Introduction to Chemical Equilibrium

Understand reversible reactions, dynamic equilibrium, and equilibrium constant expressions (K_c and K_p).

Introduction to Chemical Equilibrium

What is Chemical Equilibrium?

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:

$\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$

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:

$\ce{aA + bB <=> cC + dD}$

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:

CaCO₃(s) ⇌ CaO(s) + CO₂(g)

$K_c = [\ce{CO2}]$

(Solids omitted)

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

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

(Water omitted - solvent)

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

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

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:

| K value | Meaning | |---------|---------| | K >> 1 (>10³) | Products favored, equilibrium far right | | K ≈ 1 (10⁻³ to 10³) | Significant amounts of both | | K << 1 (<10⁻³) | Reactants favored, equilibrium far left |

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:

Identify products and reactants
Omit pure solids, pure liquids, solvents
Products in numerator
Reactants in denominator
Use coefficients as exponents

📚 Practice Problems

1Problem 1easy

❓ Question:

Write the equilibrium constant expression (K_c) for: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

💡 Show Solution

Reaction: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

Identify components:

Products: SO₃ (coefficient: 2)
Reactants: SO₂ (coefficient: 2), O₂ (coefficient: 1)
All are gases → all included

K_c expression:

$K_c = \frac{[\text{products}]}{[\text{reactants}]}$

Apply coefficients as exponents:

$K_c = \frac{[\ce{SO3}]^2}{[\ce{SO2}]^2[\ce{O2}]}$

Answer:

$K_c = \frac{[\ce{SO3}]^2}{[\ce{SO2}]^2[\ce{O2}]}$

Key points:

Products (SO₃) in numerator
Reactants (SO₂, O₂) in denominator
Coefficient 2 becomes exponent 2
All concentrations at equilibrium
Unitless (convention)

2Problem 2medium

❓ Question:

For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), at equilibrium the concentrations are: [N₂] = 0.50 M, [H₂] = 0.20 M, and [NH₃] = 0.40 M. Calculate the equilibrium constant Kc.

💡 Show Solution

Solution:

Equilibrium expression: K_c = [NH₃]² / ([N₂][H₂]³)

Calculate K_c: K_c = (0.40)² / [(0.50)(0.20)³] K_c = 0.16 / [(0.50)(0.008)] K_c = 0.16 / 0.004 K_c = 40

Interpretation: K_c > 1 means the equilibrium favors products (ammonia formation).

3Problem 3medium

❓ Question:

For the reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), K_c = 0.500 at 400°C. Calculate K_p at the same temperature.

💡 Show Solution

Given:

Reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
K_c = 0.500
T = 400°C = 673 K
R = 0.08206 L·atm/(mol·K)

Relationship:

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

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)

4Problem 4medium

❓ Question:

For the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), K_p = 3.0 × 10⁴ at 700 K. Calculate K_c for this reaction at the same temperature. (R = 0.08206 L·atm/(mol·K))

💡 Show Solution

Solution:

Relationship between K_p and K_c: K_p = K_c(RT)^Δn

where Δn = (moles of gaseous products) - (moles of gaseous reactants)

Calculate Δn: Products: 2 moles SO₃ Reactants: 2 moles SO₂ + 1 mole O₂ = 3 moles Δn = 2 - 3 = -1

Solve for K_c: K_c = K_p / (RT)^Δn K_c = K_p × (RT)^(-Δn) K_c = 3.0 × 10⁴ × [(0.08206)(700)]^(1) K_c = 3.0 × 10⁴ × 57.4 K_c = 1.7 × 10⁶

Note: K_c > K_p when Δn < 0

5Problem 5hard

❓ Question:

At equilibrium at 500 K: H₂(g) + I₂(g) ⇌ 2HI(g), [H₂] = 0.20 M, [I₂] = 0.20 M, [HI] = 1.60 M. (a) Calculate K_c. (b) If this reaction is reversed, what is the new K_c?

💡 Show Solution

Given:

Reaction: H₂(g) + I₂(g) ⇌ 2HI(g)
At equilibrium:
- [H₂] = 0.20 M
- [I₂] = 0.20 M
- [HI] = 1.60 M
T = 500 K

(a) Calculate K_c

K_c expression:

$K_c = \frac{[\ce{HI}]^2}{[\ce{H2}][\ce{I2}]}$

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 ✓

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