The Equilibrium Constant Kc — AQA A-level Chemistry

SPEC 3.1.6.2 • CHEMICAL EQUILIBRIA

Equilibrium Constant Kc

Writing Kc expressions, calculating Kc values, and understanding what Kc tells us about equilibrium position

Why Kc Matters

In Lesson 6.01, we used Le Chatelier's principle to make qualitative predictions about equilibrium — which way will it shift? Now we take the next step: quantitative analysis.

The equilibrium constant Kc allows us to:

Calculate exact concentrations at equilibrium

Predict whether products or reactants are favoured

Determine how far a reaction will proceed

Compare equilibrium positions at different temperatures

Key Concept

Kc is a number that tells us the ratio of product concentrations to reactant concentrations at equilibrium. It transforms our qualitative understanding ("products favoured") into precise, calculable values.

Link to 6.01

Remember: only temperature changes the value of Kc. Changes in concentration, pressure, or adding a catalyst shift the position but do NOT change Kc itself.

Homogeneous Systems

Definition

A homogeneous system is one in which all reactants and products are in the same phase (e.g., all gases or all in aqueous solution).

For Kc calculations at A-Level, we work exclusively with homogeneous systems. This means every species in the equilibrium can be expressed as a concentration in mol dm⁻³.

System TypeExamplePhases

Homogeneous (gas)N₂(g) + 3H₂(g) ⇌ 2NH₃(g)All gases

Homogeneous (aq)CH₃COOH(aq) + C₂H₅OH(aq) ⇌ CH₃COOC₂H₅(aq) + H₂O(l)All in solution

HeterogeneousCaCO₃(s) ⇌ CaO(s) + CO₂(g)Mixed phases

Important

Solids are ignored in equilibrium constant expressions because their concentrations are effectively constant. Pure liquids are also omitted for the same reason.

Example with Solid

For the reaction: Ag⁺(aq) + Fe²⁺(aq) ⇌ Ag(s) + Fe³⁺(aq)

K_{c} = \frac{[ Fe ^{3 +} ]}{[ Fe ^{2 +} ] [ Ag ^{+} ]}

Notice that [Ag(s)] is not included in the expression because it is a solid.

Writing Kc Expressions

For any equilibrium reaction, we can write an expression for Kc. The general rule is:

General Form

For the reaction: aA + bB ⇌ cC + dD

K_{c} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}

The key rules for writing Kc expressions:

Rules for Kc

1. Products on TOP (numerator), reactants on BOTTOM (denominator)

2. Powers = stoichiometric coefficients from the balanced equation

3. Square brackets [ ] mean concentration in mol dm⁻³

4. Only use EQUILIBRIUM concentrations — not initial concentrations!

Memory Aid

"Products over Reactants" — the products of the reaction go on top. Think of products as the "result" that rises to the top!

Important

The equilibrium constant Kc is specific to a given equation. If you double all the coefficients, the Kc expression changes — and so does the numerical value of Kc.

For example: H₂ + I₂ ⇌ 2HI has a different Kc value than ½H₂ + ½I₂ ⇌ HI

01Worked Example

Write the Kc expression for each of the following equilibria:

(a) N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

(b) 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

(c) H₂(g) + I₂(g) ⇌ 2HI(g)

(a)

Products: NH₃ (coefficient 2) → [NH₃]²

Reactants: N₂ (coefficient 1), H₂ (coefficient 3) → [N₂][H₂]³

K_{c} = \frac{[ NH _{3} ] ^{2}}{[ N _{2} ] [ H _{2} ] ^{3}}

(b)

Products: SO₃ (coefficient 2) → [SO₃]²

Reactants: SO₂ (coefficient 2), O₂ (coefficient 1) → [SO₂]²[O₂]

K_{c} = \frac{[ SO _{3} ] ^{2}}{[ SO _{2} ] ^{2} [ O _{2} ]}

(c)

Products: HI (coefficient 2) → [HI]²

Reactants: H₂ (coefficient 1), I₂ (coefficient 1) → [H₂][I₂]

K_{c} = \frac{[ HI ] ^{2}}{[ H _{2} ] [ I _{2} ]}

What Does Kc Tell Us?

The magnitude of Kc tells us whether the equilibrium position favours products or reactants:

Kc ValueMeaningExample

Kc >> 1 (e.g., 10⁶)Products strongly favoured; reaction goes nearly to completionCombustion reactions

Kc ≈ 1 (e.g., 0.1 to 10)Significant amounts of both reactants and productsEsterification

Kc << 1 (e.g., 10⁻⁶)Reactants strongly favoured; very little product formsN₂ + O₂ ⇌ 2NO at low T

Key Insight

Kc = [products]/[reactants]. If products are high, numerator is large → Kc is large. If reactants dominate, denominator is large → Kc is small.

Calculating Kc

To calculate Kc, we need the equilibrium concentrations of all species. The process is:

Method

Step 1: Write the balanced equation

Step 2: Write the Kc expression (products over reactants)

Step 3: Substitute the equilibrium concentrations

Step 4: Calculate the value

Step 5: Determine the units

Remember: concentration = moles ÷ volume

[X] = \frac{n}{V}

where n = amount in moles, V = volume in dm³

Common Error

Students often use initial concentrations instead of equilibrium concentrations. The question will specify which values are at equilibrium — read carefully!

Significant Figures

The smallest number of significant figures used in the question determines the number of significant figures in your answer. If the data has 3 s.f., give your Kc value to 3 s.f.

02Worked Example

For the equilibrium H₂(g) + I₂(g) ⇌ 2HI(g), the equilibrium concentrations at 700 K are:

[H₂] = 0.10 mol dm⁻³, [I₂] = 0.10 mol dm⁻³, [HI] = 0.80 mol dm⁻³

Calculate Kc and state its units.

1

Write the Kc expression:

K_{c} = \frac{[ HI ] ^{2}}{[ H _{2} ] [ I _{2} ]}

2

Substitute equilibrium concentrations:

K_{c} = \frac{( 0.80 ) ^{2}}{( 0.10 ) ( 0.10 )} = \frac{0.64}{0.01} = 64

3

Determine units:

\frac{( mol dm ^{- 3} ) ^{2}}{( mol dm ^{- 3} ) ( mol dm ^{- 3} )} = \frac{( mol dm ^{- 3} ) ^{2}}{( mol dm ^{- 3} ) ^{2}} =

no units

Answer: Kc = 64 (no units)

Note: Kc > 1, so products (HI) are favoured at this temperature.

ICE Tables

When we're given initial amounts and need to find equilibrium concentrations, we use an ICE table to track the changes:

ICE Table

I = Initial concentration (or moles)

C = Change during reaction

E = Equilibrium concentration (or moles)

The change row uses the stoichiometric ratios from the balanced equation. If the reaction proceeds forward, reactants decrease (−) and products increase (+).

Example setup for: A + 2B ⇌ 3C

A2B3C

I (Initial)a₀b₀0

C (Change)−x−2x+3x

E (Equilibrium)a₀ − xb₀ − 2x3x

Key Point

The coefficients in the balanced equation determine the ratio of the changes. If 1 mole of A reacts, 2 moles of B react and 3 moles of C form.

Using Molar Ratios

Use the molar ratio of reactants and products in the stoichiometric equation to determine the change row. If you know one change value, you can calculate all the others from the ratios.

Example: If 0.10 mol of Cl₂ forms in PCl₅ ⇌ PCl₃ + Cl₂, then 0.10 mol of PCl₃ also forms (1:1 ratio) and 0.10 mol of PCl₅ decomposes.

03Worked Example

0.50 mol of PCl₅ is placed in a 2.0 dm³ vessel and heated to 500 K. At equilibrium, 0.20 mol of Cl₂ is present.

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)

Calculate Kc for this reaction.

1

Set up ICE table (in moles):

2

Convert to concentrations (÷ 2.0 dm³):

[PCl₅] = 0.30 ÷ 2.0 = 0.15 mol dm⁻³

[PCl₃] = 0.20 ÷ 2.0 = 0.10 mol dm⁻³

[Cl₂] = 0.20 ÷ 2.0 = 0.10 mol dm⁻³

3

Calculate Kc:

K_{c} = \frac{[ PCl _{3} ] [ Cl _{2} ]}{[ PCl _{5} ]} = \frac{( 0.10 ) ( 0.10 )}{0.15} = \frac{0.01}{0.15} = 0.067

4

Units:

\frac{( mol dm ^{- 3} ) ^{2}}{mol dm ^{- 3}} =

mol dm⁻³

Answer: Kc = 0.067 mol dm⁻³

Units of Kc

The units of Kc depend on the stoichiometry of the reaction. We must work them out for each reaction.

Method for Units

Step 1: Write the Kc expression

Step 2: Replace each concentration with (mol dm⁻³)

Step 3: Simplify the powers

Step 4: Cancel where possible

ReactionKc ExpressionUnits CalculationUnits

H₂ + I₂ ⇌ 2HI

\frac{[ HI ] ^{2}}{[ H _{2} ] [ I _{2} ]}

\frac{( mol dm ^{- 3} ) ^{2}}{( mol dm ^{- 3} ) ^{2}}

No units

PCl₅ ⇌ PCl₃ + Cl₂

\frac{[ PCl _{3} ] [ Cl _{2} ]}{[ PCl _{5} ]}

\frac{( mol dm ^{- 3} ) ^{2}}{mol dm ^{- 3}}

mol dm⁻³

N₂ + 3H₂ ⇌ 2NH₃

\frac{[ NH _{3} ] ^{2}}{[ N _{2} ] [ H _{2} ] ^{3}}

\frac{( mol dm ^{- 3} ) ^{2}}{( mol dm ^{- 3} ) ^{4}}

mol⁻² dm⁶

Quick Check

Count the powers: (products) − (reactants). If the result is 0, no units. If positive, mol dm⁻³ to that power. If negative, the inverse.

Finding Unknown Concentrations

If we know Kc and some equilibrium concentrations, we can calculate the unknown concentration by rearranging the Kc expression.

Method

Step 1: Write the Kc expression

Step 2: Substitute known values

Step 3: Rearrange to find the unknown

Step 4: Solve (take square roots if needed)

Example: For H₂ + I₂ ⇌ 2HI, if Kc = 49, [H₂] = 0.20, [I₂] = 0.10, find [HI].

K_{c} = \frac{[ HI ] ^{2}}{[ H _{2} ] [ I _{2} ]}

Rearrange:

[HI]^{2} = K_{c} \times [H_{2}] \times [I_{2}]

[HI]^{2} = 49 \times 0.20 \times 0.10 = 0.98

[HI] = 0.98 = 0.99 mol dm^{- 3}

Watch Out

When taking square roots, only take the positive root — concentrations cannot be negative!

04Worked Example

For the equilibrium N₂O₄(g) ⇌ 2NO₂(g), Kc = 0.36 mol dm⁻³ at 400 K.

At equilibrium, [N₂O₄] = 0.25 mol dm⁻³. Calculate [NO₂].

1

Write Kc expression:

K_{c} = \frac{[ NO _{2} ] ^{2}}{[ N _{2} O _{4} ]}

2

Rearrange for [NO₂]²:

[NO_{2}]^{2} = K_{c} \times [N_{2} O_{4}]

3

Substitute values:

[NO_{2}]^{2} = 0.36 \times 0.25 = 0.090

4

Take square root:

[NO_{2}] = 0.090 = 0.30 mol dm^{- 3}

Answer: [NO₂] = 0.30 mol dm⁻³

What Affects Kc?

This is a crucial concept that links back to Lesson 6.01:

Key Rule

ONLY TEMPERATURE changes the value of Kc

ChangeEffect on PositionEffect on Kc

Change concentrationShifts to oppose changeNO CHANGE

Change pressureShifts to oppose changeNO CHANGE

Add catalystNo effect (faster both ways)NO CHANGE

Change temperatureShifts to oppose changeCHANGES

Why Concentration Doesn't Change Kc

Consider: 2HI(g) ⇌ H₂(g) + I₂(g) where Kc = 6.25 × 10⁻³

If we add more HI:

Initially, the ratio [H₂][I₂]/[HI]² becomes smaller than Kc

The system shifts RIGHT to restore equilibrium

[H₂] and [I₂] increase, [HI] decreases

The ratio returns to exactly 6.25 × 10⁻³

The system adjusts concentrations to restore the same Kc value — the value itself doesn't change.

Catalysts

A catalyst speeds up both forward and reverse reactions at the same rate. This means equilibrium is reached faster, but the ratio of [products]/[reactants] remains unchanged. Therefore, Kc is unaffected.

Pressure Changes

A change in pressure only changes the position of the equilibrium (via Le Chatelier's principle). If all other conditions stay the same, Kc is not affected by any changes in pressure — the system simply adjusts concentrations to maintain the same Kc ratio.

Temperature Effects on Kc

The relationship between temperature and Kc depends on whether the forward reaction is exothermic or endothermic:

Understanding the Pattern

When temperature increases, the system shifts in the endothermic direction to absorb the extra heat:

Exothermic forward: reverse is endothermic → shifts LEFT → [products] decreases → Kc decreases

Endothermic forward: forward is endothermic → shifts RIGHT → [products] increases → Kc increases

Exam Tip

Questions may give you Kc values at two temperatures and ask you to deduce whether the reaction is exothermic or endothermic. If Kc increases with temperature → endothermic. If Kc decreases with temperature → exothermic.

05Worked Example

For the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), the following data was obtained:

Temperature (K)Kc

5002.5 × 10¹⁰

7003.0 × 10⁴

(a) Is the forward reaction exothermic or endothermic?

(b) Explain your answer.

(a)

Exothermic

(b)

As temperature increases (500 K → 700 K), Kc decreases (2.5 × 10¹⁰ → 3.0 × 10⁴).

A decrease in Kc means the equilibrium shifts to favour reactants.

According to Le Chatelier's principle, increasing temperature favours the endothermic direction.

Since the reverse reaction is favoured, the reverse must be endothermic.

Therefore, the forward reaction is exothermic.

Answer: (a) Exothermic (b) Kc decreases as T increases, so forward reaction releases heat.

Practice

Question 1:

[CO] = 0.040 mol dm⁻³, [H₂] = 0.080 mol dm⁻³, [CH₃OH] = 0.020 mol dm⁻³

Calculate Kc and state its units.

1

Kc expression:

K_{c} = \frac{[ CH _{3} OH ]}{[ CO ] [ H _{2} ] ^{2}}

2

Substitute:

K_{c} = \frac{0.020}{( 0.040 ) ( 0.080 ) ^{2}} = \frac{0.020}{( 0.040 ) ( 0.0064 )} = \frac{0.020}{0.000256}

3

Kc = 78.1 (or 78)

4

Units:

\frac{mol dm ^{- 3}}{( mol dm ^{- 3} ) ^{3}} = mol^{- 2} dm^{6}

Answer: Kc = 78 mol⁻² dm⁶

Practice

Question 2:

CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O

Calculate Kc for this reaction.

1

ICE table (in moles, V = 1.0 dm³ so conc = moles):

2

K_{c} = \frac{[ Ester ] [ H _{2} O ]}{[ Acid ] [ Alcohol ]} = \frac{( 0.67 ) ( 0.67 )}{( 0.33 ) ( 0.33 )} = \frac{0.449}{0.109} = 4.1

3

Units: All cancel → no units

Answer: Kc = 4.1 (no units)

Practice

Question 3:

At equilibrium, [N₂] = 2.0 mol dm⁻³ and [H₂] = 3.0 mol dm⁻³.

Calculate the equilibrium concentration of ammonia.

1

Kc expression:

K_{c} = \frac{[ NH _{3} ] ^{2}}{[ N _{2} ] [ H _{2} ] ^{3}}

2

Rearrange:

[NH_{3}]^{2} = K_{c} \times [N_{2}] \times [H_{2}]^{3}

3

Substitute:

[NH_{3}]^{2} = 0.50 \times 2.0 \times (3.0)^{3} = 0.50 \times 2.0 \times 27 = 27

4

[NH_{3}] = 27 = 5.2 mol dm^{- 3}

Answer: [NH₃] = 5.2 mol dm⁻³

Summary

Key Formulas

ConceptFormula/Expression

General Kc expression

K_{c} = \frac{[ products ] ^{coefficients}}{[ reactants ] ^{coefficients}}

Concentration

[X] = \frac{n}{V}

(mol dm⁻³)

Kc >> 1Products favoured

Kc << 1Reactants favoured

Key Exam Points

Products on top, reactants on bottom — never reverse this!

Use equilibrium concentrations — not initial values

Work out units every time — they depend on stoichiometry

Only temperature changes Kc — concentration, pressure, catalyst do NOT

Exothermic: ↑T → ↓Kc | Endothermic: ↑T → ↑Kc

Common Mistakes

1. Writing reactants on top (inverted expression)

2. Using initial instead of equilibrium concentrations

3. Forgetting to convert moles to concentrations (÷ volume)

4. Not working out units systematically

5. Thinking pressure or catalyst changes Kc