium" — 9 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 11 Chemistry chapter "Equilibrium" — 9 important questions with detailed answers for CBSE board exam prepa…
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NCERT Solutions for Class 11 Chemistry — Equilibrium. Step-by-step answers to all 9 textbook questions from this chapter, written for CBSE board preparation and free to use.
Equilibrium — All 9 Questions Solved
Q1. Define equilibrium constant Kc. How does it differ from Kp?
Equilibrium constant Kc is the ratio of concentrations of products to reactants, each raised to their stoichiometric coefficients at equilibrium.
For reaction: aA + bB ⇌ cC + dD
Kc = [C]^c [D]^d / [A]^a [B]^b
Difference from Kp:
- Kc uses molar concentrations (mol/L)
- Kp uses partial pressures (atm or bar)
- For reactions involving gases, Kp = Kc(RT)^Δn where Δn = moles of gaseous products - moles of gaseous reactants
- R = 0.0821 L·atm·K^-1·mol^-1, T = temperature in Kelvin
Example: N2 + 3H2 ⇌ 2NH3
Kp = Kc(RT)^(2-1-3) = Kc(RT)^-2 = Kc/(RT)^2
Q2. Calculate Kc for the reaction: H2 + I2 ⇌ 2HI at equilibrium when [H2] = 0.5 M, [I2] = 0.5 M, [HI] = 1 M
Given reaction: H2 + I2 ⇌ 2HI
Equilibrium concentrations:
[H2] = 0.5 M
[I2] = 0.5 M
[HI] = 1 M
Kc = [HI]^2 / ([H2][I2])
Kc = (1)^2 / (0.5 × 0.5)
Kc = 1 / 0.25
Kc = 4
Therefore, the equilibrium constant Kc = 4 (dimensionless for this reaction)
Q3. What does Le Chatelier's principle state? Give an example with the Haber process.
Le Chatelier's Principle states: When a system in equilibrium is disturbed, it shifts to counteract the disturbance and restore equilibrium.
Haber Process: N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = -92.4 kJ/mol
Applications:
1. Increase pressure: Shifts RIGHT (fewer moles of gas on products side: 4 mol -> 2 mol). Favors NH3 formation.
2. Increase temperature: Shifts LEFT (reaction is exothermic). Decreases yield but increases reaction rate. Optimum ~450°C.
3. Increase [N2] or [H2]: Shifts RIGHT, increases NH3 formation.
4. Remove NH3: Shifts RIGHT, increases NH3 formation.
Industrial optimization: High pressure (150-300 atm), moderate temperature (400-500°C), and continuous removal of NH3 maximize yield.
Q4. At 25°C, for the reaction PCl5 ⇌ PCl3 + Cl2, if initial PCl5 = 0.5 M and Kc = 1.8 × 10^-7, find equilibrium concentrations.
Reaction: PCl5(g) ⇌ PCl3(g) + Cl2(g)
ICE Table:
PCl5 PCl3 Cl2
I 0.5 0 0
C -x +x +x
E 0.5-x x x
Kc = [PCl3][Cl2] / [PCl5]
1.8 × 10^-7 = x · x / (0.5 - x)
Since Kc is very small, x << 0.5, so 0.5 - x ≈ 0.5
1.8 × 10^-7 = x^2 / 0.5
x^2 = 1.8 × 10^-7 × 0.5 = 9 × 10^-8
x = 3 × 10^-4 M
Equilibrium concentrations:
[PCl5] = 0.5 - 0.0003 ≈ 0.4997 M
[PCl3] = 3 × 10^-4 M
[Cl2] = 3 × 10^-4 M
Q5. What is the relationship between ΔG° and Kc? Explain spontaneity.
Relationship: ΔG° = -RT ln(Kc)
Where:
R = 8.314 J·mol^-1·K^-1 (or 8.314 / 1000 kJ·mol^-1·K^-1)
T = temperature in Kelvin
Kc = equilibrium constant
Spontaneity Criteria:
1. If Kc > 1: ln(Kc) > 0, so ΔG° < 0
Forward reaction is spontaneous. Products favored at equilibrium.
2. If Kc < 1: ln(Kc) < 0, so ΔG° > 0
Forward reaction is non-spontaneous. Reactants favored at equilibrium.
3. If Kc = 1: ln(Kc) = 0, so ΔG° = 0
System is at equilibrium.
Example: For reaction with ΔG° = -20 kJ/mol at 298 K
-20,000 = -(8.314)(298) ln(Kc)
ln(Kc) = 20,000 / 2477.6 ≈ 8.07
Kc = e^8.07 ≈ 3160 (reaction strongly favored)
Q6. Calculate pH of a solution if [H+] = 2.5 × 10^-4 M
pH is defined as: pH = -log[H+]
Given: [H+] = 2.5 × 10^-4 M
pH = -log(2.5 × 10^-4)
pH = -[log(2.5) + log(10^-4)]
pH = -[log(2.5) - 4]
pH = -[0.398 - 4]
pH = -[-3.602]
pH = 3.602
pH ≈ 3.6
This is an acidic solution (pH < 7).
Alternative calculation:
log(2.5) ≈ 0.398
log(2.5 × 10^-4) = 0.398 - 4 = -3.602
pH = 3.602
Q7. For weak acid HA, Ka = 1.8 × 10^-5. Calculate pH of 0.1 M HA solution.
Weak acid dissociation: HA ⇌ H+ + A-
Ka = [H+][A-] / [HA] = 1.8 × 10^-5
ICE Table:
HA H+ A-
I 0.1 0 0
C -x +x +x
E 0.1-x x x
Ka = x^2 / (0.1 - x)
Since Ka is small, assume x << 0.1, so 0.1 - x ≈ 0.1
1.8 × 10^-5 = x^2 / 0.1
x^2 = 1.8 × 10^-6
x = 1.34 × 10^-3 M = [H+]
Verify assumption: 1.34 × 10^-3 / 0.1 = 0.0134 = 1.34% < 5%, so assumption valid.
pH = -log(1.34 × 10^-3)
pH = -(-2.87)
pH ≈ 2.87 ≈ 2.9
Q8. What is ionic product of water Kw? Calculate [OH-] if [H+] = 5 × 10^-9 M.
Ionic product of water Kw is defined as:
Kw = [H+][OH-] = 1.0 × 10^-14 at 25°C
Given: [H+] = 5 × 10^-9 M
Using Kw = [H+][OH-]:
1.0 × 10^-14 = (5 × 10^-9) × [OH-]
[OH-] = 1.0 × 10^-14 / (5 × 10^-9)
[OH-] = 0.2 × 10^-5
[OH-] = 2 × 10^-6 M
pOH = -log(2 × 10^-6) = -(-5.7) = 5.7
pH = 14 - pOH = 14 - 5.7 = 8.3
This is a basic solution since pH > 7.
Relationship: pH + pOH = 14 at 25°C
Q9. Explain common ion effect with example of acetate buffer containing CH3COOH and CH3COONa
Common Ion Effect: Addition of a common ion suppresses the dissociation of a weak electrolyte.
Example: Acetate buffer (CH3COOH + CH3COONa)
CH3COOH ⇌ CH3COO- + H+
Ka = [H+][CH3COO-] / [CH3COOH] = 1.8 × 10^-5
Without CH3COONa:
If only 0.1 M CH3COOH, [H+] ≈ 1.34 × 10^-3 M (from previous problem)
With CH3COONa (adding common ion CH3COO-):
Na+ is spectator. [CH3COO-] increases significantly from 0.1 M CH3COONa dissociation.
Let's say mixture has 0.1 M CH3COOH + 0.1 M CH3COONa:
Ka = [H+][0.1 + x] / [0.1 - x] ≈ [H+](0.1) / (0.1)
1.8 × 10^-5 = [H+] × 1
[H+] = 1.8 × 10^-5 M
pH = -log(1.8 × 10^-5) ≈ 4.74
Effect: [H+] decreased from 1.34 × 10^-3 to 1.8 × 10^-5 M due to common ion shift of equilibrium left.
Equilibrium — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. What is the defining characteristic of a chemical equilibrium?
- A. The reaction stops.
- B. The concentrations of reactants and products become equal.
- C. The rate of the forward reaction equals the rate of the reverse reaction.
- D. The change in Gibbs free energy is positive.
Answer: C. The rate of the forward reaction equals the rate of the reverse reaction. — At equilibrium, the forward and reverse reaction rates are equal, leading to no net change in concentrations of reactants and products. It is a dynamic state.
MCQ 2. For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), what is the relationship between Kp and Kc?
- A. Kp = Kc (RT)⁻²
- B. Kp = Kc (RT)²
- C. Kp = Kc (RT)⁻¹
- D. Kp = Kc (RT)
Answer: A. Kp = Kc (RT)⁻² — The relationship is Kp = Kc(RT)Δn, where Δn = (moles of gaseous products) - (moles of gaseous reactants). Here, Δn = 2 - (1+3) = 2 - 4 = -2, so Kp = Kc(RT)⁻².
MCQ 3. Consider the reaction: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). If the pressure of the system at equilibrium is increased, what will happen?
- A. The equilibrium will shift to the right, favoring product formation.
- B. The equilibrium will shift to the left, favoring reactant formation.
- C. The equilibrium will remain unchanged.
- D. The value of Kp will decrease.
Answer: B. The equilibrium will shift to the left, favoring reactant formation. — According to Le Chatelier's principle, increasing pressure favors the side with fewer moles of gas. Here, there are 2 moles of gas on the product side and 1 mole on the reactant side, so the equilibrium shifts to the left.
MCQ 4. According to Brønsted-Lowry theory, an acid is a substance that:
- A. Produces H⁺ ions in water.
- B. Accepts a pair of electrons.
- C. Donates a proton.
- D. Accepts a proton.
Answer: C. Donates a proton. — A Brønsted-Lowry acid is defined as a proton (H⁺) donor, while a Brønsted-Lowry base is a proton acceptor.
MCQ 5. What is the pH of a neutral aqueous solution at 25°C?
- A. 0
- B. 7
- C. 14
- D. Between 0 and 7
Answer: B. 7 — At 25°C, the ionic product of water (Kw) is 1.0 x 10⁻¹⁴. In a neutral solution, [H⁺] = [OH⁻] = 1.0 x 10⁻⁷ M, so pH = -log[H⁺] = 7.
MCQ 6. What is the pH of a 0.001 M HCl solution?
- A. 1
- B. 2
- C. 3
- D. 4
Answer: C. 3 — HCl is a strong acid, so it dissociates completely. Thus, [H⁺] = 0.001 M = 10⁻³ M. The pH is -log[H⁺] = -log(10⁻³) = 3.
MCQ 7. For the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), at 1000 K, the value of Kp is 3.5 atm⁻¹. Calculate Kc at the same temperature. (R = 0.0821 L atm mol⁻¹ K⁻¹)
- A. 2.87 x 10⁻² M⁻¹
- B. 2.87 x 10² M⁻¹
- C. 2.87 M⁻¹
- D. 28.7 M⁻¹
Answer: B. 2.87 x 10² M⁻¹ — Δn = (2) - (2+1) = -1. The relationship is Kp = Kc(RT)Δn, so Kc = Kp / (RT)Δn = Kp * (RT)⁻Δn = Kp * (RT)¹. Kc = 3.5 atm⁻¹ * (0.0821 L atm mol⁻¹ K⁻¹ * 1000 K) = 3.5 * 82.1 = 287.35 M⁻¹ or 2.87 x 10² M⁻¹.
MCQ 8. For an endothermic reaction, increasing the temperature will:
- A. Shift the equilibrium to the left.
- B. Shift the equilibrium to the right.
- C. Not affect the equilibrium position.
- D. Decrease the value of the equilibrium constant.
Answer: B. Shift the equilibrium to the right. — For an endothermic reaction, heat is a reactant. Increasing the temperature adds heat, which shifts the equilibrium to the right to consume the added heat and favor product formation, thereby increasing the equilibrium constant.
MCQ 9. The solubility product (Ksp) of AgCl is 1.6 x 10⁻¹⁰ at 25°C. What is the solubility of AgCl in 0.01 M NaCl solution?
- A. 1.6 x 10⁻⁸ M
- B. 1.6 x 10⁻¹⁰ M
- C. 1.6 x 10⁻¹² M
- D. 1.6 x 10⁻⁵ M
Answer: A. 1.6 x 10⁻⁸ M — Let 's' be the solubility of AgCl. In 0.01 M NaCl, [Cl⁻] is effectively 0.01 M due to the common ion effect. Ksp = [Ag⁺][Cl⁻] = s * 0.01 = 1.6 x 10⁻¹⁰. Thus, s = 1.6 x 10⁻¹⁰ / 0.01 = 1.6 x 10⁻⁸ M.
MCQ 10. Which statement best describes the nature of chemical equilibrium?
- A. All chemical activity stops.
- B. Reactants are completely converted to products.
- C. The forward and reverse reactions continue at equal rates.
- D. The concentrations of reactants are always greater than products.
Answer: C. The forward and reverse reactions continue at equal rates. — Chemical equilibrium is a dynamic state where the forward and reverse reactions occur continuously at the same rate, resulting in no net change in the concentrations of reactants and products.
MCQ 11. What is the effect of adding a catalyst to a reversible reaction at equilibrium?
- A. It shifts the equilibrium to the right.
- B. It shifts the equilibrium to the left.
- C. It increases the rate of the forward reaction only.
- D. It increases the rates of both forward and reverse reactions equally, reaching equilibrium faster.
Answer: D. It increases the rates of both forward and reverse reactions equally, reaching equilibrium faster. — A catalyst speeds up both the forward and reverse reactions equally by lowering the activation energy, allowing the system to reach equilibrium faster without changing the equilibrium position or the value of the equilibrium constant.
MCQ 12. If the Kw for water at 60°C is 9.6 x 10⁻¹⁴, what is the pH of a neutral solution at this temperature?
- A. 7.00
- B. 6.51
- C. 7.49
- D. 13.02
Answer: B. 6.51 — For a neutral solution, [H⁺] = [OH⁻] = √(Kw). So, [H⁺] = √(9.6 x 10⁻¹⁴) = 3.098 x 10⁻⁷ M. The pH = -log[H⁺] = -log(3.098 x 10⁻⁷) ≈ 6.51.
MCQ 13. The dissociation constant of a weak acid HA is Ka. Its conjugate base is A⁻. What is the relationship between Ka, Kb (dissociation constant of A⁻), and Kw?
- A. Ka / Kb = Kw
- B. Ka * Kb = Kw
- C. Ka + Kb = Kw
- D. Ka - Kb = Kw
Answer: B. Ka * Kb = Kw — For a conjugate acid-base pair in an aqueous solution, the product of their dissociation constants (Ka for the acid and Kb for its conjugate base) is equal to the ionic product of water, Kw.
MCQ 14. Which of the following mixtures will form a buffer solution?
- A. HCl and NaOH (strong acid, strong base)
- B. CH₃COOH and CH₃COONa (weak acid, salt of its conjugate base)
- C. NH₄Cl and NaOH (salt of weak base, strong base)
- D. HCl and NaCl (strong acid, salt of strong acid)
Answer: B. CH₃COOH and CH₃COONa (weak acid, salt of its conjugate base) — A buffer solution consists of a weak acid and its conjugate base, or a weak base and its conjugate acid. Acetic acid (CH₃COOH) is a weak acid, and sodium acetate (CH₃COONa) provides its conjugate base (acetate ion, CH₃COO⁻).
MCQ 15. For the reaction aA + bB ⇌ cC + dD, the equilibrium constant Kc is expressed as:
- A. Kc = ([A]ᵃ[B]ᵇ) / ([C]ᶜ[D]ᵈ)
- B. Kc = ([C]ᶜ[D]ᵈ) / ([A]ᵃ[B]ᵇ)
- C. Kc = [C]ᶜ[D]ᵈ[A]ᵃ[B]ᵇ
- D. Kc = ([A]ᵃ + [B]ᵇ) / ([C]ᶜ + [D]ᵈ)
Answer: B. Kc = ([C]ᶜ[D]ᵈ) / ([A]ᵃ[B]ᵇ) — The equilibrium constant Kc is the ratio of the product of the concentrations of products to the product of the concentrations of reactants, with each concentration raised to the power of its stoichiometric coefficient.
MCQ 16. A 0.1 M solution of a weak base B has a Kb of 1.0 x 10⁻⁵. What is the pH of the solution?
- A. 5.0
- B. 9.0
- C. 11.0
- D. 3.0
Answer: C. 11.0 — For B + H₂O ⇌ BH⁺ + OH⁻, Kb = [BH⁺][OH⁻]/[B] = x²/(0.1-x). Assuming x << 0.1, x² = 0.1 * 1.0 x 10⁻⁵ = 1.0 x 10⁻⁶. So, [OH⁻] = x = 1.0 x 10⁻³ M. pOH = -log(1.0 x 10⁻³) = 3. pH = 14 - pOH = 14 - 3 = 11.
MCQ 17. Which of the following statements about the equilibrium constant K is INCORRECT?
- A. It is independent of the initial concentrations of reactants.
- B. It depends on temperature.
- C. It indicates the extent of the reaction.
- D. It changes with the addition of a catalyst.
Answer: D. It changes with the addition of a catalyst. — The equilibrium constant K is independent of initial concentrations and catalysts. It does depend on temperature and indicates the relative amounts of products and reactants at equilibrium.
MCQ 18. Which of the following represents a physical equilibrium?
- A. 2NO₂(g) ⇌ N₂O₄(g)
- B. H₂O(l) ⇌ H₂O(g)
- C. N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
- D. CaCO₃(s) ⇌ CaO(s) + CO₂(g)
Answer: B. H₂O(l) ⇌ H₂O(g) — Physical equilibrium involves a reversible phase change of a substance, such as the evaporation and condensation of water, where the chemical identity remains unchanged.
MCQ 19. For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), what happens if an inert gas like Argon is added to the system at constant volume?
- A. The equilibrium shifts to the right.
- B. The equilibrium shifts to the left.
- C. The equilibrium remains unchanged.
- D. The equilibrium constant decreases.
Answer: C. The equilibrium remains unchanged. — Adding an inert gas at constant volume increases the total pressure but does not change the partial pressures (or concentrations) of the reacting gases. Since equilibrium depends on partial pressures, the equilibrium position remains unchanged.
MCQ 20. A 0.01 M solution of a weak acid HA has a pH of 4. Calculate the percentage dissociation of the acid.
- A. 1%
- B. 10%
- C. 0.1%
- D. 5%
Answer: A. 1% — pH = 4 implies [H⁺] = 10⁻⁴ M. For a weak acid, the degree of dissociation (α) is [H⁺] / C, where C is the initial concentration. α = 10⁻⁴ M / 0.01 M = 10⁻⁴ / 10⁻² = 10⁻² = 0.01. Percentage dissociation = α * 100% = 0.01 * 100% = 1%.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.