AQA A-Level Chemistry: Chemical equilibria, Le Chatelier's principle and Kc (7405) – Revision Notes

These notes are for the final weeks before your exams. For full explanations and longer worked examples, use the chemical equilibria study guide.

They follow section 3.1.6 of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2 (July 2026), for AS and A-level exams from June 2016 onwards: 3.1.6.1 Chemical equilibria and Le Chatelier’s principle, and 3.1.6.2 Equilibrium constant Kc for homogeneous systems. The whole section is AS content and is also part of the full A-level; none of it is “A-level only”. Kp (section 3.1.10, A-level only) is not covered here. The topic is listed for AS Papers 1 and 2 and A-level Papers 1 and 2; A-level Paper 3 can test any content.

Course links: AQA A-Level Chemistry hub | printable checklist | practice questions | relative mass and the mole revision notes

Definitions (3.1.6.1)

• Reversible reaction: products can react to re-form reactants (⇌).
• Dynamic equilibrium (closed system): forward and reverse reactions proceed at equal rates, and the concentrations of reactants and products remain constant.
• Homogeneous system: all species in the same phase.
• Le Chatelier’s principle: if a factor affecting a system at equilibrium is changed, the position of equilibrium shifts to oppose the change.
• Catalyst: speeds up forward and reverse reactions equally; equilibrium is reached faster; the position does not change.

Le Chatelier at a glance (3.1.6.1)

Change	Shift in position	Why
Add a reactant / remove a product	to the right	uses up added reactant / replaces product
Add a product / remove a reactant	to the left	opposes the change
Increase pressure	to the side with fewer moles of gas	lowers the pressure
Decrease pressure	to the side with more moles of gas	raises the pressure
Same moles of gas each side	no shift with pressure	no side can lower it
Increase temperature	endothermic direction	absorbs heat
Decrease temperature	exothermic direction	releases heat
Add a catalyst	no shift	both rates increase equally

Method in steps: a Le Chatelier explanation

1. Identify the change (concentration, pressure or temperature).
2. Find the relevant information: which species is added or removed; moles of gas on each side; sign of ΔH for the forward reaction.
3. State the direction of shift (left or right).
4. Give the reason, using “to oppose the change” in specific terms (for example “to the side with fewer moles of gas, reducing the pressure”).
5. State the effect on the yield or on what you would observe.

Compromise conditions (3.1.6.1)

For an exothermic forward reaction with fewer moles of gas on the product side (for example N₂ + 3H₂ ⇌ 2NH₃ in the Haber process, which uses an iron catalyst):

Condition	Best for yield	Problem	Compromise
Temperature	low	rate too slow	moderate temperature: acceptable yield at acceptable rate
Pressure	high	costly equipment, energy to compress, safety	high enough for a good yield and rate, limited by cost
Catalyst	no effect on yield	–	used so equilibrium is reached quickly at a lower temperature

A full compromise answer always mentions yield and rate for temperature, and yield and cost or safety for pressure.

Kc: the formulas (3.1.6.2)

Item	Rule
Reaction	aA + bB ⇌ cC + dD (homogeneous)
Expression	Kc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ)
[X]	equilibrium concentration of X in mol dm⁻³
Concentration	moles ÷ volume in dm³
Units	substitute mol dm⁻³ and cancel
Same total moles each side	no units; volume cancels, so moles can be used
What changes Kc	temperature only
What does not change Kc	concentration, pressure, catalyst

Unit patterns

(moles on top) minus (moles on bottom)	Units of Kc
0	none
+1	mol dm⁻³
−1	mol⁻¹ dm³
−2	mol⁻² dm⁶

Method in steps: Kc from initial amounts

1. Write the balanced equation and the Kc expression.
2. Draw a moles table: initial, change, equilibrium.
3. Use the one equilibrium amount you are given to find the change, then use the equation’s ratios for every other species.
4. Divide each equilibrium amount by the volume (skip this only if the volume cancels).
5. Substitute, calculate, and work out the units.
6. Round to the least accurate data.

Small worked reminder

2A(g) ⇌ B(g). Start with 1.00 mol A in 2.00 dm³. At equilibrium there is 0.200 mol B.

A used = 2 x 0.200 = 0.400 mol  ->  A left = 0.600 mol
[A] = 0.600 / 2.00 = 0.300 mol dm-3
[B] = 0.200 / 2.00 = 0.100 mol dm-3
Kc = [B] / [A]^2 = 0.100 / 0.300^2 = 1.11 mol-1 dm3

Method in steps: finding an unknown concentration

1. Write the Kc expression.
2. Rearrange so the unknown is the subject before you substitute.
3. Substitute the known equilibrium concentrations and the value of Kc.
4. If the unknown is squared, take the square root at the end.
5. Check the units of your answer are mol dm⁻³.

Example: for X(g) ⇌ 2Y(g), Kc = 0.0500 mol dm⁻³ and [X] = 0.200 mol dm⁻³. Then [Y]² = 0.0500 × 0.200 = 0.0100, so [Y] = 0.100 mol dm⁻³.

Temperature and Kc (3.1.6.2)

Forward ΔH	Temperature rises	Kc	Product yield
negative (exothermic)	shifts left	decreases	decreases
positive (endothermic)	shifts right	increases	increases

Working backwards: Kc falls as temperature rises → forward reaction exothermic.

Practical links

• Test-tube equilibrium shifts (suggested by the specification): for example [Cu(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CuCl₄]²⁻ + 6H₂O. Adding concentrated HCl shifts it right (blue towards green-yellow); adding water shifts it back. Warming shows which direction is endothermic.
• Kc for ethyl ethanoate (suggested): equilibrium mixture of ethanoic acid, ethanol, ester and water with a strong acid catalyst. Titrate with NaOH, subtract the catalyst blank to get moles of ethanoic acid at equilibrium. Include water from the catalyst solution as an initial amount. Volume cancels.

Must-know distinctions

• Constant concentrations vs equal concentrations: only “constant” is correct.
• Position of equilibrium vs value of Kc: concentration and pressure change the position but not Kc; temperature changes both.
• Rate vs yield: a catalyst and higher temperature both raise rate; only Le Chatelier shifts affect yield.
• Moles of gas vs total moles: pressure arguments count gas moles only.
• Moles vs concentrations in Kc: moles are allowed only when the volume cancels.

Quick self-test

1. State the two features of a reaction at dynamic equilibrium.
2. Write the Kc expression for N₂(g) + 3H₂(g) ⇌ 2NH₃(g) and give its units.
3. Give the units of Kc for PCl₅(g) ⇌ PCl₃(g) + Cl₂(g).
4. For A + B ⇌ C, [A] = 0.50, [B] = 0.20 and [C] = 0.30 mol dm⁻³. Calculate Kc.
5. What happens to the position of equilibrium in H₂(g) + I₂(g) ⇌ 2HI(g) when the pressure is increased?
6. The forward reaction is exothermic. What happens to Kc when the temperature is raised?
7. What is the effect of a catalyst on the value of Kc?
8. For 2A ⇌ B, 1.00 mol A reacts until 0.30 mol B is present. How many moles of A remain?
9. Concentrations are given to 2 s.f. and 3 s.f. To how many s.f. should you give Kc?
10. Kc for a reaction increases as temperature increases. Is the forward reaction exothermic or endothermic?
11. More of a reactant is added at constant temperature. What happens to Kc and to the position of equilibrium?

Answers

1. Forward and reverse rates are equal; concentrations of reactants and products remain constant.
2. Kc = [NH₃]² / ([N₂][H₂]³); units mol⁻² dm⁶.
3. mol dm⁻³.
4. Kc = 0.30 / (0.50 × 0.20) = 3.0 mol⁻¹ dm³.
5. No change: 2 moles of gas on each side.
6. Kc decreases (equilibrium shifts left, in the endothermic direction).
7. None.
8. 1.00 − 2 × 0.30 = 0.40 mol.
9. 2 s.f.
10. Endothermic.
11. Kc is unchanged; the position shifts to the right.

Where marks are usually lost

• Writing “concentrations are equal” instead of “remain constant” in the definition of dynamic equilibrium.
• Giving a direction of shift with no reason, or a reason that just repeats “Le Chatelier’s principle” without saying what is opposed.
• Counting liquid water or other non-gases when arguing about pressure.
• Claiming a catalyst increases yield or shifts the equilibrium.
• Writing that adding a reactant “increases Kc”.
• Using moles directly in Kc when the volume does not cancel.
• Forgetting the 2:1 or 3:1 ratio when filling in the change row of a moles table.
• Omitting units, or giving units when they cancel.
• Rounding mid-calculation and giving the final answer to the wrong number of significant figures.
• Compromise answers that mention only yield, with nothing about rate, cost or safety.

Official syllabus

AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards, published by AQA. Section 3.1.6: Chemical equilibria, Le Chatelier’s principle and Kc.