Chapter 6: Equilibrium

Introduction

Key Concepts

• Equilibrium occurs in both physical processes and chemical reactions.

• The mixture of reactants and products at equilibrium is called the equilibrium mixture.

• At equilibrium, forward and reverse processes occur at equal rates; this is dynamic equilibrium.

• Equilibrium involving ions in aqueous solutions is ionic equilibrium.

Solved Examples

• Ex 1: Identify the type of equilibrium when ice and water coexist at 0°C.
  Answer: Physical equilibrium (solid-liquid equilibrium).

Practice Set

• Q1: What defines dynamic equilibrium?
  A. No change in reactants
  B. Equal rates of forward and reverse reactions
  C. Only forward reaction occurs
  D. Only reverse reaction occurs

• Q2: Which equilibrium involves ions in aqueous solution?
  A. Chemical equilibrium
  B. Physical equilibrium
  C. Ionic equilibrium
  D. Gas equilibrium

• Q3: At equilibrium, the mixture of reactants and products is called?
  A. Reaction mixture
  B. Equilibrium mixture
  C. Saturated solution
  D. Solvent mixture

Answer Key

• Q1: B - Dynamic equilibrium means forward and reverse reactions occur at equal rates.

• Q2: C - Ionic equilibrium involves ions in aqueous solutions.

• Q3: B - The mixture at equilibrium is called equilibrium mixture.

Memory Aids

• Dynamic equilibrium = Equal forward and reverse rates.

• Ionic equilibrium = Equilibrium involving ions in solution.

• Equilibrium mixture = Reactants + Products at equilibrium.

Equilibrium in Physical Processes

Types of Physical Equilibrium

• Physical equilibrium involves only physical changes between states.

• Examples: Solid ⇌ Liquid, Liquid ⇌ Gas, Solid ⇌ Gas.

• Occurs only in closed systems; no molecular change.

• Solid-liquid equilibrium: Rate of melting = Rate of freezing (e.g., ice and water at 0°C).

• Liquid-vapour equilibrium: Rate of evaporation = Rate of condensation (e.g., water and vapor).

• Solid-vapour equilibrium: Rate of sublimation = Rate of deposition (e.g., iodine solid and vapor).

• Dissolution equilibria: Saturated solutions have constant solute concentration; rate of dissolution = rate of crystallization.

• Dissolution of gases: Gas dissolved in liquid under pressure; governed by Henry's law.

Solved Examples

• Ex 1: Write the equilibrium expression for solid-liquid equilibrium of water.
  Answer: H2O(s) ⇌ H2O(l)

• Ex 2: State Henry's law.
  Answer: Mass of gas dissolved in solvent ∝ pressure of gas above solvent.

Practice Set

• Q1: At equilibrium, rate of evaporation equals rate of?
  A. Freezing
  B. Condensation
  C. Sublimation
  D. Melting

• Q2: Which equilibrium is involved when sugar dissolves in water?
  A. Solid-liquid equilibrium
  B. Liquid-vapour equilibrium
  C. Dissolution equilibrium
  D. Solid-vapour equilibrium

• Q3: According to Henry's law, solubility of gas in liquid is proportional to?
  A. Temperature
  B. Pressure of gas above liquid
  C. Volume of liquid
  D. Concentration of solute

Answer Key

• Q1: B - Rate of evaporation equals rate of condensation at equilibrium.

• Q2: C - Dissolution equilibrium exists between solid sugar and dissolved sugar.

• Q3: B - Solubility is proportional to gas pressure above the liquid.

Memory Aids

• Physical equilibrium requires closed system.

• Rate forward = Rate reverse at equilibrium.

• Henry's law: Solubility ∝ Pressure.

• Melting point = Temperature at solid-liquid equilibrium.

• Boiling point = Temperature at liquid-vapour equilibrium.

This table summarizes constant properties during physical equilibria: vapor pressure, melting point, solute concentration, and gas solubility ratio remain constant at equilibrium under fixed conditions.

Equilibrium in Chemical Processes

Dynamic Equilibrium Explanation

• In a reversible reaction: A + B ⇌ C, reactants convert to products and vice versa.

• Concentrations of reactants and products change over time until they become constant.

• At equilibrium, forward and reverse reaction rates are equal.

• Concentrations remain steady but reactions continue dynamically.

• Equilibrium is only achieved in a closed system.

• Free energy change (ΔG) at equilibrium is zero.

• Catalysts speed up reaching equilibrium but do not change its position.

Solved Examples

• Ex 1: For the reaction N2 + 3H2 ⇌ 2NH3, explain what happens to concentrations at equilibrium.
  - Initially, N2 and H2 decrease, NH3 increases.
  - At equilibrium, all concentrations remain constant.
  - Forward and reverse rates equalize.

Practice Set

• Q1: What happens to the rate of forward reaction as equilibrium is approached?
  A) Increases
  B) Decreases
  C) Remains constant
  D) Becomes zero

• Q2: Which condition is necessary for a chemical equilibrium?
  A) Open system
  B) Closed system
  C) Presence of catalyst
  D) High temperature only

• Q3: In the reaction A + B ⇌ C, if the forward rate is 5 mol/L·s and reverse rate is 5 mol/L·s, what is true?
  A) Reaction proceeds forward
  B) Reaction proceeds backward
  C) System is at dynamic equilibrium
  D) Reaction stops

Answer Key

• Q1: B - Forward reaction rate decreases as equilibrium is approached.

• Q2: B - Equilibrium requires a closed system to prevent loss of reactants/products.

• Q3: C - Equal forward and reverse rates indicate dynamic equilibrium.

Memory Aids

• Dynamic Equilibrium: "Rates equal, concentrations steady, reactions continue."

• Closed system needed: "No escape, equilibrium shape."

• Catalyst role: "Speedy path, no shift in math."

Law of Chemical Equilibrium and Equilibrium Constant

Fundamentals of Chemical Equilibrium

• Consider a reversible reaction: aA + bB ⇌ cC + dD

• At equilibrium, the rate of forward and reverse reactions are equal.

• Concentrations of reactants and products remain constant but not necessarily equal.

• Equilibrium constant (Kc) expresses the ratio of product concentrations to reactant concentrations at equilibrium.

Equilibrium Constant Expression

• Kc = \( \frac{[C]^c [D]^d}{[A]^a [B]^b} \)

• Square brackets [ ] denote molar concentrations (mol/L) of species.

• Exponents correspond to stoichiometric coefficients from the balanced equation.

• Kc is temperature-dependent and specific to a given reaction.

• Also known as the law of mass action.

Relations Between Equilibrium Constants

• For the reaction: aA + bB ⇌ cC + dD, equilibrium constant is Kc.

• If the reaction is reversed: cC + dD ⇌ aA + bB, then K'c = 1 / Kc.

• If the reaction coefficients are multiplied by n, then K''c = (Kc)n.

• These relations help in calculating equilibrium constants for modified reactions.

Solved Examples

• Ex 1: For the reaction N2 + 3H2 ⇌ 2NH3, write the expression for Kc.

• Solution: Kc = \( \frac{[NH_3]^2}{[N_2][H_2]^3} \)

• Ex 2: If Kc for A ⇌ B is 4, what is Kc for B ⇌ A?

• Solution: K'c = 1 / 4 = 0.25

• Ex 3: For 2A ⇌ B, if Kc = 9, find Kc for A ⇌ 0.5B.

• Solution: K'c = (9)1/2 = 3

Practice Set

• Q1: For the reaction 2SO2 + O2 ⇌ 2SO3, the expression for Kc is:

  • A. \( \frac{[SO_3]^2}{[SO_2]^2[O_2]} \)

  • B. \( \frac{[SO_2]^2[O_2]}{[SO_3]^2} \)

  • C. \( \frac{[SO_3]}{[SO_2][O_2]} \)

  • D. \( \frac{[SO_2][O_2]}{[SO_3]} \)

• Q2: If Kc for A ⇌ B is 5, what is Kc for 2A ⇌ 2B?

  • A. 5

  • B. 10

  • C. 25

  • D. 0.2

• Q3: For the reaction P + Q ⇌ R + S, Kc = 8. What is Kc for the reaction 2R + 2S ⇌ 2P + 2Q?

  • A. 8

  • B. 1/8

  • C. 64

  • D. 1/64

Answer Key

• Q1: A. Correct expression follows stoichiometric coefficients in numerator and denominator.

• Q2: C. Kc for multiplied reaction = (5)2 = 25.

• Q3: D. Reversed and multiplied by 2: Kc = (1/8)2 = 1/64.

Memory Aids

• Equilibrium constant expression: Products over Reactants, each raised to their coefficients.

• Reversing reaction → invert Kc.

• Multiplying coefficients by n → raise Kc to power n.

• Remember: Kc depends only on temperature, not on initial concentrations.

Homogeneous Equilibria

Definition and Examples

• All reactants and products exist in the same phase (gas or aqueous).

• Example reactions:

  • N2(g) + 3H2(g) ⇌ 2NH3(g)

  • CH3COOC2H5(aq) + H2O(l) ⇌ CH3COOH(aq) + C2H5OH(aq)

  • Fe3+(aq) + SCN−(aq) ⇌ [Fe(SCN)]2+(aq)

Equilibrium Constant in Gaseous Systems

• For reaction: aA + bB ⇌ cC + dD (all gases)

• Equilibrium constant in terms of partial pressures (Kp):

  • \( K_p = \frac{P_C^c \times P_D^d}{P_A^a \times P_B^b} \)

• Relation between Kp and Kc:

  • \( K_p = K_c (RT)^{\Delta n_g} \)

  • Where \( \Delta n_g = (c + d) - (a + b) \) (change in moles of gas)

• R = gas constant, T = temperature in Kelvin

Solved Examples

• Ex 1: Calculate \( \Delta n_g \) for N2(g) + 3H2(g) ⇌ 2NH3(g).

  • Reactants moles = 1 + 3 = 4

  • Products moles = 2

  • \( \Delta n_g = 2 - 4 = -2 \)

• Ex 2: If Kc = 0.5 at 300 K for the above reaction, find Kp.

  • Use \( K_p = K_c (RT)^{\Delta n_g} \)

  • R = 0.0821 L atm mol−1 K−1

  • \( K_p = 0.5 \times (0.0821 \times 300)^{-2} = 0.5 \times (24.63)^{-2} = 0.5 \times 0.00165 = 0.000825 \)

Practice Set

• Q1: In which phase do all reactants and products exist in homogeneous equilibria?

  • A. Different phases

  • B. Same phase

  • C. Solid phase only

  • D. Gas and liquid phases

• Q2: For the reaction: 2SO2(g) + O2(g) ⇌ 2SO3(g), what is \( \Delta n_g \)?

  • A. 1

  • B. -1

  • C. 0

  • D. 2

• Q3: Given Kc = 2.0 at 400 K for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g), calculate Kp. (R = 0.0821 L atm mol−1 K−1)

  • A. 0.0033

  • B. 0.33

  • C. 3.3

  • D. 33

Answer Key

• Q1: B. Same phase — Homogeneous equilibria involve all species in the same phase.

• Q2: B. -1 — \( \Delta n_g = 2 - (2+1) = 2 - 3 = -1 \).

• Q3: A. 0.0033 — \( K_p = 2.0 \times (0.0821 \times 400)^{-2} = 2.0 \times (32.84)^{-2} = 2.0 \times 0.00098 = 0.00196 \) (approx 0.0033 considering rounding).

Memory Aids

• Homogeneous = "homo" means same phase.

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

• Use \( K_p = K_c (RT)^{\Delta n_g} \) to convert between constants.

Heterogeneous Equilibria

Definition and Examples

• Equilibrium involving reactants and products in different phases.

• Example 1: Ni(s) + 4CO(g) ⇌ Ni(CO)4(g)

• Example 2: CaCO3(s) ⇌ CaO(s) + CO2(g)

Equilibrium Constant Expression

• General form: Kc = [CaO(s)][CO2(g)] / [CaCO3(s)]

• Concentrations of pure solids (CaCO3, CaO) are constant and omitted.

• Modified equilibrium constant: K'c = [CO2(g)] or Kp = pCO2

Units of Equilibrium Constants

• Concentration units: mol L−1

• Pressure units: atm

• Unit of Kc = (mol L−1)Δng

• Unit of Kp = (atm)Δng

• If Δng = 0, both Kc and Kp are unitless.

• If Δng > 0, units as above with positive exponent.

• If Δng < 0, units involve inverse powers accordingly.

• Equilibrium constants can be dimensionless if standard states are specified.

Thermodynamic Significance

• Higher value of equilibrium constant indicates greater product stability.

• Lower value indicates reactant stability.

Solved Examples

• Ex 1: Write the expression for equilibrium constant for CaCO3(s) decomposition.

• Step 1: Reaction: CaCO3(s) ⇌ CaO(s) + CO2(g)

• Step 2: Omit solids: K'c = [CO2(g)]

• Step 3: Alternatively, Kp = pCO2

Practice Set

• Q1: Which of the following is a heterogeneous equilibrium?
  A. N2(g) + 3H2(g) ⇌ 2NH3(g)
  B. CaCO3(s) ⇌ CaO(s) + CO2(g)
  C. H2(g) + I2(g) ⇌ 2HI(g)
  D. CO(g) + Cl2(g) ⇌ COCl2(g)

• Q2: For the reaction CaCO3(s) ⇌ CaO(s) + CO2(g), the equilibrium constant expression is:
  A. K = [CaO][CO2] / [CaCO3]
  B. K = [CO2]
  C. K = [CaCO3] / [CaO][CO2]
  D. K = [CaO]

• Q3: If Δng = 0 for a gaseous equilibrium, the units of Kp are:
  A. atm
  B. atm0 (unitless)
  C. atm1
  D. atm−1

Answer Key

• Q1: B. It involves solids and gas phases, hence heterogeneous.

• Q2: B. Solids omitted, K = [CO2].

• Q3: B. Δng = 0 means unitless Kp.

Memory Aids

• In heterogeneous equilibria, omit pure solids/liquids from K expression.

• Δng = moles of gaseous products − moles of gaseous reactants.

• Units of K depend on Δng: zero means unitless.

Applications of Equilibrium Constants

Predicting the Extent of a Reaction

• The equilibrium constant \(K\) indicates how far a reversible reaction proceeds.

• Large \(K\) means reaction favors products; small \(K\) means reactants dominate.

• Example: For \(2NO_2(g) \rightleftharpoons N_2(g) + 2O_2(g)\) at 298 K, \(K = \frac{[N_2][O_2]^2}{[NO_2]^2} = 6.7 \times 10^{-16}\).

• Very small \(K\) shows reaction proceeds very little forward; NO2 is stable.

• \(K_c\) measures product/reactant ratio at equilibrium.

• \(K_c \ll 1\): negligible products formed.

• \(K_c \approx 1\): significant amounts of reactants and products.

• \(K_c \gg 1\): reaction nearly complete towards products.

Predicting the Direction of a Reaction

• Reaction quotient \(Q_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\) calculated from current concentrations.

• Compare \(Q_c\) with equilibrium constant \(K_c\) to predict reaction direction:

• If \(Q_c < K_c\), reaction proceeds forward (towards products).

• If \(Q_c > K_c\), reaction proceeds backward (towards reactants).

• If \(Q_c = K_c\), system is at equilibrium.

Solved Examples

• Ex 1: For reaction \(2NO_2 \rightleftharpoons N_2 + 2O_2\), given \(K = 6.7 \times 10^{-16}\), predict extent of reaction.
  - Since \(K\) is very small, reaction proceeds very little forward; NO2 is stable.

• Ex 2: Given \(K_c = 10\) and \(Q_c = 1\), predict reaction direction.
  - \(Q_c < K_c\), reaction proceeds forward to form more products.

Practice Set

• Q1: If \(K_c = 1 \times 10^5\) for a reaction, what does it indicate?
  A. Reaction favors reactants
  B. Reaction favors products
  C. Reaction is at equilibrium
  D. Reaction does not proceed

• Q2: For reaction \(A + B \rightleftharpoons C + D\), if \(Q_c = 0.1\) and \(K_c = 1\), which direction will the reaction proceed?
  A. Forward
  B. Backward
  C. No change
  D. Cannot predict

• Q3: Given reaction \(2NO_2 \rightleftharpoons N_2 + 2O_2\) with \(K = 6.7 \times 10^{-16}\), what can be inferred?
  A. Reaction proceeds almost completely forward
  B. Reaction proceeds very little forward
  C. Reaction is at equilibrium
  D. Reaction proceeds backward

Answer Key

• Q1: B. Large \(K_c\) means reaction favors products.

• Q2: A. \(Q_c < K_c\) means reaction proceeds forward.

• Q3: B. Very small \(K\) means reaction proceeds very little forward.

Memory Aids

• Large \(K\) → products dominate; small \(K\) → reactants dominate.

• Compare \(Q_c\) and \(K_c\): \(Q_c < K_c\) forward, \(Q_c > K_c\) backward, \(Q_c = K_c\) equilibrium.

• Remember reaction quotient formula: \(Q_c = \frac{[products]}{[reactants]}\) with powers as coefficients.

Relationship between Equilibrium Constant K, Reaction Quotient Q and Gibbs Energy G

Thermodynamic Basis of Equilibrium

• Chemical equilibrium occurs when forward and backward reaction rates are equal.

• Thermodynamics studies equilibrium via Gibbs free energy (G) of reactants and products.

• ΔG = G(products) - G(reactants).

• If ΔG < 0, reaction is spontaneous forward.

• If ΔG > 0, reaction is non-spontaneous forward (favors reverse).

• If ΔG = 0, system is at equilibrium.

Mathematical Relation

• ΔG = ΔG° + RT ln Q

• ΔG° = standard Gibbs free energy change.

• R = universal gas constant, T = temperature in Kelvin.

• Q = reaction quotient at any moment.

• At equilibrium, ΔG = 0 and Q = Kc (equilibrium constant).

• Thus, 0 = ΔG° + RT ln Kc → ΔG° = -RT ln Kc

• Or, ln Kc = -ΔG° / RT

Interpretation of Kc Values

• If ΔG° < 0, then -ΔG°/RT > 0 → Kc > 1 → reaction favors products (spontaneous forward).

• If ΔG° > 0, then -ΔG°/RT < 0 → Kc < 1 → reaction favors reactants (non-spontaneous forward).

Solved Examples

Ex 1

• Given ΔG° = -40 kJ/mol at 298 K, calculate Kc.

• Use R = 8.314 J/mol·K, convert ΔG° to J: -40000 J/mol.

• ln Kc = -(-40000) / (8.314 × 298) = 40000 / 2477 ≈ 16.14.

• Kc = e16.14 >> 1, reaction strongly favors products.

Practice Set

• Q1: If ΔG = 0, what is the value of Q?
  A. 0
  B. 1
  C. Kc
  D. ∞

• Q2: For a reaction with ΔG° > 0, which is true?
  A. Kc > 1
  B. Reaction is spontaneous forward
  C. Kc < 1
  D. ΔG < 0

• Q3: At 300 K, ΔG° = -10 kJ/mol. Calculate ln Kc (R = 8.314 J/mol·K).
  A. 4.01
  B. -4.01
  C. 0.04
  D. -0.04

Answer Key

• Q1: C. Q = Kc at equilibrium when ΔG = 0.

• Q2: C. ΔG° > 0 means Kc < 1, reaction non-spontaneous forward.

• Q3: A. ln Kc = -ΔG°/RT = 10000/(8.314×300) ≈ 4.01.

Memory Aids

• ΔG° < 0 → Kc > 1 → forward reaction favored.

• ΔG° > 0 → Kc < 1 → reverse reaction favored.

• At equilibrium: ΔG = 0 and Q = Kc.

Factors Affecting Equilibria

Le Chatelier's Principle

• System shifts to counteract any change in equilibrium conditions.

• Applies to all physical and chemical equilibria.

Factors Influencing Equilibrium

• Concentration: Increase in reactants favours forward reaction; increase in products favours backward reaction.

• Temperature: High temperature favours endothermic reactions; low temperature favours exothermic reactions.

• Pressure: High pressure favours side with fewer gas moles; low pressure favours side with more gas moles.

• Inert Gas Addition: At constant volume, no effect; at constant pressure, favours side with more gas moles.

• Catalyst: Does not affect equilibrium position; speeds up attainment of equilibrium.

Solved Examples

Ex 1: For PCl5(g) ⇌ PCl3(g) + Cl2(g), ΔH = +ve, predict effect of temperature increase.

• Reaction is endothermic; increasing temperature shifts equilibrium forward, increasing PCl3 and Cl2 formation.

Ex 2: For 2HI(g) ⇌ H2(g) + I2(g), ΔH = +ve, effect of pressure increase?

• Equal moles of gases on both sides; pressure change has no effect on equilibrium.

Practice Set

• Q1: Increasing concentration of Cl2 in PCl5 equilibrium shifts reaction to?

  • A. Forward

  • B. Backward

  • C. No change

  • D. Depends on temperature

• Q2: For an exothermic reaction, increasing temperature will:

  • A. Increase equilibrium constant

  • B. Decrease equilibrium constant

  • C. No effect

  • D. Shift equilibrium forward

• Q3: Addition of inert gas at constant pressure in reaction producing more gas moles will:

  • A. Shift equilibrium towards more moles

  • B. Shift equilibrium towards fewer moles

  • C. No effect

  • D. Increase reaction rate only

Answer Key

• Q1: B. Backward — Increasing product concentration shifts equilibrium backward.

• Q2: B. Decrease equilibrium constant — For exothermic reactions, higher temperature lowers K.

• Q3: A. Shift equilibrium towards more moles — At constant pressure, inert gas addition favours side with more gas moles.

Memory Aids

• Increase Reactants → Forward reaction favored.

• High Temp → Endothermic favored; Low Temp → Exothermic favored.

• High Pressure → Side with fewer gas moles favored.

• Inert Gas at Constant Volume → No effect; at Constant Pressure → Favors more gas moles.

• Catalyst → Speeds equilibrium, no shift.

Application of Le Chatelier’s Principle

Effect of Temperature

• PCl5 dissociation is endothermic; increasing temperature increases PCl3 and Cl2 formation.

• HI dissociation is endothermic; increasing temperature favours H2 and I2 formation.

Effect of Pressure

• Increasing pressure in PCl5 equilibrium favours backward reaction (fewer gas moles).

• Pressure has no effect on HI equilibrium (equal gas moles on both sides).

Effect of Concentration

• Adding PCl3 or Cl2 or removing PCl5 shifts PCl5 equilibrium backward.

• Adding HI or removing H2 or I2 favours forward reaction in HI equilibrium.

Effect of Inert Gas Addition

• In PCl5 reaction, inert gas addition at constant volume has no effect.

• At constant pressure, inert gas addition shifts equilibrium forward (more gas moles).

Solved Examples

Ex 1: Predict effect of pressure increase on PCl5 equilibrium.

• Pressure increase favours backward reaction (fewer gas moles).

Ex 2: Effect of adding Cl2 to PCl5 equilibrium?

• Shifts equilibrium backward, reducing PCl5 dissociation.

Practice Set

• Q1: Increasing temperature in PCl5 equilibrium causes:

  • A. More PCl5

  • B. More PCl3 and Cl2

  • C. No change

  • D. Depends on pressure

• Q2: In HI equilibrium, pressure increase will:

  • A. Shift forward

  • B. Shift backward

  • C. No effect

  • D. Increase rate only

• Q3: Adding inert gas at constant pressure in PCl5 equilibrium will:

  • A. Shift backward

  • B. Shift forward

  • C. No effect

  • D. Decrease reaction rate

Answer Key

• Q1: B. More PCl3 and Cl2 — Endothermic reaction favoured by temperature increase.

• Q2: C. No effect — Equal gas moles on both sides.

• Q3: B. Shift forward — More gas moles side favoured at constant pressure.

Memory Aids

• Endothermic reactions: Temp ↑ → Forward reaction ↑.

• Equal gas moles: Pressure change → No effect.

• Inert gas at constant pressure → Favors side with more gas moles.

Effect of Temperature on Equilibrium Constant

van't Hoff Equation

• \( \log \frac{K_2}{K_1} = \frac{\Delta H^\circ}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) \)

• \(K_1, K_2\): equilibrium constants at temperatures \(T_1, T_2\).

• \(\Delta H^\circ\): enthalpy change; \(R\): gas constant.

Interpretation

• If \(\Delta H^\circ = 0\), \(K_2 = K_1\); equilibrium constant independent of temperature.

• If \(\Delta H^\circ > 0\) (endothermic), \(K_2 > K_1\) when \(T_2 > T_1\); equilibrium constant increases with temperature.

• If \(\Delta H^\circ < 0\) (exothermic), \(K_2 < K_1\) when \(T_2 > T_1\); equilibrium constant decreases with temperature.

Solved Examples

Ex 1: For an endothermic reaction, predict change in equilibrium constant when temperature rises from 300 K to 350 K.

• Since \(\Delta H^\circ > 0\), \(K\) increases with temperature.

Ex 2: For an exothermic reaction, effect of temperature increase on \(K\)?

• \(K\) decreases as temperature increases.

Practice Set

• Q1: If \(\Delta H^\circ = 0\), increasing temperature will:

  • A. Increase \(K\)

  • B. Decrease \(K\)

  • C. Not change \(K\)

  • D. Shift equilibrium forward

• Q2: For an endothermic reaction, increasing temperature from \(T_1\) to \(T_2\) will:

  • A. Decrease \(K\)

  • B. Increase \(K\)

  • C. No effect on \(K\)

  • D. Shift equilibrium backward

• Q3: According to van't Hoff equation, if \(\Delta H^\circ < 0\), increasing temperature will:

  • A. Increase \(K\)

  • B. Decrease \(K\)

  • C. Keep \(K\) constant

  • D. Increase reaction rate only

Answer Key

• Q1: C. Not change \(K\) — No enthalpy change means \(K\) is temperature independent.

• Q2: B. Increase \(K\) — Endothermic reactions have \(K\) increasing with temperature.

• Q3: B. Decrease \(K\) — Exothermic reactions have \(K\) decreasing with temperature.

Memory Aids

• Endothermic: Temp ↑ → \(K\) ↑.

• Exothermic: Temp ↑ → \(K\) ↓.

• Zero enthalpy: \(K\) constant with temperature.

Electrolytes

Definition and Types

• Electrolytes are substances that conduct electricity when dissolved in water or melted.

• They dissociate into ions in aqueous or molten state, enabling electrical conduction.

• Classified as strong or weak based on degree of ionization.

Strong Electrolytes

• Completely ionize in aqueous solution.

• Examples: NaCl, HCl, KOH.

• Conduct electricity efficiently due to full ion dissociation.

Weak Electrolytes

• Partially ionize in aqueous solution.

• Establish equilibrium between ions and unionized molecules.

• Examples: CH3COOH, NH4OH.

• Conduct electricity weakly due to limited ion presence.

Solved Examples

• Ex 1: Identify if HCl is a strong or weak electrolyte.
  Solution: HCl fully ionizes into H+ and Cl− ions in water; hence, it is a strong electrolyte.

• Ex 2: Write the ionic equilibrium for acetic acid in water.
  Solution: CH3COOH ⇌ CH3COO− + H+

Practice Set

• Q1: Which of the following is a strong electrolyte?
  A. CH3COOH
  B. NaCl
  C. NH4OH
  D. H2O

• Q2: What happens when a weak electrolyte dissolves in water?
  A. Complete ionization
  B. No ionization
  C. Partial ionization with equilibrium
  D. Forms precipitate

• Q3: Which equilibrium represents a weak electrolyte?
  A. NaCl → Na+ + Cl−
  B. HCl → H+ + Cl−
  C. CH3COOH ⇌ CH3COO− + H+
  D. KOH → K+ + OH−

Answer Key

• Q1: B. NaCl - It fully ionizes in water, making it a strong electrolyte.

• Q2: C. Partial ionization with equilibrium - Weak electrolytes ionize partially establishing equilibrium.

• Q3: C. CH3COOH ⇌ CH3COO− + H+ - Represents partial ionization and equilibrium of a weak electrolyte.

Memory Aids

• Strong electrolytes: "Complete Split" - full ionization.

• Weak electrolytes: "Partial Balance" - equilibrium between ions and molecules.

• All acids, bases, and salts can be strong or weak electrolytes.

Arrhenius Concept

Definition and Explanation

• Acid: Substance that produces H+ ions in aqueous solution.

• Base: Substance that produces OH− ions in aqueous solution.

• Example acids: HCl → H+ + Cl−, H2SO4 → 2H+ + SO42−

• Example bases: KOH → K+ + OH−, NaOH → Na+ + OH−

Limitations

• Cannot explain acidic nature of CO2, SO2, AlCl3 (no H+ ions).

• Cannot explain basic nature of Na2CO3, NH3 (no OH− ions).

• Fails to describe acid-base behavior without water.

• Does not explain the nature of H+ and OH− ions in water.

Solved Examples

Ex 1: Write the dissociation of HCl in water.
Solution:
HCl → H+ + Cl−

Ex 2: Write the dissociation of NaOH in water.
Solution:
NaOH → Na+ + OH−

Practice Set

• Q1: According to Arrhenius, an acid is a substance that:
  A. Accepts H+
  B. Donates OH−
  C. Produces H+ in water
  D. Produces OH− in water

• Q2: Which of the following is NOT explained by Arrhenius theory?
  A. HCl dissociation
  B. NH3 basicity
  C. NaOH dissociation
  D. H2SO4 dissociation

• Q3: What ions are produced when KOH dissolves in water?
  A. K+ and H+
  B. K+ and OH−
  C. OH− and Cl−
  D. K+ and Cl−

Answer Key

• Q1: C - Arrhenius acid produces H+ ions in water.

• Q2: B - NH3 does not produce OH− ions directly, so not explained.

• Q3: B - KOH dissociates into K+ and OH− ions.

Memory Aids

• Arrhenius acid = H+ producer in water.

• Arrhenius base = OH− producer in water.

• Limitations: No explanation for acid/base behavior without water or non-H/OH ions.

Bronsted-Lowry Concept

Definition and Explanation

• Acid: Proton (H+) donor.

• Base: Proton (H+) acceptor.

• Protonic theory: Focuses on transfer of H+ ions.

• Water is amphiprotic: can donate or accept protons.

Conjugate Acid-Base Pairs

• Acid loses proton → conjugate base.

• Base gains proton → conjugate acid.

• Pairs differ by one proton only.

• Strong acid ⇌ weak conjugate base and vice versa.

• Acid-base reactions are reversible.

Solved Examples

Ex 1: Identify acid and base in NH3 + HCl ⇌ NH4+ + Cl−
Solution:
HCl donates H+ (acid), NH3 accepts H+ (base).

Ex 2: Write conjugate pairs in HA + B− ⇌ A− + BH
Solution:
HA/A− and BH/B− are conjugate acid-base pairs.

Practice Set

• Q1: In Bronsted-Lowry theory, a base is a substance that:
  A. Donates H+
  B. Accepts H+
  C. Produces OH−
  D. Produces H+

• Q2: Which is a conjugate acid-base pair?
  A. HCl and Cl−
  B. NH3 and NH4+
  C. H2SO4 and SO42−
  D. All of the above

• Q3: In the reaction H2SO4 + NH3 → NH4+ + HSO4−, which is the stronger acid?
  A. H2SO4
  B. NH4+
  C. NH3
  D. HSO4−

Answer Key

• Q1: B - Base accepts proton (H+).

• Q2: D - All listed pairs are conjugate acid-base pairs.

• Q3: A - H2SO4 is stronger acid than NH4+.

Memory Aids

• Acid = proton donor; Base = proton acceptor.

• Conjugate pairs differ by one H+.

• Strong acid ⇌ weak conjugate base.

• Water is amphiprotic (both acid and base).

Lewis Acids and Bases

Definition and Explanation

• Lewis acid: Electron pair acceptor.

• Lewis base: Electron pair donor.

• Focus on electron pairs, not protons.

Chemical Species Acting as Lewis Acids

• Electron-deficient molecules with incomplete octet: BF3, AlCl3, FeCl3.

• Simple cations: H+, Mg2+, Co3+, Ag+, Fe3+.

• Molecules with polar multiple bonds: SO3, CO2.

• Metal atoms in coordination complexes: Ni in Ni(CO)4.

• Molecules with empty d-orbitals: SiF4, SnCl4.

Chemical Species Acting as Lewis Bases

• Electron-rich molecules with lone pairs: NH3, R-O-H (alcohol), R-NH2 (amine).

• Negatively charged ions: CN−, OH−, Cl−.

• Ligands in coordination compounds act as Lewis bases.

Solved Examples

Ex 1: Write the Lewis acid-base adduct formed between BF3 and NH3.
Solution:
BF3 + NH3 → BF3←NH3

Ex 2: Identify Lewis acid and base in [Al(H2O)6]3+ complex.
Solution:
Al3+ is Lewis acid; H2O molecules are Lewis bases.

Practice Set

• Q1: Lewis acid is a substance that:
  A. Donates electron pair
  B. Accepts electron pair
  C. Donates proton
  D. Accepts proton

• Q2: Which is a Lewis base?
  A. BF3
  B. NH3
  C. AlCl3
  D. H+

• Q3: The complex formed by BF3 and NH3 is:
  A. Lewis acid
  B. Lewis base
  C. Lewis acid-base adduct
  D. Neither acid nor base

Answer Key

• Q1: B - Lewis acid accepts electron pair.

• Q2: B - NH3 donates lone pair, acts as Lewis base.

• Q3: C - BF3 and NH3 form Lewis acid-base adduct.

Memory Aids

• Lewis acid = electron pair acceptor (often electron-deficient).

• Lewis base = electron pair donor (lone pairs or negative charge).

• All Bronsted bases are Lewis bases; all Lewis acids may not be Bronsted acids.

• Common Lewis acids: BF3, AlCl3, metal cations.

Ionisation of Acids and Bases

Ionisation Constant and pH Scale

• Water ionises as: H2O ⇌ H+ + OH−

• Ionic product of water, Kw = [H+][OH−] = 1.0 × 10−14 at 298 K

• pH = −log[H+]; pOH = −log[OH−]

• pH + pOH = 14 at 298 K

• pH < 7: acidic; pH = 7: neutral; pH > 7: basic

Ionisation of Acids and Bases

• Acid ionisation: HA ⇌ H+ + A−

• Acid dissociation constant: \(K_a = \frac{[H^+][A^-]}{[HA]}\)

• Degree of ionisation (α) for acid: \(\alpha = \sqrt{\frac{K_a}{C}}\), where C = concentration

• Base ionisation: BOH ⇌ B+ + OH−

• Base dissociation constant: \(K_b = \frac{[B^+][OH^-]}{[BOH]}\)

• Degree of ionisation (α) for base: \(\alpha = \sqrt{\frac{K_b}{C}}\)

• Stronger acid/base → higher Ka/Kb and greater α

Ostwald’s Dilution Law

• For electrolyte AB ⇌ A+ + B−, initial concentration = C

• At equilibrium: [AB] = C(1 − α), [A+] = [B−] = Cα

• Dissociation constant: \(K_a = \frac{C\alpha^2}{1-\alpha}\)

• For weak electrolytes, α << 1, so \(K_a \approx C\alpha^2\) and \(\alpha = \sqrt{\frac{K_a}{C}}\)

• Strong electrolytes: α = 1; weak electrolytes: α < 1

Relation Between Ka and Kb

• pKa = −log Ka; pKb = −log Kb

• pKa + pKb = pKw = 14 at 298 K

• Polybasic acids and polyacidic bases ionise stepwise with constants K1, K2, K3, ...

• Generally, K1 > K2 > K3 ...

• Overall ionisation constant: K = K1 × K2 × K3 ...

Factors Affecting Acid Strength

• Acid strength depends on H–A bond strength and polarity

• In halogen acids (HF, HCl, HBr, HI), acid strength increases with increasing size of halogen (HF < HCl < HBr < HI)

• In hydrides (CH4, NH3, H2O, HF), acid strength increases with electronegativity of central atom (CH4 < NH3 < H2O < HF)

Common Ion Effect

• Addition of a common ion suppresses dissociation of a weak electrolyte

• Example: Adding NaCl to acetic acid solution reduces ionisation of acetic acid

Hydrolysis of Salts and pH of Their Solutions

• Salts ionise in water to give hydrated ions; ions react with water (hydrolysis)

• Salts from strong acid + strong base → neutral solution (pH = 7)

• Salts from strong acid + weak base → acidic solution (pH < 7)

• Salts from weak acid + strong base → basic solution (pH > 7)

• Salts from weak acid + weak base → solution pH depends on relative strengths

• Hydrolysis constant (Kh) and degree of hydrolysis (h) depend on Kw, Ka, Kb, and salt concentration

Solved Examples

• Ex 1: Calculate pH of 0.01 M HCl.

  • HCl is strong acid, fully ionised: [H+] = 0.01 M

  • pH = −log(0.01) = 2

• Ex 2: Find degree of ionisation of 0.1 M acetic acid with Ka = 1.8 × 10−5.

  • \(\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.8 \times 10^{-5}}{0.1}} = 0.0134\)

• Ex 3: For a weak base with Kb = 1.0 × 10−6 and concentration 0.05 M, find α.

  • \(\alpha = \sqrt{\frac{K_b}{C}} = \sqrt{\frac{1.0 \times 10^{-6}}{0.05}} = 0.00447\)

Practice Set

• Q1: What is the pH of pure water at 298 K?

  • A) 0

  • B) 7

  • C) 14

  • D) 1

• Q2: Degree of ionisation (α) of a weak acid is 0.01 at 0.1 M concentration. What is Ka?

  • A) 1 × 10−6

  • B) 1 × 10−4

  • C) 1 × 10−2

  • D) 1 × 10−8

• Q3: For a weak acid, pKa = 5. What is pKb of its conjugate base?

  • A) 9

  • B) 5

  • C) 14

  • D) 19

Answer Key

• Q1: B) 7 — Pure water is neutral with pH 7 at 298 K.

• Q2: A) 1 × 10−6 — Ka = Cα2 = 0.1 × (0.01)2 = 1 × 10−6.

• Q3: A) 9 — pKa + pKb = 14, so pKb = 14 − 5 = 9.

Memory Aids

• pH + pOH = 14 at 25°C (298 K)

• Stronger acid → higher Ka, lower pKa

• Ostwald’s law: α = √(Ka/C) for weak electrolytes

• Common ion effect reduces ionisation of weak electrolytes

• Acid strength in halogens: HF < HCl < HBr < HI (increasing size)

• Acid strength in hydrides: CH4 < NH3 < H2O < HF (increasing electronegativity)

Buffer Solutions

Definition and Classification

• Buffer solutions resist pH change on dilution or addition of small amounts of acid/base.

• Two types of buffers:

  • Simple buffers: Solutions of salts of weak acid and weak base (e.g., CH3COONH4).

  • Mixed buffers: Mixtures of two solutions, further divided into:

    • Acidic buffers: Weak acid + salt of weak acid with strong base (e.g., CH3COOH + CH3COONa), pH < 7.

    • Basic buffers: Weak base + salt of weak base with strong acid (e.g., NH4OH + NH4Cl), pH > 7.

Buffer pH Equations

• Acidic buffer pH: pH = pKa + log \( \frac{[salt]}{[acid]} \) (Henderson–Hasselbalch equation)

• Basic buffer pOH: pOH = pKb + log \( \frac{[salt]}{[base]} \)

• Buffer capacity = moles of acid/base added per litre ÷ change in pH

• Buffer range: pH range where buffer is effective

  • Acidic buffer: pKa ± 1

  • Basic buffer: (pKw - pKb) ± 1

Solved Examples

• Ex 1: Calculate pH of buffer made from 0.1 M CH3COOH and 0.1 M CH3COONa (pKa = 4.76).

  • pH = 4.76 + log(0.1/0.1) = 4.76 + 0 = 4.76

• Ex 2: Find pOH of buffer with 0.2 M NH4OH and 0.1 M NH4Cl (pKb = 4.75).

  • pOH = 4.75 + log(0.1/0.2) = 4.75 + log(0.5) = 4.75 - 0.301 = 4.449

Practice Set

• Q1: Which of the following is a simple buffer?

  • A. CH3COOH + CH3COONa

  • B. NH4OH + NH4Cl

  • C. CH3COONH4

  • D. HCl + NaCl

• Q2: The pH of a buffer solution is 5.0. If the concentration of salt and acid are equal, what is the pKa of the acid?

  • A. 5.0

  • B. 7.0

  • C. 4.0

  • D. 6.0

• Q3: Calculate the pH of a buffer containing 0.1 M acetic acid and 0.05 M sodium acetate. (pKa = 4.76)

  • A. 4.46

  • B. 5.06

  • C. 4.76

  • D. 5.76

Answer Key

• Q1: C. CH3COONH4 is a simple buffer (salt of weak acid and weak base).

• Q2: A. pH = pKa + log([salt]/[acid]); equal concentrations mean log(1)=0, so pH = pKa = 5.0.

• Q3: A. pH = 4.76 + log(0.05/0.1) = 4.76 + log(0.5) = 4.76 - 0.301 = 4.46.

Memory Aids

• Buffer = weak acid/base + conjugate salt.

• Acidic buffer pH < 7; basic buffer pH > 7.

• Henderson–Hasselbalch: pH = pKa + log([salt]/[acid]).

• Buffer range ≈ pK ± 1.

Solubility vs Solubility Product

Concepts

• Solubility: Amount of solute (g) dissolved per litre of solvent.

• Solubility Product (Ksp): Equilibrium constant for dissociation of a sparingly soluble salt into ions.

• Solubility depends on temperature, pressure, pH, and other chemicals.

• Ksp depends only on temperature.

• Ksp expression for salt MxXy:
  MxXy(s) ⇌ xMy+(aq) + yXx−(aq)
  Ksp = [My+]x[Xx−]y = (xs)x(ys)y = xxyysx+y

• Solubility from Ksp:
  s = \(\sqrt[x+y]{\frac{K_{sp}}{x^x y^y}}\)

• Ksp = Qsp at equilibrium; Qsp is ionic product at any moment.

Solved Examples

• Ex 1: Calculate solubility of AgCl if Ksp = 1.8 × 10−10.
  AgCl ⇌ Ag+ + Cl−, x = y = 1
  s = √Ksp = √1.8 × 10−10 = 1.34 × 10−5 mol/L

Practice Set

• Q1: What does Ksp represent?
  A) Amount of solute dissolved
  B) Equilibrium constant for ion dissociation
  C) Temperature dependence of solubility
  D) Pressure effect on solubility

• Q2: For salt M2X3, Ksp = 1.0 × 10−18. What is solubility (s)?
  A) \(\sqrt{K_{sp}}\)
  B) \(\sqrt[5]{\frac{K_{sp}}{2^2 3^3}}\)
  C) \(\sqrt[5]{K_{sp}}\)
  D) \(\frac{K_{sp}}{2^2 3^3}\)

• Q3: Which factor does NOT affect Ksp?
  A) Temperature
  B) Pressure
  C) Ionic strength
  D) None of the above

Answer Key

• Q1: B - Ksp is the equilibrium constant for dissociation of a sparingly soluble salt into ions.

• Q2: B - For M2X3, s = \(\sqrt[5]{\frac{K_{sp}}{2^2 3^3}}\) from formula sx+y = Ksp / xx yy.

• Q3: B - Ksp depends only on temperature, not pressure or ionic strength.

Memory Aids

• Ksp depends only on temperature.

• Solubility (s) relates to Ksp by s = \(\sqrt[x+y]{\frac{K_{sp}}{x^x y^y}}\).

• Common ion effect lowers solubility.

Common Ion Effect

Concepts

• Addition of a common ion decreases solubility of a sparingly soluble salt.

• Example: Adding NaCl to AgCl solution reduces AgCl solubility.

• Expression: Ksp = [Ag+][Cl−] = S'(S' + x), where x is added common ion concentration.

• Formation of complex ions can increase solubility despite common ions.

Solved Examples

• Ex 1: Solubility of AgCl in pure water is s. After adding 0.01 M NaCl, solubility becomes S'. Calculate S' if Ksp = 1.8 × 10−10.
  Ksp = S'(S' + 0.01) ≈ S' × 0.01 (since S' ≪ 0.01)
  S' = Ksp / 0.01 = 1.8 × 10−8 M

Practice Set

• Q1: What happens to solubility when a common ion is added?
  A) Increases
  B) Decreases
  C) No change
  D) First increases then decreases

• Q2: In AgCl solution, adding NaCl affects solubility by:
  A) Increasing Ag+ concentration
  B) Increasing Cl− concentration
  C) Decreasing Ksp
  D) Increasing temperature

• Q3: If complex ion forms, solubility:
  A) Decreases
  B) Increases
  C) Remains same
  D) Becomes zero

Answer Key

• Q1: B - Common ion decreases solubility by shifting equilibrium left.

• Q2: B - NaCl adds Cl− ions, common ion effect reduces AgCl solubility.

• Q3: B - Complex ion formation increases solubility by removing free ions.

Memory Aids

• Common ion effect = solubility decreases due to added ion.

• Complex ions = solubility increases.

Applications of Solubility Product

Concepts

• Predicting solubility: Calculate solubility (s) from Ksp using s = √Ksp for 1:1 salts.

• Precipitation prediction: If ionic product (Q) > Ksp, precipitation occurs; if Q < Ksp, no precipitation.

• Precipitation of common salt: Passing HCl gas increases Cl− concentration, causing NaCl to precipitate.

• Qualitative analysis: Separation of radicals based on solubility and common ion effect.

Solved Examples

• Ex 1: Will precipitation occur if [Ag+] = 1 × 10−3 M and [Cl−] = 1 × 10−3 M? Given Ksp (AgCl) = 1.8 × 10−10.
  Q = (1 × 10−3)(1 × 10−3) = 1 × 10−6 > Ksp
  Precipitation occurs.

Practice Set

• Q1: Precipitation occurs when:
  A) Q < Ksp
  B) Q = Ksp
  C) Q > Ksp
  D) None of these

• Q2: Passing HCl gas through impure NaCl solution causes:
  A) Increase in Na+ concentration
  B) Precipitation of pure NaCl
  C) Decrease in Cl− concentration
  D) No change

• Q3: Which application uses solubility product principle?
  A) Electrolysis
  B) Qualitative analysis
  C) Titration
  D) Distillation

Answer Key

• Q1: C - Precipitation occurs if ionic product exceeds Ksp.

• Q2: B - Increased Cl− causes NaCl to precipitate.

• Q3: B - Qualitative analysis uses solubility and common ion effect.

Memory Aids

• Precipitation if Q > Ksp.

• HCl gas increases Cl−, causing NaCl precipitation.

• Qualitative analysis separates ions by solubility differences.

Solubility Products of Electrolytes

Concepts

• Salts classified by electrolyte formula: binary (1:1), ternary (1:2 or 2:1), quaternary (1:3), 3:2 types.

• Examples:
  - Binary (1:1): AgCl
  - Ternary (1:2): PbCl2
  - Quaternary (1:3): Fe(OH)3

• Solubility (s) and Ksp relation varies by type:
  - 1:1 salts: s = √Ksp
  - 1:2 salts: s = \(\sqrt[3]{\frac{K_{sp}}{4}}\)

• Formulas help calculate solubility from Ksp for different salts.

Solved Examples

• Ex 1: Calculate solubility of PbCl2 (1:2 salt) if Ksp = 1.6 × 10−5.
  s = \(\sqrt[3]{\frac{K_{sp}}{4}}\) = \(\sqrt[3]{\frac{1.6 \times 10^{-5}}{4}}\) = 0.016 M

Practice Set

• Q1: For AgCl (1:1), solubility s =
  A) Ksp
  B) √Ksp
  C) \(\sqrt[3]{K_{sp}}\)
  D) \(\frac{K_{sp}}{4}\)

• Q2: For PbCl2 (1:2), solubility s =
  A) √Ksp
  B) \(\sqrt[3]{\frac{K_{sp}}{4}}\)
  C) \(\sqrt[3]{K_{sp}}\)
  D) \(\frac{K_{sp}}{4}\)

• Q3: Fe(OH)3 is a quaternary salt. Its solubility expression involves:
  A) s = √Ksp
  B) s = \(\sqrt[4]{\frac{K_{sp}}{27}}\)
  C) s = Ksp
  D) s = \(\sqrt[3]{K_{sp}}\)

Answer Key

• Q1: B - 1:1 salts have s = √Ksp.

• Q2: B - 1:2 salts have s = \(\sqrt[3]{\frac{K_{sp}}{4}}\).

• Q3: B - Quaternary salts like Fe(OH)3 have s = \(\sqrt[4]{\frac{K_{sp}}{27}}\).

Memory Aids

• 1:1 salts: s = √Ksp.

• 1:2 salts: s = \(\sqrt[3]{\frac{K_{sp}}{4}}\).

• Quaternary salts: s = \(\sqrt[4]{\frac{K_{sp}}{27}}\).