1. a) Equilibrium requires no net exchange of matter with the surroundings; in an open system, reactants or products can escape, preventing a stable equilibrium state from being reached.
b) At equilibrium, forward and reverse reactions continue at equal rates, meaning molecules are constantly reacting and reforming, maintaining constant macroscopic properties while microscopic processes persist.
2. Let x = moles of CO reacted at equilibrium.
Initial: [CO] = 2 M, [H₂O] = 2 M, [CO₂] = 0, [H₂] = 0
Change: -x, -x, +x, +x
Equilibrium: [CO] = 2 - x, [H₂O] = 2 - x, [CO₂] = x, [H₂] = x
Kc = [CO₂][H₂] / [CO][H₂O] = (x)(x) / (2 - x)(2 - x) = x² / (2 - x)² = 6.2
Take square root: x / (2 - x) = √6.2 ≈ 2.49
x = 2.49(2 - x) → x = 4.98 - 2.49x → 3.49x = 4.98 → x ≈ 1.427
[CO] = [H₂O] = 2 - 1.427 = 0.573 M
[CO₂] = [H₂] = 1.427 M
3. The given reaction is: SO₂(g) + ½ O₂(g) ⇌ SO₃(g), Kp = 72.5
The target reaction is: 2SO₃(g) ⇌ 2SO₂(g) + O₂(g)
This is the reverse of twice the given reaction.
For reversing: Kp' = 1/Kp
For doubling coefficients: Kp'' = (Kp')² = (1/Kp)²
So Kp for target = (1/72.5)² = 1 / 5256.25 ≈ 1.90 × 10⁻⁴
4. Reaction: CO₂(g) + C(s) ⇌ 2CO(g)
Kp = P_CO² / P_CO₂ = 3 at 1000 K
Relationship between Kp and Kc: Kp = Kc(RT)^Δn
Δn = moles gaseous products - moles gaseous reactants = 2 - 1 = 1
R = 0.0821 L·atm·mol⁻¹·K⁻¹, T = 1000 K
Kp = Kc(RT)¹ → Kc = Kp / (RT) = 3 / (0.0821 × 1000) = 3 / 82.1 ≈ 0.0365
5. Reaction: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
Qc = [PCl₃][Cl₂] / [PCl₅] = (0.29)(0.32) / 0.15 = 0.0928 / 0.15 ≈ 0.619
Kc = 3.5
Since Qc < Kc, the reaction will proceed to the right (forward direction) to reach equilibrium.
6. Reaction: 2NO₂(g) ⇌ N₂O₄(g)
Increasing pressure favors the side with fewer moles of gas.
Reactants: 2 moles NO₂, Products: 1 mole N₂O₄
Fewer moles on product side → equilibrium shifts right → concentration of products increases.
7. HCN(aq) ⇌ H⁺(aq) + CN⁻(aq), Ka = 4.9 × 10⁻¹⁰, initial [HCN] = 0.05 M
Let x = [H⁺] = [CN⁻] at equilibrium, [HCN] ≈ 0.05 - x ≈ 0.05 (since Ka is small)
Ka = x² / 0.05 = 4.9 × 10⁻¹⁰ → x² = 2.45 × 10⁻¹¹ → x ≈ √(2.45 × 10⁻¹¹) ≈ 4.95 × 10⁻⁶ M
[H⁺] = [CN⁻] = 4.95 × 10⁻⁶ M
[HCN] = 0.05 - 4.95×10⁻⁶ ≈ 0.05 M
pH = -log(4.95×10⁻⁶) ≈ 5.305
Degree of dissociation α = x / initial [HCN] = (4.95×10⁻⁶) / 0.05 = 9.9 × 10⁻⁵ or 0.0099%
8. For conjugate acid-base pair: Ka × Kb = Kw = 1.0 × 10⁻¹⁴
Ka (niacin) = 1.5 × 10⁻⁵
Kb (conjugate base) = Kw / Ka = (1.0 × 10⁻¹⁴) / (1.5 × 10⁻⁵) = 6.67 × 10⁻¹⁰
Parent Tip: Review the logic above to help your child master the concept of chemical equilibrium worksheet.
