Partial Pressures and Kp Expressions
A concise revision guide to gas equilibria: mole fraction, partial pressure as mole fraction × total pressure, how to write the expression for Kp from the balanced equation and its units, and why solids and liquids are left out of Kp for a heterogeneous system.
- 11.1
- 11.2
What these spec points say
- 11.1 be able to deduce an expression for Kp, for homogeneous and heterogeneous systems, in terms of equilibrium partial pressures in atm
- 11.2 be able to calculate a value, with units where appropriate, for the equilibrium constant (Kc and Kp) for homogeneous and heterogeneous reactions, from experimental data
Why Gases Have Their Own Constant
Pages 1 and 2 wrote Kc in terms of concentrations. For gases it is far easier to measure a pressure than a concentration.
So for an equilibrium in which the reacting substances are gases a second constant is used: Kp, the equilibrium constant in terms of partial pressures.
- Kp is built in the same way as Kc, with products over reactants and powers from the equation.
- Each concentration is replaced by the partial pressure of that gas.
- Like Kc, the value of Kp is constant at a given temperature.
Two ideas are needed before the expression can be used: the mole fraction of a gas and its partial pressure.
Definition: The partial pressure of a gas in a mixture is the pressure that the gas would exert if it alone occupied the whole container at the same temperature.
Mole Fraction and Partial Pressure
The mole fraction of a gas, x, is its share of the total number of moles of gas in the mixture:
mole fraction of A, x(A) = (moles of A) / (total moles of gas)
- A mole fraction has no units.
- The mole fractions of all the gases in a mixture add up to 1.
Each gas contributes to the total pressure in proportion to its mole fraction:
partial pressure of A, p(A) = x(A) × total pressure, P
- A partial pressure has the same units as the total pressure.
- The sum of the partial pressures equals the total pressure: P = p(A) + p(B) + p(C) + …
Worked example: from moles to partial pressures
An equilibrium mixture contains 2.0 mol of N₂, 6.0 mol of H₂ and 2.0 mol of NH₃ at a total pressure of 200 atm.
Step 1. Total moles of gas = 2.0 + 6.0 + 2.0 = 10.0 mol.
| Gas | Moles | Mole fraction | Partial pressure / atm |
|---|---|---|---|
| N₂ | 2.0 | (2.0) / (10.0) = 0.20 | 0.20 × 200 = 40 |
| H₂ | 6.0 | (6.0) / (10.0) = 0.60 | 0.60 × 200 = 120 |
| NH₃ | 2.0 | (2.0) / (10.0) = 0.20 | 0.20 × 200 = 40 |
| Total | 10.0 | 1.00 | 200 |
Check. The mole fractions add up to 1.00 and the partial pressures add up to 200 atm.
Worked example: using the sum
A mixture of N₂O₄ and NO₂ has a total pressure of 150 atm. The partial pressure of NO₂ is 90 atm.
Step 1. p(N₂O₄) = 150 − 90 = 60 atm.
Step 2. Mole fraction of NO₂ = (partial pressure) / (total pressure) = (90) / (150) = 0.60.
From moles to mole fractions to partial pressures for a mixture of three gases: the mole fractions add up to 1 and the partial pressures add up to the total pressure.
Key idea: Partial pressure = mole fraction × total pressure. Mole fractions add up to 1; partial pressures add up to the total pressure. Use both sums as a check on every calculation.
Check: Mole Fractions and Partial Pressures
Calculate mole fractions and partial pressures, and use their sums, for gas mixtures not used on this page.
Writing a Kp Expression
For the gas equilibrium aA(g) + bB(g) ⇌ cC(g) + dD(g):
Kp = (p(C)ᶜ × p(D)ᵈ) / (p(A)ᵃ × p(B)ᵇ)
- Products go over reactants, as in Kc.
- Each partial pressure is raised to the power of its balancing number.
- Partial pressures are written with a p and no square brackets. Square brackets mean concentration and are marked wrong in Kp.
- The partial pressures are equilibrium values.
| Equation | Kp expression | Units of Kp |
|---|---|---|
| N₂(g) + 3H₂(g) ⇌ 2NH₃(g) | (p(NH₃)²) / (p(N₂) × p(H₂)³) | atm⁻² |
| 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) | (p(SO₃)²) / (p(SO₂)² × p(O₂)) | atm⁻¹ |
| N₂O₄(g) ⇌ 2NO₂(g) | (p(NO₂)²) / (p(N₂O₄)) | atm |
| H₂(g) + I₂(g) ⇌ 2HI(g) | (p(HI)²) / (p(H₂) × p(I₂)) | none |
Units of Kp
The units are found by cancelling, exactly as for Kc, with the pressure unit in place of mol dm⁻³.
Step 1. For the ammonia equilibrium, units = (atm²) / (atm × atm³) = (atm²) / (atm⁴).
Answer. Units of Kp = atm⁻².
Pressures in these questions are usually given in atm. If a question uses kPa or Pa instead, keep that unit throughout and do not convert.
Worked example: substituting into Kp
Use the partial pressures from the ammonia mixture above: p(N₂) = 40, p(H₂) = 120 and p(NH₃) = 40 atm.
Step 1. Kp = (p(NH₃)²) / (p(N₂) × p(H₂)³) = (40²) / (40 × 120³).
Step 2. Kp = (1600) / (40 × 1 728 000) = (1600) / (69 120 000).
Answer. Kp = 2.31 × 10⁻⁵ atm⁻².
Kc and Kp for the same gas equilibrium side by side, with partial pressures written p and no square brackets, and four gas equilibria with their Kp expressions and units.
Exam focus: In a Kp expression write p(NH₃)², never [NH₃]² or [pNH₃]². Use only partial pressures, never moles or mole fractions, and state the units you have derived.
Check: Kp Expressions and Units
Write Kp expressions, derive their units and substitute partial pressures for gas equilibria not used on this page.
Quick Check: Work Through a New Mixture
Work out each step on paper, then flip the card to compare. The data are invented for this exercise.
Heterogeneous Kp
Some equilibria involve gases together with solids or liquids. The rule from Kc carries over: only gases appear in Kp.
- A solid or a pure liquid does not have a partial pressure that changes, so it is left out of the expression.
- It is also left out when the total moles of gas are counted for mole fractions.
| Equation | Left out | Kp expression | Units |
|---|---|---|---|
| CaCO₃(s) ⇌ CaO(s) + CO₂(g) | both solids | Kp = p(CO₂) | atm |
| C(s) + H₂O(g) ⇌ CO(g) + H₂(g) | carbon | Kp = (p(CO) × p(H₂)) / (p(H₂O)) | atm |
For calcium carbonate in a closed container, Kp = p(CO₂). At a given temperature the pressure of carbon dioxide at equilibrium is fixed, and it is the same whether there is a little solid or a lot.
- Heating raises Kp for this endothermic reaction, so the equilibrium pressure of carbon dioxide rises.
- In an open container the gas escapes, the equilibrium pressure is never reached, and the solid decomposes completely.
Remember: In a heterogeneous equilibrium, write partial pressures for the species marked (g) only. Solids and liquids are omitted from Kp and from the total moles of gas.
Quick Check: Solids and Gases Together
Pick the one accurate statement in each set, using equilibria that are new to this page.
Common Exam Points
Say
- “Mole fraction = (moles of the gas) / (total moles of gas).”
- “Partial pressure = mole fraction × total pressure.”
- “Kp = (p(SO₃)²) / (p(SO₂)² × p(O₂)).”
- “Units = (atm²) / (atm³) = atm⁻¹.”
Do not say
- “Kp = ([SO₃]²) / ([SO₂]²[O₂])” (square brackets mean concentration).
- “Partial pressure = moles × total pressure” (use the mole fraction).
- “Kp = (p(CaO) × p(CO₂)) / (p(CaCO₃))” (solids have no place in Kp).
- “Kp has the units of pressure” for every reaction (they depend on the equation).
Watch for
- Total moles: add up the moles of every gas at equilibrium, including any that did not react.
- A missing partial pressure: subtract the others from the total pressure.
- Percentages by volume of a gas mixture: these are mole fractions × 100.
- Rounding: keep mole fractions to at least three significant figures until the end.
Check: Partial Pressures and Kp Together
Combine mole fractions, partial pressures and Kp expressions for gas mixtures not used on this page.
FAQs
Use these quick answers to check partial pressures and Kp expressions.
What is the difference between Kc and Kp?
Kc is written with equilibrium concentrations in mol dm⁻³ and Kp with equilibrium partial pressures. Kp is used only for gaseous equilibria, and each partial pressure is written as p(X), never with square brackets.
What is a partial pressure?
The pressure that one gas in a mixture would exert if it occupied the container alone. It equals the mole fraction of that gas multiplied by the total pressure, and all the partial pressures add up to the total pressure.
How do I work out a mole fraction?
Divide the moles of that gas at equilibrium by the total moles of gas at equilibrium. Mole fractions have no units and must add up to 1, which is a quick check on your working.
Why are solids and liquids left out of Kp?
Kp contains only gases. A pure solid or liquid does not contribute a partial pressure that varies, so it is omitted. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), the expression is simply Kp = p(CO₂).
What units does Kp have?
Substitute atm for each partial pressure and cancel. Equal powers top and bottom give no units. For N₂O₄ ⇌ 2NO₂ the units are atm; for N₂ + 3H₂ ⇌ 2NH₃ they are atm⁻².
Copyright notice: This OLS revision content, including the explanations, layout, diagrams, tables and embedded learning structure, is authored for Online Learning System by Dr. Mohammed Al-Fatah. It may not be copied, reproduced, redistributed or adapted without written permission.
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