Chemical equilibrium: Kc, ICE tables and graphs — Grade 12 Chemical equilibrium

Definitions and the Kc expression

📖 Definitions

Chemical equilibrium: a dynamic equilibrium in a closed system when the rate of the forward reaction equals the rate of the reverse reaction.

Le Chatelier’s principle: when the equilibrium in a closed system is disturbed, the system will re-instate a new equilibrium by favouring the reaction that will oppose the disturbance.

For aA + bB ⇌ cC + dD: Kc = [C]ᶜ[D]ᵈ ÷ ([A]ᵃ[B]ᵇ)

Use equilibrium concentrations in mol·dm⁻³, not moles.
Leave out solids and pure liquids.
Kc depends only on temperature. A large Kc (> 1) means a high yield of products.

The ICE table method

Work in moles in the table (Initial, Change, Equilibrium), then divide by the volume to get concentrations. The Change row must follow the mole ratio of the balanced equation.

✏️ Worked example 1: H₂(g) + I₂(g) ⇌ 2HI(g)

10,50 mol H₂ and 0,50 mol I₂ are sealed in a 2,0 dm³ flask. At equilibrium there is 0,80 mol HI.

2Change: HI +0,80 mol, so H₂ and I₂ each −0,40 mol (ratio 1 : 1 : 2).

3Equilibrium moles: H₂ 0,10; I₂ 0,10; HI 0,80. Concentrations (÷ 2,0): 0,05; 0,05; 0,40 mol·dm⁻³.

4Kc = (0,40)² ÷ (0,05 × 0,05) = 64.

	H₂	I₂	HI
Initial (mol)	0,50	0,50	0
Change (mol)	−0,40	−0,40	+0,80
Equilibrium (mol)	0,10	0,10	0,80
Concentration (mol·dm⁻³)	0,05	0,05	0,40

✏️ Worked example 2: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

14 mol SO₂ and 2 mol O₂ in 2 dm³; at equilibrium there is 3 mol SO₃.

2Change: SO₃ +3, SO₂ −3, O₂ −1,5. Equilibrium: SO₂ 1; O₂ 0,5; SO₃ 3 mol → 0,5; 0,25; 1,5 mol·dm⁻³.

3Kc = 1,5² ÷ (0,5² × 0,25) = 36.

Le Chatelier and Kc

Disturbance	System responds by	Kc
Add reactant / remove product	Favouring the forward reaction	No change
Increase pressure (decrease volume)	Favouring the side with fewer moles of gas	No change
Increase temperature	Favouring the endothermic reaction	Increases if forward is endothermic; decreases if forward is exothermic
Add a catalyst	Speeding up both reactions equally; no shift	No change

Reading the graphs

Rate–time graphs (forward and reverse rates plotted together):

Reactant added: the forward rate jumps up suddenly; the reverse rate rises gradually; they meet at a new, higher equal rate.
Temperature increased: both rates jump up at once; the endothermic reaction’s rate jumps more; they become equal again.
Pressure increased: both rates jump; the reaction producing fewer gas moles jumps more.
Catalyst added: both rates jump by the same amount and stay equal — a single higher horizontal line.

Concentration–time graphs: a substance that is added shows a vertical jump followed by a gradual decrease; the others change gradually (curves) until new horizontal lines show the new equilibrium. A change in temperature or pressure shows no vertical jump for added substances — every concentration changes gradually (unless the volume is changed, which makes all concentrations jump together).

Exam tip: horizontal lines on either graph mean equilibrium has been reached. When explaining a shift, state the disturbance, the reaction that is favoured, and why it opposes the disturbance.

Answer in your exercise book, then open each memo to mark your work.

Question 1 · (2 marks)

State Le Chatelier’s principle.

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When the equilibrium in a closed system is disturbed, the system will re-instate a new equilibrium ✓ by favouring the reaction that will oppose the disturbance ✓.

Question 2 · (7 marks)

2 mol of NO₂ is placed in a sealed 1 dm³ container: 2NO₂(g) ⇌ N₂O₄(g). At equilibrium, 0,4 mol of NO₂ remains. Calculate Kc using an ICE table.

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NO₂ reacted = 2 − 0,4 = 1,6 mol ✓; N₂O₄ formed = 0,8 mol (ratio 2 : 1) ✓✓; concentrations: [NO₂] = 0,4, [N₂O₄] = 0,8 mol·dm⁻³ ✓; Kc = [N₂O₄]/[NO₂]² ✓ = 0,8 ÷ 0,4² ✓ = 5 ✓

Question 3 · (4 marks)

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH < 0, the temperature is increased. (a) How will the yield of NH₃ change? (b) How will Kc change? Explain.

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(a) Yield decreases ✓. An increase in temperature favours the endothermic (reverse) reaction ✓.
(b) Kc decreases ✓ because [NH₃] decreases while [N₂] and [H₂] increase ✓.

Question 4 · (4 marks)

Describe the rate–time graph when a catalyst is added to a system at equilibrium, and explain why the yield does not change.

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Both forward and reverse rates increase suddenly ✓ by the same amount ✓ and remain equal ✓. A catalyst lowers the activation energy for both reactions equally, so equilibrium does not shift and the yield stays the same ✓.