CBSE Class 11 Chemistry Chapter 6 Equilibrium Worksheet with Answers

Quick Chapter Recap: Equilibrium Concepts

• Equilibrium constant expression: Kc = [Products]ⁿ/[Reactants]ᵐ at constant temperature
• Relationship between Kp and Kc: Kp = Kc(RT)^Δn where Δn = moles of gaseous products - moles of gaseous reactants
• Henderson-Hasselbalch equation for buffers: pH = pKa + log([Salt]/[Acid])
• Common ion effect reduces ionization of weak electrolytes when a salt with common ion is added
• Solubility product (Ksp) governs precipitation and dissolution of sparingly soluble salts

Section A: Multiple Choice Questions (MCQs)

• Q1. For the equilibrium 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), the equilibrium constant Kp is related to Kc by: (A) Kp = Kc(RT) (B) Kp = Kc/(RT) (C) Kp = Kc(RT)² (D) Kp = Kc
• Q2. The pH of 0.001 M HCl solution is: (A) 1 (B) 2 (C) 3 (D) 4
• Q3. A buffer solution is prepared by mixing CH₃COOH and CH₃COONa. Which of the following will NOT change its pH significantly? (A) Adding small amount of NaOH (B) Adding small amount of HCl (C) Dilution (D) All of these
• Q4. According to Le Chatelier's principle, increasing pressure will favor the formation of products in: (A) N₂ + O₂ ⇌ 2NO (B) H₂ + I₂ ⇌ 2HI (C) N₂ + 3H₂ ⇌ 2NH₃ (D) PCl₅ ⇌ PCl₃ + Cl₂
• Q5. The ionic product of water (Kw) at 25°C is: (A) 1×10⁻⁷ (B) 1×10⁻¹⁴ (C) 1×10⁻¹² (D) 1×10⁻¹⁰
• Q6. Which of the following salts will give basic solution on hydrolysis? (A) NH₄Cl (B) NaCl (C) CH₃COONa (D) Na₂SO₄

Section B: Fill in the Blanks

• Q7. The state in which the rate of forward reaction equals the rate of backward reaction is called __________.
• Q8. For a general reaction aA + bB ⇌ cC + dD, the equilibrium constant expression is Kc = __________.
• Q9. The pH of a neutral solution at 25°C is __________.
• Q10. A solution that resists change in pH on addition of small amounts of acid or base is called a __________ solution.
• Q11. The common ion effect is based on __________ principle.
• Q12. The degree of ionization of a weak acid __________ (increases/decreases) on addition of a salt having a common ion.

Section C: Match the Following

• Column A: (13) Kp = Kc(RT)^Δn | (14) Henderson-Hasselbalch equation | (15) Le Chatelier's principle | (16) Ionic product of water | (17) Common ion effect
• Column B: (A) Kw = [H⁺][OH⁻] | (B) Suppression of ionization | (C) pH = pKa + log([Salt]/[Acid]) | (D) Relation between Kp and Kc | (E) Predicts shift in equilibrium

Section D: Short Answer Questions (2-3 marks each)

• Q18. (2 marks) State the law of chemical equilibrium. Write the expression for Kc for the reaction: 2NO(g) + O₂(g) ⇌ 2NO₂(g)
• Q19. (3 marks) Calculate the pH of 0.01 M HNO₃ solution. If this solution is diluted 100 times, what will be the pH of the resulting solution?
• Q20. (2 marks) What is a buffer solution? Give one example each of acidic and basic buffer.
• Q21. (3 marks) For the reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g); ΔH = -92 kJ. State and explain the effect of (i) increasing temperature and (ii) increasing pressure on the equilibrium.
• Q22. (2 marks) Explain the common ion effect with an example. How does it affect the ionization of a weak acid?

Section E: Long Answer and HOTS Questions (5 marks each)

• Q23. (5 marks) Derive the relationship between Kp and Kc for a general gaseous equilibrium reaction. For the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), Kc = 100 at 727°C. Calculate Kp. (R = 0.082 L atm K⁻¹ mol⁻¹)
• Q24. (5 marks) A buffer solution contains 0.2 M CH₃COOH and 0.2 M CH₃COONa. Calculate the pH of this buffer. (Ka for CH₃COOH = 1.8×10⁻⁵). What will be the pH after addition of 0.01 mole of HCl to 1 litre of this buffer? Explain why buffer solutions resist pH change.
• Q25. (5 marks) The solubility product of AgCl is 1.8×10⁻¹⁰ at 25°C. (a) Calculate the solubility of AgCl in pure water. (b) Calculate the solubility of AgCl in 0.01 M NaCl solution. (c) Explain the difference in solubility values based on common ion effect.

Section F: Case Study Question

Complete Answer Key with Explanations

• Section A Answers: Q1-(B) Kp=Kc/(RT) because Δn=-1 for this reaction. Q2-(C) pH=3 as [H⁺]=0.001=10⁻³. Q3-(C) Dilution changes pH slightly; buffers resist change to acid/base addition. Q4-(C) N₂+3H₂⇌2NH₃ has Δn=-2, high pressure favors fewer moles. Q5-(B) Kw=1×10⁻¹⁴ at 25°C. Q6-(C) CH₃COONa is salt of weak acid-strong base, undergoes hydrolysis to give basic solution.
• Section B Answers: Q7-chemical equilibrium or dynamic equilibrium. Q8-[C]^c[D]^d/[A]^a[B]^b. Q9-7. Q10-buffer. Q11-Le Chatelier's. Q12-decreases (common ion suppresses ionization).
• Section C Answers: Q13-D, Q14-C, Q15-E, Q16-A, Q17-B.
• Q18 Answer: Law of chemical equilibrium states that at constant temperature, the ratio of product of concentrations of products to product of concentrations of reactants, each raised to their stoichiometric coefficients, is constant. Kc=[NO₂]²/[NO]²[O₂].
• Q19 Answer: For 0.01 M HNO₃, [H⁺]=0.01=10⁻², pH=2. After 100-fold dilution, [H⁺]=10⁻⁴ M, pH=4. Note: pH changes by 2 units on 100-fold dilution.
• Q20 Answer: Buffer solution resists pH change on addition of small amounts of acid or base. Acidic buffer example: CH₃COOH + CH₃COONa. Basic buffer example: NH₄OH + NH₄Cl.
• Q21 Answer: (i) Increasing temperature shifts equilibrium backward (endothermic direction) as reaction is exothermic, decreasing NH₃ yield. (ii) Increasing pressure shifts equilibrium forward (toward fewer moles) as Δn=-2, increasing NH₃ yield per Le Chatelier principle.
• Q22 Answer: Common ion effect: suppression of ionization of weak electrolyte by adding a strong electrolyte with common ion. Example: ionization of CH₃COOH decreases when CH₃COONa is added because CH₃COO⁻ ion is common. This shifts equilibrium CH₃COOH⇌CH₃COO⁻+H⁺ backward, reducing [H⁺].
• Q23 Answer: Derivation shown in example above. For 2SO₂+O₂⇌2SO₃, Δn=2-3=-1. T=1000K. Kp=Kc(RT)⁻¹=100/(0.082×1000)=1.22 atm⁻¹.
• Q24 Answer: Using Henderson-Hasselbalch equation: pH=pKa+log([Salt]/[Acid])=-log(1.8×10⁻⁵)+log(0.2/0.2)=4.74+0=4.74. After adding 0.01 mol HCl to 1L: HCl reacts with CH₃COO⁻ forming CH₃COOH. New [CH₃COOH]=0.2+0.01=0.21M, [CH₃COO⁻]=0.2-0.01=0.19M. New pH=4.74+log(0.19/0.21)=4.74-0.044=4.70. pH change is only 0.04, demonstrating buffer action. Buffers resist pH change because added H⁺ is consumed by conjugate base (CH₃COO⁻) and added OH⁻ by weak acid (CH₃COOH).
• Q25 Answer: (a) In pure water: AgCl⇌Ag⁺+Cl⁻. If solubility=s, then Ksp=[Ag⁺][Cl⁻]=s×s=s². s=√(1.8×10⁻¹⁰)=1.34×10⁻⁵ mol/L. (b) In 0.01M NaCl: [Cl⁻]=0.01 (from NaCl, common ion). Ksp=[Ag⁺](0.01)=1.8×10⁻¹⁰. [Ag⁺]=1.8×10⁻⁸ mol/L = solubility. (c) Solubility decreases from 1.34×10⁻⁵ to 1.8×10⁻⁸ due to common ion effect; excess Cl⁻ shifts equilibrium backward, reducing AgCl dissolution.
• Case Study Answers: Q26(i)-High pressure (200-300 atm) shifts equilibrium toward product side as product has fewer moles (Δn=-2), increasing ammonia yield. Q26(ii)-High temperature increases reaction rate and catalyst efficiency; although equilibrium shifts backward, faster attainment of equilibrium compensates. Economic balance between yield and rate. Q26(iii)-Continuous removal of NH₃ decreases product concentration, shifting equilibrium forward to produce more NH₃ per Le Chatelier principle, increasing overall yield. Q26(iv)-Iron catalyst increases both forward and backward reaction rates equally, helping system reach equilibrium faster without changing Kc value or equilibrium position.

How to Use This Worksheet Effectively

• Attempt worksheet in one 90-minute sitting without breaks to simulate exam pressure
• Cover answer key section before starting; no peeking during the test
• Show all working for numerical questions even if final answer is correct
• For HOTS questions, write explanations in complete sentences with proper chemical terminology
• After self-assessment, maintain an error log noting concepts that need revision
• Redo incorrect numerical problems after reviewing NCERT solved examples
• Time yourself on case-study questions; these require quick comprehension and application
• Target minimum 75 percent score before moving to next chapter practice

Why Equilibrium Matters for CBSE Class 11 and Beyond

• Equilibrium concepts directly link to Class 12 chapters: electrochemistry (Nernst equation), kinetics (rate-equilibrium relation), solutions (colligative properties)
• JEE Main 2024 had 2-3 questions (8-12 marks) from equilibrium, especially pH and Ksp calculations
• NEET 2024 included buffer solution and Le Chatelier principle application questions in Chemistry section
• CBSE 2024 Class 11 term exams saw 8-10 marks from Equilibrium chapter across paper patterns
• Industrial chemistry case studies (Haber, Contact process) are high-scoring if concepts are clear
• Strong foundation in ionic equilibrium simplifies Class 12 topics like salt hydrolysis and electrochemical cells