OCR A Level Chemistry A: Chemical equilibrium (H432)

This study guide teaches section 3.2.3, Chemical equilibrium, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. This section belongs to Module 3, Periodic table and energy; all seven lettered outcomes, 3.2.3(a) to (g), are covered below. H432 is the full A level. The specification states that Module 3 content is assessed in Periodic table, elements and physical chemistry (01) and Unified chemistry (03). The specification links no PAG to 3.2.3, although outcome (d) is practical.

Every H432 topic is on the OCR chemistry hub; tick off outcomes on the printable checklist. Afterwards, use the Chemical equilibrium revision notes and test yourself with the Chemical equilibrium practice questions. Check other topics with the free diagnostics.

Outcomes in 3.2.3

Outcome	Your task	Additional guidance (specification)
3.2.3(a)	Explain that a dynamic equilibrium exists in a closed system when forward and reverse rates are equal and concentrations do not change	None given
3.2.3(b)	Apply le Chatelier’s principle to homogeneous equilibria to deduce, qualitatively, the effect of temperature, pressure or concentration on the position of equilibrium	Definition of le Chatelier’s principle not required; HSW1,2,5
3.2.3(c)	Explain why a catalyst, which speeds up both directions equally, leaves the equilibrium position where it was	None given
3.2.3(d)	Describe techniques used to investigate changes to the position of equilibrium when concentration and temperature change	Qualitative effects only; HSW4
3.2.3(e)	Explain why industry compromises between equilibrium and rate when choosing operating conditions	HSW6: balancing equilibrium, rate, safety and economics, e.g. the Haber process
3.2.3(f)	Write Kc expressions for homogeneous reactions and calculate Kc from equilibrium concentrations you are given	M0.2, M1.1, M2.3, M2.4; you will not need to determine the units for Kc
3.2.3(g)	Estimate the position of equilibrium from the size of Kc	M0.3; qualitative estimation only

Heterogeneous Kc, Kp, Kc units and equilibrium amounts from starting amounts come later, in 5.1.2 (see the How far? study guide).

3.2.3(a): Dynamic equilibrium

A reversible reaction can go both ways, shown by the ⇌ sign. Suppose you seal some reactant in a container and leave it at a fixed temperature.

• At the start, only reactant is present. The forward rate is highest and the reverse rate is zero.
• As reactant is used up, the forward rate falls. As product builds up, the reverse rate rises.
• Eventually the rates are equal, and product is made as fast as it is used up.

That state is a dynamic equilibrium. Learn its features together:

1. Forward and reverse reactions go at the same rate.
2. No concentration in the mixture changes.
3. It exists only in a closed system, where nothing can enter or leave.

“Dynamic” means both reactions are still going on, but the totals stay the same. “Constant concentrations” does not mean “equal concentrations”: the mixture might be mostly products or mostly reactants.

Worked example 1: why a closed system?

Calcium carbonate is heated in an open crucible. Explain why the mixture never reaches equilibrium.

CaCO₃(s) ⇌ CaO(s) + CO₂(g)

Carbon dioxide escapes as it forms, so the reverse reaction, which needs CO₂, barely happens. The two rates never become equal and the carbonate keeps decomposing. In a sealed vessel the CO₂ would stay and equilibrium could be reached.

3.2.3(b): le Chatelier’s principle

The specification does not ask you to recite a definition, but you need the idea: if a system at equilibrium is disturbed, the position of equilibrium shifts to oppose the change. “Position of equilibrium” just means the relative amounts of reactants and products. A shift to the right gives more products. A shift to the left gives more reactants.

This outcome covers homogeneous equilibria: all species share one phase.

Change made	Shift in position	Reason
Add a reactant, or remove a product	Right	Uses up some added reactant, or replaces lost product
Add a product, or remove a reactant	Left	Opposes the change the same way
Raise total pressure (gases)	Side with fewer gas moles	Fewer molecules lowers the pressure
Lower total pressure	Side with more gas moles	More molecules raises the pressure
Raise temperature	Endothermic direction	Absorbs some of the added heat
Lower temperature	Exothermic direction	Releases heat to oppose the cooling

The shift only partly opposes the change: after adding reactant, there is still more of it than before.

If both sides have the same number of gas moles, a pressure change has no effect on the position.

Worked example 2: steam reforming of methane

CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g)     ΔH = +206 kJ mol⁻¹

Predict the effect on the equilibrium yield of hydrogen of each change.

(i) Raising the temperature. The forward reaction is endothermic. The position shifts to the right to absorb the extra heat. The yield of hydrogen increases.

(ii) Raising the total pressure. Count gas moles: 1 + 1 = 2 on the left, 1 + 3 = 4 on the right. The position shifts left, to fewer gas molecules. The yield of hydrogen decreases.

(iii) Adding more steam. The position shifts to the right to use up some of the added steam. The yield of hydrogen increases.

Worked example 3: when pressure does nothing

N₂(g) + O₂(g) ⇌ 2NO(g)     ΔH = +180 kJ mol⁻¹

There are 2 gas moles on each side, so changing the pressure does not shift the position. Raising the temperature shifts it right, the endothermic direction, which is why nitrogen monoxide forms in very hot places such as engine cylinders.

3.2.3(c): Catalysts and equilibrium

A catalyst opens a different pathway whose activation energy is lower (see 3.2.2 and the Reaction rates study guide). The reverse reaction uses the same route backwards, so it speeds up too. A catalyst increases the forward and reverse rates by the same amount. The rates stay equal, so:

• the position of equilibrium is unchanged and so is the equilibrium yield
• equilibrium is reached in less time.

That second point is why industry still uses catalysts: more product is made per hour, even though the equilibrium yield is no higher.

3.2.3(d): Investigating changes to equilibrium in the laboratory

The specification asks for qualitative effects only. Coloured equilibria work best, because you can watch the position shift.

Concentration: chromate and dichromate

2CrO₄²⁻(aq) + 2H⁺(aq) ⇌ Cr₂O₇²⁻(aq) + H₂O(l)
   yellow                    orange

1. Put about 2 cm³ of potassium chromate(VI) solution in each of two test tubes. Keep one as a reference.
2. Add dilute sulfuric acid dropwise to the other. It turns orange: added H⁺ shifts the position right.
3. Add dilute sodium hydroxide dropwise. OH⁻ removes H⁺, so the position shifts left and yellow returns.

Temperature: the cobalt(II) chloride system

[Co(H₂O)₆]²⁺(aq) + 4Cl⁻(aq) ⇌ [CoCl₄]²⁻(aq) + 6H₂O(l)
     pink                         blue

1. Add concentrated hydrochloric acid to pink cobalt(II) chloride solution until it turns violet.
2. Pour equal volumes into three test tubes. Keep one at room temperature as a control.
3. Stand one tube in hot water and one in iced water for a few minutes.
4. Compare each with the control against a white background.

The hot tube turns blue and the cold tube turns pink. Heating shifts the position right, so the forward reaction must be endothermic.

Safety: chromate(VI) compounds are toxic and cobalt(II) chloride is harmful, so wear gloves and eye protection. Use concentrated hydrochloric acid, which is corrosive, in a fume cupboard.

Worked example 4: interpreting an observation

A sealed tube of brown nitrogen dioxide is cooled in ice and becomes paler.

2NO₂(g) ⇌ N₂O₄(g)
 brown     colourless

Deduce the sign of ΔH for the forward reaction.

Paler means less NO₂, so cooling shifted the position right. Cooling favours the exothermic direction. The forward reaction is exothermic and ΔH is negative.

3.2.3(e): Compromise in industry

A chemical company wants a good yield, but it also needs a fast rate, safe working and low costs. The best conditions for yield are often poor for these other aims, so plants run at a compromise. The Haber process is the classic case.

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)     ΔH = −92 kJ mol⁻¹

Condition	Effect on equilibrium yield	Other factors	Typical choice
Temperature	Exothermic, so low temperature gives a higher yield	Low temperature makes the rate too slow	About 450 °C
Pressure	4 gas moles go to 2, so high pressure gives a higher yield and a faster rate	Very high pressure needs thick-walled plant, costs more energy and is a greater safety hazard	About 2 × 10⁴ kPa
Catalyst	No effect on yield	Iron catalyst lets equilibrium be reached quickly at a workable temperature	Finely divided iron

Ammonia is condensed out and the leftover nitrogen and hydrogen go back through the reactor, so each pass need only give a modest yield.

Contrast steam reforming (Worked example 2). Its forward reaction is endothermic, so a high temperature favours both yield and rate. There the limit on temperature comes from cost and plant materials, not from the equilibrium.

3.2.3(f): The equilibrium constant, Kc

For a homogeneous reaction:

aA + bB ⇌ cC + dD

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

Square brackets mean equilibrium concentration in mol dm⁻³. Products go on top, each concentration raised to the power of its balancing number. Kc is fixed at a given temperature. In 3.2.3 you will not be asked for its units.

Worked example 5: Kc from equilibrium concentrations

At a high temperature, an equilibrium mixture for N₂(g) + O₂(g) ⇌ 2NO(g) contains:

[N₂] = 0.0360 mol dm⁻³, [O₂] = 0.00900 mol dm⁻³, [NO] = 4.00 × 10⁻⁴ mol dm⁻³

Kc = [NO]² / ([N₂][O₂])
   = (4.00 × 10⁻⁴)² / (0.0360 × 0.00900)
   = 1.60 × 10⁻⁷ / 3.24 × 10⁻⁴
   = 4.94 × 10⁻⁴

Match the significant figures of the data, here 3 (M1.1).

Worked example 6: Kc with a cubed term

For CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g), equilibrium concentrations at one temperature are [CH₄] = 0.120, [H₂O] = 0.150, [CO] = 0.0500 and [H₂] = 0.160 mol dm⁻³.

Kc = [CO][H₂]³ / ([CH₄][H₂O])
   = 0.0500 × 0.160³ / (0.120 × 0.150)
   = 0.0500 × 0.004096 / 0.0180
   = 1.14 × 10⁻²

Worked example 7: finding an unknown concentration (M2.4)

At a different temperature, Kc for the same reaction is 0.0216. An equilibrium mixture has [CH₄] = 0.200, [H₂O] = 0.250 and [CO] = 0.0400 mol dm⁻³. Find [H₂].

[H₂]³ = Kc × [CH₄][H₂O] / [CO]
      = 0.0216 × 0.200 × 0.250 / 0.0400
      = 0.0270
[H₂]  = ∛0.0270 = 0.300 mol dm⁻³

Rearrange first, then substitute, and take the cube root at the end.

3.2.3(g): What the size of Kc tells you

Kc compares products (top) with reactants (bottom), so its size gives a quick estimate of the position.

Value of Kc	Position of equilibrium
Much greater than 1	Far to the right: mostly products
About 1	Similar amounts of reactants and products
Much less than 1	Far to the left: mostly reactants

In Worked example 5, Kc = 4.94 × 10⁻⁴, much less than 1, so the mixture is mostly nitrogen and oxygen with very little nitrogen monoxide. Only a qualitative estimate is needed; do not try to work out percentages from Kc.

Common errors

• Writing “concentrations are equal” instead of “concentrations do not change” for dynamic equilibrium.
• Saying a catalyst “shifts the equilibrium to the right” or “increases the yield”.
• Counting all moles, including liquids, when deciding the effect of pressure. Count gas moles only.
• Stating a shift without the reason, such as “towards the endothermic direction” or “fewer gas moles”.
• Putting reactants on top in Kc, or forgetting to raise a concentration to its balancing number.
• Discussing only yield in a compromise answer. Rate, and often cost or safety, are needed too.

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 3.2.3 Chemical equilibrium. Published by OCR.