Kc, Kp and Calculations
Section: 7 Equilibria | Syllabus: Cambridge AS Level Chemistry (9701)
The Equilibrium Constant Kc
For the general homogeneous equilibrium: aA + bB ⇌ cC + dD
Kc (Equilibrium Constant in terms of concentrations) Kc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ)Concentrations are equilibrium values in mol dm⁻³. Products appear in the numerator; reactants in the denominator. Each concentration is raised to the power of its stoichiometric coefficient. Writing Kc Expressions
Example 1: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
Kc = [NH₃]² / ([N₂][H₂]³)
Example 2: H₂(g) + I₂(g) ⇌ 2HI(g)
Kc = [HI]² / ([H₂][I₂])
Example 3: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
Kc = [SO₃]² / ([SO₂]²[O₂])
Heterogeneous Equilibria
Rule Pure solids and pure liquids are not included in the Kc expression because their concentrations are effectively constant and do not change during the reaction. Heterogeneous ExamplesCaCO₃(s) ⇌ CaO(s) + CO₂(g)
Kc = [CO₂] (both solids omitted)
Fe₃O₄(s) + 4H₂(g) ⇌ 3Fe(s) + 4H₂O(g)
Kc = [H₂O]⁴ / [H₂]⁴ (solids omitted)
Units of Kc
Units are found by substituting mol dm⁻³ for each concentration term and simplifying.
Finding Units of KcN₂(g) + 3H₂(g) ⇌ 2NH₃(g)
Kc = [NH₃]² / ([N₂][H₂]³)
Units = (mol dm⁻³)² / [(mol dm⁻³)(mol dm⁻³)³]
= mol² dm⁻⁶ / mol⁴ dm⁻¹² = mol⁻² dm⁶
Tip If the total moles of gaseous/aqueous products equals the total moles of gaseous/aqueous reactants (Δn = 0), Kc has no units. Always check and state units explicitly.Mole Fractions and Partial Pressures
Mole Fraction (x) The fraction of the total moles in a mixture that belongs to one component:x(A) = n(A) / n(total)
Mole fractions are dimensionless and must sum to 1. Partial Pressure (p) The pressure that a gas would exert if it alone occupied the entire volume. For a component A in a gas mixture:
p(A) = x(A) × P(total)
The sum of all partial pressures equals the total pressure (Dalton's law). Worked Example - Mole Fractions and Partial Pressures
An equilibrium mixture contains: 1.0 mol N₂, 3.0 mol H₂, 0.5 mol NH₃. Total pressure = 200 atm.
Total moles = 1.0 + 3.0 + 0.5 = 4.5 mol
x(N₂) = 1.0 / 4.5 = 0.222 → p(N₂) = 0.222 × 200 = 44.4 atm
x(H₂) = 3.0 / 4.5 = 0.667 → p(H₂) = 0.667 × 200 = 133.4 atm
x(NH₃) = 0.5 / 4.5 = 0.111 → p(NH₃) = 0.111 × 200 = 22.2 atm
Check: 44.4 + 133.4 + 22.2 = 200 atm ✓
Fig 7.5 - Partial Pressures in a Gas MixtureDiagram: A horizontal bar representing the total pressure (200 atm) divided into three coloured segments proportional to the partial pressures of N₂ (44.4 atm), H₂ (133.4 atm), and NH₃ (22.2 atm). Below, a pie chart shows the corresponding mole fractions. Labels indicate that p(N₂) + p(H₂) + p(NH₃) = P(total).
The Equilibrium Constant Kp
For gas-phase equilibria, Kp is expressed in terms of partial pressures rather than concentrations.
Kp (Equilibrium Constant in terms of partial pressures) For aA(g) + bB(g) ⇌ cC(g) + dD(g):Kp = [p(C)]ᶜ × [p(D)]ᵈ / ([p(A)]ᵃ × [p(B)]ᵇ)
Pure solids are excluded (same rule as Kc). The units of Kp depend on the pressure units used and the value of Δn(gas). Writing Kp Expressions
Example 1: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
Kp = p(NH₃)² / [p(N₂) × p(H₂)³]
Example 2: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
Kp = p(SO₃)² / [p(SO₂)² × p(O₂)]
Units for Example 2: atm² / (atm² × atm) = atm⁻¹
Equilibrium Calculations - The ICE Table Method
ICE stands for Initial → Change → Equilibrium. It is a systematic way to track concentrations (or moles) from the start of a reaction to equilibrium.
Fig 7.6 - The ICE Table StructureDiagram: A generic ICE table with columns for each species (A, B, C, D) and three rows labelled Initial, Change (showing −ax, −bx, +cx, +dx where x is the extent of reaction), and Equilibrium (Initial ± Change). Annotations explain: "Substitute the Equilibrium row into the Kc expression and solve for x." An arrow shows that once x is found, all equilibrium values can be calculated.
Worked Example 1 - Calculating Kc from Equilibrium DataReaction: H₂(g) + I₂(g) ⇌ 2HI(g) in a 1.0 dm³ flask
0.50 mol H₂ and 0.50 mol I₂ are mixed. At equilibrium, 0.40 mol HI is present. Calculate Kc.
| H₂ | I₂ | HI | |
|---|---|---|---|
| Initial (mol dm⁻³) | 0.50 | 0.50 | 0 |
| Change | −0.20 | −0.20 | +0.40 |
| Equilibrium | 0.30 | 0.30 | 0.40 |
Kc = [HI]² / ([H₂][I₂]) = (0.40)² / (0.30 × 0.30) = 0.16 / 0.090 = 1.78 (no units)
Worked Example 2 - Calculating Quantities at Equilibrium Given KcReaction: CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)
Kc = 4.0. 1.0 mol ethanoic acid is mixed with 1.0 mol ethanol. Volume = 1.0 dm³. Find moles at equilibrium.
| CH₃COOH | C₂H₅OH | CH₃COOC₂H₅ | H₂O | |
|---|---|---|---|---|
| Initial | 1.0 | 1.0 | 0 | 0 |
| Change | −x | −x | +x | +x |
| Equilibrium | 1.0−x | 1.0−x | x | x |
Kc = x² / (1.0 − x)² = 4.0
Taking square roots of both sides: x / (1.0 − x) = 2.0
x = 2.0 − 2.0x → 3.0x = 2.0 → x = 0.667 mol
At equilibrium: CH₃COOH = C₂H₅OH = 0.333 mol; CH₃COOC₂H₅ = H₂O = 0.667 mol
Worked Example 3 - Calculating KpReaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). Total pressure = 200 atm.
At equilibrium: x(N₂) = 0.20, x(H₂) = 0.60, x(NH₃) = 0.20.
p(N₂) = 0.20 × 200 = 40 atm
p(H₂) = 0.60 × 200 = 120 atm
p(NH₃) = 0.20 × 200 = 40 atm
Kp = p(NH₃)² / [p(N₂) × p(H₂)³] = (40)² / (40 × 120³)
= 1600 / (40 × 1 728 000) = 1600 / 69 120 000 = 2.31 × 10⁻⁵ atm⁻²
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