OCR A Level Chemistry A: Chemical equilibrium (H432) – Revision Notes

These notes condense 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. All seven lettered outcomes, 3.2.3(a) to (g), from Module 3, Periodic table and energy, are here. H432 is the full A level, and the specification states that Module 3 is assessed in Periodic table, elements and physical chemistry (01) and Unified chemistry (03).

For full explanations and longer worked examples, read the Chemical equilibrium study guide. When you are ready, try the Chemical equilibrium practice questions. The OCR chemistry hub lists every H432 topic, and the printable checklist helps you see what is left to revise. For a quick check of other topics, use the free diagnostics.

Key terms (3.2.3(a))

• Reversible reaction: a reaction that can go in both directions, shown by ⇌.
• Closed system: nothing can enter or leave. Equilibrium can only be reached in one.
• Dynamic equilibrium: forward and reverse reactions run at the same rate, so no concentration in the mixture changes.
• Position of equilibrium: the relative amounts of reactants and products in the equilibrium mixture.
• Homogeneous equilibrium: all reactants and products share one phase, such as a gas mixture or one solution.

Picture it on a graph of rate against time. The forward rate starts high and falls. The reverse rate starts at zero and rises. Where the two lines meet and level off together, equilibrium has been reached.

Shifting the position (3.2.3(b))

The specification does not require a definition of le Chatelier’s principle. Use the working idea: the position of equilibrium shifts to oppose the change you make.

You do this	The system responds by	So the position moves
Add more of a reactant	using some of it up	right
Take away some product	making more of it	right
Add more of a product	using some of it up	left
Raise the total pressure	forming fewer gas molecules	to the side with fewer gas moles
Lower the total pressure	forming more gas molecules	to the side with more gas moles
Heat the mixture	taking in heat	in the endothermic direction
Cool the mixture	giving out heat	in the exothermic direction
Add a catalyst	nothing changes in the balance	nowhere

Method: predicting a shift

1. Write the balanced equation with state symbols and the sign of ΔH for the forward reaction.
2. For pressure, count gas moles on each side. Ignore liquids and solids.
3. For temperature, identify which direction is endothermic.
4. State the direction of shift and the reason.
5. Say what happens to the amount (or yield) of the product you were asked about.

Small worked reminder

2CO(g) + O₂(g) ⇌ 2CO₂(g), forward reaction exothermic.

• Increase pressure: 3 gas moles on the left, 2 on the right. Shifts right. More CO₂.
• Increase temperature: shifts left, the endothermic direction. Less CO₂.

Catalysts (3.2.3(c))

• A catalyst speeds up both directions equally, by the same factor.
• The position of equilibrium is unchanged, so the equilibrium yield is unchanged.
• Equilibrium is reached sooner, so more product is made in a given time.

Practical work on equilibrium (3.2.3(d))

The specification asks for qualitative effects only. Choose a system with a colour change so you can see the shift.

System	Colours	Change made	What you see	Explanation
2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O	yellow ⇌ orange	Add dilute acid	Turns orange	Added H⁺ shifts the position right
(same)	(same)	Then add dilute alkali	Turns yellow	OH⁻ removes H⁺, shift left
[Co(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CoCl₄]²⁻ + 6H₂O	pink ⇌ blue	Add concentrated HCl	Turns blue	Added Cl⁻ shifts the position right
(same)	(same)	Warm in hot water	Turns blue	Forward reaction is endothermic
(same)	(same)	Cool in iced water	Turns pink	Exothermic direction (reverse) favoured

Good practice in a temperature investigation:

• Use equal volumes of the same solution in each tube.
• Keep one tube at room temperature as a control for colour comparison.
• Use a hot water bath and an ice bath, and leave the tubes long enough for the colour to stop changing.
• View the tubes against a white background.

Reading the result: if heating gives more of the product colour, the forward reaction is endothermic. If cooling gives more product colour, the forward reaction is exothermic.

Industrial compromise (3.2.3(e))

Industry balances four things: equilibrium yield, rate, safety and cost.

Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ mol⁻¹.

Condition	Equilibrium says	Rate, safety, cost say	Outcome
Temperature	Low (exothermic)	Too slow if low	Compromise, about 450 °C
Pressure	High (4 → 2 gas moles)	Expensive, energy-hungry plant; safety risk	High but limited, about 2 × 10⁴ kPa
Catalyst	No effect	Faster at a lower temperature	Iron catalyst

Unreacted gases are recycled, so the overall conversion is high even though a single pass through the reactor is not.

The equilibrium constant Kc (3.2.3(f))

For the homogeneous reaction wW + xX ⇌ yY + zZ:

Kc = [Y]ʸ [Z]ᶻ / ([W]ʷ [X]ˣ)

• Each square bracket is an equilibrium concentration in mol dm⁻³.
• Products on top; every concentration carries its balancing number as a power.
• You do not need Kc units in 3.2.3.
• At a fixed temperature Kc has one value.

Method: calculating Kc

1. Write the Kc expression from the balanced equation.
2. Check every value you have is an equilibrium concentration.
3. Substitute, raising each concentration to its power.
4. Work out the top and bottom separately, then divide.
5. Round to the significant figures of the data.

Small worked reminder

2NOCl(g) ⇌ 2NO(g) + Cl₂(g). At equilibrium, [NOCl] = 0.250, [NO] = 0.0500 and [Cl₂] = 0.0250 mol dm⁻³.

Kc = [NO]²[Cl₂] / [NOCl]²
   = 0.0500² × 0.0250 / 0.250²
   = 6.25 × 10⁻⁵ / 0.0625
   = 1.00 × 10⁻³

To find an unknown concentration, rearrange the expression first, then substitute. When the unknown appears squared, finish with a square root.

Size of Kc (3.2.3(g))

Kc	Position
≫ 1	Far right, mostly products
≈ 1	Similar amounts of both
≪ 1	Far left, mostly reactants

Only a qualitative estimate is required.

Must-know distinctions

• Constant concentrations at equilibrium, not equal concentrations.
• Position of equilibrium (moved by concentration, pressure and temperature) versus the value of Kc (changed only by temperature).
• Rate (how fast equilibrium is reached) versus yield (how much product at equilibrium). A catalyst changes the first, not the second.
• Total moles versus gas moles when you judge a pressure change.

Quick self-test

1. Give two features of a mixture at dynamic equilibrium.
2. Why can a reaction in an open beaker that gives off a gas not reach equilibrium?
3. For H₂(g) + Cl₂(g) ⇌ 2HCl(g), what is the effect of raising the pressure on the position?
4. A forward reaction is exothermic. What happens to the yield of product when the temperature is raised?
5. What does a catalyst do to the time taken to reach equilibrium, and to the position?
6. What colour change do you see when dilute acid is added to yellow chromate(VI) solution?
7. A blue solution turns pink when cooled in ice: [Co(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CoCl₄]²⁻ + 6H₂O. Is the forward reaction exothermic or endothermic?
8. Write the Kc expression for 2CO(g) + O₂(g) ⇌ 2CO₂(g).
9. For CO(g) + Cl₂(g) ⇌ COCl₂(g), equilibrium concentrations are [CO] = 0.0200, [Cl₂] = 0.0150 and [COCl₂] = 0.120 mol dm⁻³. Calculate Kc.
10. For 2H₂S(g) ⇌ 2H₂(g) + S₂(g), Kc = 2.00 × 10⁻⁴. At equilibrium [H₂S] = 0.100 and [H₂] = 0.0100 mol dm⁻³. Calculate [S₂].
11. A reaction has Kc = 3 × 10¹⁸. Estimate the position of equilibrium.
12. Why is the Haber process not run at a much lower temperature, even though that would raise the yield?

Answers

1. Forward and reverse rates are equal, and no concentration changes.
2. It is not a closed system. The gas escapes, so the reverse reaction cannot happen at an equal rate.
3. No effect: 2 gas moles on each side.
4. The yield falls. The position shifts left, the endothermic direction.
5. It shortens the time taken; the position is unchanged.
6. Yellow to orange.
7. Cooling shifts it left, so the reverse is exothermic and the forward reaction is endothermic.
8. Kc = [CO₂]² / ([CO]²[O₂]).
9. Kc = 0.120 / (0.0200 × 0.0150) = 0.120 / 3.00 × 10⁻⁴ = 400.
10. [S₂] = Kc × [H₂S]² / [H₂]² = 2.00 × 10⁻⁴ × 0.0100 / 1.00 × 10⁻⁴ = 0.0200 mol dm⁻³.
11. Far to the right: almost all products.
12. The rate would be too slow, so too little ammonia would be made per hour.

Where marks are usually lost

• Writing “the concentrations are equal” instead of “do not change” in a definition of dynamic equilibrium.
• Stating a shift with no reason. “Shifts right” alone rarely earns full credit; add “fewer gas moles” or “endothermic direction”.
• Counting water as a gas mole when it is a liquid in the equation, or counting solids.
• Saying a catalyst increases yield, or that it speeds up only the forward reaction.
• Describing a practical with no control tube, or with tubes that differ in volume or starting colour.
• Concluding the wrong sign of ΔH from a colour change. Decide which way the position moved first, then link it to heating or cooling.
• Inverting the Kc expression, or leaving out a power such as the 2 in [NO]².
• Substituting starting concentrations instead of equilibrium concentrations.
• In a compromise question, writing only about yield and forgetting rate, cost and safety.

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.