Kc, Kp and Calculations
Section: 7 Equilibria | Syllabus: Cambridge AS Level Physics 9702
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 Examples CaCO₃(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 Kc N₂(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 Mixture Diagram: 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 I nitial → C hange → E quilibrium.
It is a systematic way to track concentrations (or moles) from the start of a reaction to equilibrium. Fig 7.6 - The ICE Table Structure Diagram: 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 Data Reaction: 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 Kc Reaction: 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 Kp Reaction: 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 …
Interactive revision notes, videos and practice questions load below.