How to Use This Page
This page provides complete exercise solutions for Chapter 8 — Physical and Chemical Equilibrium of the Samacheer Kalvi Class 11 Chemistry textbook. Work through systematically:
1. Key Concepts — revise formulas and definitions. 2. MCQs through Match the Following — test your conceptual understanding. 3. Brief and Detailed Answers — practice structured responses. 4. ICE Table Numericals — follow each step carefully. 5. Common Mistakes & Exam Checklist — final review before exams.
Key Concepts (Quick Recap)
| Concept | Formula / Statement |
|---|
| Equilibrium | Rate of forward reaction = Rate of backward reaction |
| Kc (concentration) | Kc = [products]^coefficients / [reactants]^coefficients |
| Kp (pressure) | Kp = (partial pressures of products)^coefficients / (partial pressures of reactants)^coefficients |
| Kp and Kc relation | Kp = Kc(RT)^Δn, where Δn = (moles of gaseous products) − (moles of gaseous reactants) |
| Reaction Quotient Q | Same expression as Kc but using initial (non-equilibrium) concentrations |
| Q vs Kc | Q < Kc → forward reaction favoured; Q > Kc → backward reaction favoured; Q = Kc → at equilibrium |
| Le Chatelier's Principle | A system at equilibrium, when disturbed, shifts to counteract the change |
| Ksp | Solubility product = product of ion concentrations raised to stoichiometric powers |
| Ka (acid) | Ka = [H+][A-]/[HA]; stronger acid = larger Ka |
| Kw | Kw = [H+][OH-] = 1 × 10⁻¹⁴ at 25°C |
| pH | pH = −log[H+]; pOH = −log[OH-]; pH + pOH = 14 |
Important Notes on Equilibrium Constants:
- Pure solids and pure liquids do NOT appear in Kc or Kp expressions
- Kc is dimensionless (but concentrations must be in mol/L)
- If Δn = 0, then Kp = Kc
- Larger Kc means products are favoured at equilibrium
- Kc depends only on temperature, NOT on concentration or pressure
I. Choose the Best Answer (MCQs)
1. For the reaction H₂(g) + I₂(g) ⇌ 2HI(g), if the equilibrium concentrations are [H₂] = 0.2 M, [I₂] = 0.2 M, [HI] = 1.6 M, what is Kc?
Answer
(a) 8 (b) 16 (c) 64 (d) 32
Answer(c)
Kc = [HI]² / ([H₂][I₂])
Kc = (1.6)² / (0.2 × 0.2)
Kc = 2.56 / 0.04
Kc = 64
2. If Kp = Kc for a reaction, then:
Answer
(a) Δn > 0 (b) Δn < 0 (c) Δn = 0 (d) Temperature is zero
Answer(c) Kp = Kc(RT)^Δn. If Kp = Kc, then (RT)^Δn = 1, which means Δn = 0. The number of moles of gaseous products equals the number of moles of gaseous reactants.
3. According to Le Chatelier's principle, adding an inert gas at constant volume to a system at equilibrium:
Answer
(a) Shifts equilibrium to the right (b) Shifts equilibrium to the left (c) Has no effect on equilibrium (d) Increases the value of Kc
Answer(c) Adding an inert gas at constant volume does not change the partial pressures or concentrations of the reacting species. Hence, it has no effect on the equilibrium position or Kc. (Note: At constant pressure, adding an inert gas would increase volume and shift equilibrium toward the side with more moles.)
4. The solubility product of AgCl is 1.8 × 10⁻¹⁰. The solubility of AgCl in water is:
Answer
(a) 1.34 × 10⁻⁵ M (b) 3.24 × 10⁻¹⁰ M (c) 1.8 × 10⁻¹⁰ M (d) 1.8 × 10⁻⁵ M
Answer(a)
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
Ksp = [Ag⁺][Cl⁻] = s × s = s²
s² = 1.8 × 10⁻¹⁰
s = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ M
5. Which of the following changes will shift the equilibrium N₂(g) + 3H₂(g) ⇌ 2NH₃(g) (ΔH = −92 kJ) to the right?
Answer
(a) Increase temperature (b) Decrease pressure (c) Remove NH₃ (d) Add an inert gas at constant volume
Answer(c) Removing NH₃ (a product) reduces its concentration. By Le Chatelier's principle, the system shifts to the right to produce more NH₃. Increasing temperature would shift left (reaction is exothermic). Decreasing pressure shifts toward more moles (left side, 4 moles vs 2). Inert gas at constant volume has no effect.
6. For the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), if the initial moles of PCl₅ is 1 and α is the degree of dissociation, the total moles at equilibrium is:
Answer
(a) 1 + α (b) 1 − α (c) 1 + 2α (d) 1 − 2α
Answer(a)
PCl₅ ⇌ PCl₃ + Cl₂
Initial: 1 0 0
Change: −α +α +α
Equilibrium: 1−α α α
Total moles at equilibrium = (1−α) + α + α = 1 + α
7. If the pH of a solution is 3, the concentration of H⁺ ions is:
Answer
(a) 3 × 10⁻⁷ M (b) 1 × 10⁻³ M (c) 3 M (d) 1 × 10¹¹ M
Answer(b)
pH = −log[H⁺]
3 = −log[H⁺]
log[H⁺] = −3
[H⁺] = 10⁻³ = 1 × 10⁻³ M
II. Fill in the Blanks
1. At equilibrium, the rate of the forward reaction equals the rate of the backward (reverse) reaction. 2. The equilibrium constant Kc depends only on temperature. 3. If Q < Kc, the reaction proceeds in the forward direction. 4. For the reaction N₂(g) + O₂(g) ⇌ 2NO(g), Δn = 0, so Kp = Kc. 5. The solubility product is applicable only for sparingly soluble salts. 6. Le Chatelier's principle states that a system at equilibrium counteracts the change imposed on it. 7. An increase in pressure shifts equilibrium toward the side with fewer moles of gas. 8. In the expression for Kc, pure solids and pure liquids are not included (have unit activity). 9. If Kc >> 1, the equilibrium favours products. 10. The ionic product of water Kw = 1 × 10⁻¹⁴ at 25°C. 11. For an exothermic reaction, increasing temperature shifts equilibrium toward reactants (left). 12. The degree of dissociation is represented by the symbol α. 13. If Ksp > ionic product, the solution is unsaturated and more solute can dissolve. 14. The relationship between Ka, Kb, and Kw for a conjugate acid-base pair is Ka × Kb = Kw. 15. Adding a catalyst at equilibrium does not shift the equilibrium position; it only speeds up attainment of equilibrium.
III. True or False (with Corrections)
1. At equilibrium, the concentrations of reactants and products are always equal.
False. At equilibrium, the concentrations of reactants and products are constant (unchanging), but they are not necessarily equal. Their ratio is determined by Kc.
2. A catalyst increases the value of the equilibrium constant.
False. A catalyst speeds up both forward and backward reactions equally. It helps reach equilibrium faster but does not change the equilibrium constant or the equilibrium position. Kc depends only on temperature.
3. For the reaction 2SO₃(g) ⇌ 2SO₂(g) + O₂(g), increasing pressure will shift equilibrium toward SO₃.
True. There are 3 moles of gas on the right (2SO₂ + O₂) and 2 moles on the left (2SO₃). Increasing pressure shifts equilibrium toward fewer moles, i.e., toward SO₃ (left side).
4. Kp = Kc for all reactions involving gases.
False. Kp = Kc(RT)^Δn. Only when Δn = 0 (equal moles of gas on both sides) does Kp = Kc. If Δn is not 0, Kp does not equal Kc.
5. The solubility of a salt decreases when a common ion is added to the solution.
True. This is the common ion effect. Adding an ion that is already present in the equilibrium shifts the equilibrium backward, reducing the solubility of the salt. For example, adding NaCl to a saturated AgCl solution reduces AgCl solubility.
IV. Match the Following
| Column A | Column B |
|---|
| 1. Le Chatelier's principle | (a) [H+][A-]/[HA] |
| 2. Solubility product (Ksp) | (b) System counteracts change |
| 3. Equilibrium constant (Kc) | (c) Q < Kc |
| 4. Acid dissociation constant (Ka) | (d) Ratio of products to reactants at equilibrium |
| 5. Forward reaction favoured | (e) [Ag+][Cl-] |
| 6. pH | (f) −log[H+] |
Answer
1 → (b) Le Chatelier's principle: system counteracts the imposed change
2 → (e) Ksp for AgCl = [Ag+][Cl-]
3 → (d) Kc = ratio of products to reactants at equilibrium
4 → (a) Ka = [H+][A-]/[HA]
5 → (c) Forward reaction favoured when Q < Kc
6 → (f) pH = −log[H+]
V. Answer Briefly (2-mark)
1. What is dynamic equilibrium? Give one example.
Answer
Dynamic equilibrium is a state in which the rate of the forward reaction equals the rate of the backward reaction, and the concentrations of reactants and products remain constant. Both reactions continue to occur, but there is no net change.
Example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
At dynamic equilibrium, N₂ and H₂ combine to form NH₃ at the same
rate as NH₃ decomposes back to N₂ and H₂.
2. Write the expression for Kc and Kp for the reaction: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
Kc = [SO₃]² / ([SO₂]² × [O₂])
Kp = (P_SO₃)² / ((P_SO₂)² × P_O₂)
Relation: Kp = Kc(RT)^Δn
where Δn = 2 − 3 = −1
Kp = Kc(RT)⁻¹ = Kc / (RT)
3. What is the common ion effect? Give one example.
Answer
The common ion effect is the suppression of the dissociation (or solubility) of a weak electrolyte by the addition of a strong electrolyte containing a common ion.
Example: Adding NaCl (strong electrolyte) to a solution of AgCl
increases [Cl⁻], which shifts the equilibrium backward:
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
More Cl⁻ forces the equilibrium left, reducing AgCl solubility.
4. State Le Chatelier's principle. What happens when temperature is increased for an exothermic reaction?
Answer
Le Chatelier's principle states that if a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the system shifts in the direction that counteracts the change.
For an exothermic reaction (ΔH < 0), heat is treated as a product. Increasing temperature adds more "product" (heat), so the system shifts toward reactants (left) to absorb the excess heat.
VI. Answer in Detail (5-mark)
1. Explain Le Chatelier's principle with the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ. Show the effect of (i) adding N₂, (ii) increasing pressure, (iii) increasing temperature, (iv) adding a catalyst, (v) removing NH₃.
Answer
Le Chatelier's principle states that when a system at equilibrium is disturbed, it shifts in the direction that partially counteracts the disturbance to restore a new equilibrium.
For the reaction:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ (exothermic)
Moles of gas: Left = 4 (1 N₂ + 3 H₂), Right = 2 (2 NH₃)
(i) Adding N₂ (increasing [N₂]):
Adding more N₂ increases reactant concentration.
System shifts RIGHT (forward) to consume the added N₂.
Result: More NH₃ is produced. Equilibrium shifts toward products.
(ii) Increasing pressure:
4 moles of gas (left) vs 2 moles of gas (right).
Increasing pressure favours the side with fewer gas moles.
System shifts RIGHT (forward) to reduce total moles.
Result: More NH₃ is produced. This is why the Haber process
uses high pressure (200-300 atm).
(iii) Increasing temperature:
The reaction is exothermic (ΔH = −92 kJ). Heat is a product.
Increasing temperature adds heat, so system shifts LEFT (backward)
to absorb the excess heat.
Result: Less NH₃ is produced. Equilibrium favours reactants.
This is why the Haber process uses a compromise temperature (~450°C)
rather than low temperature.
(iv) Adding a catalyst:
A catalyst speeds up both forward and backward reactions equally.
It does NOT shift the equilibrium position.
It only helps the system reach equilibrium faster.
Result: No change in equilibrium concentrations. Same Kc value.
(v) Removing NH₃:
Removing the product NH₃ decreases its concentration.
System shifts RIGHT (forward) to replace the removed NH₃.
Result: More NH₃ is produced. This is why in the Haber process,
NH₃ is continuously removed by condensation.
Summary table:
| Change | Direction of shift | Effect on NH₃ yield |
|---|
| Add N₂ | Right (forward) | Increases |
| Increase pressure | Right (forward) | Increases |
| Increase temperature | Left (backward) | Decreases |
| Add catalyst | No shift | No change |
| Remove NH₃ | Right (forward) | Increases |
2. Derive the relationship between Kp and Kc. Calculate Kp for the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) given Kc = 1.2 × 10² at 600 K.
Answer
For a general gaseous reaction:
aA(g) + bB(g) ⇌ cC(g) + dD(g)
Kp = (P_C)^c × (P_D)^d / ((P_A)^a × (P_B)^b)
Kc = [C]^c × [D]^d / ([A]^a × [B]^b)
Using ideal gas equation: P = (n/V)RT = CRT (where C is concentration)
P_A = [A]RT, P_B = [B]RT, P_C = [C]RT, P_D = [D]RT
Substituting into Kp expression:
Kp = ([C]RT)^c × ([D]RT)^d / (([A]RT)^a × ([B]RT)^b)
Kp = ([C]^c × [D]^d) / ([A]^a × [B]^b) × (RT)^(c+d−a−b)
Kp = Kc × (RT)^Δn
where Δn = (c + d) − (a + b)
= moles of gaseous products − moles of gaseous reactants
Numerical calculation:
Given: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
Kc = 1.2 × 10², T = 600 K
R = 0.0821 L·atm mol⁻¹ K⁻¹
Δn = n_products(g) − n_reactants(g) = 2 − 3 = −1
Kp = Kc × (RT)^Δn
Kp = Kc × (RT)⁻¹
Kp = Kc / (RT)
Kp = (1.2 × 10²) / (0.0821 × 600)
Kp = 120 / 49.26
Kp = 2.44
So Kp = 2.44 atm⁻¹ (since Δn = −1, the unit is atm⁻¹)
3. Explain ICE table method with a worked example. At 700 K, 2.0 moles of PCl₅ are taken in a 2 L flask. At equilibrium, 0.4 moles of PCl₃ are formed. Calculate Kc and the degree of dissociation (α).
Answer
The ICE table method is a systematic way to solve equilibrium problems:
- I = Initial concentrations
- C = Change in concentrations
- E = Equilibrium concentrations
Step-by-step solution:
Reaction: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
Initial moles: PCl₅ = 2.0, PCl₃ = 0, Cl₂ = 0
Volume = 2 L
ICE Table (in moles):
PCl₅ ⇌ PCl₃ + Cl₂
Initial: 2.0 0 0
Change: −x +x +x
Equilibrium: 2.0−x x x
Given: At equilibrium, moles of PCl₃ = 0.4
So: x = 0.4
Equilibrium moles:
PCl₅ = 2.0 − 0.4 = 1.6
PCl₃ = 0.4
Cl₂ = 0.4
Equilibrium concentrations (divide by volume = 2 L):
[PCl₅] = 1.6/2 = 0.8 M
[PCl₃] = 0.4/2 = 0.2 M
[Cl₂] = 0.4/2 = 0.2 M
Kc = [PCl₃][Cl₂] / [PCl₅]
Kc = (0.2 × 0.2) / 0.8
Kc = 0.04 / 0.8
Kc = 0.05
Degree of dissociation (α):
α = (moles dissociated) / (initial moles) = x / 2.0 = 0.4 / 2.0 = 0.2
α = 0.2 or 20%
Verification using Kc formula with α:
Kc = (α² × C) / (1 − α)
where C = initial concentration = 2.0/2.0 = 1.0 M
Kc = (0.04 × 1.0) / (1 − 0.2) = 0.04/0.8 = 0.05 ✓
Common Mistakes to Avoid
1. Including pure solids and liquids in Kc expressions — pure solids (e.g., CaCO₃(s)) and pure liquids (e.g., H₂O(l) as solvent) have unit activity and do NOT appear in equilibrium constant expressions.
2. Confusing Kc with Q (reaction quotient) — Kc uses equilibrium concentrations; Q uses initial or any concentrations. They are different unless the system is at equilibrium.
3. Forgetting to convert moles to concentrations — Kc requires molar concentrations (mol/L), not moles. Always divide moles by volume.
4. Getting the direction of shift wrong with Le Chatelier's principle — remember: the system shifts to OPPOSE the change. Adding reactant shifts toward products; adding product shifts toward reactants.
5. Using wrong Δn in Kp-Kc relation — Δn counts only GASEOUS species. Solid and liquid moles are excluded. Δn = (moles of gaseous products) − (moles of gaseous reactants).
6. Assuming catalyst shifts equilibrium — a catalyst only speeds up reaching equilibrium; it does NOT change Kc or the equilibrium position.
7. Confusing solubility (s) with solubility product (Ksp) — Ksp is the product of ion concentrations at saturation. Solubility 's' is the molar concentration that dissolves. They are related but different quantities.
8. Not squaring/cubing concentrations in Kc — if the stoichiometric coefficient is 2, the concentration must be squared; if 3, it must be cubed. [A]² is not equal to [2A].
Exam Checklist
- [ ] Can you write Kc and Kp expressions for any reversible reaction?
- [ ] Can you convert between Kp and Kc using Kp = Kc(RT)^Δn?
- [ ] Can you set up and solve an ICE table for equilibrium problems?
- [ ] Can you apply Le Chatelier's principle to predict equilibrium shifts?
- [ ] Do you know the effect of concentration, pressure, temperature, and catalyst on equilibrium?
- [ ] Can you calculate Ksp and solubility from each other?
- [ ] Can you use the reaction quotient Q to predict direction of reaction?
- [ ] Can you explain the common ion effect with examples?
- [ ] Do you understand the relationship between Ka, Kb, and Kw?
- [ ] Can you calculate pH, pOH, [H+], and [OH-] from given data?
