CHEMICAL EQUILIBRIUM MASTERY

Definition: Change in concentration per unit time

Rate = Δ[concentration]/Δtime

Units: mol·dm⁻³·s⁻¹

Forward Rate (Rf): Rate at which reactants form products

Reverse Rate (Rr): Rate at which products reform reactants

At Equilibrium: Rf = Rr ≠ 0

Graphical Representation: Concentration vs time shows plateau at equilibrium

Key Features:

1. Closed system required
2. Macroscopically static, microscopically dynamic
3. Forward and reverse reactions continue at equal rates
4. Concentrations remain constant over time
5. Can be approached from either direction

Visual Analogy: Like people moving between two rooms at equal rates – room populations remain constant

Experimental Evidence: Use of radioactive tracers shows continuous exchange

Formulated by: C.M. Guldberg and P. Waage (1864)

Statement: Rate of reaction is proportional to product of active masses (concentrations) of reactants

For reaction: aA + bB ⇌ cC + dD

Rf = kf[A]ᵃ[B]ᵇ

Rr = kr[C]ᶜ[D]ᵈ

At equilibrium: kf[A]ᵃ[B]ᵇ = kr[C]ᶜ[D]ᵈ

Thus: Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ = kf/kr

Definition: Concentration in mol·dm⁻³ that alters during reaction

Important Points:

• For solids: Active mass = 1 (constant)
• For liquids: Use concentration if pure, otherwise active mass varies
• For gases: Use partial pressures or concentrations
• For solutions: Use molar concentrations

Example: In CaCO₃(s) ⇌ CaO(s) + CO₂(g)

Kc = [CO₂] only (solids have constant active mass = 1)

Kp = P_CO₂ only

Definition: Ratio of product concentrations to reactant concentrations at equilibrium

For: aA + bB ⇌ cC + dD

Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ

Characteristics:

• Constant at constant temperature
• Independent of initial concentrations
• Independent of pressure/volume changes
• No units if Δn = 0
• Units = (mol·dm⁻³)^Δn if Δn ≠ 0

For Gaseous Reactions: Use partial pressures instead of concentrations

Kp = (P_C)ᶜ(P_D)ᵈ / (P_A)ᵃ(P_B)ᵇ

Relation with Kc:

Kp = Kc(RT)^Δn

where Δn = (c+d) – (a+b)

Cases:

• Δn = 0 → Kp = Kc
• Δn > 0 → Kp > Kc
• Δn < 0 → Kp < Kc

Example 1: N₂ + 3H₂ ⇌ 2NH₃

Kc = [NH₃]² / [N₂][H₂]³ (units: dm⁶·mol⁻²)

Kp = (P_NH₃)² / (P_N₂)(P_H₂)³

Example 2: PCl₅ ⇌ PCl₃ + Cl₂

Kc = [PCl₃][Cl₂] / [PCl₅] (units: mol·dm⁻³)

Example 3: 2HI ⇌ H₂ + I₂

Kc = [H₂][I₂] / [HI]² (no units, Δn=0)

Depends on: Temperature only

Independent of:

• Initial concentrations
• Pressure/volume changes
• Presence of catalyst
• Direction from which equilibrium is established

Temperature Effect:

• Endothermic reactions: K increases with T
• Exothermic reactions: K decreases with T

Memory Aid: “Only Temperature Changes K”

Statement: If a stress is applied to a system at equilibrium, the system adjusts to minimize that stress.

Key Concept: Equilibrium position shifts but K remains constant (except for temperature changes).

Types of Stress:

1. Change in concentration
2. Change in pressure/volume
3. Change in temperature
4. Addition of catalyst

Memory Aid: “Equilibrium fights back against change”

Rule: Increase concentration of reactant → shift right (toward products)

Increase concentration of product → shift left (toward reactants)

Examples:

• N₂ + 3H₂ ⇌ 2NH₃
• Add N₂ → more NH₃ produced
• Remove NH₃ → more NH₃ produced

Practical Application: Continuous removal of product drives reaction forward

Note: K remains unchanged, only position shifts

Rule: Increase pressure → shift toward fewer gas moles

Decrease pressure → shift toward more gas moles

Only affects gaseous reactions where Δn ≠ 0

Examples:

• N₂ + 3H₂ ⇌ 2NH₃ (Δn = -2)
• Increase P → more NH₃ (shift right)
• PCl₅ ⇌ PCl₃ + Cl₂ (Δn = +1)
• Increase P → less PCl₃, Cl₂ (shift left)

Volume: Decrease volume = increase pressure

Rule: Increase T → favor endothermic direction

Decrease T → favor exothermic direction

Changes K value!

For Endothermic (ΔH = +ve):

• Increase T → K increases
• Decrease T → K decreases

For Exothermic (ΔH = -ve):

• Increase T → K decreases
• Decrease T → K increases

Memory Aid: “Heat is like a reactant in endothermic, product in exothermic”

Reaction: N₂ + 3H₂ ⇌ 2NH₃ (ΔH = -92.46 kJ)

Optimal Conditions:

• Pressure: 200-300 atm
• Temperature: 400-450°C (673K)
• Catalyst: Fe with K₂O, Al₂O₃ promoters

Why these conditions?

• High P: Favors forward reaction (Δn = -2)
• Moderate T: Compromise between rate and yield
• Catalyst: Increases rate without affecting K
• Continuous NH₃ removal: Drives reaction forward

Yield: ~35% NH₃ at equilibrium

Key Step: 2SO₂ + O₂ ⇌ 2SO₃ (ΔH = -194 kJ)

Optimal Conditions:

• Pressure: 1-2 atm (Δn = -1, but equipment costs limit high P)
• Temperature: 400-450°C
• Catalyst: V₂O₅ or Pt
• Excess O₂: Drives reaction forward

Why moderate conditions?

• Low T: Favors exothermic forward reaction
• Excess O₂: Increases SO₂ conversion
• Catalyst: Allows lower T operation
• Economic balance between yield and rate

Yield: ~98% conversion

Key Step: 4NH₃ + 5O₂ ⇌ 4NO + 6H₂O (ΔH = -905 kJ)

Conditions:

• Pressure: 4-10 atm
• Temperature: 850-900°C
• Catalyst: Pt-Rh gauze
• Excess air: Provides O₂, dilutes mixture

Features:

• High T despite exothermic nature (for rate)
• Catalyst allows reaction at lower T than uncatalyzed
• Quick cooling of products prevents decomposition
• Multiple oxidation steps after NO formation

Common Strategies:

1. Use optimal pressure (high for Δn<0)
2. Choose moderate temperature (balance yield vs rate)
3. Use effective catalysts
4. Remove products continuously
5. Use excess cheaper reactant
6. Recycle unreacted materials

Economic Considerations:

• Equipment costs vs yield
• Energy requirements
• Catalyst life and cost
• Safety considerations
• Environmental impact

Memory Aid: “PTC” – Pressure, Temperature, Catalyst

Definition: Solutions that resist pH change when small amounts of acid/base are added.

Composition: Weak acid + its salt OR weak base + its salt

Types:

• Acidic Buffer: Weak acid + salt of weak acid (e.g., CH₃COOH + CH₃COONa)
• Basic Buffer: Weak base + salt of weak base (e.g., NH₄OH + NH₄Cl)

Buffer Range: pH = pKa ± 1 (most effective)

Buffer Capacity: Maximum acid/base that can be added without significant pH change

For Acidic Buffer:

pH = pKa + log([salt]/[acid])

For Basic Buffer:

pOH = pKb + log([salt]/[base])

or pH = 14 – pOH

Special Cases:

• [salt] = [acid] → pH = pKa
• [salt] > [acid] → pH > pKa
• [salt] < [acid] → pH < pKa

Example: CH₃COOH (0.1M) + CH₃COONa (0.1M)

pKa = 4.74, so pH = 4.74 + log(0.1/0.1) = 4.74

Acidic Buffer Example: CH₃COOH/CH₃COO⁻

Added Acid (H⁺): H⁺ + CH₃COO⁻ → CH₃COOH

The acetate ions neutralize added H⁺

Added Base (OH⁻): OH⁻ + CH₃COOH → CH₃COO⁻ + H₂O

The acetic acid neutralizes added OH⁻

Basic Buffer Example: NH₄⁺/NH₃

Added Acid: H⁺ + NH₃ → NH₄⁺

Added Base: OH⁻ + NH₄⁺ → NH₃ + H₂O

Key: Reservoir of both conjugate acid and base

Blood Buffer System (pH ~7.4):

• Bicarbonate: H₂CO₃/HCO₃⁻ (pKa = 6.1)
• Phosphate: H₂PO₄⁻/HPO₄²⁻ (pKa = 7.2)
• Proteins: Histidine residues

Other Biological Buffers:

• Intracellular: Phosphate, proteins
• Urine: Phosphate, ammonia
• Stomach: Mucous layer protects

Buffer Capacity in Blood:

Can handle 12-15 mEq/L of acid or base without significant pH change

Acidosis/Alkalosis: When buffers overwhelmed, pH changes dangerously

Definition: Product of ion concentrations raised to stoichiometric coefficients

For: AₐBₓ(s) ⇌ aAⁿ⁺(aq) + xBᵐ⁻(aq)

Ksp = [Aⁿ⁺]ᵃ[Bᵐ⁻]ˣ

Applicability: Only for sparingly soluble salts (solubility < 0.01M)

Examples:

• AgCl: Ksp = [Ag⁺][Cl⁻] = 1.8×10⁻¹⁰
• PbCl₂: Ksp = [Pb²⁺][Cl⁻]² = 1.6×10⁻⁵
• Al(OH)₃: Ksp = [Al³⁺][OH⁻]³ = 1.3×10⁻³³

Note: Pure solids don’t appear in Ksp expression

Relationship: Convert between Ksp and molar solubility (s)

AgCl (1:1 salt): AgCl ⇌ Ag⁺ + Cl⁻

[Ag⁺] = s, [Cl⁻] = s

Ksp = s × s = s² → s = √Ksp

PbCl₂ (1:2 salt): PbCl₂ ⇌ Pb²⁺ + 2Cl⁻

[Pb²⁺] = s, [Cl⁻] = 2s

Ksp = s × (2s)² = 4s³ → s = ³√(Ksp/4)

Al(OH)₃ (1:3 salt): Al(OH)₃ ⇌ Al³⁺ + 3OH⁻

[Al³⁺] = s, [OH⁻] = 3s

Ksp = s × (3s)³ = 27s⁴ → s = ⁴√(Ksp/27)

General: For AₐBₓ: Ksp = aᵃxˣs⁽ᵃ⁺ˣ⁾

Definition: Solubility decreases when common ion added

Example 1: AgCl in NaCl solution

Normally: s = √(1.8×10⁻¹⁰) = 1.34×10⁻⁵ M

In 0.1M NaCl: Ksp = [Ag⁺][0.1] = 1.8×10⁻¹⁰

[Ag⁺] = 1.8×10⁻⁹ M (solubility reduced 7400×!)

Example 2: Purification of NaCl

Pass HCl gas through saturated NaCl

Common Cl⁻ reduces NaCl solubility → precipitation

Applications:

• Qualitative analysis
• Salt purification
• Controlling precipitation
• Buffer preparation

Ion Product (IP): = [Aⁿ⁺]ᵃ[Bᵐ⁻]ˣ (actual concentrations)

Rules:

• IP < Ksp → Unsaturated, no precipitate
• IP = Ksp → Saturated, at equilibrium
• IP > Ksp → Supersaturated, precipitation occurs

Example: Will AgCl precipitate when equal volumes of 2×10⁻⁴M AgNO₃ and 2×10⁻⁴M NaCl mixed?

After mixing: [Ag⁺] = 1×10⁻⁴M, [Cl⁻] = 1×10⁻⁴M

IP = (1×10⁻⁴)(1×10⁻⁴) = 1×10⁻⁸

Ksp(AgCl) = 1.8×10⁻¹⁰

IP > Ksp → Yes, precipitate forms

Selective Precipitation: Using different Ksp values to separate ions