Course Topics

Introduction to Chemical Equilibrium

Chemical Equilibrium: A state in reversible reactions where forward and reverse rates are equal, and concentrations remain constant.

A
B
→
C
D
←

Reversible Reactions

  • Proceed in both directions
  • Don’t go to completion
  • Show ⇌ symbol
  • Dynamic equilibrium exists
  • Example: N₂ + 3H₂ ⇌ 2NH₃

Irreversible Reactions

  • Proceed in one direction
  • Go to completion
  • Show → symbol
  • No equilibrium
  • Example: 2Na + 2H₂O → 2NaOH + H₂
Memory Tip

Dynamic ≠ Static! Equilibrium is dynamic – molecules still react, but rates are equal so concentrations stay constant!

Law of Mass Action & Equilibrium Constant

Law of Mass Action (Guldberg & Waage, 1864): Rate of reaction ∝ product of active masses (concentrations) of reactants.

For: aA + bB ⇌ cC + dD
Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ
CRITICAL CONCEPT – Characteristics of Kc:
  • Depends on: Temperature only
  • Independent of: Initial concentration, pressure, volume, catalyst, direction
  • Units: (mol dm⁻³)^Δn where Δn = moles products – moles reactants
Reaction Kc Expression Δn
PCl₅ ⇌ PCl₃ + Cl₂ Kc = [PCl₃][Cl₂]/[PCl₅] +1
N₂O₄ ⇌ 2NO₂ Kc = [NO₂]²/[N₂O₄] +1
N₂ + 3H₂ ⇌ 2NH₃ Kc = [NH₃]²/[N₂][H₂]³ -2
2SO₂ + O₂ ⇌ 2SO₃ Kc = [SO₃]²/[SO₂]²[O₂] -1
Memory Tip

Kc = Products over Reactants, raised to their coefficients! Temperature changes Kc, but concentration/pressure don’t!

Kp and Relationship with Kc

Kp: Equilibrium constant in terms of partial pressures for gaseous reactions.

Kp = Kc(RT)^Δn
where Δn = moles of products – moles of reactants
Δn Value Relationship Example
Δn = 0 Kp = Kc H₂ + I₂ ⇌ 2HI
Δn > 0 Kp > Kc PCl₅ ⇌ PCl₃ + Cl₂
Δn < 0 Kp < Kc N₂ + 3H₂ ⇌ 2NH₃
EXAMPLE CALCULATION:
For: 2HI ⇌ H₂ + I₂
At equilibrium: P(HI) = 100 torr, P(H₂) = 50 torr, P(I₂) = 50 torr
Kp = (50 × 50) / (100)² = 0.25
Memory Tip

Kp = Kc × (RT)^Δn! When Δn=0, Kp=Kc. Remember R=0.0821 L·atm/mol·K for calculations!

Applications of Equilibrium Constant

1. Direction of Reaction:

  • If Q < Kc → Forward reaction
  • If Q > Kc → Reverse reaction
  • If Q = Kc → At equilibrium

2. Extent of Reaction:

Kc Value Extent Example
Very large (>10³) Almost complete 2H₂ + O₂ ⇌ 2H₂O (Kc=10⁵⁵)
Very small (<10⁻³) Very little forward reaction N₂ + O₂ ⇌ 2NO (Kc=10⁻¹³)
Moderate (~1) Significant amounts of both H₂ + I₂ ⇌ 2HI
CRITICAL CONCEPT: Kc indicates product stability. Large Kc = products are very stable. Small Kc = products are unstable relative to reactants.
Memory Tip

Q vs Kc: QK → reverse, Q=K → equilibrium! Large K = products favored, Small K = reactants favored!

Le Chatelier’s Principle

Principle: When stress is applied to a system at equilibrium, the system shifts to counteract the stress.

Stress Applied Shift Direction Effect on Kc
Increase [Reactant] → Forward No change
Increase [Product] ← Backward No change
Increase Pressure (Δn≠0) Toward fewer moles No change
Increase Temperature Toward endothermic side Changes
Add Catalyst No shift No change
TEMPERATURE EFFECTS:
  • Endothermic (ΔH>0): Increase T → Forward shift, Kc increases
  • Exothermic (ΔH<0): Increase T → Backward shift, Kc decreases
Memory Tip

Le Chatelier: “Fight the stress!” Add reactant → make product. Increase pressure → go to fewer moles. Increase T → favor endothermic!

Industrial Applications

Haber Process (NH₃)

  • N₂ + 3H₂ ⇌ 2NH₃, ΔH = -92.46 kJ
  • Conditions: 200-300 atm, 400°C
  • Catalyst: Fe with promoters
  • 35% NH₃ at equilibrium
  • Remove NH₃ by liquefaction

Contact Process (SO₃)

  • 2SO₂ + O₂ ⇌ 2SO₃, ΔH = -194 kJ
  • Conditions: 1 atm, 400-500°C
  • Catalyst: V₂O₅ or Pt
  • Continuous O₂ supply
  • Increase pressure favors forward
OPTIMIZATION: Industrial processes balance rate and yield. Lower T increases yield but slows rate. Higher P increases yield but is expensive. Catalysts increase rate without affecting equilibrium.
Memory Tip

Haber process: High pressure (fewer moles), moderate T (compromise), catalyst (Fe). Contact process: Excess O₂, moderate T, catalyst (V₂O₅)!

Common Ion Effect

Definition: Suppression of ionization of a weak electrolyte by adding a strong electrolyte with a common ion.

Example: CH₃COOH ⇌ CH₃COO⁻ + H⁺
Add CH₃COONa → [CH₃COO⁻] increases
Equilibrium shifts ←, [H⁺] decreases, pH increases

Applications:

  • Purification of NaCl: Pass HCl gas → NaCl precipitates
  • Suppress H₂S ionization: Add HCl → [H⁺] increases, shifts ←
  • Decrease NH₄OH dissociation: Add NH₄Cl → [NH₄⁺] increases
EFFECTS:
  • Increases: Crystallization, Association
  • Decreases: Ionization, Solubility
Memory Tip

Common ion effect = Add common ion → suppress ionization! Like adding salt to weak acid → makes it even weaker!

Buffer Solutions

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

CH₃COOH
+
CH₃COONa
=
Buffer

Types:

  • Acidic Buffer: Weak acid + its salt (e.g., CH₃COOH + CH₃COONa)
  • Basic Buffer: Weak base + its salt (e.g., NH₄OH + NH₄Cl)
Henderson-Hasselbalch Equations:
Acidic: pH = pKa + log([salt]/[acid])
Basic: pOH = pKb + log([salt]/[base])
[CH₃COOH] M [CH₃COO⁻] M % Dissociation pH
0.10 0.00 1.3 2.89
0.10 0.05 0.036 4.44
0.10 0.10 0.018 4.74
Memory Tip

Buffer = Weak acid + its salt OR weak base + its salt. pH = pKa + log(salt/acid). When [salt]=[acid], pH=pKa!

Solubility Product (Ksp)

Definition: Product of ion concentrations raised to their coefficients for saturated solutions of sparingly soluble salts.

For: AₓBᵧ(s) ⇌ xA⁺(aq) + yB⁻(aq)
Ksp = [A⁺]ˣ[B⁻]ʸ
Salt Dissolution Ksp Expression Relation to s
AgCl AgCl ⇌ Ag⁺ + Cl⁻ Ksp = [Ag⁺][Cl⁻] Ksp = s²
PbCl₂ PbCl₂ ⇌ Pb²⁺ + 2Cl⁻ Ksp = [Pb²⁺][Cl⁻]² Ksp = 4s³
Al(OH)₃ Al(OH)₃ ⇌ Al³⁺ + 3OH⁻ Ksp = [Al³⁺][OH⁻]³ Ksp = 27s⁴
Ca₃(PO₄)₂ Ca₃(PO₄)₂ ⇌ 3Ca²⁺ + 2PO₄³⁻ Ksp = [Ca²⁺]³[PO₄³⁻]² Ksp = 108s⁵
APPLICATIONS:
  • Determining solubility from Ksp
  • Predicting precipitation (Qsp vs Ksp)
  • Effect of common ion on solubility
  • Purification by fractional precipitation
Memory Tip

Ksp = Ion product at saturation! Qsp < Ksp = unsaturated, Qsp = Ksp = saturated, Qsp > Ksp = precipitate forms!

Ksp Calculations

Example 1: Solubility of PbF₂ = 2×10⁻³ M. Calculate Ksp.

PbF₂ ⇌ Pb²⁺ + 2F⁻
s = 2×10⁻³ M
[Pb²⁺] = s = 2×10⁻³ M
[F⁻] = 2s = 4×10⁻³ M
Ksp = [Pb²⁺][F⁻]² = (2×10⁻³)(4×10⁻³)² = 3.2×10⁻⁸

Example 2: Ksp of Ca(OH)₂ = 3.6×10⁻⁸. Calculate solubility.

Ca(OH)₂ ⇌ Ca²⁺ + 2OH⁻
Ksp = [Ca²⁺][OH⁻]² = s(2s)² = 4s³
4s³ = 3.6×10⁻⁸
s³ = 9×10⁻⁹
s = 2.08×10⁻³ M
PRECIPITATION PREDICTION:
  • If Qsp < Ksp: Unsaturated, no precipitate
  • If Qsp = Ksp: Saturated, at equilibrium
  • If Qsp > Ksp: Supersaturated, precipitate forms
Memory Tip

For AB₂ type: Ksp = 4s³. For A₂B type: Ksp = 4s³. For AB type: Ksp = s². For AₓBᵧ type: Ksp = (xˣyʵ)sˣ⁺ʸ!

Important Distinctions

Homogeneous Equilibrium

  • Same phase reactants & products
  • Example: H₂ + I₂ ⇌ 2HI (all gases)
  • Example: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O (all liquids)
  • Kc = [products]/[reactants]

Heterogeneous Equilibrium

  • Different phases reactants & products
  • Example: CaCO₃(s) ⇌ CaO(s) + CO₂(g)
  • Example: NH₄Cl(s) ⇌ NH₃(g) + HCl(g)
  • Pure solids/liquids not included in Kc
KEY POINTS:
  • Kc changes with temperature for all reactions
  • Catalyst affects rate but not equilibrium position or Kc
  • Common ion decreases solubility and ionization
  • Buffer capacity maximum when [acid]=[salt] or [base]=[salt]
Memory Tip

Homogeneous = same phase, Heterogeneous = different phases! Pure solids/liquids = constant concentration = not in K expression!