4.6 The rate and extent of chemical change — notes on one page

Every AQA GCSE Chemistry 8462 specification point in 4.6, read straight through: 11 points with definitions, explanations, mark-scheme wording and worked examples. Practice questions are in the printable question pack.

Checked against AQA 8462 section 4.6. Review basis: the qualification registry sourced from the AQA GCSE Chemistry (8462) specification; registry verification recorded 17 July 2026.

How this checking works

4.6.1.1 · Calculating rates of reactions

Reaction rate measures how quickly a reactant is used or a product is formed.
Calculate mean rate using $\text{mean rate}=\dfrac{\text{change in quantity}}{\text{time taken}}$ ; quantity may be mass in grams or gas volume in $\text{cm}^3$ , giving units such as $\text{g s}^{-1}$ or $\text{cm}^3\text{ s}^{-1}$ .
On a quantity–time graph, gradient represents rate: a steeper section is faster and a horizontal section has zero rate.
A tangent gives the rate at one instant.
Higher tier: AQA restricts "calculate the gradient of a tangent to the curve on these graphs as a measure of rate of reaction at a specific time" to Higher tier, which also uses quantity of reactants in moles and rate in mol/s; calculating a mean rate and drawing tangents are examined on both tiers.

A product–time curve with a tangent whose gradient gives the instantaneous rate.

Worked example

The mass of a flask decreases from $92.80\,\text{g}$ to $91.96\,\text{g}$ in $35\,\text{s}$ . Calculate the mean rate of mass loss.

1.Find the mass used: $92.80-91.96=0.84\,\text{g}$ .
2.Substitute into the rate equation: $\text{rate}=0.84\div35$ .
3.Evaluate and include the correct compound unit: $0.024\,\text{g s}^{-1}$ .

Answer: $0.024\,\text{g s}^{-1}$

Common mistakes

Don't fall into the trap of dividing the final balance reading by the time instead of first finding the change in mass.
Don't fall into the trap of using the whole curve for an instantaneous rate instead of drawing a tangent at the stated time.

Exam tip. For a ‘calculate the rate’ question, show change ÷ time and give a quantity-per-second unit.

4.6.1.2 · Factors which affect the rates of chemical reactions

Definition. The factors that affect the rate of a chemical reaction are the concentrations of reactants in solution, the pressure of reacting gases, the surface area of solid reactants, the temperature, and the presence of a catalyst.

Four of the five factors act through collision frequency, and temperature acts twice over; a catalyst works differently. Increasing concentration puts more reactant particles in the same volume, so the particles are closer together and collide more frequently.
Increasing the pressure of a gas does the same thing by another route: the same number of particles occupies a smaller volume. Breaking a solid into smaller pieces increases its surface area to volume ratio, so more of the solid is exposed and collisions are more frequent.
Increasing the temperature gives the particles more kinetic energy so they collide more frequently, and it also means a greater proportion of collisions carry at least the activation energy — the mark scheme wants both halves.
A catalyst increases the rate without being used up, by providing a different reaction pathway with a lower activation energy.
Rate also falls as a reaction proceeds, because reactant is used up and collision frequency falls.

Worked example

Describe how to investigate the effect of sodium thiosulfate concentration on reaction rate using a disappearing cross.

1.Place measured sodium thiosulfate solution over a marked cross and add a fixed volume and concentration of acid.
2.Start the timer on mixing and stop it when the cross is no longer visible through the sulfur precipitate.
3.Repeat for several thiosulfate concentrations while keeping total volume, acid, temperature, viewing position and cross constant.
4.Repeat each concentration and compare mean $1\div\text{time}$ values.

Answer: A controlled concentration series with repeated disappearance times provides comparable relative-rate data.

Mark-scheme wording to use

“particles are closer together. 1 mark: particles collide more frequently” — AQA 8462, June 2024 Paper 2F, Q05.6
“rate increases. 1 mark: particles have more energy/move faster. 1 mark: collision frequency increases or a greater proportion of collisions have enough energy. Ignore merely 'successful collisions'” — AQA 8462, June 2024 Paper 2H, Q09.5
“used up, concentration/surface area falls, collision frequency falls, rate falls” — AQA 8462, Specimen Paper 2H, Q09.4

What the examiner wants. The word 'frequency', not just 'more collisions'. The scheme explicitly ignores a bare 'more successful collisions', so say collisions happen more often per second AND that more of them reach the activation energy.

Common mistakes

Don't fall into the trap of changing acid concentration by changing its volume, so concentration and total volume both become independent variables.
Don't fall into the trap of calling disappearance time the reaction rate without using a comparable measure such as $1\div\text{time}$ .

Exam tip. In a practical-method answer, name the independent, dependent and at least two controlled variables before describing repeats.

4.6.1.3 · Collision theory and activation energy

Definition. Collision theory states that a chemical reaction can occur only when the reacting particles collide with each other and with sufficient energy. The activation energy is the minimum amount of energy that colliding particles must have in order to react.

Collision theory is the mechanism behind every rate factor, so a full-mark answer always links the change in conditions to the collisions and then to the rate. Increasing the concentration of reactants in solution, the pressure of reacting gases, or the surface area of a solid all increase the frequency of collisions and so increase the rate;
none of them changes how energetic the collisions are.
Increasing the temperature does both: it increases the frequency of collisions and makes them more energetic, so a greater proportion meet the activation energy. Lowering the temperature reverses both effects.
For a solid, reason in terms of the surface area to volume ratio, since smaller pieces of the same total mass expose more surface.
Catalysts act differently again: they are not used up, and they increase the rate by providing a different reaction pathway with a lower activation energy, so a greater proportion of collisions are successful.

Worked example

A $27\,\text{cm}^3$ cube is cut into $1\,\text{cm}$ cubes. Compare the total surface area before and after cutting and explain the rate change.

1.The original cube has side $3\,\text{cm}$ , so area $=6(3^2)=54\,\text{cm}^2$ .
2.There are $27$ small cubes, each with area $6(1^2)=6\,\text{cm}^2$ .
3.Total area after cutting $=27\times6=162\,\text{cm}^2$ , three times larger.
4.The greater exposed area gives more frequent collisions at the solid surface, so reaction is faster.

Answer: Surface area increases from $54$ to $162\,\text{cm}^2$ , so the smaller cubes react faster.

Mark-scheme wording to use

“(a catalyst) provides a different pathway for the reaction. 1 mark: (which has a) lower activation energy” — AQA 8462, catalyst two-marker, June 2023 Paper 2H, Q07.1
“fewer particles have activation energy. 1 mark: particles move more slowly” — AQA 8462, June 2022 Paper 2F, Q06.7
“there are more particle collisions each second. 1 mark: there are more particles in the same volume” — AQA 8462, June 2018 Paper 2F, Q07.7

What the examiner wants. Activation energy named as the minimum energy for a reaction, and the catalyst answer phrased as a different pathway of lower activation energy — never as 'the catalyst lowers the energy of the particles'.

Common mistakes

Don't fall into the trap of saying smaller pieces contain more particles even though the mass and number of particles are unchanged.
Don't fall into the trap of explaining temperature only through more frequent collisions and omitting that collisions are more energetic.

Exam tip. For an ‘explain using collision theory’ question, finish with ‘more successful collisions per second’.

4.6.1.4 · Catalysts

A catalyst changes the rate of a reaction but is not used up overall; different reactions need different catalysts, and enzymes are biological catalysts. It provides an alternative pathway with a lower activation energy.
At the same temperature, a greater fraction of collisions therefore has enough energy to react, so the reaction is faster.
On a reaction profile the catalysed pathway has a lower peak, while the reactant and product energy levels remain fixed.
A catalyst can be identified because rate increases and it is not included in the overall chemical equation.
It does not give particles extra energy or change the reaction’s overall energy transfer.

Catalysed and uncatalysed pathways have different activation energies but the same energy change.

Worked example

A substance is recovered with unchanged mass after making a reaction faster. Explain why it may be a catalyst.

1.State the rate evidence: the reaction is faster when the substance is present.
2.Use the recovery evidence: unchanged mass shows it was not used up overall.
3.Explain the action: it provides an alternative pathway with lower activation energy.

Answer: It fits the definition of a catalyst because it increases rate without being used up overall.

Common mistakes

Don't fall into the trap of writing that a catalyst supplies energy to reactant particles.
Don't fall into the trap of drawing different reactant or product energy levels for the catalysed pathway.

Exam tip. To explain catalytic action, state ‘alternative pathway’ and ‘lower activation energy’ for both marks.

4.6.2.1 · Reversible reactions

In a reversible reaction, products can react to produce the original reactants.
The equation uses $\rightleftharpoons$ : for $\mathrm{A+B\rightleftharpoons C+D}$ , A and B form C and D in the forward direction, while C and D reform A and B in the reverse direction.
Changing the conditions can change which direction is favoured.
A familiar example is hydrated copper sulfate: heating blue hydrated copper sulfate produces white anhydrous copper sulfate and water, while adding water reverses the change.
Reversible describes the ability to proceed both ways; it does not by itself mean that equilibrium has been reached.

Worked example

Heating blue hydrated copper sulfate makes a white solid. Adding water makes it blue again. Explain what this shows.

1.Heating changes hydrated copper sulfate into anhydrous copper sulfate and water.
2.Adding water makes the products reform the original hydrated copper sulfate.
3.Products reforming the original reactant is evidence that the reaction is reversible.

Answer: The observations show a reversible reaction whose direction changes with the conditions.

Common mistakes

Don't fall into the trap of reading the right-hand substances as products in both directions instead of reactants for the reverse reaction.
Don't fall into the trap of claiming the reversible arrow means the forward and reverse rates are already equal.

Exam tip. When asked for evidence of reversibility, state that the products reform the original reactants under changed conditions.

4.6.2.2 · Energy changes and reversible reactions

The two directions of a reversible reaction have opposite energy transfers.
If the forward reaction is exothermic, it transfers energy to the surroundings; the reverse reaction must take the same amount of energy from the surroundings and is endothermic.
Conversely, an endothermic forward reaction has an exothermic reverse.
On an energy-level diagram the reactant and product levels simply exchange roles when the direction is reversed, so an energy change of $-q$ forwards becomes $+q$ backwards.
Examiners expect both the opposite classification and the fact that the magnitude of the energy transfer is unchanged.

Worked example

The forward direction of a reversible reaction transfers $38\,\text{kJ}$ to the surroundings. Describe the reverse reaction.

1.Transferring energy to the surroundings makes the forward reaction exothermic.
2.The reverse reaction has the opposite energy transfer, so it is endothermic.
3.The same energy magnitude is involved, so the reverse takes in $38\,\text{kJ}$ .

Answer: The reverse reaction is endothermic and takes in $38\,\text{kJ}$ .

Common mistakes

Don't fall into the trap of writing that the reverse reaction transfers a different amount of energy.
Don't fall into the trap of changing the sign of the energy change without stating whether energy enters or leaves the surroundings.

Exam tip. For a reverse-reaction energy question, keep the numerical magnitude, reverse the sign and name the opposite energy-transfer type.

4.6.2.3 · Equilibrium

Dynamic equilibrium is reached when a reversible reaction occurs in a closed system, so reactants and products cannot escape, and the forward and reverse reactions happen at exactly the same rate. Both reactions continue: ‘dynamic’ means particles still react in each direction.
Equal rates make the observable amounts or concentrations of every substance remain constant.
Those amounts need not be equal; an equilibrium mixture may contain much more reactant than product.
If material escapes from the apparatus, a stable equilibrium cannot be maintained.
In definitions, AQA expects the closed-system condition and equality of forward and reverse rates.

Inside one closed container, separate A and B particles and joined AB pairs are intermixed. Equal forward and reverse arrows show A and B joining as AB and AB separating back into A and B.

Worked example

A sealed reaction mixture has constant concentrations, but particles continue changing between reactants and products. Explain this observation.

1.The sealed apparatus provides the required closed system.
2.Constant concentrations are consistent with equal forward and reverse reaction rates.
3.Particles still reacting in both directions shows that the equilibrium is dynamic.

Answer: The mixture is at dynamic equilibrium: forward and reverse reactions continue at equal rates.

Common mistakes

Don't fall into the trap of stating that both reactions stop when equilibrium is reached.
Don't fall into the trap of stating that equilibrium requires equal amounts of reactants and products rather than equal reaction rates.

Exam tip. A two-mark definition normally needs both ‘closed system’ and ‘forward and reverse reactions at the same rate’.

4.6.2.4 · The effect of changing conditions on equilibrium (HT only)

Higher tier: the relative amounts of reactants and products at equilibrium depend on the conditions.
Le Chatelier’s principle states that when a condition changes, the equilibrium system responds in the direction that counteracts that change.
A prediction therefore needs three parts: identify the imposed change, decide which direction opposes it, then state whether the relative amount of product increases or decreases.
Concentration, temperature and gas pressure can change equilibrium position, but their rules differ.
The response only partially counteracts the disturbance and produces a new equilibrium; it does not necessarily restore the original composition.

Worked example

Higher tier: $\mathrm{N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)}$ is at equilibrium. Predict and explain what happens to the equilibrium position and the relative amount of ammonia when the pressure is increased at constant temperature.

1.Count the gaseous molecules from the balanced equation: there are $1+3=4$ on the left and $2$ on the right.
2.Increasing pressure favours the side with fewer gas molecules, which counteracts the pressure increase.
3.The equilibrium therefore shifts to the right, so the relative amount of ammonia increases as a new equilibrium is reached.

Answer: The equilibrium shifts to the right and the relative amount of $\mathrm{NH_3}$ increases because the product side has fewer gas molecules: $2$ compared with $4$ on the reactant side.

Common mistakes

Don't fall into the trap of writing that every condition change shifts equilibrium towards products.
Don't fall into the trap of describing a faster reaction rate without stating the new relative amount of product at equilibrium.

Exam tip. For a ‘predict and explain’ question, name the direction of shift, the counteracted change and the effect on product amount.

4.6.2.5 · The effect of changing concentration (HT only)

Higher tier: changing the concentration of a reactant or product disturbs equilibrium, so all concentrations change until a new equilibrium is reached.
If a reactant concentration is increased, the system shifts towards products and uses some of the added reactant.
If a product concentration is decreased, the system also shifts towards products and replaces some of what was removed.
Apply the same counteracting logic to any named substance: an added substance is consumed and a removed substance is replaced.
The shift changes relative amounts but does not completely undo the imposed change, and the concentrations eventually become constant at new values.

Worked example

For $\mathrm{A+B\rightleftharpoons C+D}$ , some C is removed from an equilibrium mixture. Predict the effect.

1.C is a product, so removing it decreases a product concentration.
2.The system counteracts the removal by favouring the forward reaction.
3.More A and B react, producing more C and D until a new equilibrium is reached.

Answer: Equilibrium shifts to the right and the relative amounts of C and D increase.

Common mistakes

Don't fall into the trap of assuming increasing any concentration must increase the amount of product.
Don't fall into the trap of saying the system fully restores the original concentration rather than only counteracting the change.

Exam tip. Circle whether the changed substance is a reactant or product, then choose the direction that consumes an addition or replaces a removal.

4.6.2.6 · The effect of temperature changes on equilibrium (HT only)

Higher tier: treat energy as part of the equilibrium: the endothermic direction takes in energy and the exothermic direction releases it.
Increasing temperature shifts equilibrium in the endothermic direction, counteracting the added heat; decreasing temperature shifts it in the exothermic direction, counteracting the cooling.
Therefore, for an endothermic forward reaction, heating increases the relative amount of products and cooling decreases it.
For an exothermic forward reaction, the pattern is reversed.
Temperature also affects reaction rates, but an equilibrium-position answer must use the direction of energy transfer to predict the final relative amounts.

Worked example

The forward reaction in $\mathrm{X\rightleftharpoons Y}$ is exothermic. Predict the effect of increasing temperature.

1.The forward direction is exothermic, so the reverse direction is endothermic.
2.Increasing temperature favours the endothermic direction to counteract the heating.
3.The equilibrium shifts towards X, so the relative amount of Y decreases.

Answer: The equilibrium shifts left and the equilibrium yield of Y decreases.

Common mistakes

Don't fall into the trap of claiming higher temperature always increases product yield because reactions happen faster.
Don't fall into the trap of using the sign of the forward reaction without first identifying which direction is endothermic.

Exam tip. Annotate the equation with ‘exo’ and ‘endo’ before deciding which direction counteracts the temperature change.

4.6.2.7 · The effect of pressure changes on equilibrium (HT only)

Higher tier: pressure changes affect equilibria involving gases.
Increasing pressure shifts equilibrium towards the side with the smaller number of gas molecules, which counteracts the increase; decreasing pressure favours the side with more gas molecules.
Use the balanced equation and add the coefficients of gaseous species only.
For $\mathrm{N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)}$ , there are four gas molecules on the left and two on the right, so increasing pressure favours ammonia.
If both sides contain equal numbers of gas molecules, pressure does not change the equilibrium position, although it may still affect reaction rates.

Worked example

Predict the effect of increasing pressure on $\mathrm{2SO_2(g)+O_2(g)\rightleftharpoons2SO_3(g)}$ .

1.Count gaseous molecules from the coefficients: $2+1=3$ on the left and $2$ on the right.
2.Increasing pressure favours the side with fewer gas molecules.
3.The equilibrium shifts right, increasing the relative amount of sulfur trioxide.

Answer: The equilibrium shifts towards $\mathrm{SO_3}$ because the product side has fewer gas molecules.

Common mistakes

Don't fall into the trap of counting the number of different gas formulae instead of adding their balanced coefficients.
Don't fall into the trap of including solids or liquids when comparing the numbers of gas molecules.

Exam tip. Write the gas-molecule totals under both sides of the equation before stating the pressure shift.