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For gas-phase equilibria, the equilibrium constant is most usefully expressed not in terms of concentrations but in terms of partial pressures. The reason is practical: in an industrial reactor, in a sealed laboratory vessel, or in the atmosphere, the partial pressure of each gas is the directly measurable quantity — concentration of a gas requires you to know V and T as well. The constant Kp is built from equilibrium partial pressures in exactly the same way that Kc is built from equilibrium concentrations. This lesson develops the partial-pressure definition through Dalton's law, mole fractions, the Kp expression and its units, the conversion Kp = Kc(RT)^Δn, and a sequence of worked examples covering Δn = 0, +1, −1, and +2 cases — including the Haber synthesis and the Contact process. Lesson 5 (Kc) is the prerequisite; lesson 4 (Le Chatelier) supplies the qualitative shift rules that Kp formalises quantitatively.
Spec mapping (AQA 7405): This lesson maps to §3.1.10 (equilibrium constant Kp for homogeneous gaseous systems — A2). It builds on lesson 5 of this course (Kc, §3.1.6 + §3.1.10) and on lesson 4 (Le Chatelier's principle — qualitative, §3.1.6). The conversion Kp = Kc(RT)^Δn relies on the ideal gas equation pV = nRT introduced in §3.1.2.5 (covered in lesson 9 of the AQA A-Level Chemistry — Atomic Structure course). Refer to the official AQA specification document for the exact wording of each section.
Assessment objectives: Definition of partial pressure, mole fraction, and the Kp expression for a given gaseous equilibrium are AO1 recall items. Computing Kp from partial pressures, from total pressure and mole fractions, from an ICE table of equilibrium moles, or by conversion from Kc, tests AO2. Interpreting the magnitude of Kp, predicting the direction of shift in Kp on a temperature change using the sign of ΔH, and reasoning about why pressure and catalyst do not change Kp, are AO3 items that anchor the longer essay-style and multi-step calculation questions on Paper 2 of the AQA A-Level Chemistry assessment.
In a mixture of gases that do not react with one another, each gas behaves — to a very good approximation — as if it alone occupied the container. The partial pressure p_i of gas i is the pressure that gas i would exert in the absence of the other gases:
p_i = (n_i / n_total) × P_total = x_i × P_total
where x_i = n_i / Σn is the mole fraction of gas i. The sum of all mole fractions in a mixture is exactly 1.
Dalton's law of partial pressures states that the total pressure of a mixture of (ideal) gases is the sum of the partial pressures of the components:
P_total = p_A + p_B + p_C + …
Two routes to a partial pressure are useful in equilibrium problems:
Worked check: A sealed 2.00 dm³ vessel at 400 K contains 0.150 mol N₂ and 0.0500 mol O₂. Find the partial pressure of each gas and the total pressure.
Route 1: total moles = 0.200 mol. Use pV = nRT: P_total = (0.200 × 8.314 × 400) / (2.00 × 10⁻³) = 332.6 kPa. Then p(N₂) = (0.150/0.200) × 332.6 = 249.4 kPa and p(O₂) = (0.0500/0.200) × 332.6 = 83.2 kPa. Sum = 332.6 kPa as expected by Dalton's law.
For a homogeneous gaseous equilibrium aA(g) + bB(g) ⇌ cC(g) + dD(g) at constant temperature, the equilibrium constant Kp is defined as
Kp = (p_C^c × p_D^d) / (p_A^a × p_B^b)
where each p is the equilibrium partial pressure of that species. The Kp expression follows three structural rules:
Common Misconception: "Kp is just Kc with the letter changed." Not quite. Kp uses partial pressures (units of pressure, e.g. kPa or atm or Pa); Kc uses concentrations (units of mol dm⁻³). Numerically Kp ≠ Kc in general — they are related by the formula Kp = Kc × (RT)^Δn, derived below. Confusing them in a calculation is one of the most common Paper 2 errors.
The units of Kp are obtained by substituting the chosen pressure unit (commonly kPa, sometimes atm or Pa) into the expression and simplifying. Define Δn(gas) = (c + d) − (a + b) — the change in the total stoichiometric coefficient of gaseous species across the reaction. (Solids and liquids contribute nothing to Δn.) Then the units of Kp are (pressure unit)^Δn:
| Reaction | Δn(gas) | Units of Kp (in kPa) |
|---|---|---|
| H₂(g) + I₂(g) ⇌ 2HI(g) | 0 | dimensionless |
| 2NO₂(g) ⇌ N₂O₄(g) | −1 | kPa⁻¹ |
| PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) | +1 | kPa |
| N₂(g) + 3H₂(g) ⇌ 2NH₃(g) | −2 | kPa⁻² |
| 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) | −1 | kPa⁻¹ |
| 2H₂O(g) ⇌ 2H₂(g) + O₂(g) | +1 | kPa |
The numerical value of Kp depends on the pressure unit chosen (a Kp expressed in atm will be numerically different from the same equilibrium's Kp expressed in kPa or Pa, except when Δn = 0). The predictions — direction of shift on a temperature change, position of equilibrium — do not depend on the unit choice.
Exam Tip: On AQA papers, the convention is to use kPa unless the question explicitly directs otherwise. Always state the units of Kp; a numerical answer without units is normally penalised by 1 mark.
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). A 1.00 mol sample of PCl₅ is sealed in a vessel and heated to 500 K. At equilibrium 40.0% has dissociated and the total equilibrium pressure is 200 kPa. Calculate Kp.
Step 1 — Equilibrium moles (ICE in moles).
| Species | PCl₅ | PCl₃ | Cl₂ |
|---|---|---|---|
| Initial / mol | 1.00 | 0 | 0 |
| Change / mol | −0.400 | +0.400 | +0.400 |
| Equilibrium / mol | 0.600 | 0.400 | 0.400 |
Total moles at equilibrium = 0.600 + 0.400 + 0.400 = 1.400.
Step 2 — Mole fractions.
x(PCl₅) = 0.600/1.400 = 0.4286 x(PCl₃) = 0.400/1.400 = 0.2857 x(Cl₂) = 0.400/1.400 = 0.2857
(Check: 0.4286 + 0.2857 + 0.2857 = 1.000 ✓)
Step 3 — Partial pressures (p_i = x_i × P_total).
p(PCl₅) = 0.4286 × 200 = 85.7 kPa p(PCl₃) = 0.2857 × 200 = 57.1 kPa p(Cl₂) = 0.2857 × 200 = 57.1 kPa
Step 4 — Kp.
Kp = p(PCl₃) × p(Cl₂) / p(PCl₅) = (57.1 × 57.1) / 85.7 = 3261 / 85.7 = 38.1 kPa
Δn = (1+1) − 1 = +1, so the units are (kPa)^+1 = kPa, as expected.
N₂(g) + 3H₂(g) ⇌ 2NH₃(g). In an industrial Haber reactor at 700 K and a total pressure of 10 000 kPa (≈ 100 atm), the equilibrium mole fractions are x(N₂) = 0.200, x(H₂) = 0.600, x(NH₃) = 0.200. Calculate Kp.
Partial pressures:
Kp = [p(NH₃)]² / {[p(N₂)] × [p(H₂)]³}
Numerator: (2000)² = 4.00 × 10⁶ kPa² Denominator: 2000 × (6000)³ = 2000 × 2.16 × 10¹¹ = 4.32 × 10¹⁴ kPa⁴
Kp = 4.00 × 10⁶ / 4.32 × 10¹⁴ = 9.26 × 10⁻⁹ kPa⁻²
Δn = 2 − (1 + 3) = −2, so units are kPa⁻².
Industrial commentary: The very small Kp at 700 K shows that, at this temperature, the equilibrium lies well to the left — the equilibrium yield of NH₃ is modest (~20% mole fraction here). The Haber process compensates by operating at very high total pressure (which shifts the position of equilibrium right but does not change Kp) and by recycling unreacted N₂/H₂. The temperature is a compromise: lowering T would raise Kp (forward reaction is exothermic, ΔH = −92 kJ mol⁻¹) but slows the rate to industrial impossibility.
2SO₂(g) + O₂(g) ⇌ 2SO₃(g). A reaction vessel at 700 K contains 0.0800 mol SO₂, 0.0400 mol O₂ and 0.180 mol SO₃ at equilibrium. The total pressure is 150 kPa. Calculate Kp.
Total moles = 0.0800 + 0.0400 + 0.180 = 0.300 mol.
Mole fractions: x(SO₂) = 0.0800/0.300 = 0.2667; x(O₂) = 0.0400/0.300 = 0.1333; x(SO₃) = 0.180/0.300 = 0.600.
Partial pressures: p(SO₂) = 0.2667 × 150 = 40.0 kPa; p(O₂) = 0.1333 × 150 = 20.0 kPa; p(SO₃) = 0.600 × 150 = 90.0 kPa.
Kp = [p(SO₃)]² / {[p(SO₂)]² × p(O₂)} = (90.0)² / [(40.0)² × 20.0] = 8100 / 32 000 = 0.253 kPa⁻¹
Δn = 2 − 3 = −1, so units are kPa⁻¹. The Contact process is operated near 700 K and ~200 kPa total pressure; this Kp is consistent with the high SO₃ yields (>95%) observed industrially.
2NO₂(g) ⇌ N₂O₄(g). 0.500 mol NO₂ is sealed in a 1.00 dm³ flask at 350 K. At equilibrium 80.0% of the NO₂ has dimerised. Calculate the equilibrium partial pressures and Kp.
ICE in moles (with 2NO₂ → 1 N₂O₄, so 0.400 mol NO₂ consumed produces 0.200 mol N₂O₄):
| Species | NO₂ | N₂O₄ |
|---|---|---|
| Initial / mol | 0.500 | 0 |
| Change / mol | −0.400 | +0.200 |
| Equilibrium / mol | 0.100 | 0.200 |
Total equilibrium moles = 0.300. Using pV = nRT with V = 1.00 × 10⁻³ m³ and T = 350 K:
P_total = nRT/V = (0.300 × 8.314 × 350) / 1.00 × 10⁻³ = 873 kPa.
Mole fractions: x(NO₂) = 0.100/0.300 = 0.3333; x(N₂O₄) = 0.200/0.300 = 0.6667.
Partial pressures: p(NO₂) = 0.3333 × 873 = 291 kPa; p(N₂O₄) = 0.6667 × 873 = 582 kPa.
Kp = p(N₂O₄) / [p(NO₂)]² = 582 / (291)² = 582 / 84 681 = 6.87 × 10⁻³ kPa⁻¹
H₂(g) + I₂(g) ⇌ 2HI(g). At 700 K, Kp = 54 (dimensionless, since Δn = 0). 1.00 mol H₂ and 1.00 mol I₂ are mixed in a sealed vessel; find the equilibrium mole fractions.
Let α mol of H₂ react. ICE in moles:
| Species | H₂ | I₂ | HI |
|---|---|---|---|
| Equilibrium / mol | 1.00 − α | 1.00 − α | 2α |
Total moles = 2.00 mol (constant because Δn = 0 — moles of gas are conserved).
When Δn = 0 the total pressure cancels in the Kp expression, so we can substitute mole fractions (or just moles divided by the same total) directly:
Kp = [p(HI)]² / [p(H₂) × p(I₂)] = (2α)² / [(1.00 − α)(1.00 − α)] = [2α/(1.00 − α)]² = 54
Taking the (positive) square root: 2α/(1.00 − α) = √54 = 7.348
2α = 7.348(1.00 − α) ⟹ 2α + 7.348α = 7.348 ⟹ α = 7.348/9.348 = 0.786
Mole fractions: x(HI) = 2 × 0.786 / 2.00 = 0.786; x(H₂) = x(I₂) = (1.00 − 0.786) / 2.00 = 0.107.
Δn = 0 shortcut: When Δn = 0, you can use mole fractions, or even just equilibrium moles, directly in the Kp expression — the total pressure (or total moles) cancels. This is a frequent A-Level shortcut for the H₂/I₂/HI system and the CO/H₂O/CO₂/H₂ water-gas shift.
NH₂CO₂NH₄(s) ⇌ 2NH₃(g) + CO₂(g) (ammonium carbamate decomposition). At a particular temperature the total equilibrium pressure above the solid is 30.0 kPa. Calculate Kp.
The Kp expression includes only the gases (the solid is omitted): Kp = [p(NH₃)]² × p(CO₂).
Because the gases arise from a single solid in fixed stoichiometric ratio, the mole ratio of NH₃ : CO₂ is 2 : 1, so x(NH₃) = 2/3 and x(CO₂) = 1/3.
p(NH₃) = (2/3) × 30.0 = 20.0 kPa; p(CO₂) = (1/3) × 30.0 = 10.0 kPa.
Kp = (20.0)² × 10.0 = 400 × 10.0 = 4.00 × 10³ kPa³
Δn = 3 (three moles of gas from zero on the reactant gas side), so units are kPa³.
For gas-phase systems, ICE tables are most cleanly constructed in moles (rather than concentrations or partial pressures directly). After finding equilibrium moles, convert to partial pressures via one of two routes:
A common trap: when Δn ≠ 0, the total moles change as the equilibrium shifts. If 0.400 mol of PCl₅ decomposes, total moles rise from 1.00 to 1.40, so the partial pressures cannot simply be read off the initial mole fractions. Always recompute mole fractions from equilibrium moles.
Kp and Kc are related through the ideal gas equation. For each gaseous species: p_i V = n_i RT, so p_i = (n_i/V) RT = [i] × RT, where [i] is the molar concentration. Substituting into the Kp expression for aA(g) + bB(g) ⇌ cC(g) + dD(g):
Kp = ([C] RT)^c × ([D] RT)^d / ([A] RT)^a × ([B] RT)^b = {[C]^c [D]^d / [A]^a [B]^b} × (RT)^{(c+d) − (a+b)} = Kc × (RT)^Δn
Kp = Kc × (RT)^Δn(gas)
Two cautions on units:
For 2HI(g) ⇌ H₂(g) + I₂(g) at 700 K, Kc = 1.85 × 10⁻². Calculate Kp.
Δn = (1 + 1) − 2 = 0, so Kp = Kc = 1.85 × 10⁻² (dimensionless).
For N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at 700 K, Kp = 9.26 × 10⁻⁹ kPa⁻². Calculate Kc.
Δn = −2. So Kp = Kc × (RT)^−2, hence Kc = Kp × (RT)².
Use R = 8.314 J K⁻¹ mol⁻¹, T = 700 K, with p in kPa and [ ] in mol dm⁻³. The product RT in compatible units is: R = 8.314 kPa dm³ K⁻¹ mol⁻¹, so RT = 8.314 × 700 = 5820 kPa dm³ mol⁻¹.
Kc = 9.26 × 10⁻⁹ × (5820)² = 9.26 × 10⁻⁹ × 3.39 × 10⁷ = 0.314 dm⁶ mol⁻² (= (mol dm⁻³)⁻²)
This is consistent with the small but non-negligible equilibrium constant observed for the Haber synthesis at this temperature.
The rules for Kp are identical to those for Kc (see lesson 5 and lesson 4):
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