Partial Pressures and Mole Fractions — AQA A-level Chemistry

SPEC 3.1.10 - EQUILIBRIUM CONSTANT Kp

Partial Pressure, Mole Fractions & Kp

Quantifying Gaseous Equilibria

Why Kp?

In AS Chemistry, you learned about

K_{c}

- the equilibrium constant expressed in terms of concentrations. But when dealing with gaseous equilibria, there's a more practical approach: measuring pressures instead.

The Problem

Measuring the concentration of a gas requires knowing its amount in a specific volume. But for gases in industrial processes (like the Haber process), it's much easier to measure pressure directly using pressure gauges.

This is where

K_{p}

comes in - the equilibrium constant expressed in terms of partial pressures. It applies to homogeneous gaseous equilibria, where all reactants and products are gases.

Key Insight

K_{p}

and

K_{c}

describe the same equilibrium position - they're just different ways of quantifying it. For gases,

K_{p}

is often more convenient because pressures are easier to measure experimentally. Many industrial processes involve reactions between gases, so understanding

K_{p}

has important consequences for controlling reaction conditions.

Before we can understand

K_{p}

, we need to understand two fundamental concepts: partial pressure and mole fraction.

Partial Pressure

Definition

Partial pressure is the pressure a gas exerts in a mixture of gases if it occupied the container on its own at the same temperature.

Imagine a container with a mixture of nitrogen and oxygen. Each gas contributes to the total pressure based on how many molecules of that gas are present. The nitrogen molecules collide with the walls creating "nitrogen's partial pressure", while oxygen molecules create "oxygen's partial pressure". Partial pressure is given the symbol

p

, so for a gas X, it is written as

p_{X}

.

Dalton's Law of Partial Pressures

Formula

P_{total} = p_{A} + p_{B} + p_{C} + \dots

The total pressure is the sum of the partial pressures - this is known as Dalton's Law. It works because gas molecules act independently of each other (assuming ideal behaviour).

Notation

Partial pressure uses lowercase

p

with the gas symbol:

p_{N_{2}}

,

p_{O_{2}}

,

p_{NH_{3}}

. Total pressure uses uppercase

P

or

P_{total}

.

Visualising Partial Pressure

Each gas in a mixture contributes its own partial pressure. If we could separate the gases, each would exert its partial pressure independently:

Check Your Understanding

The partial pressures always add up to the total pressure: 180 + 120 = 300 kPa. This is a useful check in calculations.

Mole Fraction

Definition

Mole fraction (

χ

, the Greek letter "chi") is the ratio of the number of moles of a particular component to the total number of moles in the mixture.

Formula

χ_{A} = \frac{n _{A}}{n _{total}} = \frac{n _{A}}{n _{A} + n _{B} + n _{C} + \dots}

The mole fraction tells us what proportion of the mixture is made up of a particular gas. If a mixture contains 2 mol of nitrogen and 3 mol of oxygen, the mole fraction of nitrogen is

\frac{2}{5} = 0.4

- meaning 40% of the molecules are nitrogen.

Critical Property

The sum of all mole fractions in a mixture always equals exactly 1:

χ_{A} + χ_{B} + χ_{C} + \dots = 1

This is a useful check in calculations - if your mole fractions don't sum to 1, you've made an error.

Common Error

Mole fractions are dimensionless (no units) - they are pure numbers between 0 and 1. Don't include units like "mol" in your final answer for mole fraction.

Worked Example

1

Calculating Mole Fractions

A mixture of gases contains 0.51 mol N₂, 0.28 mol H₂, and 0.52 mol NH₃. Calculate the mole fraction of each gas.

Step 1: Calculate total moles

n_{total} = 0.51 + 0.28 + 0.52 = 1.31 mol

Step 2: Calculate mole fraction of each gas

χ_{N_{2}} = \frac{0.51}{1.31} = 0.389

χ_{H_{2}} = \frac{0.28}{1.31} = 0.214

χ_{NH_{3}} = \frac{0.52}{1.31} = 0.397

Check:

0.389 + 0.214 + 0.397 = 1.000

✓

Examiner Tip

You can check you have the mole fractions correct by adding them up and making sure they come to 1. This is a quick way to catch calculation errors before moving on.

The Key Relationship

Here's the crucial connection: for an ideal gas mixture, the partial pressure of each component is directly proportional to its mole fraction. To find the partial pressure of a gas, you need two pieces of information: the total pressure in the container and the mole fraction.

Fundamental Equation

p_{A} = χ_{A} \times P_{total}

Partial pressure of gas A = Mole fraction of A × Total pressure

Why Does This Work?

Consider the ideal gas equation:

P V = n RT

. For a gas mixture at constant temperature and volume:

Total pressure:

P_{total} = \frac{n _{total} RT}{V}

Partial pressure of A:

p_{A} = \frac{n _{A} RT}{V}

Taking the ratio:

\frac{p _{A}}{P _{total}} = \frac{n _{A}}{n _{total}} = χ_{A}

Therefore:

p_{A} = χ_{A} \times P_{total}

Exam Strategy

This equation is your workhorse for

K_{p}

calculations. The typical workflow is: Equilibrium moles → Mole fractions → Partial pressures → K_p

Worked Example

2

Calculating Partial Pressures from Mole Fractions

The equilibrium mixture below contains 1.75 moles each of PCl₃ and Cl₂, and 1.25 moles of PCl₅ at equilibrium. If the total pressure is 714 kPa, calculate the partial pressure of PCl₃.

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)

Step 1: Calculate total moles

n_{total} = 1.75 + 1.75 + 1.25 = 4.75 mol

Step 2: Calculate mole fraction of PCl₃

χ_{PCl_{3}} = \frac{1.75}{4.75} = 0.368

Step 3: Calculate partial pressure using the equation

p_{PCl_{3}} = χ_{PCl_{3}} \times P_{total}

p_{PCl_{3}} = 0.368 \times 714 = 263 kPa

Answer: $p (PCl_{3}) = 263$ kPa (3 s.f.)

The Kp Expression

Definition

K_p is the equilibrium constant calculated from the partial pressures of gaseous reactants and products at a constant temperature.

For a general gaseous equilibrium:

a A (g) + b B (g) ⇌ c C (g) + d D (g)

Formula

K_{p} = \frac{( p _{C} ) ^{c} \times ( p _{D} ) ^{d}}{( p _{A} ) ^{a} \times ( p _{B} ) ^{b}}

p^{a} A

and

p^{b} B

= equilibrium reactant partial pressures (kPa)

p^{c} C

and

p^{d} D

= equilibrium product partial pressures (kPa)

a, b, c, d

= coefficients in the balanced equation

This follows the same pattern as

K_{c}

: products over reactants, each raised to the power of its stoichiometric coefficient. Solids and liquids are ignored in

K_{p}

equilibrium expressions.

Critical Notation

Use round brackets for partial pressures:

(p_{NO_{2}})^{2}

NOT square brackets:

[p_{NO_{2}}]^{2}

✗

Square brackets [ ] mean concentration - using them for partial pressures is a common exam error that loses marks.

Writing Kp Expressions

Let's practise writing

K_{p}

expressions for common equilibria:

EquilibriumK_p Expression

N_{2} O_{4} (g) ⇌ 2 NO_{2} (g)

K_{p} = \frac{( p _{NO_{2}} ) ^{2}}{( p _{N_{2} O_{4}} )}

N_{2} (g) + 3 H_{2} (g) ⇌ 2 NH_{3} (g)

K_{p} = \frac{( p _{NH_{3}} ) ^{2}}{( p _{N_{2}} ) ( p _{H_{2}} ) ^{3}}

PCl_{5} (g) ⇌ PCl_{3} (g) + Cl_{2} (g)

K_{p} = \frac{( p _{PCl_{3}} ) ( p _{Cl_{2}} )}{( p _{PCl_{5}} )}

2 SO_{2} (g) + O_{2} (g) ⇌ 2 SO_{3} (g)

K_{p} = \frac{( p _{SO_{3}} ) ^{2}}{( p _{SO_{2}} ) ^{2} ( p _{O_{2}} )}

Homogeneous Systems Only

You are only required to know how to write

K_{p}

expressions for homogeneous systems - that is, where all the substances present are gases. If any species is a solid or liquid, it is not included in the

K_{p}

expression.

Units of Kp

Unlike

K_{c}

which uses concentration units (mol dm⁻³),

K_{p}

uses pressure units. The units depend on the stoichiometry of the reaction.

Determining Units

The units of

K_{p}

are: (pressure unit)^Δn

where

Δ n

= (total moles of gaseous products) − (total moles of gaseous reactants)

EquilibriumΔnUnits of K_p

N_{2} O_{4} ⇌ 2 NO_{2}

2 − 1 = +1kPa (or Pa, atm)

N_{2} + 3 H_{2} ⇌ 2 NH_{3}

2 − 4 = −2kPa⁻² (or Pa⁻²)

H_{2} + I_{2} ⇌ 2 HI

2 − 2 = 0No units (dimensionless)

PCl_{5} ⇌ PCl_{3} + Cl_{2}

2 − 1 = +1kPa (or Pa, atm)

Unit Consistency

Use the same pressure unit throughout your calculation. If partial pressures are in kPa, your

K_{p}

units will involve kPa. Don't mix Pa and kPa - convert first!

Pressure Units

Partial pressures are commonly measured in Pascals (Pa) or kilopascals (kPa), but can occasionally be measured in atmospheres (atm). Always check which unit is used in the question and maintain consistency throughout your calculation.

Full Kp Calculation

Most exam questions give you equilibrium amounts and total pressure.

K_{p}

calculations are a step-by-step process:

The 5-Step Method

Write the balanced equation with state symbols

Calculate total moles at equilibrium

Calculate mole fractions for each gas

Calculate partial pressures using

p = χ \times P_{total}

Substitute into the $K_{p}$ expression and solve

Exam Tip

Set up a clear table with columns for: Species | Moles | Mole Fraction | Partial Pressure

This organises your working and makes it easy for the examiner to award method marks even if your final answer is wrong.

Many questions also involve calculating equilibrium amounts from initial amounts and a percentage conversion. We'll see this in the next worked example.

Worked Example

3

Complete Kp Calculation (Haber Process)

For the Haber process:

N_{2} (g) + 3 H_{2} (g) ⇌ 2 NH_{3} (g)

At equilibrium, a mixture contains 0.10 mol N₂, 0.30 mol H₂, and 0.20 mol NH₃ at a total pressure of 20.0 MPa. Calculate

K_{p}

and state its units.

Step 1: Calculate total moles

n_{total} = 0.10 + 0.30 + 0.20 = 0.60 mol

Step 2: Calculate mole fractions

χ_{N_{2}} = \frac{0.10}{0.60} = 0.167 χ_{H_{2}} = \frac{0.30}{0.60} = 0.500 χ_{NH_{3}} = \frac{0.20}{0.60} = 0.333

Step 3: Calculate partial pressures

p_{N_{2}} = 0.167 \times 20.0 = 3.33 MPa

p_{H_{2}} = 0.500 \times 20.0 = 10.0 MPa

p_{NH_{3}} = 0.333 \times 20.0 = 6.67 MPa

Step 4: Substitute into

K_{p}

expression

K_{p} = \frac{( p _{NH_{3}} ) ^{2}}{( p _{N_{2}} ) ( p _{H_{2}} ) ^{3}} = \frac{( 6.67 ) ^{2}}{( 3.33 ) ( 10.0 ) ^{3}} = \frac{44.5}{3330} = 0.0134

Units: Δn = 2 − 4 = −2, so units = MPa⁻²

$K_{p} = 0.013$ MPa⁻² (2 s.f.)

Worked Example

4

Kp from Percentage Dissociation

1.00 mol of PCl₅ is heated in a sealed container. At equilibrium, 40% has dissociated. The total pressure is 200 kPa.

PCl_{5} (g) ⇌ PCl_{3} (g) + Cl_{2} (g)

Calculate

K_{p}

.

Step 1: Calculate equilibrium moles

PCl₅ dissociated = 40% of 1.00 = 0.40 mol

Step 2: Total moles = 0.60 + 0.40 + 0.40 = 1.40 mol

Step 3: Partial pressures (combining mole fraction and pressure steps)

p_{PCl_{5}} = \frac{0.60}{1.40} \times 200 = 85.7 kPa

p_{PCl_{3}} = p_{Cl_{2}} = \frac{0.40}{1.40} \times 200 = 57.1 kPa

Step 4: Calculate

K_{p}

K_{p} = \frac{( p _{PCl_{3}} ) ( p _{Cl_{2}} )}{( p _{PCl_{5}} )} = \frac{57.1 \times 57.1}{85.7} = \frac{3266}{85.7} = 38.1 kPa

$K_{p} = 38$ kPa (2 s.f.)

Worked Example

5

Finding Partial Pressure from Kp

K_{p}

for the dissociation equilibrium reaction of nitrogen monoxide is 0.0021 (no units).

2 NO (g) ⇌ N_{2} (g) + O_{2} (g)

If pure NO is introduced into a reaction flask at an initial pressure of 100 kPa, what is the equilibrium partial pressure of nitrogen?

Step 1: Let

x

= equilibrium partial pressure of N₂

Step 2: Write the

K_{p}

expression and substitute

K_{p} = \frac{( p _{N_{2}} ) ( p _{O_{2}} )}{( p _{NO} ) ^{2}} = \frac{x \times x}{( 100 - 2 x ) ^{2}} = 0.0021

Step 3: Taking square roots of both sides:

\frac{x}{100 - 2 x} = 0.0021 = 0.04583

x = 0.04583 \times (100 - 2 x)

x = 4.583 - 0.09166 x

1.09166 x = 4.583

x = \frac{4.583}{1.09166} = 4.2 kPa

The partial pressure of N₂ = 4.2 kPa (to 2 s.f.)

Note:

p_{O_{2}}

= 4.2 kPa and

p_{NO}

= 100 − (2 × 4.2) = 91.6 kPa

Worked Example

6

Finding Partial Pressure by Rearranging Kp

The reaction

N_{2} (g) + 3 H_{2} (g) ⇌ 2 NH_{3} (g)

has an equilibrium constant

K_{p} = 0.50

atm⁻² at 25°C.

At equilibrium:

p (N_{2}) = 0.30

atm and

p (H_{2}) = 0.60

atm.

Determine

p (NH_{3})

. Give your answer to 2 significant figures.

Step 1: Write the equilibrium constant (

K_{p}

) expression

K_{p} = \frac{( p _{NH_{3}} ) ^{2}}{( p _{N_{2}} ) ( p _{H_{2}} ) ^{3}}

Step 2: Rearrange the equilibrium constant (

K_{p}

) expression

(p_{NH_{3}})^{2} = K_{p} \times p_{N_{2}} \times (p_{H_{2}})^{3}

p_{NH_{3}} = K_{p} \times p_{N_{2}} \times (p_{H_{2}})^{3}

Step 3: Substitution and correct evaluation

p_{NH_{3}} = 0.50 \times 0.30 \times (0.60)^{3}

p_{NH_{3}} = 0.50 \times 0.30 \times 0.216

p_{NH_{3}} = 0.0324 = 0.18 atm

$p (NH_{3}) = 0.18$ atm

Exam Tip

If the equilibrium constant

K_{p}

and some equilibrium partial pressures are known, an unknown partial pressure can be calculated by rearranging the

K_{p}

expression. Always check your rearrangement is correct before substituting values.

Interpreting Kp Values

Just like

K_{c}

, the magnitude of

K_{p}

tells us about the position of equilibrium. The

K_{p}

of a reaction is constant and only changes if the temperature of the reaction changes.

Large K_p (>> 1)

Products dominate at equilibrium

Equilibrium lies to the right

Reaction "goes to completion"

Example: $K_{p} = 1 0^{6}$

Small K_p (<< 1)

Reactants dominate at equilibrium

Equilibrium lies to the left

Little product formed

Example: $K_{p} = 1 0^{- 6}$

Key Point

We can apply Le Chatelier's Principle to gaseous equilibria in the same way it is applied to aqueous systems. Remember: Le Chatelier's principle says that if a change is made to a system in dynamic equilibrium, the position of equilibrium moves to counteract this change.

Effect of Temperature on Kp

The equilibrium constant

K_{p}

changes with temperature in the same way as

K_{c}

. Temperature is the only factor that changes the value of

K_{p}

.

Key Rules

Increasing temperature causes

K_{p}

to increase for endothermic reactions and decrease for exothermic reactions

Decreasing temperature has the opposite effects on

K_{p}

The effects of temperature changes on exothermic and endothermic reactions can be summarised in the following table:

Type of ReactionTemperature ChangeImpact on Equilibrium PositionImpact on K_p

Exothermic reactionIncreaseShifts leftDecreases

Exothermic reactionDecreaseShifts rightIncreases

Endothermic reactionIncreaseShifts rightIncreases

Endothermic reactionDecreaseShifts leftDecreases

Important

Changes in pressure and adding a catalyst do not affect the value of

K_{p}

, only shifting the equilibrium position. The value of

K_{p}

itself depends only on temperature, just like

K_{c}

.

Effect of Pressure on Kp

Changes in pressure only affect reactions where the reactants or products are gases. Here's how the equilibrium shifts:

ChangeHow the Equilibrium Shifts

Increase in pressureEquilibrium shifts in the direction that produces the SMALLER number of molecules of gas to decrease the pressure again

Decrease in pressureEquilibrium shifts in the direction that produces the LARGER number of molecules of gas to increase the pressure again

Effect on the Value of K_p

Critical Point

The value of

K_{p}

is NOT affected by any changes in pressure.

Changes in pressure cause a shift in the position of equilibrium to a new position which restores the value of

K_{p}

. This is analogous to what happens to

K_{c}

when you change concentration in an aqueous equilibrium.

Why Doesn't Kp Change?

When pressure changes, the system adjusts the amounts of products and reactants until the ratio of partial pressures returns to the equilibrium constant value. The ratio is restored, not changed.

Pressure Changes: Position vs Value

This diagram shows how pressure affects position but NOT the value of

K_{p}

:

Key Point

For

N_{2} O_{4} ⇌ 2 NO_{2}

: The system shifts left (towards fewer gas moles) to counteract the pressure increase, but the ratio

\frac{( p _{NO_{2}} ) ^{2}}{p _{N_{2} O_{4}}}

returns to the same

K_{p}

value once equilibrium is re-established.

Effect of a Catalyst

Key Principle

If all other conditions stay the same, the equilibrium constant

K_{p}

is NOT affected by the presence of a catalyst.

A catalyst speeds up both the forward and reverse reactions at the same rate, so the ratio of [products] to [reactants] remains unchanged. The catalyst provides an alternative reaction pathway with lower activation energy, but this affects both directions equally.

What Catalysts Do

Catalysts only cause a reaction to reach equilibrium faster

Catalysts have no effect on the position of the equilibrium once this is reached

Catalysts do not change the value of

K_{p}

Catalysts do not change the equilibrium yield of products

Industrial Importance

In industrial processes like the Haber process, catalysts are used to make the reaction reach equilibrium quickly enough to be economically viable. Without a catalyst, equilibrium might take too long to establish, even if the yield would be the same.

Practice

Question 1: For the equilibrium H₂(g) + I₂(g) ⇌ 2HI(g), an equilibrium mixture contains 0.50 mol H₂, 0.50 mol I₂, and 3.00 mol HI at a total pressure of 500 kPa. Calculate K_p.

Total moles: 0.50 + 0.50 + 3.00 = 4.00 mol

Mole fractions:

χ_{H_{2}} = χ_{I_{2}} = \frac{0.50}{4.00} = 0.125 χ_{HI} = \frac{3.00}{4.00} = 0.75

Partial pressures:

p_{H_{2}} = p_{I_{2}} = 0.125 \times 500 = 62.5 kPa

p_{HI} = 0.75 \times 500 = 375 kPa

K_p:

K_{p} = \frac{( p _{HI} ) ^{2}}{( p _{H_{2}} ) ( p _{I_{2}} )} = \frac{( 375 ) ^{2}}{62.5 \times 62.5} = \frac{140625}{3906.25} = 36

Units: Δn = 2 − 2 = 0, so K_p is dimensionless (no units)

K_p = 36

Practice

Question 2: Hydrogen and bromine were mixed in a flask in a 1:1 ratio and allowed to reach equilibrium at 450 K. When equilibrium had been achieved, the total pressure in the flask was 140 kPa and the mole fraction of bromine was 0.35.

H₂(g) + Br₂(g) ⇌ 2HBr(g)

Determine the partial pressures for each gas at equilibrium and the value of K_p.

Since H₂ and Br₂ are in 1:1 ratio:

Mole fraction of Br₂ = Mole fraction of H₂ = 0.35

Mole fraction of HBr = 1 − (0.35 × 2) = 0.30

Partial pressures:

p_{Br_{2}} = p_{H_{2}} = 0.35 \times 140 = 49 kPa

p_{HBr} = 0.30 \times 140 = 42 kPa

K_p:

K_{p} = \frac{( p _{HBr} ) ^{2}}{( p _{H_{2}} ) ( p _{Br_{2}} )} = \frac{( 42 ) ^{2}}{49 \times 49} = \frac{1764}{2401} = 0.735

Units: Δn = 2 − 2 = 0, so K_p has no units

K_p = 0.74 (2 s.f.)

Practice

Question 3: 2.00 mol of SO₃ is placed in a sealed container and heated. At equilibrium, 25% has dissociated:

2SO₃(g) ⇌ 2SO₂(g) + O₂(g)

The total pressure is 800 kPa. Calculate K_p.

Step 1: Calculate equilibrium moles

SO₃ dissociated = 25% of 2.00 = 0.50 mol

Step 2: Total moles = 1.50 + 0.50 + 0.25 = 2.25 mol

Step 3: Partial pressures

p_{SO_{3}} = \frac{1.50}{2.25} \times 800 = 533.3 kPa

p_{SO_{2}} = \frac{0.50}{2.25} \times 800 = 177.8 kPa

p_{O_{2}} = \frac{0.25}{2.25} \times 800 = 88.9 kPa

Step 4: Calculate K_p

K_{p} = \frac{( p _{SO_{2}} ) ^{2} ( p _{O_{2}} )}{( p _{SO_{3}} ) ^{2}} = \frac{( 177.8 ) ^{2} \times 88.9}{( 533.3 ) ^{2}} = 9.9 kPa

K_p = 9.9 kPa

Summary

Key Concepts

ConceptFormula/Definition

Mole Fraction

χ_{A} = \frac{n _{A}}{n _{total}}

(sum of all = 1)

Partial Pressure

p_{A} = χ_{A} \times P_{total}

Dalton's Law

P_{total} = p_{A} + p_{B} + p_{C} + \dots

K_p ExpressionProducts over reactants (partial pressures), raised to stoichiometric powers

K_p Units(pressure)^Δn where Δn = moles products − moles reactants

Exam Checklist

Use round brackets for partial pressures in K_p expressions

Only include gaseous species in K_p (homogeneous systems only)

Check mole fractions sum to 1 and partial pressures sum to P_total

Only temperature changes the value of K_p

Pressure and catalysts do NOT change K_p

Show clear working: moles → mole fractions → partial pressures → K_p

SPEC 3.1.10 - EQUILIBRIUM CONSTANT Kp

Partial Pressure, Mole Fractions & Kp

Quantifying Gaseous Equilibria

Why Kp?

Partial Pressure

Dalton's Law of Partial Pressures

Visualising Partial Pressure

Mole Fraction

Worked Example

The Key Relationship

Worked Example

The Kp Expression

Writing Kp Expressions

Units of Kp

Full Kp Calculation

Worked Example

Worked Example

Worked Example

Worked Example

Interpreting Kp Values

Large K_p (>> 1)

Products dominate at equilibrium

Reaction "goes to completion"

Example: Kp​=106

Small K_p (<< 1)

Reactants dominate at equilibrium

Little product formed

Example: Kp​=10−6

Effect of Temperature on Kp

Effect of Pressure on Kp

Effect on the Value of K_p

Pressure Changes: Position vs Value

Effect of a Catalyst

Practice

Practice

Practice

Summary

Example: $K_{p} = 1 0^{6}$

Example: $K_{p} = 1 0^{- 6}$