Unit 7 Explanation 1

Unit 7: Equilibrium ⚖️

7.1 Dynamic Equilibrium and the Equilibrium Constant (K)

Equilibrium in chemistry occurs when opposing reactions happen at the same rate, leading to no net change in concentrations of reactants and products over time.

At dynamic equilibrium, forward and reverse reactions continue, but their rates are equal. This balance can be disturbed by changing conditions (temperature, pressure, concentration).

aA + bB ⇌ cC + dD
K_c = [C]^c[D]^d / [A]^a[B]^b

This ratio defines the position of equilibrium. The square brackets [ ] mean molar concentration (mol/L). The exponents match the coefficients in the balanced equation because the rate laws for elementary steps depend on molecular collisions and stoichiometry.

💡 Solids and liquids are excluded from K expressions – they have constant "activity."

This is CRUCIAL. Solids (like metals or precipitates) and pure liquids (like water in many reactions) don’t change their concentration during the reaction. Their activity is considered 1. Including them would incorrectly affect the value of K.

Example 1: CaCO₃(s) ⇌ CaO(s) + CO₂(g)
K_p = P_CO₂ (only gas appears in the expression)

Example 2: H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)
K_w = [H⁺][OH⁻] (water is excluded)

Manipulating Equilibrium Constants (K)

1. If the equation is reversed:

Then the new K is the reciprocal.

Example 1:
Original: N₂ + 3H₂ ⇌ 2NH₃, K = 6.0 × 10⁵
Reversed: 2NH₃ ⇌ N₂ + 3H₂
New K = 1 / (6.0 × 10⁵) = 1.67 × 10⁻⁶

Example 2:
CO + H₂O ⇌ CO₂ + H₂, K = 5.4 → Reverse: K = 1 / 5.4 = 0.185

2. If coefficients are multiplied by n:

Raise K to the power n.

Example 1:
Original: N₂ + 3H₂ ⇌ 2NH₃, K = 6.0 × 10⁵
Multiplied by 2: 2N₂ + 6H₂ ⇌ 4NH₃, K = (6.0 × 10⁵)² = 3.6 × 10¹¹

Example 2:
Half reaction: 1/2 N₂ + 3/2 H₂ ⇌ NH₃ → K = √(6.0 × 10⁵) ≈ 775

3. If equations are added:

Multiply their K values.

Example 1:
Rxn 1: A ⇌ B, K₁ = 4
Rxn 2: B ⇌ C, K₂ = 3
Combined: A ⇌ C, K = 4 × 3 = 12

Example 2:
Step 1: NO₂ ⇌ NO + O, K₁ = 2.0
Step 2: O + O₂ ⇌ O₃, K₂ = 1.5
Total: NO₂ + O₂ ⇌ NO + O₃ → K = 2.0 × 1.5 = 3.0

K_p vs. K_c: What if we want to convert?

For gas-phase reactions:

K_p = K_c(RT)^Δn

• R = 0.0821 L·atm/mol·K
• T = temperature in Kelvin
• Δn = moles of gaseous products – moles of gaseous reactants

This allows you to convert between pressure-based and concentration-based constants.

Example 1:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
Δn = 2 – 4 = –2 → K_p = K_c(RT)^–2

Example 2:
H₂(g) + I₂(g) ⇌ 2HI(g)
Δn = 2 – 2 = 0 → K_p = K_c

Reaction Quotient (Q) – Detailed

Q has the same form as K but is calculated using concentrations or pressures at ANY moment — not just at equilibrium.

• Q < K: Too few products → system shifts right to form more products
• Q > K: Too many products → system shifts left to form more reactants
• Q = K: Already at equilibrium → no shift

Example 1:
K = 10
Current: [P] = 1, [R] = 2 → Q = 1 / 2 = 0.5 → Q < K → shift right

Example 2:
K = 0.1
Current: [P] = 0.3, [R] = 0.1 → Q = 0.3 / 0.1 = 3 → Q > K → shift left

Example 3:
K = 5
Current: [P] = 2.5, [R] = 0.5 → Q = 5 → Q = K → equilibrium

Quick Summary – Everything You Must Know

1. K expressions exclude solids and liquids – they have constant activity.
2. Dynamic equilibrium = forward and reverse rates are equal.
3. K_c uses molar concentrations; K_p uses partial pressures.
4. Magnitude of K tells you direction (≫1 = products favored, ≪1 = reactants favored).
5. Reversing a reaction → 1/K; multiplying → Kⁿ; adding → multiply Ks.
6. K_p and K_c connected via Δn: K_p = K_c(RT)^Δn
7. Reaction quotient Q helps predict direction of shift.
8. If Q < K → shift right. If Q > K → shift left. If Q = K → equilibrium.

🎯 Ready to test your understanding? Take the Unit 7 Mini Quiz Here