Introduction to Chemical Equilibrium
Understand reversible reactions, dynamic equilibrium, and equilibrium constant expressions (K_c and K_p).
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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:
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:
Equilibrium expression:
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)
(Solids omitted)
- CH₃COOH(aq) + H₂O(l) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq)
(Water omitted - solvent)
- N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
Equilibrium Constant (K_p)
For gas-phase reactions:
Use partial pressures (atm) instead of concentrations
Relationship between K_c and K_p:
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:
Apply coefficients as exponents:
Answer:
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), 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:
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:
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)
3Problem 3hard
❓ 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:
Substitute values:
Answer (a): K_c = 64
(b) Reversed reaction K_c
Reversed reaction: 2HI(g) ⇌ H₂(g) + I₂(g)
Rule: K_reverse = 1/K_forward
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 ✓