Equilibrium: State where forward and reverse reaction rates are equal

Dynamic equilibrium:

• Reactions still occurring
• No net change in concentrations
• Forward rate = Reverse rate

Example:

$N2\text{(g)} + 3H2\text{(g)} \rightleftharpoons 2NH3\text{(g)}$

At equilibrium: [N₂], [H₂], [NH₃] constant (but reactions ongoing)

Reversible Reactions

Notation:

• ⇌ or <=> indicates reversible
• Single arrow → indicates irreversible

Characteristics:

• Can proceed in both directions
• Eventually reach equilibrium
• Position depends on conditions

Equilibrium Constant (K_c)

For general reaction:

$aA + bB \rightleftharpoons cC + dD$

Equilibrium expression:

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

Key points:

• Products in numerator
• Reactants in denominator
• Coefficients become exponents
• Concentrations at equilibrium (mol/L)
• Temperature dependent

Rules for K Expressions

What to Include:

Include: Gases and aqueous solutions

• Use [ ] for molarity (mol/L)

Exclude:

• Pure solids
• Pure liquids
• Solvents (usually water)

Examples:

1. CaCO₃(s) ⇌ CaO(s) + CO₂(g)

$K_c = [CO2]$

(Solids omitted)

1. CH₃COOH(aq) + H₂O(l) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq)

$K_c = \frac{[CH3COO-][H3O+]}{[CH3COOH]}$

(Water omitted - solvent)

1. N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

$K_c = \frac{[NH3]^2}{[N2][H2]^3}$

Equilibrium Constant (K_p)

For gas-phase reactions:

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

Use partial pressures (atm) instead of concentrations

Relationship between K_c and K_p:

$K_p = K_c(RT)^{\Delta n}$

Where:

• R = 0.08206 L·atm/(mol·K)
• T = temperature (K)
• Δn = moles gas products - moles gas reactants

If Δn = 0: K_p = K_c

Magnitude of K

Interpretation:

Examples:

• K = 1.0 × 10⁵: Products dominate
• K = 1.0 × 10⁻⁵: Reactants dominate
• K = 5.0: Comparable amounts

Manipulating Equilibrium Expressions

Reverse Reaction:

If K_forward = x, then K_reverse = 1/x

Example: If K = 100 for A ⇌ B, then K = 0.01 for B ⇌ A

Multiply Equation:

If multiply by n, then K_new = (K_original)^n

Example: If K = 10 for A ⇌ B, then K = 100 for 2A ⇌ 2B

Add Equations:

If add reactions, multiply K values

K_overall = K₁ × K₂ × K₃...

Heterogeneous vs Homogeneous Equilibria

Homogeneous: All species in same phase

• Example: All gases or all aqueous

Heterogeneous: Multiple phases present

• Example: Solid + gas, solid + aqueous
• Remember: Omit pure solids and liquids from K expression

Writing K Expressions - Practice

General strategy:

1. Identify products and reactants
2. Omit pure solids, pure liquids, solvents
3. Products in numerator
4. Reactants in denominator
5. Use coefficients as exponents

Calculate Δn:

Δn = moles gas products - moles gas reactants

Products: 2 mol NH₃ Reactants: 1 mol N₂ + 3 mol H₂ = 4 mol

Δn = 2 - 4 = -2

Calculate RT:

RT = (0.08206)(673) = 55.2 L·atm/mol

Calculate K_p:

$K_p = K_c(RT)^{\Delta n}$

$K_p = (0.500)(55.2)^{-2}$

$K_p = (0.500) \times \frac{1}{(55.2)^2}$

$K_p = (0.500) \times \frac{1}{3047}$

$K_p = 1.64 \times 10^{-4}$

Answer: K_p = 1.64 × 10⁻⁴

Interpretation:

K_p < K_c because Δn < 0

• Decreasing moles of gas
• Pressure-based K is smaller
• Both indicate reactants favored (K < 1)

Substitute values:

$K_c = \frac{(1.60)^2}{(0.20)(0.20)}$

$K_c = \frac{2.56}{0.040}$

$K_c = 64$

Answer (a): K_c = 64

(b) Reversed reaction K_c

Reversed reaction: 2HI(g) ⇌ H₂(g) + I₂(g)

Rule: K_reverse = 1/K_forward

$K_{c,\text{reverse}} = \frac{1}{K_{c,\text{forward}}}$

$K_{c,\text{reverse}} = \frac{1}{64}$

$K_{c,\text{reverse}} = 0.0156$

Answer (b): K_c = 0.0156 or 1.56 × 10⁻²

Interpretation:

Forward reaction (K = 64):

• Products strongly favored
• Equilibrium lies far right
• HI formation favored

Reverse reaction (K = 0.0156):

• Reactants strongly favored
• Equilibrium lies far left
• HI decomposition unfavored

Note: K_forward × K_reverse = 64 × 0.0156 = 1 ✓