Chapter 6: Equilibrium

6.1 Equilibrium in Physical Processes

Physical equilibrium is the equilibrium established in physical processes such as:

• Solid ⇌ Liquid (melting/freezing at melting point)
• Liquid ⇌ Gas (evaporation/condensation at boiling point)
• Solute in Solid ⇌ Solute in Solution (dissolution/crystallization at saturation)
• Gas ⇌ Gas dissolved in liquid (Henry’s law)

At equilibrium, the rate of the forward process equals the rate of the reverse process.

6.2 Equilibrium in Chemical Processes — Dynamic Equilibrium

In a reversible reaction at equilibrium:

\[ \text{Rate of forward reaction} = \text{Rate of backward reaction} \]

Characteristics of chemical equilibrium:

1. Equilibrium is dynamic (both forward and reverse reactions continue)
2. Equilibrium can be reached from either direction
3. A catalyst does not change the equilibrium position, only speeds up attainment
4. At equilibrium, the concentrations remain constant

6.3 Law of Mass Action and Equilibrium Constant

For a general reversible reaction:

\[ a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D} \]

The equilibrium constant expression (in terms of concentrations):

\[ K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b} \]

In terms of partial pressures (for gaseous reactions):

\[ K_p = \frac{(P_\text{C})^c(P_\text{D})^d}{(P_\text{A})^a(P_\text{B})^b} \]

Relationship between Kp and Kc

\[ K_p = K_c(RT)^{\Delta n_g} \]

where \(\Delta n_g\) = (moles of gaseous products) − (moles of gaseous reactants)

Important Relationships

6.4 Factors Affecting Equilibrium — Le Chatelier’s Principle

“If a system at equilibrium is subjected to a change, the equilibrium will shift in the direction that tends to counteract the change.”

Example: Haber’s process:

\[ \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \quad \Delta H = -92.4;\text{kJ/mol} \]

Favourable conditions: High pressure (forward has fewer moles of gas), Low temperature (exothermic), use of Fe catalyst (for faster rate).

6.5 Ionic Equilibrium in Solution

Electrolytes

• Strong electrolytes: Completely ionize in solution (e.g., NaCl, HCl, NaOH, KNO₃)
• Weak electrolytes: Partially ionize in solution (e.g., CH₃COOH, NH₃, H₂CO₃)

Degree of Ionization (\(\alpha\))

\[ \alpha = \frac{\text{Number of moles ionized}}{\text{Total moles of electrolyte}} \]

6.6 Ionization of Acids and Bases

Arrhenius Concept

• Acid: Produces H⁺ in water
• Base: Produces OH⁻ in water

Brønsted-Lowry Concept

• Acid: Proton (H⁺) donor
• Base: Proton (H⁺) acceptor

Conjugate acid-base pairs:

\[ \text{CH}_3\text{COOH} + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}_3\text{O}^+ \]

Here, CH₃COOH/CH₃COO⁻ is a conjugate acid-base pair.

Lewis Concept

• Acid: Electron pair acceptor (e.g., BF₃, AlCl₃)
• Base: Electron pair donor (e.g., NH₃, H₂O)

6.7 Ionization Constant of Weak Acids (Ka)

For a weak acid HA:

\[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \]

\[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \]

If \(c\) = initial concentration and \(\alpha\) = degree of ionization:

\[ K_a = \frac{c\alpha^2}{1 - \alpha} \approx c\alpha^2 \quad (\text{if } \alpha \ll 1) \]

\[ \alpha = \sqrt{\frac{K_a}{c}} \]

Polybasic Acids

For acids with more than one ionizable hydrogen (e.g., H₃PO₄):

\[ K_{a_1} \gg K_{a_2} \gg K_{a_3} \]

Each successive proton is harder to remove.

6.8 Ionization Constant of Weak Bases (Kb)

\[ \text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \]

\[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} \]

Relationship between Ka and Kb for a conjugate pair:

\[ K_a \times K_b = K_w = 1.0 \times 10^{-14} \quad (\text{at 25°C}) \]

6.9 Concept of pH

Ionic Product of Water (Kw)

\[ \text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^- \]

\[ K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14} \quad (\text{at 25°C}) \]

pH Scale

\[ \text{pH} = -\log[\text{H}^+] \]

\[ \text{pOH} = -\log[\text{OH}^-] \]

\[ \text{pH} + \text{pOH} = 14 \quad (\text{at 25°C}) \]

6.10 Hydrolysis of Salts

Hydrolysis is the reaction of a salt with water, producing acidic or basic solutions.

For a salt of weak acid + strong base (e.g., CH₃COONa):

\[ \text{pH} = 7 + \frac{1}{2}\text{p}K_a + \frac{1}{2}\log c \]

For a salt of strong acid + weak base (e.g., NH₄Cl):

\[ \text{pH} = 7 - \frac{1}{2}\text{p}K_b - \frac{1}{2}\log c \]

6.11 Buffer Solutions

A buffer solution resists changes in pH on the addition of small amounts of acid or base.

Types:

1. Acidic buffer: Weak acid + its conjugate base (salt), e.g., CH₃COOH + CH₃COONa
2. Basic buffer: Weak base + its conjugate acid (salt), e.g., NH₃ + NH₄Cl

Henderson-Hasselbalch Equation

For acidic buffer:

\[ \text{pH} = \text{p}K_a + \log\frac{[\text{Salt}]}{[\text{Acid}]} \]

For basic buffer:

\[ \text{pOH} = \text{p}K_b + \log\frac{[\text{Salt}]}{[\text{Base}]} \]

6.12 Solubility Equilibria — Solubility Product (Ksp)

For a sparingly soluble salt \(\text{A}_x\text{B}_y\):

\[ \text{A}_x\text{B}_y(s) \rightleftharpoons x\text{A}^{y+}(aq) + y\text{B}^{x-}(aq) \]

\[ K_{sp} = [\text{A}^{y+}]^x[\text{B}^{x-}]^y \]

Relationship between \(K_{sp}\) and solubility (\(s\)):

For AB type (e.g., AgCl): \(K_{sp} = s^2\)

For AB₂ type (e.g., PbCl₂): \(K_{sp} = 4s^3\)

Common Ion Effect

The solubility of a sparingly soluble salt decreases in the presence of a common ion.

Example: Solubility of AgCl decreases in NaCl solution (Cl⁻ is the common ion).

Practice Questions

Multiple Choice Questions (MCQs)

1. The equilibrium constant for the reaction \(\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3\) is \(K\). The equilibrium constant for \(\text{NH}_3 \rightleftharpoons \frac{1}{2}\text{N}_2 + \frac{3}{2}\text{H}_2\) is:

 (a) \(\frac{1}{K}\)

 (b) \(K^{1/2}\)

 (c) \(\frac{1}{K^{1/2}}\)

 (d) \(K^2\)

2. The pH of a 0.001 M HCl solution is:

 (a) 1

 (b) 2

 (c) 3

 (d) 4

3. A buffer solution can be prepared by mixing:

 (a) NaCl + HCl

 (b) CH₃COOH + CH₃COONa

 (c) NaOH + NaCl

 (d) HCl + NaOH

4. The solubility product of AgCl is \(1.6 \times 10^{-10}\). Its solubility in mol/L is:

 (a) \(1.26 \times 10^{-5}\)

 (b) \(1.6 \times 10^{-5}\)

 (c) \(4.0 \times 10^{-5}\)

 (d) \(1.6 \times 10^{-10}\)

5. For which reaction does \(K_p = K_c\)?

 (a) \(\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)\)

 (b) \(\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)\)

 (c) \(\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)\)

 (d) \(2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)\)

Short Answer Questions (2–3 Marks)

6. State Le Chatelier’s principle. Predict the direction of shift for the reaction \(\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3 + \text{heat}\) when (i) pressure is increased, (ii) temperature is increased.

7. Calculate the pH of a 0.01 M Ba(OH)₂ solution.

8. The \(K_a\) for HF is \(6.8 \times 10^{-4}\). Calculate the degree of ionization of HF in its 0.1 M solution.

9. Write the Henderson-Hasselbalch equation. Calculate the pH of a buffer made by mixing 0.1 M CH₃COOH and 0.1 M CH₃COONa. (\(K_a = 1.8 \times 10^{-5}\))

10. What is the common ion effect? How does it affect the solubility of AgCl in NaCl solution?

Long Answer Questions (5 Marks)

11. (a) Derive the relationship between \(K_p\) and \(K_c\).

 (b) For the reaction \(\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)\), the value of \(K_c = 0.36\) at 100°C. If the initial concentration of N₂O₄ is 0.1 M, calculate the concentrations at equilibrium.

12. (a) Define ionic product of water. What is its value at 25°C? How does it vary with temperature?

 (b) The pH of gastric juice is 1.2. Calculate the concentration of HCl in gastric juice.

 (c) What is the pH of a 0.002 M NaOH solution?

13. (a) Explain hydrolysis of salts with examples. Why does a solution of sodium acetate have a pH > 7?

 (b) Calculate the pH of 0.1 M NH₄Cl solution. (\(K_b\) for NH₃ = \(1.8 \times 10^{-5}\))

Assertion-Reason Questions

14. Assertion (A): Addition of an inert gas at constant volume does not affect the equilibrium.

Reason (R): The concentrations (partial pressures) of the reactants and products remain unchanged.

15. Assertion (A): The pH of 10⁻⁸ M HCl is not 8.

Reason (R): For very dilute solutions of acids, the contribution of H⁺ from water must be considered.

Answer Key