Equilibrium Constants: Kp

For gaseous equilibria, it is often more convenient to work with partial pressures rather than concentrations. The equilibrium constant expressed in terms of partial pressures is called Kp. At A-level, Kp is used exclusively for reactions where all species are gases.

Partial Pressures

In a mixture of gases, each gas contributes to the total pressure. The partial pressure of a gas is the pressure it would exert if it alone occupied the entire container at the same temperature.

Dalton's Law: The total pressure equals the sum of all partial pressures:

P_total = p_A + p_B + p_C + ...

The partial pressure of a gas is related to its mole fraction (x):

p_A = x_A × P_total

Where the mole fraction of A is:

x_A = moles of A / total moles of gas

The mole fractions of all gases in the mixture must sum to 1.

Worked Example: Calculating Partial Pressures

A mixture at equilibrium contains 2.0 mol N₂, 6.0 mol H₂, and 4.0 mol NH₃ at a total pressure of 200 atm.

Total moles = 2.0 + 6.0 + 4.0 = 12.0 mol

x(N₂) = 2.0/12.0 = 0.167 x(H₂) = 6.0/12.0 = 0.500 x(NH₃) = 4.0/12.0 = 0.333

Check: 0.167 + 0.500 + 0.333 = 1.000 ✓

p(N₂) = 0.167 × 200 = 33.3 atm p(H₂) = 0.500 × 200 = 100.0 atm p(NH₃) = 0.333 × 200 = 66.7 atm

Check: 33.3 + 100.0 + 66.7 = 200.0 atm ✓

Writing the Kp Expression

For the general gaseous equilibrium: aA(g) + bB(g) ⇌ cC(g) + dD(g)

Kp = p(C)^c × p(D)^d / (p(A)^a × p(B)^b)

The same rules apply as for Kc: products over reactants, each raised to the power of the stoichiometric coefficient.

Examples

Equilibrium	Kp expression	Units (atm)
N₂ + 3H₂ ⇌ 2NH₃	p(NH₃)² / (p(N₂) × p(H₂)³)	atm⁻²
2SO₂ + O₂ ⇌ 2SO₃	p(SO₃)² / (p(SO₂)² × p(O₂))	atm⁻¹
N₂O₄ ⇌ 2NO₂	p(NO₂)² / p(N₂O₄)	atm
H₂ + I₂ ⇌ 2HI	p(HI)² / (p(H₂) × p(I₂))	No units
PCl₅ ⇌ PCl₃ + Cl₂	p(PCl₃) × p(Cl₂) / p(PCl₅)	atm

Units of Kp

As with Kc, the units depend on the expression. Substitute the pressure unit (atm, Pa, or kPa) for each partial pressure term.

For Kp = p(NH₃)² / (p(N₂) × p(H₂)³):

Units = atm² / (atm × atm³) = atm² / atm⁴ = atm⁻²

If the total powers on top and bottom are equal, Kp has no units.

Step-by-Step Method for Kp Calculations

Every Kp calculation follows the same steps:

ICE table in moles -- find equilibrium moles of each species.
Total moles -- add up all equilibrium moles.
Mole fractions -- divide each species' moles by total moles. CHECK: they must sum to 1.
Partial pressures -- multiply each mole fraction by total pressure. CHECK: they must sum to P_total.
Substitute into Kp -- plug partial pressures into the expression and evaluate.

Calculating Kp: Full Worked Example 1

Problem: 1.00 mol of N₂O₄ is allowed to reach equilibrium at 400 K and a total pressure of 1.00 atm. At equilibrium, 50% of the N₂O₄ has dissociated. Calculate Kp for:

N₂O₄(g) ⇌ 2NO₂(g)

Step 1: Find equilibrium moles.

	N₂O₄	2NO₂
Initial	1.00	0
Change	−0.50	+1.00
Equilibrium	0.50	1.00

Total moles at equilibrium = 0.50 + 1.00 = 1.50 mol

Step 2: Calculate mole fractions.

x(N₂O₄) = 0.50/1.50 = 1/3 x(NO₂) = 1.00/1.50 = 2/3

Check: 1/3 + 2/3 = 1 ✓

Step 3: Calculate partial pressures.

p(N₂O₄) = (1/3) × 1.00 = 0.333 atm p(NO₂) = (2/3) × 1.00 = 0.667 atm

Check: 0.333 + 0.667 = 1.000 atm ✓

Step 4: Calculate Kp.

Kp = p(NO₂)² / p(N₂O₄) = (0.667)² / 0.333 = 0.445 / 0.333 = 1.33 atm

Units: atm² / atm = atm ✓

Kp Calculation from Percentage Conversion: Worked Example 2

Problem: In the Haber process, N₂(g) + 3H₂(g) ⇌ 2NH₃(g), a 1:3 mixture of N₂:H₂ is used at 200 atm. At equilibrium, 15% of the N₂ has reacted. Calculate Kp.

Step 1: Start with convenient moles. Take 1.00 mol N₂ and 3.00 mol H₂.

	N₂	3H₂	2NH₃
Initial	1.00	3.00	0
Change	−0.15	−0.45	+0.30
Equilibrium	0.85	2.55	0.30

(15% of 1.00 = 0.15; 3 × 0.15 = 0.45 H₂ reacted; 2 × 0.15 = 0.30 NH₃ formed)

Step 2: Total moles and mole fractions.

Total moles = 0.85 + 2.55 + 0.30 = 3.70 mol

x(N₂) = 0.85/3.70 = 0.2297 x(H₂) = 2.55/3.70 = 0.6892 x(NH₃) = 0.30/3.70 = 0.08108

Check: 0.2297 + 0.6892 + 0.08108 = 1.000 ✓

Step 3: Partial pressures.

p(N₂) = 0.2297 × 200 = 45.95 atm p(H₂) = 0.6892 × 200 = 137.84 atm p(NH₃) = 0.08108 × 200 = 16.22 atm

Check: 45.95 + 137.84 + 16.22 = 200.01 ≈ 200 ✓

Step 4: Calculate Kp.

Kp = p(NH₃)² / (p(N₂) × p(H₂)³) Kp = (16.22)² / (45.95 × (137.84)³) Kp = 263.1 / (45.95 × 2.620 × 10⁶) Kp = 263.1 / (1.204 × 10⁸) Kp = 2.19 × 10⁻⁶ atm⁻²

Kp Calculation: Worked Example 3 (Different Stoichiometry)

Problem: 4.00 mol of SO₃ is heated at 2.00 atm total pressure. At equilibrium, 25% has decomposed:

2SO₃(g) ⇌ 2SO₂(g) + O₂(g)

25% of 4.00 = 1.00 mol SO₃ decomposed.

	2SO₃	2SO₂	O₂
Initial	4.00	0	0
Change	−1.00	+1.00	+0.50
Equilibrium	3.00	1.00	0.50

(Ratio 2:2:1 means 1.00 mol SO₂ and 0.50 mol O₂ formed per 1.00 mol SO₃ decomposed)

Total moles = 3.00 + 1.00 + 0.50 = 4.50

x(SO₃) = 3.00/4.50 = 0.667; p(SO₃) = 0.667 × 2.00 = 1.333 atm x(SO₂) = 1.00/4.50 = 0.222; p(SO₂) = 0.222 × 2.00 = 0.444 atm x(O₂) = 0.50/4.50 = 0.111; p(O₂) = 0.111 × 2.00 = 0.222 atm

Kp = p(SO₂)² × p(O₂) / p(SO₃)²

Kp = (0.444)² × 0.222 / (1.333)²

Kp = 0.197 × 0.222 / 1.777

Kp = 0.0438 / 1.777 = 0.0246 atm

Units: atm² × atm / atm² = atm ✓

Relationship Between Kc and Kp

For an ideal gas, pV = nRT, so p = (n/V)RT = cRT, where c is the molar concentration.

Substituting into the Kp expression leads to:

Kp = Kc × (RT)^Δn

Where Δn = (moles of gaseous products) − (moles of gaseous reactants) in the balanced equation.

Example	Δn	Relationship
H₂ + I₂ ⇌ 2HI	2 − 2 = 0	Kp = Kc
N₂O₄ ⇌ 2NO₂	2 − 1 = +1	Kp = Kc × RT
N₂ + 3H₂ ⇌ 2NH₃	2 − 4 = −2	Kp = Kc / (RT)²
PCl₅ ⇌ PCl₃ + Cl₂	2 − 1 = +1	Kp = Kc × RT

Common Mistakes to Avoid

Forgetting that total moles change. In dissociation equilibria, the total moles at equilibrium are NOT the same as the initial moles. Always recalculate.
Mole fractions not summing to 1. This is the most reliable check for arithmetic errors. If they don't sum to 1, go back to your ICE table.
Partial pressures not summing to P_total. Same principle -- always verify this before calculating Kp.
Using wrong stoichiometric ratios in the ICE table. For 2SO₃ ⇌ 2SO₂ + O₂, if x mol of SO₃ decomposes, x mol of SO₂ forms and x/2 mol of O₂ forms (not x).
Confusing Kp with Kc. They are only numerically equal when Δn = 0. For all other cases, you need the Kp = Kc(RT)^Δn relationship.
Inconsistent pressure units. If the question gives pressure in kPa, your Kp units will be in kPa (not atm). Stay consistent throughout.

Summary

Kp uses partial pressures instead of concentrations for gaseous equilibria. Partial pressures are calculated from mole fractions and total pressure. The method involves: finding equilibrium moles → mole fractions → partial pressures → substituting into the Kp expression. Units depend on the expression. Kp and Kc are related by Kp = Kc(RT)^Δn. Like Kc, only temperature changes the value of Kp.

A-Level Deep Dive: Equilibrium Constants: Kp

Spec mapping

Edexcel 9CH0 specification, Topic 11 — Equilibrium II introduces the equilibrium constant $K_p$ for gas-phase equilibria, expressed in terms of partial pressures of gaseous species rather than concentrations (refer to the official specification document for exact wording). Candidates must be able to: compute partial pressures from mole fractions and total pressure (Dalton's law: $p_i = x_i P$ ), construct $K_p$ expressions including correct stoichiometric exponents, derive units of $K_p$ from the partial-pressure form, and relate $K_p$ to $K_c$ via $K_p = K_c (RT)^{\Delta n}$ . Examined principally on Paper 2 (Topics 11–19) and Paper 3 (synoptic with industrial chemistry, especially Haber and Contact). The data booklet does not list $K_p = K_c (RT)^{\Delta n}$ — this conversion must be derived if needed. Like $K_c$ , $K_p$ depends only on temperature.

Worked example with full mark scheme

Question (8 marks):

For the equilibrium $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ at 500 K and 200 atm total pressure, the mole fractions at equilibrium are $x(\text{N}_2) = 0.20$ , $x(\text{H}_2) = 0.60$ , $x(\text{NH}_3) = 0.20$ .

(a) Write the expression for $K_p$ in terms of partial pressures. (1)

(b) Calculate the partial pressures of each gas. (3)

(d) State, with reasoning, why $K_p$ for this reaction has units of pressure raised to a negative power. (1)

Solution with mark scheme:

(a) M1 — $K_p = p(\text{NH}_3)^2 / [p(\text{N}_2) \cdot p(\text{H}_2)^3]$ .

(b) M1 — Dalton's law: $p_i = x_i \cdot P_{\text{total}}$ . M1 — substitute: $p(\text{N}_2) = 0.20 \times 200 = 40$ atm; $p(\text{H}_2) = 0.60 \times 200 = 120$ atm; $p(\text{NH}_3) = 0.20 \times 200 = 40$ atm. A1 — sanity check: sum = 40 + 120 + 40 = 200 atm = $P_{\text{total}}$ .

(c) M1 — substitute into $K_p$ : $K_p = (40)^2 / [40 \times (120)^3] = 1600 / [40 \times 1.728 × 10^6]$ . M1 — $K_p = 1600 / (6.912 × 10^7) = 2.31 × 10^{-5}$ . A1 — units: atm² / (atm × atm³) = atm² / atm⁴ = atm⁻², so $K_p = 2.31 × 10^{-5}$ atm⁻².

(d) M1 — $\Delta n = 2 - (1 + 3) = -2$ (fewer moles of gas products than reactants). Units of $K_p$ are pressure raised to $\Delta n$ , so atm⁻² for this reaction.

Total: 8 marks (M5 A3, split as shown).

Specimen question modelled on the Edexcel 9CH0 Paper 2 format

Question (6 marks): The dissociation of N₂O₄: $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$ at 350 K. A 0.100 mol sample of pure N₂O₄ in a 4.00 dm³ vessel reaches equilibrium at total pressure 1.20 atm.

(a) Calculate the moles of each gas at equilibrium. (3)

(b) Calculate $K_p$ . (3)

Mark scheme decomposition by AO:

Part	Mark	AO	Earned by
(a) M1	1	AO2	Use ideal gas law: $n_{\text{total}} = PV/RT = 1.20 \times 4.00 / (0.0821 \times 350) = 4.80 / 28.74 = 0.167$ mol total.
(a) M1	1	AO2	If $x$ mol of N₂O₄ dissociates: equilibrium has $(0.100 - x)$ mol N₂O₄ + $2x$ mol NO₂; total = $0.100 + x = 0.167$ , so $x = 0.067$ mol.
(a) A1	1	AO2	Equilibrium: 0.033 mol N₂O₄, 0.134 mol NO₂; mole fractions 0.198 and 0.802.
(b) M1	1	AO2	Partial pressures: $p(\text{N}_2\text{O}_4) = 0.198 × 1.20 = 0.238$ atm; $p(\text{NO}_2) = 0.802 × 1.20 = 0.962$ atm.
(b) M1	1	AO1	$K_p = p(\text{NO}_2)^2 / p(\text{N}_2\text{O}_4)$ .
(b) A1	1	AO2	$K_p = (0.962)^2 / 0.238 = 0.925 / 0.238 = 3.89$ atm.