Equilibrium — Previous Year Questions (Class 11 Chemistry)
Chemical equilibrium explores dynamic reversible reactions at constant composition. Master Le Chatelier's principle and equilibrium constants for exam exce
TL;DR: Chemical equilibrium explores dynamic reversible reactions at constant composition. Master Le Chatelier's principle and equilibrium constants for exam…
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Chemical equilibrium explores dynamic reversible reactions at constant composition. Master Le Chatelier's principle and equilibrium constants for exam exce
Equilibrium — Previous Year Questions with Solutions
Q (2023, 5 marks): For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), if 1 mole of N₂ and 3 moles of H₂ are taken and 0.5 mole of NH₃ is formed at equilibrium, calculate Kc and Kp at 400 K.
Answer: Initial moles: N₂ = 1, H₂ = 3, NH₃ = 0
Change: N₂ = -0.25, H₂ = -0.75, NH₃ = +0.5
Equilibrium: N₂ = 0.75, H₂ = 2.25, NH₃ = 0.5
Total moles at equilibrium = 0.75 + 2.25 + 0.5 = 3.5
Assuming volume = 1 L:
[N₂] = 0.75 M, [H₂] = 2.25 M, [NH₃] = 0.5 M
Kc = [NH₃]²/([N₂][H₂]³)
Kc = (0.5)²/((0.75)(2.25)³)
Kc = 0.25/(0.75 × 11.39)
Kc = 0.25/8.54 ≈ 0.029
For Kp:
Δn_g = 2 - (1 + 3) = -2
Kp = Kc(RT)^Δn_g = 0.029 × (0.082 × 400)^(-2)
Kp = 0.029 × (32.8)^(-2) = 0.029/1075 ≈ 2.7 × 10^(-5)
Final Answer: Kc ≈ 0.029, Kp ≈ 2.7 × 10^(-5)
Q (2022, 3 marks): State Le Chatelier's principle and explain how pressure change affects the following equilibrium: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
Answer: Le Chatelier's principle: When a stress (change in pressure, temperature, or concentration) is applied to a system at equilibrium, the system shifts to counteract the stress and restore equilibrium.
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
Δn_g = 2 - (2 + 1) = -1 (reaction produces fewer moles of gas)
Effect of pressure increase:
Increased pressure favors the side with fewer gas molecules (right side).
Equilibrium shifts right, producing more SO₃.
Effect of pressure decrease:
Decreased pressure favors the side with more gas molecules (left side).
Equilibrium shifts left, producing more SO₂ and O₂.
Conclusion: Increased pressure shifts equilibrium right (toward products); decreased pressure shifts left.
Final Answer: Pressure increase shifts equilibrium right (forward); pressure decrease shifts left (backward)
Q (2023, 3 marks): Define degree of dissociation (α) and derive the relation between Kc and α for the reaction AB ⇌ A + B if initial concentration is C₀.
Answer: Degree of dissociation (α): Fraction of the initial amount of substance that dissociates/reacts.
For reaction AB ⇌ A + B:
Initial concentration of AB = C₀
At equilibrium: AB dissociates by fraction α
Initial: AB = C₀, A = 0, B = 0
Change: AB = -C₀α, A = +C₀α, B = +C₀α
Equilibrium: AB = C₀(1-α), A = C₀α, B = C₀α
Kc = [A][B]/[AB]
Kc = (C₀α)(C₀α)/(C₀(1-α))
Kc = (C₀α²)/(1-α)
Rearranging: α² + (Kc/C₀)α - Kc/C₀ = 0
Final Answer: Kc = (C₀α²)/(1-α) or C₀α²/(1-α) = Kc
Q (2021, 3 marks): For the reaction PCl₅ ⇌ PCl₃ + Cl₂, if Kc = 0.04 at 500 K, calculate the degree of dissociation if initial concentration of PCl₅ is 0.1 M.
Answer: Using Kc = (C₀α²)/(1-α) with Kc = 0.04, C₀ = 0.1 M:
0.04 = (0.1 × α²)/(1-α)
0.04(1-α) = 0.1α²
0.04 - 0.04α = 0.1α²
0.1α² + 0.04α - 0.04 = 0
Dividing by 0.1: α² + 0.4α - 0.4 = 0
Using quadratic formula:
α = [-0.4 ± √(0.16 + 1.6)]/2
α = [-0.4 ± √1.76]/2
α = [-0.4 ± 1.327]/2
α = 0.927/2 ≈ 0.464 (taking positive root)
Therefore, degree of dissociation = 0.464 or 46.4%
Final Answer: α ≈ 0.464 or 46.4%
Q (2022, 3 marks): Explain the effect of temperature change on equilibrium for an exothermic reaction. Use N₂O₄ ⇌ 2NO₂ as an example (ΔH = +58 kJ/mol).
Answer: Effect of temperature on equilibrium depends on whether the reaction is exothermic or endothermic.
N₂O₄ ⇌ 2NO₂ is endothermic (ΔH = +58 kJ/mol, energy absorbed).
Increasing temperature:
According to Le Chatelier's principle, increased temperature favors the endothermic direction (forward).
Equilibrium shifts right, producing more NO₂ (brown gas).
Kc increases.
Decreasing temperature:
Decreased temperature favors the exothermic direction (backward).
Equilibrium shifts left, producing more N₂O₄ (colorless).
Kc decreases.
Observation: At higher temperatures, the mixture appears more brown (more NO₂); at lower temperatures, more colorless (more N₂O₄).
Final Answer: Temperature increase shifts endothermic equilibrium right, increasing Kc; decrease shifts left, decreasing Kc
Q (2023, 2 marks): Calculate the solubility product (Ksp) of AgCl if the solubility of AgCl is 1.9 × 10^(-5) mol/L at 25°C.
Answer: For AgCl dissolving: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
If solubility = s = 1.9 × 10^(-5) mol/L, then:
[Ag⁺] = s = 1.9 × 10^(-5) M
[Cl⁻] = s = 1.9 × 10^(-5) M
Ksp = [Ag⁺][Cl⁻] = s × s = s²
Ksp = (1.9 × 10^(-5))²
Ksp = 3.61 × 10^(-10)
Final Answer: Ksp = 3.61 × 10^(-10)
Frequently Asked Questions
What is the significance of the equilibrium constant?
The equilibrium constant (K) is a numerical value indicating the extent of a reaction at equilibrium. A large K (>1000) favors products; a small K (<0.001) favors reactants. K is temperature-dependent but independent of pressure, volume, or concentration changes. It allows prediction of the reaction direction and calculation of equilibrium concentrations.
How does a catalyst affect chemical equilibrium?
A catalyst speeds up both forward and backward reactions equally, allowing the system to reach equilibrium faster. However, it does not change the equilibrium position or the equilibrium constant. The concentrations of reactants and products at equilibrium remain the same; only the time to reach equilibrium is reduced.
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