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

This study guide covers 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. It teaches both parts of the section: 3.1.6.1 Chemical equilibria and Le Chatelier’s principle, and 3.1.6.2 Equilibrium constant Kc for homogeneous systems. All of this content is AS content, so it is also part of the full A-level; nothing in section 3.1.6 is marked “A-level only”. The partial-pressure constant Kp is a separate section (3.1.10, A-level only) and is not covered here.

Kc calculations rely on moles and concentrations, so revise the relative mass and the mole study guide first if needed. The AQA A-Level Chemistry hub lists every topic, and the printable checklist lets you tick off each outcome.

What this topic covers

Spec ref	What you must be able to do	Level
3.1.6.1	State that many reactions are reversible; describe a reaction at equilibrium (equal forward and reverse rates, constant concentrations)	AS and A-level
3.1.6.1	Use Le Chatelier’s principle to predict qualitatively the effect of temperature, pressure and concentration on the position of equilibrium in homogeneous reactions	AS and A-level
3.1.6.1	State that a catalyst does not affect the position of equilibrium	AS and A-level
3.1.6.1	Explain why a compromise temperature and pressure may be used for a reversible industrial reaction	AS and A-level
3.1.6.2	Construct an expression for Kc for a homogeneous system, using [X] for concentration in mol dm⁻³	AS and A-level
3.1.6.2	Calculate Kc from equilibrium concentrations at constant temperature, and perform other calculations involving Kc	AS and A-level
3.1.6.2	State that Kc is not affected by changes in concentration or by a catalyst; predict qualitatively the effect of temperature on Kc	AS and A-level

Where it is examined: the AS assessment overview lists section 3.1.6 under both AS Paper 1 and AS Paper 2. For the A-level, it is listed under Paper 1 and Paper 2, and Paper 3 can test any content. The specification lists no required practical in section 3.1.6, but it suggests two practical activities, which are covered below.

Dynamic equilibrium (3.1.6.1)

Many chemical reactions are reversible, shown by the sign ⇌. In a closed system, a reversible reaction reaches a state of dynamic equilibrium. The specification gives two features you must be able to state:

• the forward and reverse reactions proceed at equal rates
• the concentrations of reactants and products remain constant

“Dynamic” means both reactions are still happening. Concentrations are constant, not necessarily equal.

A homogeneous system is one where every species is in the same phase, for example all gases, or all dissolved in the same solution. Both parts of section 3.1.6 deal with homogeneous systems.

Le Chatelier’s principle (3.1.6.1)

Le Chatelier’s principle: if a factor affecting a system in dynamic equilibrium is changed, the position of equilibrium shifts in the direction that opposes the change.

Shifting “to the right” means more products; “to the left” means more reactants.

Concentration

• Increase the concentration of a reactant: the equilibrium shifts right, to use up some of the added reactant.
• Remove a product: the equilibrium shifts right, to replace some of it.

Pressure (reactions involving gases)

Count the moles of gas on each side of the equation.

• Increase the pressure: the equilibrium shifts to the side with fewer moles of gas, which lowers the pressure.
• Decrease the pressure: it shifts to the side with more moles of gas.
• Equal moles of gas on both sides: pressure has no effect on the position.

Temperature

You need the sign of ΔH for the forward reaction.

• Increase the temperature: the equilibrium shifts in the endothermic direction, which absorbs heat.
• Decrease the temperature: it shifts in the exothermic direction.

Catalysts

A catalyst increases the rates of the forward and reverse reactions equally. Equilibrium is reached faster, but the position of equilibrium does not change.

Worked example 1: predicting shifts

For the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), the forward reaction is exothermic. Predict the effect on the yield of SO₃ of (a) raising the pressure, (b) raising the temperature, (c) adding a catalyst.

Moles of gas: left = 2 + 1 = 3, right = 2
(a) Higher pressure -> shifts to the side with fewer moles of gas
    -> shifts right -> yield of SO3 increases
(b) Higher temperature -> shifts in the endothermic direction
    forward is exothermic, so reverse is endothermic
    -> shifts left -> yield of SO3 decreases
(c) Catalyst -> no change in position -> yield unchanged
    (equilibrium is reached faster)

Suggested practical: test-tube equilibrium shifts

The specification suggests test-tube experiments showing the effect of concentration and temperature, for example Cu(H₂O)₆²⁺ with concentrated HCl:

[Cu(H₂O)₆]²⁺(aq) + 4Cl⁻(aq) ⇌ [CuCl₄]²⁻(aq) + 6H₂O(l)

Adding concentrated hydrochloric acid increases [Cl⁻], so the equilibrium shifts right and the blue solution turns green and then yellow as [CuCl₄]²⁻ forms. Adding water shifts it back to the left. If you then warm and cool the tube, the colour change tells you which direction is endothermic: whichever way the colour moves on warming is the endothermic direction.

Compromise conditions in industry (3.1.6.1)

Industry wants a high yield, a fast rate and a low cost. These conflict, so a compromise is used.

Take the Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). The forward reaction is exothermic, and iron is used as the catalyst.

Temperature. A low temperature shifts the equilibrium right (exothermic direction) and gives a higher equilibrium yield. But at a low temperature the rate is too slow, so equilibrium takes far too long to reach. A moderate temperature gives an acceptable yield at an acceptable rate.

Pressure. There are 4 moles of gas on the left and 2 on the right. A high pressure shifts the equilibrium right and also increases the rate. But very high pressure needs expensive, strong equipment and a lot of energy to compress the gases, and it carries safety risks. The pressure chosen is high enough to give a reasonable yield, but limited by cost and safety.

Catalyst. The catalyst does not raise the yield. It lets equilibrium be reached faster, so a lower temperature can be used while keeping a useful rate.

The equilibrium constant Kc (3.1.6.2)

For a general homogeneous reaction:

aA + bB ⇌ cC + dD

         [C]^c [D]^d
  Kc  =  -----------
         [A]^a [B]^b

Products go on top, reactants on the bottom, each equilibrium concentration [X] (in mol dm⁻³) raised to the power of its coefficient.

To find the units, substitute mol dm⁻³ for every concentration and cancel.

Worked example 2: expressions and units

H2(g) + I2(g) <=> 2HI(g)
  Kc = [HI]^2 / ([H2][I2])
  units: (mol dm-3)^2 / (mol dm-3)^2 -> no units

N2O4(g) <=> 2NO2(g)
  Kc = [NO2]^2 / [N2O4]
  units: (mol dm-3)^2 / (mol dm-3) = mol dm-3

2SO2(g) + O2(g) <=> 2SO3(g)
  Kc = [SO3]^2 / ([SO2]^2 [O2])
  units: (mol dm-3)^2 / (mol dm-3)^3 = (mol dm-3)^-1 = mol-1 dm3

Worked example 3: Kc from equilibrium concentrations

At a fixed temperature, an equilibrium mixture of 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) contains [SO₂] = 0.0400 mol dm⁻³, [O₂] = 0.0250 mol dm⁻³ and [SO₃] = 0.180 mol dm⁻³. Calculate Kc.

Kc = [SO3]^2 / ([SO2]^2 [O2])
   = 0.180^2 / (0.0400^2 x 0.0250)
   = 0.0324 / 0.0000400
   = 810 mol-1 dm3

All data are given to 3 significant figures, so the answer is given to 3 s.f.

Worked example 4: from initial moles to Kc

0.250 mol of PCl₅ is placed in a sealed 5.00 dm³ container and heated. At equilibrium, 0.100 mol of Cl₂ is present. Calculate Kc for PCl₅(g) ⇌ PCl₃(g) + Cl₂(g).

Set up a moles table. Use the equation’s ratios to find how much of each species changes.

                    PCl5      PCl3      Cl2
initial / mol       0.250     0         0
change / mol       -0.100    +0.100    +0.100
equilibrium / mol   0.150     0.100     0.100
conc / mol dm-3     0.0300    0.0200    0.0200   (divide by 5.00)

Kc = [PCl3][Cl2] / [PCl5]
   = (0.0200 x 0.0200) / 0.0300
   = 0.0133 mol dm-3

The volume matters here because the moles of gas differ on each side. Forgetting to divide by 5.00 gives 0.0667, which is wrong.

Worked example 5: finding an unknown concentration

For H₂(g) + I₂(g) ⇌ 2HI(g) at a certain temperature, Kc = 49.0. An equilibrium mixture has [H₂] = [I₂] = 0.0200 mol dm⁻³. Calculate [HI].

[HI]^2 = Kc x [H2][I2] = 49.0 x 0.0200 x 0.0200 = 0.0196
[HI]   = sqrt(0.0196) = 0.140 mol dm-3

Suggested practical: Kc for ethyl ethanoate

The specification suggests determining Kc for the reaction of ethanol with ethanoic acid, using a strong acid catalyst:

CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O

Known amounts are mixed with dilute strong acid, sealed and left at constant temperature to reach equilibrium, then titrated with standard sodium hydroxide. Subtracting a blank titre for the catalyst alone gives the moles of ethanoic acid at equilibrium; the other amounts follow from the equation. Water in the catalyst solution counts as initial product.

Worked example 6: using the ester data

A mixture initially contains 0.200 mol ethanoic acid, 0.250 mol ethanol and 0.0500 mol water (from the catalyst solution). At equilibrium, 0.0700 mol ethanoic acid remains. Calculate Kc.

                  CH3COOH   C2H5OH    ester     H2O
initial / mol     0.200     0.250     0         0.0500
change / mol     -0.130    -0.130    +0.130    +0.130
equilibrium/mol   0.0700    0.120     0.130     0.180

Kc = [ester][H2O] / ([CH3COOH][C2H5OH])
The volume V cancels (2 moles on each side), so use moles:
Kc = (0.130 x 0.180) / (0.0700 x 0.120)
   = 0.0234 / 0.00840
   = 2.79   (no units)

What changes Kc, and what does not (3.1.6.2)

The value of Kc depends only on temperature. It is not affected by changes in concentration or by adding a catalyst.

Concentration. Adding a reactant makes the fraction (products over reactants) briefly smaller than Kc, so more products form until it equals Kc again. The position moves; Kc does not.

Catalyst. Both rates increase equally, so neither the position nor Kc changes.

Temperature. This is the only change that alters Kc:

Forward reaction	Temperature increased	Effect on Kc	Effect on yield of products
Exothermic (ΔH negative)	equilibrium shifts left	Kc decreases	decreases
Endothermic (ΔH positive)	equilibrium shifts right	Kc increases	increases

You can also work backwards: if Kc gets smaller as the temperature rises, the forward reaction is exothermic.

Significant figures. Report Kc to the limit of the least accurate data: if data are given to 2 and 3 s.f., answer to 2 s.f.

Common errors

• Writing “the concentrations of reactants and products are equal” at equilibrium. They are constant, not equal.
• Saying a catalyst “shifts the equilibrium to the right”. It only speeds up the approach to equilibrium.
• Counting all moles in the equation for a pressure argument, including solids or liquids, instead of moles of gas only.
• Using moles instead of concentrations in Kc when the moles of particles on each side are not equal (Worked example 4).
• Forgetting initial water from a catalyst solution in the ester calculation.
• Saying Kc changes when the concentration or pressure changes. At constant temperature, it does not.
• Writing a compromise answer about yield only. Marks also need rate, cost and, for pressure, safety or equipment.

Next steps

Condense this with the revision notes, then test yourself with the practice questions. For paper formats, see the exam preparation guide, and try the free diagnostics.

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.