4.6 The rate and extent of chemical change — revision question pack

11 specification points · notes, questions, answers and worked methods

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

Loading your tier…

Answer all questions in the spaces provided.

4.6.1.1 · Calculating rates of reactions

Explanation

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: calculate the tangent's gradient and also express quantity in moles, with units $\text{mol s}^{-1}$ .

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.

Tier 1 · Easy

A reaction produces $72\,\text{cm}^3$ of gas in $40\,\text{s}$ . Calculate its mean rate.
[2 marks]
Total for this question: 2
A gas volume rises from $18\,\text{cm}^3$ to $66\,\text{cm}^3$ between $15\,\text{s}$ and $39\,\text{s}$ . Calculate the mean rate of gas production during this interval.
[2 marks]
Total for this question: 2

Tier 2 · Standard

The mass of a reaction flask falls from $83.40\,\text{g}$ to $82.56\,\text{g}$ during the first $35\,\text{s}$ . Calculate the mean rate of mass loss.
[3 marks]
Total for this question: 3
A reaction produces $168\,\text{cm}^3$ of gas in $2.0$ minutes. Calculate the mean rate of gas production in $\text{cm}^3\text{ s}^{-1}$ .
[3 marks]
Total for this question: 3
A reaction has $20.0\,\text{g}$ of reactant available, which is used at a constant mean rate. The reaction uses $5.4\,\text{g}$ in $45\,\text{s}$ . Calculate the mass used in $2.0$ minutes.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Reaction R forms $84\,\text{cm}^3$ of gas in $70\,\text{s}$ . Reaction S forms $99\,\text{cm}^3$ in $55\,\text{s}$ . Calculate both mean rates and determine the percentage by which S is faster than R.
[5 marks]
Total for this question: 5
A reaction has produced $24\,\text{cm}^3$ of gas after $20\,\text{s}$ , $69\,\text{cm}^3$ after $50\,\text{s}$ and $99\,\text{cm}^3$ after $90\,\text{s}$ . Calculate the mean rate from $20$ to $50\,\text{s}$ and from $50$ to $90\,\text{s}$ . Calculate the percentage decrease in mean rate.
[5 marks]
Total for this question: 5
The mass of a reactant remaining is $12.0\,\text{g}$ at $0\,\text{s}$ , $9.5\,\text{g}$ at $25\,\text{s}$ , $7.1\,\text{g}$ at $55\,\text{s}$ , $6.2\,\text{g}$ at $85\,\text{s}$ and $6.2\,\text{g}$ at $115\,\text{s}$ . Calculate the mean rate at which reactant is used between $25$ and $85\,\text{s}$ . Determine the first recorded time by which the reaction has finished and describe how the data show that the rate decreases.
[5 marks]
Total for this question: 5
A reaction flask has a mass of $74.62\,\text{g}$ at $0\,\text{s}$ . Its mean rate of mass loss during the first $25\,\text{s}$ is $0.032\,\text{g s}^{-1}$ . From the reading at $25\,\text{s}$ , the mass falls to $73.12\,\text{g}$ during the next $50\,\text{s}$ . Calculate the missing $25\,\text{s}$ reading, the mean rate during the next $50\,\text{s}$ , and the factor by which the first rate is greater.
[5 marks]
Total for this question: 5
A reaction eventually produces $180\,\text{cm}^3$ of gas. It produces $35\%$ of this total in the first $30\,\text{s}$ , a further $45\%$ during the next $60\,\text{s}$ , and the remainder during the following $90\,\text{s}$ . Calculate the mean rate during each stage and describe how the gradient of a gas-volume graph changes.
[5 marks]
Total for this question: 5

4.6.1.2 · Factors which affect the rates of chemical reactions

Explanation

Reaction rate is affected by reactant concentration in solution, gas pressure, the surface area of a solid, temperature and catalysts.
Increasing concentration, pressure, surface area or temperature increases rate; a suitable catalyst also increases it.
Required practical 5 investigates concentration in two ways: measuring gas volume produced over time and timing a colour or turbidity change.
Develop a hypothesis, vary only concentration and keep temperature, total liquid volume, amount and surface area of solid, and apparatus controlled.
Start timing when reactants mix, collect quantitative results and repeat trials so a mean can be calculated and anomalies identified.

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.

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.

Tier 1 · Easy

A solid reactant is crushed into smaller pieces without changing its mass. State the effect on the reaction rate.
[1 mark]
Total for this question: 1
State the effect on reaction rate of lowering the temperature while keeping all other conditions unchanged.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A student investigates how acid concentration affects the rate of gas production from marble chips. Describe how the student should collect suitable results while changing only the acid concentration.
[4 marks]
Total for this question: 4
A student measures the time for a cross to disappear during the reaction of sodium thiosulfate solution with dilute hydrochloric acid. Suggest two changes that would improve the repeatability of the results, and explain each change.
[4 marks]
Total for this question: 4
A student compares $1.0\,\text{g}$ of powdered marble in $40\,\text{cm}^3$ of $1.0\,\text{mol dm}^{-3}$ acid with $2.0\,\text{g}$ of marble chips in $40\,\text{cm}^3$ of $0.5\,\text{mol dm}^{-3}$ acid. Explain why the results cannot show the effect of surface area alone. Give two changes that would make the comparison valid.
[4 marks]
Total for this question: 4

Tier 3 · Hard

In a turbidity experiment, relative acid concentrations of $20\%$ , $40\%$ and $60\%$ give disappearance times of $162$ , $81$ and $54\,\text{s}$ . Describe the relationship shown and give two limitations of using the disappearance time as a rate measurement.
[5 marks]
Total for this question: 5
In a disappearing-cross experiment, $0.40\,\text{mol dm}^{-3}$ sodium thiosulfate gives times of $76$ , $74$ and $75\,\text{s}$ . At $0.80\,\text{mol dm}^{-3}$ , the times are $38$ , $92$ and $37\,\text{s}$ . Identify the anomalous result, calculate both mean times after excluding it, and compare the relative rates using $1/\text{time}$ .
[5 marks]
Total for this question: 5
Four experiments use the same marble mass, acid volume and apparatus, and measure gas formed in the first $20\,\text{s}$ . A uses large chips, $0.50\,\text{mol dm}^{-3}$ acid and $20\,^\circ\text{C}$ , producing $20\,\text{cm}^3$ . B uses powder with A's other conditions, producing $50\,\text{cm}^3$ . C uses large chips, $1.00\,\text{mol dm}^{-3}$ acid and $20\,^\circ\text{C}$ , producing $32\,\text{cm}^3$ . D uses large chips, $0.50\,\text{mol dm}^{-3}$ acid and $30\,^\circ\text{C}$ , producing $28\,\text{cm}^3$ . Calculate the rate factor for each change relative to A and determine which change has the greatest effect in these experiments.
[4 marks]
Total for this question: 4
Equal masses of the same solid react with identical acid samples. Powder, small chips and large chips produce $54$ , $36$ and $18\,\text{cm}^3$ of gas respectively in the first $15\,\text{s}$ , and all three eventually produce $90\,\text{cm}^3$ . A repeat using powder produces $51\,\text{cm}^3$ in $15\,\text{s}$ but only $72\,\text{cm}^3$ in total. Calculate the three valid initial mean rates, identify the surface-area order, and explain why the repeat should not be used in the comparison.
[6 marks]
Total for this question: 6
A fixed gas volume takes $45\,\text{s}$ to form without a catalyst and $25\,\text{s}$ with the catalyst at $20\,^\circ\text{C}$ . At $30\,^\circ\text{C}$ , it takes $36\,\text{s}$ without the catalyst and $20\,\text{s}$ with it. A student says the catalyst is less effective at $30\,^\circ\text{C}$ because it saves only $16\,\text{s}$ instead of $20\,\text{s}$ . Calculate the catalyst's relative-rate factor at each temperature and evaluate the claim.
[5 marks]
Total for this question: 5

4.6.1.3 · Collision theory and activation energy

Explanation

Collision theory states that reacting particles must collide with sufficient energy. The minimum energy needed is the activation energy.
Greater solution concentration or gas pressure places more particles in a given volume, so collisions occur more frequently.
Breaking a solid into smaller pieces increases its surface-area-to-volume ratio, exposing more particles for collisions.
Raising temperature has two effects: particles collide more often and collisions are more energetic, so a greater proportion meet or exceed the activation energy.
Examiners expect a complete chain from the changed condition to collision frequency or energy, then to more successful collisions per second and a faster rate.

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.

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’.

Tier 1 · Easy

State the two conditions needed for a collision between reactant particles to lead to a reaction.
[2 marks]
Total for this question: 2
State why a collision between two reactant particles may not produce a reaction.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Explain, using collision theory, why increasing the pressure of two reacting gases increases their reaction rate at constant temperature.
[3 marks]
Total for this question: 3
Explain, using collision theory, why warming a reaction mixture increases its reaction rate.
[4 marks]
Total for this question: 4
Solution Q contains twice as many reactant particles per unit volume as solution P. Both solutions are at the same temperature. Explain, using collision theory, why Q reacts faster without claiming that its particles move faster.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A fixed $64\,\text{cm}^3$ of solid is cut either into cubes of side $2.0\,\text{cm}$ or cubes of side $1.0\,\text{cm}$ . Calculate the total surface area for each set and explain which reacts faster with an acid.
[6 marks]
Total for this question: 6
Equal masses of limestone are reacted with acid. Experiment A uses large chips in $0.50\,\text{mol dm}^{-3}$ acid. Experiment B uses powder in $1.00\,\text{mol dm}^{-3}$ acid at the same temperature. Predict which experiment has the greater initial rate and explain both reasons using collision theory.
[5 marks]
Total for this question: 5
Change C increases concentration. Change T increases temperature. Measurements show that each change causes the same increase in the total number of collisions per second. Predict which change may produce the larger increase in reaction rate and explain using collision energy.
[4 marks]
Total for this question: 4
At one temperature, $1000$ collisions have the following energies: $250$ are below $40$ energy units, $500$ are from $40$ up to but not including $60$ energy units, and $250$ are at least $60$ energy units. Compare pathways with activation energies of $60$ and $40$ energy units. Calculate the successful collisions for each pathway, the increase and the factor change.
[5 marks]
Total for this question: 5
Before warming, reactant particles make $2400$ collisions each second and one collision in every $12$ has enough energy to react. After warming, they make $3000$ collisions each second and one collision in every $5$ has enough energy to react. Calculate the successful collisions per second before and after warming, determine the rate factor, and explain the two effects of temperature.
[5 marks]
Total for this question: 5

4.6.1.4 · Catalysts

Explanation

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.

Tier 1 · Easy

Complete the statement: a catalyst increases reaction rate by providing a different pathway with a lower what?
[1 mark]
Total for this question: 1
State the name given to a biological catalyst.
[1 mark]
Total for this question: 1

Tier 2 · Standard

An uncatalysed reaction has an activation energy of $79\,\text{kJ mol}^{-1}$ and an overall energy change of $-24\,\text{kJ mol}^{-1}$ . A catalyst lowers the activation energy to $43\,\text{kJ mol}^{-1}$ . State the reduction in activation energy and the catalysed reaction's overall energy change.
[3 marks]
Total for this question: 3
Explain why lowering the activation energy increases reaction rate even though the temperature is unchanged.
[3 marks]
Total for this question: 3
Substance K speeds up reaction 1 but has no measurable effect on reaction 2. A student claims that K therefore cannot be a catalyst. Evaluate the claim.
[3 marks]
Total for this question: 3

Tier 3 · Hard

In the first $30\,\text{s}$ , a reaction forms $48\,\text{cm}^3$ of gas with solid X and $21\,\text{cm}^3$ without X. Both tests eventually form $72\,\text{cm}^3$ , and the dry mass of X is unchanged. Use the data to explain why X is a catalyst.
[5 marks]
Total for this question: 5
At one temperature, $80$ out of every $1000$ collisions have enough energy to react by an uncatalysed pathway. With a catalyst, $240$ out of every $1000$ collisions have enough energy. Calculate the increase in successful collisions per $1000$ and the factor by which this number increases. Explain the catalyst's effect.
[5 marks]
Total for this question: 5
An uncatalysed energy profile starts at $35\,\text{kJ mol}^{-1}$ , reaches a peak at $118\,\text{kJ mol}^{-1}$ and ends at $61\,\text{kJ mol}^{-1}$ . A catalyst gives an activation energy of $46\,\text{kJ mol}^{-1}$ . Describe the start, peak and end values of the catalysed profile. Calculate the overall energy change and explain why the catalyst increases the rate.
[5 marks]
Total for this question: 5
The same reaction takes $60\,\text{s}$ with no added substance. With P it takes $60\,\text{s}$ and all $1.50\,\text{g}$ of P is recovered. With Q it takes $32\,\text{s}$ but only $1.10\,\text{g}$ of the original $1.50\,\text{g}$ is recovered. With R it takes $30\,\text{s}$ and all $1.50\,\text{g}$ is recovered. Evaluate P, Q and R and identify which has the strongest evidence of being a catalyst.
[5 marks]
Total for this question: 5
A reaction occurs in two steps: $\mathrm{A+C\rightarrow AC}$ and $\mathrm{AC+B\rightarrow AB+C}$ . Combine the steps to give the overall equation, identify the catalyst, and explain using both the equations and activation energy why it can increase the reaction rate.
[5 marks]
Total for this question: 5

4.6.2.1 · Reversible reactions

Explanation

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.

Tier 1 · Easy

State what is meant by a reversible reaction.
[1 mark]
Total for this question: 1
State one observation that would show the products of a reaction can reform the original reactants.
[1 mark]
Total for this question: 1

Tier 2 · Standard

The reaction $\mathrm{J+K\rightleftharpoons L+M}$ is reversible. State which substances react in the reverse reaction and which substances they form.
[2 marks]
Total for this question: 2
Heating solid Q forms gases R and S. Cooling R and S together forms solid Q again. Explain why this change is reversible.
[2 marks]
Total for this question: 2
Heating U forms V, and cooling V reforms U. Heating X forms Y, but cooling Y leaves Y unchanged. Compare the evidence and identify the process shown to be reversible.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Blue hydrated copper sulfate is heated and forms white anhydrous copper sulfate and water. Adding water to the white solid reforms the blue substance. Explain how these observations show a reversible reaction and identify the change of condition used in each direction.
[4 marks]
Total for this question: 4
Heating solid L produces solid M and gas N. Passing N back over M at room temperature reforms L. A student claims that cooling M by itself would be enough to prove the reaction is reversible. Evaluate the claim and explain which observation provides the evidence of reversibility.
[4 marks]
Total for this question: 4
In a sealed tube, heating solid A forms solid C and gas B. Cooling the sealed tube reforms A. A student repeats the test but opens the hot tube, allowing B to escape. Cooling now leaves C unchanged. Evaluate the claim that the second result proves the reaction is not reversible.
[4 marks]
Total for this question: 4
Light changes red substance R into colourless substances S and T. In darkness, a mixture of S and T reforms red R. Write one equation representing the complete process, identify the forward and reverse reactants, and evaluate whether these observations alone prove that the mixture reaches equilibrium.
[5 marks]
Total for this question: 5
Heating a $6.00\,\text{g}$ sample of A in a sealed tube forms B and a gas. Cooling reforms $5.76\,\text{g}$ of A. In an open tube, the gas escapes and cooling reforms only $4.20\,\text{g}$ of A. Calculate the percentage of A recovered in the sealed tube and evaluate the claim that recovering less than $100\%$ means the reaction is not reversible.
[5 marks]
Total for this question: 5

4.6.2.2 · Energy changes and reversible reactions

Explanation

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.

Tier 1 · Easy

The forward direction of a reversible reaction is exothermic. State the energy-change type of the reverse direction.
[1 mark]
Total for this question: 1
State the relationship between the magnitudes of the energy transfers in the forward and reverse directions of a reversible reaction.
[1 mark]
Total for this question: 1

Tier 2 · Standard

The forward direction of a reversible reaction transfers $67\,\text{kJ mol}^{-1}$ to the surroundings. State the energy change for the reverse direction, including its sign.
[2 marks]
Total for this question: 2
The products of a reversible reaction are $45\,\text{kJ}$ higher in energy than the reactants. State the energy-change type and energy change for the forward and reverse directions.
[3 marks]
Total for this question: 3
The surroundings warm during the forward direction of a reversible reaction. State the energy-change type of each direction and compare the amounts of energy transferred.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A reversible reaction has a forward activation energy of $145\,\text{kJ mol}^{-1}$ and a forward overall energy change of $-38\,\text{kJ mol}^{-1}$ . Calculate the reverse activation energy and state the reverse overall energy change.
[4 marks]
Total for this question: 4
On an energy profile, the reactants are at $92\,\text{kJ}$ , the products are at $137\,\text{kJ}$ and the peak is at $206\,\text{kJ}$ . Calculate the forward and reverse activation energies. State the overall energy change in each direction.
[5 marks]
Total for this question: 5
Each complete forward change of a reversible reaction transfers $24\,\text{kJ}$ to the surroundings. In a sequence, the forward change occurs seven times and the reverse change occurs four times. Calculate the net energy transfer and state its direction.
[4 marks]
Total for this question: 4
A reversible reaction has a forward activation energy of $84\,\text{kJ}$ and a reverse activation energy of $51\,\text{kJ}$ . Determine the forward and reverse overall energy changes and state whether each direction is endothermic or exothermic.
[4 marks]
Total for this question: 4
The forward direction of a reversible reaction cools the surroundings and transfers $36\,\text{kJ}$ per complete change. Four students describe the pair of energy changes: P gives $+36\,\text{kJ}$ in both directions; Q gives $-36\,\text{kJ}$ forwards and $+36\,\text{kJ}$ in reverse; R gives $-36\,\text{kJ}$ forwards and $+42\,\text{kJ}$ in reverse; S gives $+36\,\text{kJ}$ forwards and $-36\,\text{kJ}$ in reverse. Choose the correct student, justify both signs, and calculate the net energy transfer when three forward changes and one reverse change occur.
[5 marks]
Total for this question: 5

4.6.2.3 · Equilibrium

Explanation

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’.

Tier 1 · Easy

State the relationship between the forward and reverse reaction rates at equilibrium.
[1 mark]
Total for this question: 1
State why a closed system is needed for a reaction mixture to reach dynamic equilibrium.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A student says, 'The concentrations stay constant at equilibrium because both reactions have stopped.' Explain why this statement is incorrect.
[3 marks]
Total for this question: 3
At equilibrium, a sealed mixture contains $0.80\,\text{mol}$ of reactant and $1.70\,\text{mol}$ of product. Explain why the unequal amounts do not show that the mixture is not at equilibrium.
[2 marks]
Total for this question: 2
At one instant in a sealed reversible system, the forward rate is $4.6$ arbitrary units and the reverse rate is $2.9$ arbitrary units. Determine whether the system is at equilibrium and state the net effect on the amount of product.
[3 marks]
Total for this question: 3

Tier 3 · Hard

In a sealed vessel, the measured forward and reverse rates in arbitrary units are: at $0\,\text{s}$ , $5.2$ and $0.0$ ; at $20\,\text{s}$ , $3.6$ and $1.7$ ; at $40\,\text{s}$ , $2.8$ and $2.8$ ; at $60\,\text{s}$ , $2.8$ and $2.8$ . Determine when equilibrium is first reached and explain what the later readings show.
[4 marks]
Total for this question: 4
Two samples of the same reversible mixture are placed in different containers. Container S is sealed, and its concentrations eventually become constant. Container O is open, so a gaseous product escapes and its reactant concentration keeps falling. Explain why only container S can establish dynamic equilibrium.
[4 marks]
Total for this question: 4
At equilibrium, $240$ forward-reaction events and $240$ reverse-reaction events occur each second. Calculate the total number of reaction events in each direction over $15\,\text{s}$ and the net change in the amounts present. Explain why this is a dynamic equilibrium.
[5 marks]
Total for this question: 5
A sealed system initially contains $1800$ product particles. During the next $10\,\text{s}$ , $320$ forward-reaction events and $200$ reverse-reaction events occur each second; each event changes one product particle. During the following $10\,\text{s}$ , both directions have $260$ events per second. Calculate the product-particle count after the first interval and explain what the second interval shows.
[6 marks]
Total for this question: 6
Two sealed systems have constant reactant and product concentrations. In system A, both the forward and reverse rates are zero. In system B, the forward and reverse rates are both $1.7$ arbitrary units. Determine which system is at dynamic equilibrium and evaluate the claim that constant concentrations alone prove dynamic equilibrium.
[5 marks]
Total for this question: 5

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

Explanation

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.

Tier 1 · Easy

State Le Chatelier's principle for a system at equilibrium when a condition is changed.
[1 mark]
Total for this question: 1
State what is meant by saying that an equilibrium position shifts to the right.
[1 mark]
Total for this question: 1

Tier 2 · Standard

For $\mathrm{R\rightleftharpoons2P}$ at equilibrium, some P is removed. Predict the direction of shift and explain it using Le Chatelier's principle.
[3 marks]
Total for this question: 3
A catalyst is added to a reversible reaction that is already at equilibrium. Explain its effect on the equilibrium position and on the time taken to re-establish equilibrium after a later disturbance.
[3 marks]
Total for this question: 3
Give three conditions that can change the relative amounts of reactants and products in a system at equilibrium.
[3 marks]
Total for this question: 3

Tier 3 · Hard

The forward reaction $\mathrm{A+2B\rightleftharpoons C}$ is exothermic. At equilibrium, the concentration of B and the temperature are both increased. Explain why the information given is insufficient to predict the final change in the amount of C.
[4 marks]
Total for this question: 4
The forward reaction $\mathrm{2A(g)+B_2(g)\rightleftharpoons2AB(g)}$ is exothermic. Predict and explain separately the effect on the equilibrium amount of AB of increasing pressure, decreasing temperature and adding a catalyst.
[6 marks]
Total for this question: 6
After a condition is changed in a system at equilibrium, the forward rate is $5.0$ arbitrary units and the reverse rate is $2.0$ arbitrary units. Later, both rates are $3.6$ arbitrary units. Determine the direction of the equilibrium shift and the change in the relative amount of product. Explain what the later rates show and evaluate whether these data identify the condition that was changed.
[5 marks]
Total for this question: 5
Higher Tier: The forward reaction $\mathrm{2A(g)+B(g)\rightleftharpoons C(g)}$ is exothermic. Three separate disturbances give these observations. X causes only A's concentration to rise instantly, then C increases. Y causes no instant concentration jump, both reaction rates become faster, then C decreases. Z causes every gas concentration to rise instantly, then C increases further. Identify each disturbance as addition of A, increased temperature or increased pressure, and explain each equilibrium shift.
[6 marks]
Total for this question: 6
Higher Tier: For $\mathrm{P(g)+Q(g)\rightleftharpoons2R(g)}$ , the forward reaction is endothermic. P is added, the temperature is decreased and the pressure is increased at the same time. Predict the separate effect of each change on the relative amount of R, then evaluate whether the final change in R can be determined.
[5 marks]
Total for this question: 5

4.6.2.5 · The effect of changing concentration (HT only)

Explanation

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.

Tier 1 · Easy

A reactant is added to a mixture at equilibrium. State the direction in which the equilibrium shifts.
[1 mark]
Total for this question: 1
State whether an imposed concentration change is completely cancelled when a new equilibrium is reached.
[1 mark]
Total for this question: 1

Tier 2 · Standard

For $\mathrm{D+E\rightleftharpoons F}$ , some F is continuously removed from an equilibrium mixture. Explain the effect on the relative amount of F that is formed.
[3 marks]
Total for this question: 3
For $\mathrm{A_2+B_2\rightleftharpoons2AB}$ , extra AB is added to an equilibrium mixture. Predict the shift and describe the changes in the relative amounts of $\mathrm{A_2}$ and $\mathrm{B_2}$ as a new equilibrium is reached.
[4 marks]
Total for this question: 4
For $\mathrm{H_2(g)+I_2(g)\rightleftharpoons2HI(g)}$ , some $\mathrm{I_2}$ is removed from an equilibrium mixture while temperature is unchanged. Predict the equilibrium shift and the changes in the relative amounts of $\mathrm{H_2}$ and $\mathrm{HI}$ .
[3 marks]
Total for this question: 3

Tier 3 · Hard

For $\mathrm{N_2+3H_2\rightleftharpoons2NH_3}$ at equilibrium, extra hydrogen is added while the temperature is kept constant. Predict the effect of this concentration change on the amounts of all three substances as a new equilibrium is reached.
[4 marks]
Total for this question: 4
For $\mathrm{A_2+B_2\rightleftharpoons2AB}$ , equilibrium concentrations are initially $\mathrm{A_2}=0.60$ , $\mathrm{B_2}=0.50$ and $\mathrm{AB}=0.40\,\text{mol dm}^{-3}$ . Immediately after one substance is added they are $\mathrm{A_2}=1.10$ , $\mathrm{B_2}=0.50$ and $\mathrm{AB}=0.40\,\text{mol dm}^{-3}$ . At the new equilibrium they are $\mathrm{A_2}=0.82$ , $\mathrm{B_2}=0.22$ and $\mathrm{AB}=0.96\,\text{mol dm}^{-3}$ . Identify the added substance, deduce the direction of shift, and use the data to show that the imposed change is only partly counteracted.
[5 marks]
Total for this question: 5
Two sealed vessels contain $\mathrm{H_2(g)+I_2(g)\rightleftharpoons2HI(g)}$ at equilibrium. Extra $\mathrm{H_2}$ is added to vessel 1. Some $\mathrm{HI}$ is removed from vessel 2. Compare the equilibrium shifts and describe how the relative amount of $\mathrm{HI}$ changes as each new equilibrium is established. Explain both responses.
[5 marks]
Total for this question: 5
Higher Tier: In $\mathrm{2P+Q\rightleftharpoons R}$ , some P is removed and extra R is added at the same time. Predict the equilibrium shift and the changes in the relative amounts of P, Q and R as a new equilibrium is established. Explain how both disturbances support your prediction.
[5 marks]
Total for this question: 5
Higher Tier: A closed system containing $\mathrm{A+B\rightleftharpoons C+D}$ is allowed to reach equilibrium. Some D is removed and the system reaches a second equilibrium. Extra C is then added and a third equilibrium is reached. Describe the direction of each shift and the changes in all four relative amounts during each shift. Evaluate whether the final mixture must match the original mixture.
[6 marks]
Total for this question: 6

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

Explanation

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.

Tier 1 · Easy

The forward reaction is endothermic. State the effect of increasing temperature on the relative amount of products at equilibrium.
[1 mark]
Total for this question: 1
State which reaction direction is favoured when the temperature of an equilibrium mixture is increased.
[1 mark]
Total for this question: 1

Tier 2 · Standard

The forward direction of $\mathrm{X\rightleftharpoons Y}$ is exothermic. Explain the effect of decreasing temperature on the equilibrium yield of Y.
[3 marks]
Total for this question: 3
In a reversible reaction, blue substance P forms yellow substance Q in the forward direction. Heating an equilibrium mixture increases the relative amount of blue P. Deduce the energy-change type of the forward reaction and explain your answer.
[3 marks]
Total for this question: 3
The forward direction of a reversible reaction is endothermic. Predict and explain the effect of decreasing temperature on the relative amount of product at equilibrium.
[3 marks]
Total for this question: 3

Tier 3 · Hard

For the same equilibrium, product yields are $68\%$ at $300\,\text{K}$ , $49\%$ at $400\,\text{K}$ and $32\%$ at $500\,\text{K}$ . Reaction time falls as temperature rises. Deduce the energy-change type of the forward reaction and explain why an industrial process might use an intermediate temperature.
[5 marks]
Total for this question: 5
Equilibrium 1 has an endothermic forward reaction. Equilibrium 2 has an exothermic forward reaction. Predict and explain how increasing temperature affects the equilibrium amount of product and the reaction rates in both systems.
[5 marks]
Total for this question: 5
The forward direction of an equilibrium in a sealed, fixed-volume system is exothermic. The temperature is raised from $350\,\text{K}$ to $450\,\text{K}$ and, after a new equilibrium is reached, returned to $350\,\text{K}$ . Predict the equilibrium shift and relative amount of product after each temperature change. Explain the complete cycle.
[5 marks]
Total for this question: 5
Higher Tier: The forward direction of $\mathrm{X\rightleftharpoons Y}$ is endothermic. After an unknown change, both reaction rates become faster immediately. As a new equilibrium is established, the relative amount of Y increases. Identify the change and explain why the rate and equilibrium observations together rule out adding a catalyst.
[4 marks]
Total for this question: 4
Higher Tier: Each batch has a maximum possible product mass of $50\,\text{kg}$ , and reset time between batches is negligible. At $350\,\text{K}$ , equilibrium yield is $80\%$ and a batch takes $100$ minutes. At $450\,\text{K}$ , equilibrium yield is $55\%$ and a batch takes $25$ minutes. Calculate the product made at each temperature during $200$ minutes, identify the energy-change type of the forward reaction, and evaluate the temperature choice.
[6 marks]
Total for this question: 6

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

Explanation

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.

Tier 1 · Easy

A gaseous equilibrium has three gas molecules on the left of its equation and one on the right. State the direction of shift when pressure is increased.
[1 mark]
Total for this question: 1
State which substances are counted when predicting how pressure changes an equilibrium position.
[1 mark]
Total for this question: 1

Tier 2 · Standard

For $\mathrm{N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)}$ , predict and explain the effect of increasing pressure on the equilibrium yield of ammonia.
[3 marks]
Total for this question: 3
For $\mathrm{CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)}$ , predict and explain the effect of increasing pressure on the equilibrium amount of calcium carbonate.
[3 marks]
Total for this question: 3
For $\mathrm{2NO_2(g)\rightleftharpoons N_2O_4(g)}$ , the volume of the container is decreased at constant temperature, increasing the pressure. Predict and explain the effect on the relative amount of $\mathrm{N_2O_4}$ .
[3 marks]
Total for this question: 3

Tier 3 · Hard

Pressure is decreased for each equilibrium: (1) $\mathrm{2SO_2(g)+O_2(g)\rightleftharpoons2SO_3(g)}$ ; (2) $\mathrm{H_2(g)+I_2(g)\rightleftharpoons2HI(g)}$ . Predict the effect on the product amount in each case and justify both predictions.
[5 marks]
Total for this question: 5
The equilibrium $\mathrm{2XY(g)+Z_2(g)\rightleftharpoons2XYZ(g)}$ is also written with every coefficient doubled: $\mathrm{4XY(g)+2Z_2(g)\rightleftharpoons4XYZ(g)}$ . A student claims that the doubled equation changes the effect of increasing pressure. Evaluate the claim and predict the pressure effect.
[5 marks]
Total for this question: 5
For $\mathrm{NH_4Cl(s)\rightleftharpoons NH_3(g)+HCl(g)}$ , the pressure is decreased at constant temperature. Predict the equilibrium shift and the changes in the relative amounts of $\mathrm{NH_4Cl}$ , $\mathrm{NH_3}$ and $\mathrm{HCl}$ . Explain which substances are counted.
[5 marks]
Total for this question: 5
Higher Tier: For $\mathrm{A(g)+2B(g)\rightleftharpoons2C(g)}$ , an equilibrium mixture contains $40$ A particles, $70$ B particles and $30$ C particles. The pressure is increased at constant temperature, and the new equilibrium contains $46$ C particles. Predict the shift and calculate the new numbers of A and B particles.
[6 marks]
Total for this question: 6
Higher Tier: The equilibrium $\mathrm{2X(g)+Y(g)\rightleftharpoons aZ(g)}$ gives a $64\%$ equilibrium yield of Z at pressures of $1$ , $3$ and $5$ arbitrary units. The times taken to reach equilibrium are $90$ , $42$ and $28$ minutes respectively. Determine the value of a and evaluate the statement, 'Pressure has no effect on this reaction.'
[5 marks]
Total for this question: 5

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

4.6.1.1 · Calculating rates of reactions

Tier 1 · Easy

Mark scheme for 4.6.1.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	$1.8\,\text{cm}^3\text{ s}^{-1}$	Use $\text{mean rate}=\dfrac{\text{quantity formed}}{\text{time}}=\dfrac{72}{40}=1.8\,\text{cm}^3\text{ s}^{-1}$ .	2
Total Question 1			2
02.1	$2.0\,\text{cm}^3\text{ s}^{-1}$	The volume change is $66-18=48\,\text{cm}^3$ and the time interval is $39-15=24\,\text{s}$ . The mean rate is $48/24=2.0\,\text{cm}^3\text{ s}^{-1}$ .	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.6.1.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$0.024\,\text{g s}^{-1}$	The mass lost is $83.40-82.56=0.84\,\text{g}$ . Therefore the mean rate is $0.84/35=0.024\,\text{g s}^{-1}$ .	3
Total Question 1			3
02.1	$1.4\,\text{cm}^3\text{ s}^{-1}$	Convert the time: $2.0\times60=120\,\text{s}$ . The mean rate is $168/120=1.4\,\text{cm}^3\text{ s}^{-1}$ .	3
Total Question 2			3
03.1	$14.4\,\text{g}$	The mean rate is $5.4/45=0.12\,\text{g s}^{-1}$ . Convert the time: $2.0\times60=120\,\text{s}$ . The mass used is $0.12\times120=14.4\,\text{g}$ .	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.6.1.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	R: $1.2\,\text{cm}^3\text{ s}^{-1}$ S: $1.8\,\text{cm}^3\text{ s}^{-1}$ S is $50\%$ faster than R.	For R, $84/70=1.2\,\text{cm}^3\text{ s}^{-1}$ . For S, $99/55=1.8\,\text{cm}^3\text{ s}^{-1}$ . The increase relative to R is $1.8-1.2=0.6$ , so the percentage increase is $(0.6/1.2)\times100=50\%$ .	5
Total Question 1			5
02.1	$1.5\,\text{cm}^3\text{ s}^{-1}$ from $20$ to $50\,\text{s}$ $0.75\,\text{cm}^3\text{ s}^{-1}$ from $50$ to $90\,\text{s}$ The mean rate decreases by $50\%$ .	The first rate is $(69-24)/(50-20)=45/30=1.5\,\text{cm}^3\text{ s}^{-1}$ . The second is $(99-69)/(90-50)=30/40=0.75\,\text{cm}^3\text{ s}^{-1}$ . The decrease is $1.5-0.75=0.75$ , so the percentage decrease is $(0.75/1.5)\times100=50\%$ .	5
Total Question 2			5
03.1	$0.055\,\text{g s}^{-1}$ The reaction finishes by $85\,\text{s}$ . The mass used in successive $30\,\text{s}$ intervals falls from $2.4\,\text{g}$ to $0.9\,\text{g}$ and then to zero.	Between $25$ and $85\,\text{s}$ , the mass used is $9.5-6.2=3.3\,\text{g}$ over $60\,\text{s}$ , so the mean rate is $3.3/60=0.055\,\text{g s}^{-1}$ . The unchanged mass from $85$ to $115\,\text{s}$ shows that the reaction has finished by $85\,\text{s}$ . From $25$ to $55\,\text{s}$ , $2.4\,\text{g}$ is used; from $55$ to $85\,\text{s}$ , only $0.9\,\text{g}$ is used, so the rate decreases.	5
Total Question 3			5
04.1	$0.80\,\text{g}$ is lost in the first $25\,\text{s}$ . The mass at $25\,\text{s}$ is $73.82\,\text{g}$ . $0.70\,\text{g}$ is lost during the next $50\,\text{s}$ . The second mean rate is $0.014\,\text{g s}^{-1}$ . The first rate is about $2.3$ times the second rate.	The first mass loss is $0.032\times25=0.80\,\text{g}$ , so the balance reads $74.62-0.80=73.82\,\text{g}$ . The later loss is $73.82-73.12=0.70\,\text{g}$ , giving $0.70/50=0.014\,\text{g s}^{-1}$ . The rate factor is $0.032/0.014=2.2857\ldots$ , which is about $2.3$ .	5
Total Question 4			5
05.1	The first-stage mean rate is $2.1\,\text{cm}^3\text{ s}^{-1}$ . The second-stage mean rate is $1.35\,\text{cm}^3\text{ s}^{-1}$ . The third-stage mean rate is $0.40\,\text{cm}^3\text{ s}^{-1}$ . The first stage has the greatest mean rate. The graph becomes progressively less steep as the rate decreases.	The three gas volumes are $0.35\times180=63\,\text{cm}^3$ , $0.45\times180=81\,\text{cm}^3$ and $180-63-81=36\,\text{cm}^3$ . Dividing by the corresponding times gives $63/30=2.1$ , $81/60=1.35$ and $36/90=0.40\,\text{cm}^3\text{ s}^{-1}$ . A smaller rate is shown by a shallower gradient.	5
Total Question 5			5

4.6.1.2 · Factors which affect the rates of chemical reactions

Tier 1 · Easy

Mark scheme for 4.6.1.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	The reaction rate increases.	Crushing exposes a greater surface area of the same solid, which increases the reaction rate.	1
Total Question 1			1
02.1	The reaction rate decreases.	Lower temperature is a condition that reduces reaction rate when the other variables are held constant.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.6.1.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Use measured acid concentrations with equal volumes, equal masses and sizes of marble chips, and the same temperature. Collect the gas in a gas syringe and record gas volume at regular times, repeating each concentration.	Name the independent variable as acid concentration and a quantitative dependent variable such as gas volume over time. Keep acid volume, marble mass and surface area, temperature and apparatus constant. Start timing on mixing, take readings at fixed intervals and repeat so a mean or anomalies can be considered.	4
Total Question 1			4
02.1	Use the same observer and fixed lighting or a light sensor so the end point is judged consistently. Repeat every concentration and calculate a mean so anomalous results have less effect.	The disappearing-cross end point is subjective, so standardising the observer and lighting, or using a light sensor, makes the end point more consistent. Repeats reveal anomalous values and allow a representative mean to be calculated.	4
Total Question 2			4
03.1	The mass of marble and the acid concentration both change as well as surface area. Use equal masses of marble. Use acid with the same concentration and volume in both tests.	A valid comparison changes only surface area. Here both the marble mass and acid concentration are additional independent variables. Keep the marble mass equal, use the same acid concentration and volume, and also keep temperature and apparatus unchanged.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.6.1.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	The time is inversely proportional to concentration for these data: doubling concentration halves the time, and tripling it divides the time by three. Judging disappearance is subjective, and a single end-point time does not show how rate changes during the reaction.	Check concentration multiplied by time: $20(162)=3240$ , $40(81)=3240$ and $60(54)=3240$ , so time is inversely proportional to concentration here. Valid limitations include subjective visual judgement, inconsistent lighting or observer, delay in mixing and timing, and recording only one end point rather than a continuous quantity-time curve.	5
Total Question 1			5
02.1	$92\,\text{s}$ is anomalous. The mean times are $75\,\text{s}$ and $37.5\,\text{s}$ . Doubling the concentration doubles the relative rate.	$92\,\text{s}$ is far from the other two results at $0.80\,\text{mol dm}^{-3}$ . The means are $(76+74+75)/3=75\,\text{s}$ and $(38+37)/2=37.5\,\text{s}$ . The relative rates are $1/75$ and $1/37.5=2/75$ , so the second is twice the first.	5
Total Question 2			5
03.1	Powdering gives a rate factor of $2.5$ . Doubling concentration gives a rate factor of $1.6$ . Raising temperature by $10\,^\circ\text{C}$ gives a rate factor of $1.4$ . Powdering has the greatest effect in these experiments.	Each experiment runs for the same time, so the gas-volume ratio is the rate ratio. For B, $50/20=2.5$ ; for C, $32/20=1.6$ ; and for D, $28/20=1.4$ . The largest factor is $2.5$ , produced by changing from chips to powder.	4
Total Question 3			4
04.1	The powder's initial mean rate is $3.6\,\text{cm}^3\text{ s}^{-1}$ . The small chips' initial mean rate is $2.4\,\text{cm}^3\text{ s}^{-1}$ . The large chips' initial mean rate is $1.2\,\text{cm}^3\text{ s}^{-1}$ . The exposed surface area decreases in the order powder, small chips, large chips. The repeat's final volume is $18\,\text{cm}^3$ below the common $90\,\text{cm}^3$ total. This suggests gas escaped or was not fully collected, so its measured volume is not comparable.	Divide each first- $15$ -second volume by $15$ : $54/15=3.6$ , $36/15=2.4$ and $18/15=1.2\,\text{cm}^3\text{ s}^{-1}$ . The rate order matches the expected surface-area order. Because the same mass of solid should give the same final gas volume under otherwise identical conditions, the repeat's $90-72=18\,\text{cm}^3$ shortfall indicates a collection problem and invalidates its rate comparison.	6
Total Question 4			6
05.1	At $20\,^\circ\text{C}$ , the rate factor is $45/25=1.8$ . At $30\,^\circ\text{C}$ , the rate factor is $36/20=1.8$ . The proportional effect of the catalyst is the same at both temperatures. Comparing only the numbers of seconds saved is misleading because the uncatalysed times differ. The student's claim is not supported by these results.	For a fixed gas volume, rate is proportional to $1/\text{time}$ . The catalysed-to-uncatalysed rate factors are therefore $45/25=1.8$ at $20\,^\circ\text{C}$ and $36/20=1.8$ at $30\,^\circ\text{C}$ . Equal factors show an equal proportional effect, even though the absolute time reductions are different.	5
Total Question 5			5

4.6.1.3 · Collision theory and activation energy

Tier 1 · Easy

Mark scheme for 4.6.1.3 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	The particles must collide, and the collision must have energy equal to or greater than the activation energy.	Collision alone is insufficient. Award one condition for a collision occurring and one for sufficient collision energy.	2
Total Question 1			2
02.1	The collision may have less energy than the activation energy.	Only collisions with at least the activation energy can lead to reaction, so a lower-energy collision is unsuccessful.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.6.1.3 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	The same number of gas particles occupies a smaller volume, so particles are closer together and collide more frequently. This produces more successful collisions per second.	Link the macroscopic change to particles: higher pressure at constant temperature means a greater particle concentration. Collision frequency therefore rises, so the number of collisions with sufficient energy per second rises and the reaction is faster.	3
Total Question 1			3
02.1	The particles move faster and collide more frequently. A greater proportion of collisions also has energy equal to or greater than the activation energy, so there are more successful collisions each second.	Link warming to greater particle kinetic energy. This raises collision frequency and increases the fraction of collisions that meet the activation-energy requirement; both changes increase successful collisions per second.	4
Total Question 2			4
03.1	Q has more reactant particles in a given volume. Collisions occur more frequently. There are more successful collisions per second, so the rate is greater.	The higher concentration places more reacting particles in each unit volume. This increases collision frequency and therefore the number of collisions with sufficient energy per second. The temperature is unchanged, so increased particle speed is not the explanation.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.6.1.3 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	$192\,\text{cm}^2$ for the $2.0\,\text{cm}$ cubes $384\,\text{cm}^2$ for the $1.0\,\text{cm}$ cubes The $1.0\,\text{cm}$ cubes react faster because their greater exposed area gives more frequent collisions at the solid surface.	A $2.0\,\text{cm}$ cube has volume $8.0\,\text{cm}^3$ , so there are $64/8=8$ cubes. Each has area $6(2.0^2)=24\,\text{cm}^2$ , giving $8(24)=192\,\text{cm}^2$ . There are $64$ cubes of side $1.0\,\text{cm}$ , each with area $6\,\text{cm}^2$ , giving $64(6)=384\,\text{cm}^2$ . The doubled exposed area permits more acid-particle collisions per second, increasing the rate.	6
Total Question 1			6
02.1	Experiment B has the greater initial rate. The powder has a larger exposed surface area, and the more concentrated acid contains more particles per unit volume. Both changes increase collision frequency at the limestone surface, giving more successful collisions per second.	The mass is unchanged, but powder exposes more limestone particles to the acid. Doubling acid concentration also places more acid particles in each unit volume. Each factor raises the frequency of collisions at the solid surface, so B has the higher initial rate.	5
Total Question 2			5
03.1	Change T may produce the larger increase in rate. Both changes increase collision frequency by the same measured amount. Increasing temperature also makes collisions more energetic. A greater proportion of collisions meet or exceed the activation energy.	The stated collision-frequency increase is the same for C and T. Increasing concentration does not, by itself, make collisions more energetic. Increasing temperature does, so a greater proportion of collisions have sufficient energy and the number of successful collisions per second may rise by more for T.	4
Total Question 3			4
04.1	$250$ collisions can react using the $60$ -energy-unit pathway. $750$ collisions can react using the $40$ -energy-unit pathway. The number of successful collisions increases by $500$ . The number of successful collisions increases by a factor of $3$ . The lower activation energy changes the energy threshold, not the energies of the particles.	Only the $250$ collisions in the highest band meet the $60$ -energy-unit threshold. At the lower threshold, the middle and highest bands both qualify, giving $500+250=750$ . The increase is $750-250=500$ and the factor is $750/250=3$ . The temperature is unchanged, so the collision-energy distribution itself has not been raised.	5
Total Question 4			5
05.1	Before warming there are $200$ successful collisions per second. After warming there are $600$ successful collisions per second. The successful-collision rate increases by a factor of $3$ . Warming increases the collision frequency. Warming also increases the proportion of collisions that meet or exceed the activation energy.	Initially, $2400/12=200$ collisions per second are successful. After warming, $3000/5=600$ per second are successful, so the factor is $600/200=3$ . The calculation separates the two temperature effects: more collisions occur each second and a larger fraction has sufficient energy.	5
Total Question 5			5

4.6.1.4 · Catalysts

Tier 1 · Easy

Mark scheme for 4.6.1.4 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Activation energy.	The defining energy change is a lower activation energy for the alternative pathway.	1
Total Question 1			1
02.1	An enzyme.	Enzymes are catalysts produced by living organisms and are described as biological catalysts.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.6.1.4 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	The activation energy is reduced by $36\,\text{kJ mol}^{-1}$ ; the overall energy change remains $-24\,\text{kJ mol}^{-1}$ .	The reduction is $79-43=36\,\text{kJ mol}^{-1}$ . A catalyst changes the pathway but not the reactant or product energy levels, so the overall change stays $-24\,\text{kJ mol}^{-1}$ .	3
Total Question 1			3
02.1	A greater proportion of particle collisions has enough energy to react, so there are more successful collisions per second and the rate increases.	At one temperature the distribution of particle energies is unchanged. Lowering the minimum required energy means more existing collisions meet that requirement, increasing the number of successful collisions each second.	3
Total Question 2			3
03.1	The claim is not justified. Different reactions can require different catalysts. K can be a catalyst for reaction 1 if it increases its rate and is not used up.	Catalysts are reaction-specific, so failing to speed up reaction 2 does not rule out catalytic action in reaction 1. Evidence that K speeds reaction 1 and is recovered unchanged would support identifying it as a catalyst for that reaction.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.6.1.4 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	X raises the initial mean rate from $0.70$ to $1.6\,\text{cm}^3\text{ s}^{-1}$ , does not change the final gas volume, and is recovered unchanged, so it increases rate without being used up.	Without X the first- $30$ -second mean rate is $21/30=0.70\,\text{cm}^3\text{ s}^{-1}$ ; with X it is $48/30=1.6\,\text{cm}^3\text{ s}^{-1}$ . The equal final volume shows that X changes how quickly product forms rather than the final quantity. Its unchanged mass shows it was not used up overall. Together these observations identify X as a catalyst.	5
Total Question 1			5
02.1	There are $160$ additional successful collisions per $1000$ . The number is multiplied by $3$ . The catalyst provides an alternative pathway with a lower activation energy.	The increase is $240-80=160$ successful collisions per $1000$ . The factor is $240/80=3$ . A lower-activation-energy pathway allows a greater proportion of the same collisions to react; it does not raise the temperature or particle energies.	5
Total Question 2			5
03.1	The catalysed profile starts at $35\,\text{kJ mol}^{-1}$ , peaks at $81\,\text{kJ mol}^{-1}$ and ends at $61\,\text{kJ mol}^{-1}$ . The overall energy change is $+26\,\text{kJ mol}^{-1}$ . The lower activation energy means a greater proportion of collisions have enough energy to react.	A catalyst does not change reactant or product energy, so the start and end remain $35$ and $61\,\text{kJ mol}^{-1}$ . The catalysed peak is $35+46=81\,\text{kJ mol}^{-1}$ . The overall energy change is $61-35=+26\,\text{kJ mol}^{-1}$ . A lower barrier increases the proportion of successful collisions.	5
Total Question 3			5
04.1	P is not supported as a catalyst because it does not increase the rate. Q increases the rate. Q is not shown to remain chemically unchanged because $0.40\,\text{g}$ is not recovered. R doubles the rate for the fixed change because it halves the time. R has the strongest evidence because it increases the rate and is recovered unchanged.	P satisfies the recovery test but not the rate test. Q shortens the time, but $1.50-1.10=0.40\,\text{g}$ is missing, so the data do not show that it is unused overall. R changes the time from $60$ to $30\,\text{s}$ , a rate factor of $60/30=2$ , and its full mass is recovered. R therefore meets both pieces of catalyst evidence.	5
Total Question 4			5
05.1	The overall equation is $\mathrm{A+B\rightarrow AB}$ . C is used in the first step. C is re-formed in the second step, so it is not used up overall. C is the catalyst. It provides an alternative pathway with a lower activation energy.	Add the two step equations and cancel AC because it is formed and then used. Cancel C because it is used and then regenerated. This leaves $\mathrm{A+B\rightarrow AB}$ . The regeneration of C shows that it is not consumed overall, while the two-step route represents an alternative pathway whose lower activation energy allows more successful collisions.	5
Total Question 5			5

4.6.2.1 · Reversible reactions

Tier 1 · Easy

Mark scheme for 4.6.2.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	The products can react to form the original reactants.	The essential idea is that the reaction can proceed in the reverse direction, turning products back into reactants.	1
Total Question 1			1
02.1	An original reactant reappears when the conditions are changed.	Re-forming an original reactant from the products is direct evidence that the reverse reaction can occur.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.6.2.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	L and M react to form J and K.	Read the equation from right to left for the reverse direction: the right-hand substances L and M are its reactants, and J and K are its products.	2
Total Question 1			2
02.1	Q forms R and S when heated, and the products R and S react to reform Q when cooled, so the reaction can proceed in both directions.	Identify the heated change as the forward reaction and the re-formation of Q on cooling as the reverse reaction.	2
Total Question 2			2
03.1	The U to V process is shown to be reversible. Changing the condition allows V to form the original substance U. The observations for X and Y do not show Y reforming X.	A reversible process is evidenced when products form the original reactants under changed conditions. V reforms U on cooling, whereas no reverse change from Y to X is observed.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.6.2.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Heating drives the hydrated salt towards anhydrous copper sulfate and water; adding water drives the products back to hydrated copper sulfate. Products reform the original reactant, so the reaction is reversible.	Identify the forward observation under heating, then the reverse observation when water is supplied. The recovery of the original blue hydrated substance from the products is the evidence of reversibility; the conditions are heating in one direction and addition of water in the other.	4
Total Question 1			4
02.1	The claim is incorrect because both products, M and N, are needed for the reverse reaction. Cooling M alone would not show that the products reform the reactant. L reappearing when N is passed over M is the evidence that the reverse reaction occurs.	In the forward change, L forms both M and N. The reverse change therefore requires M and N to form L, so simply cooling one product is not a valid test. Observing L reform from both products demonstrates reversibility.	4
Total Question 2			4
03.1	The claim is incorrect. Gas B is a product of the forward reaction and a reactant in the reverse reaction. Opening the tube removes B, so the complete reverse reaction cannot occur. Re-forming A in the sealed tube is evidence that the reaction is reversible.	The failed second reversal is caused by loss of B, not by irreversibility. Both products, C and B, must be available to reform A. The sealed-tube observation provides the valid evidence that the products can react to make the original reactant.	4
Total Question 3			4
04.1	The process can be represented as $\mathrm{R\rightleftharpoons S+T}$ . R is the reactant in the forward reaction. S and T are the reactants in the reverse reaction. Re-forming R from S and T shows that the reaction is reversible. The observations do not prove equilibrium because equal forward and reverse rates in a closed system are not shown.	Write the observed light-driven change from left to right and the dark change from right to left using a reversible arrow. The colour return shows that both products can form the original reactant. Reversibility alone is not evidence of equilibrium; evidence of a closed system and equal opposing rates would also be required.	5
Total Question 4			5
05.1	$96\%$ of A is recovered in the sealed tube. The sealed-tube result shows that products can reform A. Complete recovery is not required to demonstrate that a reverse reaction can occur. The open tube loses a gaseous product needed for the reverse reaction. The claim is incorrect; the sealed-tube observations provide evidence of reversibility.	The sealed-tube recovery is $(5.76/6.00)\times100=96\%$ . Reversibility means that products can form the original reactants, not that every particle must reverse. In the open tube, loss of the gaseous product prevents the complete set of products from reacting back, so its lower recovery does not disprove reversibility.	5
Total Question 5			5

4.6.2.2 · Energy changes and reversible reactions

Tier 1 · Easy

Mark scheme for 4.6.2.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Endothermic.	Reversing an exothermic process reverses the direction of energy transfer, so the reverse process is endothermic.	1
Total Question 1			1
02.1	The energy transfers have the same magnitude.	Reversing a reaction changes the direction of energy transfer but not the amount transferred.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.6.2.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$+67\,\text{kJ mol}^{-1}$	Transferring energy to the surroundings makes the forward change $-67\,\text{kJ mol}^{-1}$ . The reverse reaction transfers the same amount in the opposite direction, so its change is $+67\,\text{kJ mol}^{-1}$ .	2
Total Question 1			2
02.1	The forward direction is endothermic with an energy change of $+45\,\text{kJ}$ ; the reverse direction is exothermic with an energy change of $-45\,\text{kJ}$ .	Moving to a higher energy level requires $45\,\text{kJ}$ to be taken in, so the forward change is positive and endothermic. Reversing the direction releases the same amount, giving a negative, exothermic change.	3
Total Question 2			3
03.1	The forward direction is exothermic. The reverse direction is endothermic. The same amount of energy is transferred in opposite directions.	Warming the surroundings shows that the forward reaction transfers energy to them, so it is exothermic. The reverse reaction is endothermic and transfers the same quantity of energy from the surroundings.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.6.2.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Reverse activation energy $=183\,\text{kJ mol}^{-1}$ ; reverse overall energy change $=+38\,\text{kJ mol}^{-1}$ .	The products are $38\,\text{kJ mol}^{-1}$ below the reactants. The peak is $145\,\text{kJ mol}^{-1}$ above the reactants, so it is $145+38=183\,\text{kJ mol}^{-1}$ above the products. Reversing the reaction changes the sign of the overall change, giving $+38\,\text{kJ mol}^{-1}$ .	4
Total Question 1			4
02.1	Forward activation energy $=114\,\text{kJ}$ and reverse activation energy $=69\,\text{kJ}$ . The forward energy change is $+45\,\text{kJ}$ and the reverse energy change is $-45\,\text{kJ}$ .	Forwards, the peak is $206-92=114\,\text{kJ}$ above the reactants. In reverse it is $206-137=69\,\text{kJ}$ above the products. The product level is $137-92=45\,\text{kJ}$ higher, so the forward change is $+45\,\text{kJ}$ and the reverse change is $-45\,\text{kJ}$ .	5
Total Question 2			5
03.1	A net $72\,\text{kJ}$ is transferred to the surroundings.	The seven forward changes transfer $7\times24=168\,\text{kJ}$ to the surroundings. The four reverse changes transfer $4\times24=96\,\text{kJ}$ from the surroundings. The net transfer is $168-96=72\,\text{kJ}$ to the surroundings.	4
Total Question 3			4
04.1	The products are $33\,\text{kJ}$ above the reactants. The forward energy change is $+33\,\text{kJ}$ . The forward reaction is endothermic. The reverse energy change is $-33\,\text{kJ}$ and the reverse reaction is exothermic.	Both activation energies end at the same peak. Since the peak is $84\,\text{kJ}$ above the reactants but only $51\,\text{kJ}$ above the products, the products are $84-51=33\,\text{kJ}$ higher. The forward change is therefore $+33\,\text{kJ}$ and endothermic; reversing it gives $-33\,\text{kJ}$ and exothermic.	4
Total Question 4			4
05.1	Cooling the surroundings shows that the forward reaction is endothermic. The forward energy change is $+36\,\text{kJ}$ . The reverse reaction is exothermic with an energy change of $-36\,\text{kJ}$ . Student S is correct. The net transfer is $72\,\text{kJ}$ from the surroundings.	The forward change takes $36\,\text{kJ}$ from the surroundings, whereas one reverse change returns the same amount. Three forward changes take $3\times36=108\,\text{kJ}$ and one reverse change releases $36\,\text{kJ}$ . The net is $108-36=72\,\text{kJ}$ from the surroundings, consistent with S's opposite signs and equal magnitudes.	5
Total Question 5			5

4.6.2.3 · Equilibrium

Tier 1 · Easy

Mark scheme for 4.6.2.3 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	The forward and reverse reactions occur at the same rate.	Dynamic equilibrium is defined by equality of the forward and reverse rates.	1
Total Question 1			1
02.1	Reactants and products cannot escape from a closed system.	Keeping every reacting substance in the system allows both the forward and reverse reactions to continue.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.6.2.3 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Both reactions continue. Their rates are equal, so each substance is formed at the same rate as it is used and its concentration stays constant.	Correct the word 'stopped': equilibrium is dynamic. Then connect equal opposing rates to no net change in concentration.	3
Total Question 1			3
02.1	Equilibrium requires the forward and reverse reaction rates to be equal, not the amounts of reactant and product. Unequal amounts can therefore remain constant.	Distinguish the rate condition from the composition: equal opposing rates give constant amounts, but do not require those amounts to have the same value.	2
Total Question 2			2
03.1	The system is not at equilibrium. The forward rate is greater than the reverse rate. The amount of product increases overall at that instant.	At equilibrium the forward and reverse rates must be equal. Here $4.6>2.9$ , so product is formed faster than it is changed back into reactant and its amount increases overall.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.6.2.3 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Equilibrium is first reached at $40\,\text{s}$ . The equal, non-zero rates at $40$ and $60\,\text{s}$ show dynamic equilibrium continues with no net change.	Find the first row in which the two rates are equal: both are $2.8$ at $40\,\text{s}$ . Their remaining equal and non-zero at $60\,\text{s}$ shows that both directions continue at a shared rate rather than stopping.	4
Total Question 1			4
02.1	The sealed flask is a closed system, so product remains available for the reverse reaction. The forward and reverse rates can become equal, making concentrations constant. In the open flask, escaping product prevents the reverse reaction from balancing the forward reaction.	Use the closed-system condition first. Retaining both reactants and products permits equal non-zero opposing rates. Continuous product loss from the open flask disrupts that balance, so its composition keeps changing.	4
Total Question 2			4
03.1	$3600$ reaction events occur in each direction. The net change in the amounts is zero. Both reactions continue at equal rates, so the amounts remain constant.	In each direction, $240\times15=3600$ particles change. Equal numbers change in opposite directions, giving zero net change. The equilibrium is dynamic because both reactions continue rather than stopping, while their equal rates keep the amounts constant.	5
Total Question 3			5
04.1	$3200$ forward events occur in the first interval. $2000$ reverse events occur in the first interval. The net product increase is $1200$ particles. There are $3000$ product particles after the first interval. The second interval has equal, non-zero forward and reverse rates. The system is at dynamic equilibrium during the second interval, so there is no further net change.	Over $10\,\text{s}$ , the first interval contains $320\times10=3200$ forward events and $200\times10=2000$ reverse events. The net gain is $1200$ product particles, giving $1800+1200=3000$ . In the next interval the two non-zero rates are equal at $260$ events per second, so reactions continue but their effects cancel and the amount stays constant.	6
Total Question 4			6
05.1	System B is at dynamic equilibrium. Its forward and reverse rates are equal. Its rates are non-zero, so both reactions continue. System A is not dynamic because neither reaction is occurring. Constant concentrations alone are insufficient evidence of dynamic equilibrium.	Both systems satisfy the closed-system observation, but dynamic equilibrium also requires continuing forward and reverse reactions at equal rates. B has equal non-zero rates, whereas A is static. Therefore an unchanged composition by itself cannot distinguish equilibrium from a mixture in which no reaction is occurring.	5
Total Question 5			5

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

Tier 1 · Easy

Mark scheme for 4.6.2.4 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	The system responds so as to counteract the change.	The required principle is that the equilibrium shifts in the direction that opposes the imposed change.	1
Total Question 1			1
02.1	The relative amount of products increases.	A rightward shift favours the forward reaction, so the new equilibrium mixture contains relatively more products.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.6.2.4 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	The equilibrium shifts to the right. More R reacts to replace some of the removed P, counteracting the decrease in P concentration.	Identify the change as a decrease in product concentration. The counteracting response is the forward reaction, which forms more P, so the position moves right.	3
Total Question 1			3
02.1	The catalyst does not change the equilibrium position because it speeds up the forward and reverse reactions equally. It makes equilibrium re-establish more quickly after the disturbance.	A catalyst lowers the activation energy for both directions, so neither direction is favoured at equilibrium. Both rates respond faster after a change, reducing the time needed to reach the new equal-rate state.	3
Total Question 2			3
03.1	Concentration Temperature Pressure for gaseous systems	Le Chatelier's principle is used to predict responses to changes in concentration, temperature and, for equilibria involving gases, pressure.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.6.2.4 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Increasing B favours the forward reaction and more C, but increasing temperature favours the endothermic reverse reaction and less C. The effects oppose one another, and their relative sizes are not given.	Treat each change separately. Extra B is counteracted by consuming B, shifting right and increasing C. Extra thermal energy is counteracted by the endothermic reverse direction, shifting left and decreasing C. Because no data compare the magnitudes of these shifts, the net change cannot be decided.	4
Total Question 1			4
02.1	Increasing pressure raises the equilibrium amount of AB because the right side has fewer gas molecules. Decreasing temperature raises the equilibrium amount of AB because the exothermic forward direction counteracts the cooling. A catalyst does not change the equilibrium amount of AB because it speeds up both directions equally.	There are three gas molecules on the left and one on the right, so higher pressure shifts right. Cooling favours the energy-releasing, exothermic forward direction, also shifting right. A catalyst changes the speed at which equilibrium is reached but not its position.	6
Total Question 2			6
03.1	The equilibrium shifts to the right and the relative amount of product increases. The later equal rates show that a new equilibrium has been established. The data do not identify the changed condition because concentration, temperature or pressure could produce a shift when suitable reaction information is given.	Immediately after the change, the forward rate exceeds the reverse rate, so there is net product formation and a shift to the right. When both rates become $3.6$ units, equilibrium has been re-established. Rate data alone do not state whether concentration, temperature or pressure changed, so the particular disturbance cannot be identified.	5
Total Question 3			5
04.1	X is addition of A. X shifts equilibrium right to consume some added A, so C increases. Y is increased temperature. Y shifts equilibrium left because heating favours the endothermic reverse direction, so C decreases. Z is increased pressure. Z shifts equilibrium right because the right side has one gas molecule compared with three on the left.	An instantaneous change in A alone identifies addition of A, followed by the rightward counteracting response. Faster rates in both directions without an instantaneous concentration step identify heating; because the forward reaction is exothermic, heating shifts left. An instantaneous rise in every gas concentration identifies compression and increased pressure. Counting gaseous coefficients gives three on the left and one on the right, so the pressure response is a rightward shift.	6
Total Question 4			6
05.1	Adding P shifts equilibrium right and tends to increase R. Decreasing temperature favours the exothermic reverse direction and tends to decrease R. Increasing pressure causes no equilibrium shift because each side has two gas molecules. The concentration and temperature effects oppose one another. The final change in R cannot be determined without information about the sizes of the two opposing effects.	Extra P is counteracted by the forward reaction, increasing R. Cooling favours the energy-releasing reverse direction because the stated forward direction is endothermic, decreasing R. Pressure does not alter the position because the gaseous coefficients total two on each side. Since the two effective disturbances favour opposite directions, their net result cannot be found by counting changes.	5
Total Question 5			5

4.6.2.5 · The effect of changing concentration (HT only)

Tier 1 · Easy

Mark scheme for 4.6.2.5 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Towards the products.	The system counteracts the added reactant by consuming some of it in the forward reaction, so the position shifts towards products.	1
Total Question 1			1
02.1	No. The concentration change is only partly counteracted.	The equilibrium shifts to oppose a concentration change, but it does not restore every concentration to its original value.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.6.2.5 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	The equilibrium shifts to the right, so more D and E react and more F is formed to oppose its removal.	Removing F lowers a product concentration. The forward reaction replaces some F, so reactants are consumed and the equilibrium position shifts right.	3
Total Question 1			3
02.1	The equilibrium shifts to the left. Some added AB is consumed, so the relative amounts of $\mathrm{A_2}$ and $\mathrm{B_2}$ increase as the reverse reaction is favoured.	Adding product AB is counteracted by using some AB in the reverse reaction. This forms $\mathrm{A_2}$ and $\mathrm{B_2}$ , so both reactant amounts rise while a new equilibrium is established.	4
Total Question 2			4
03.1	The equilibrium shifts to the left. The relative amount of $\mathrm{H_2}$ increases. The relative amount of $\mathrm{HI}$ decreases.	Removing $\mathrm{I_2}$ lowers its concentration. The system counteracts this by favouring the reverse reaction, which forms $\mathrm{H_2}$ and $\mathrm{I_2}$ from $\mathrm{HI}$ . Therefore $\mathrm{H_2}$ increases and $\mathrm{HI}$ decreases as the new equilibrium is established.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.6.2.5 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	The equilibrium shifts right. Some of the added hydrogen and some nitrogen are consumed, while more ammonia is formed; the hydrogen increase is partly, not completely, opposed.	The imposed change is increased $\mathrm{H_2}$ concentration. The forward reaction consumes $\mathrm{H_2}$ , so it is favoured. Consequently $\mathrm{N_2}$ is also used and $\mathrm{NH_3}$ increases until the opposing rates are equal again. Do not claim that all added hydrogen disappears.	4
Total Question 1			4
02.1	$\mathrm{A_2}$ was added and the equilibrium shifted to the right. $\mathrm{A_2}$ and $\mathrm{B_2}$ each decrease by $0.28\,\text{mol dm}^{-3}$ while AB increases by $0.56\,\text{mol dm}^{-3}$ , matching the $1:1:2$ ratio. $\mathrm{A_2}$ settles at $0.82\,\text{mol dm}^{-3}$ , still above its original $0.60\,\text{mol dm}^{-3}$ , so the addition is only partly counteracted.	Only $\mathrm{A_2}$ changes instantaneously, from $0.60$ to $1.10\,\text{mol dm}^{-3}$ , so $\mathrm{A_2}$ was added. The later fall in $\mathrm{A_2}$ and $\mathrm{B_2}$ and rise in AB show a rightward shift. The changes are $-0.28$ , $-0.28$ and $+0.56\,\text{mol dm}^{-3}$ , consistent with the equation. The final $\mathrm{A_2}$ concentration remaining above its starting value proves that the response does not completely reverse the addition.	5
Total Question 2			5
03.1	Vessel 1 shifts right and the relative amount of $\mathrm{HI}$ increases. Vessel 2 also shifts right, so some of the removed $\mathrm{HI}$ is replaced. Each system responds in a direction that partly counteracts its imposed concentration change.	Adding $\mathrm{H_2}$ makes the forward reaction favourable because it uses $\mathrm{H_2}$ , so vessel 1 shifts right and forms more $\mathrm{HI}$ . Removing $\mathrm{HI}$ also favours the forward reaction because it forms $\mathrm{HI}$ , so vessel 2 shifts right and partly replaces the removed product. Neither imposed concentration change is completely cancelled.	5
Total Question 3			5
04.1	Removing P favours the reverse reaction to replace some P. Adding R also favours the reverse reaction to consume some added R. The equilibrium shifts to the left. The relative amounts of P and Q increase during the shift. The relative amount of R decreases from its value immediately after the addition.	The system counteracts removal of reactant P by forming P in the reverse direction. It counteracts addition of product R by consuming R in that same direction. Because both concentration changes favour the reverse reaction, the position shifts left, producing P and Q while using some R.	5
Total Question 4			5
05.1	Removing D causes a shift to the right. During the first shift, A and B decrease while C and D increase. Adding C causes a shift to the left. During the second shift, A and B increase while C and D decrease from their post-addition values. Each shift only partly counteracts its own concentration change. The final mixture need not match the original because different quantities were removed and added.	Removing product D favours the forward reaction, which consumes A and B and forms C and D. Adding product C later favours the reverse reaction, which forms A and B and consumes C and D. Le Chatelier's principle predicts directions, not complete cancellation, so without quantitative information there is no basis for claiming that the third equilibrium has the original composition.	6
Total Question 5			6

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

Tier 1 · Easy

Mark scheme for 4.6.2.6 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	The relative amount of products increases.	Heating favours the direction that takes in energy. Here that is the endothermic forward reaction, so the equilibrium shifts towards products.	1
Total Question 1			1
02.1	The endothermic direction.	The system counteracts heating by favouring the direction that takes in energy.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.6.2.6 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	The yield of Y increases because the equilibrium shifts in the exothermic forward direction, which releases energy and counteracts the cooling.	A temperature decrease favours the energy-releasing direction. Since the forward direction is exothermic, the position moves right and the relative amount of Y rises.	3
Total Question 1			3
02.1	The forward reaction is exothermic. Heating shifts equilibrium towards P, so the reverse direction is endothermic and takes in the added energy.	More blue P means the equilibrium shifts left when heated. Heating favours the endothermic direction, so the reverse direction is endothermic and the forward direction must be exothermic.	3
Total Question 2			3
03.1	The equilibrium shifts to the left. The relative amount of product decreases. Lower temperature favours the exothermic reverse direction.	If the forward reaction is endothermic, the reverse reaction is exothermic. Decreasing temperature favours the exothermic direction, so equilibrium shifts left and the relative amount of product falls.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.6.2.6 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	The forward reaction is exothermic because increasing temperature lowers the product yield. An intermediate temperature compromises between a higher equilibrium yield at low temperature and a faster production rate at high temperature.	The decreasing product yield as temperature rises shows that heating favours the reverse direction. The reverse direction is therefore endothermic and the forward direction exothermic. Low temperature favours products but gives a slow reaction; high temperature gives a faster reaction but a poorer equilibrium yield. An intermediate value balances these competing effects.	5
Total Question 1			5
02.1	In equilibrium 1, the product amount increases because heating favours the endothermic forward direction. In equilibrium 2, the product amount decreases because heating favours the endothermic reverse direction. The forward and reverse reaction rates increase in both systems because particles have more energy at higher temperature.	Apply Le Chatelier's principle separately to the equilibrium positions: added heat favours whichever direction is endothermic. Separately, higher temperature increases particle energy and successful-collision frequency, so both opposing reactions become faster even though their new equilibrium compositions change in opposite ways.	5
Total Question 2			5
03.1	Heating shifts equilibrium left and decreases the relative amount of product. Cooling back to $350\,\text{K}$ shifts equilibrium right and increases the relative amount of product. In the same sealed, fixed-volume system, returning to the original temperature restores the original equilibrium relative amounts.	Raising temperature favours the endothermic reverse direction, so the first shift is left and product decreases. Lowering temperature then favours the exothermic forward direction, so the shift is right and product increases. With the same sealed mixture and the original temperature restored, the equilibrium returns to its original composition.	5
Total Question 3			5
04.1	The temperature was increased. Higher temperature makes both forward and reverse reactions faster. Heating favours the endothermic forward direction, so the relative amount of Y increases. A catalyst would speed both directions but would not change the equilibrium relative amounts.	Both heating and a catalyst could make the two reactions faster, but only temperature changes the equilibrium position. Since the forward reaction is endothermic, heating shifts the position towards Y. The increase in Y therefore identifies a temperature rise and rules out catalyst addition as the complete explanation.	4
Total Question 4			4
05.1	At $350\,\text{K}$ , each batch makes $40\,\text{kg}$ and two batches make $80\,\text{kg}$ . At $450\,\text{K}$ , each batch makes $27.5\,\text{kg}$ . Eight batches at $450\,\text{K}$ make $220\,\text{kg}$ in total. The forward reaction is exothermic because increasing temperature lowers the equilibrium yield. $350\,\text{K}$ gives the greater yield per batch. $450\,\text{K}$ gives the greater output in the stated $200$ minutes, so the choice depends on whether yield per batch or output rate is prioritised.	At $350\,\text{K}$ , $0.80\times50=40\,\text{kg}$ is made per batch and $200/100=2$ batches give $80\,\text{kg}$ . At $450\,\text{K}$ , $0.55\times50=27.5\,\text{kg}$ per batch and $200/25=8$ batches give $220\,\text{kg}$ . The lower yield at higher temperature shows that heating favours the reverse direction, so the forward reaction is exothermic. The preferred operating temperature depends on the stated objective.	6
Total Question 5			6

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

Tier 1 · Easy

Mark scheme for 4.6.2.7 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	To the right.	Increasing pressure favours the side with fewer gas molecules. One on the right is fewer than three on the left.	1
Total Question 1			1
02.1	Only substances in the gaseous state are counted.	Pressure predictions compare the balanced coefficients of gases; solids and liquids are not included in the gas-molecule totals.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.6.2.7 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	The ammonia yield increases because the equilibrium shifts right, from four gas molecules to two, counteracting the pressure increase.	Count gaseous coefficients: $1+3=4$ on the left and $2$ on the right. Higher pressure favours the side with fewer gas molecules, so the position shifts right and produces relatively more ammonia.	3
Total Question 1			3
02.1	The equilibrium shifts to the left and the amount of calcium carbonate increases because the left side has no gas molecules while the right side has one.	Count gaseous species only: the solid substances do not contribute to the gas total. Higher pressure favours the left-hand side with fewer gas molecules, forming more $\mathrm{CaCO_3}$ .	3
Total Question 2			3
03.1	The equilibrium shifts to the right. There are two gas molecules on the left and one on the right. The relative amount of $\mathrm{N_2O_4}$ increases.	Decreasing volume increases pressure. The system shifts towards the side with fewer gas molecules: one molecule of $\mathrm{N_2O_4}$ rather than two molecules of $\mathrm{NO_2}$ . Therefore the relative amount of $\mathrm{N_2O_4}$ increases.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.6.2.7 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	For (1), the equilibrium shifts left and the relative amount of $\mathrm{SO_3}$ decreases. For (2), there is no shift because each side has two gas molecules.	A pressure decrease favours more gas molecules. In (1), the left has $2+1=3$ gas molecules and the right has $2$ , so the position shifts left and product amount falls. In (2), the left has $1+1=2$ and the right has $2$ ; because the gas totals are equal, changing pressure does not alter the equilibrium position.	5
Total Question 1			5
02.1	The claim is incorrect. The first equation has three gas molecules on the left and two on the right; the doubled equation has six on the left and four on the right. Both forms therefore predict a shift to the right when pressure increases because the product side has fewer gas molecules.	Add the gaseous coefficients in each representation: $2+1=3$ versus $1+1=2$ , or $4+2=6$ versus $2+2=4$ . Multiplying every coefficient by the same factor changes both totals but not which side has fewer gas molecules, so higher pressure favours the right in either form.	5
Total Question 2			5
03.1	The equilibrium shifts to the right. The relative amount of $\mathrm{NH_4Cl}$ decreases, while $\mathrm{NH_3}$ and $\mathrm{HCl}$ increase. Only gases are counted: there are no gas molecules on the left and two on the right; solid $\mathrm{NH_4Cl}$ is not counted.	Decreased pressure favours the side with more gas molecules. The solid on the left is excluded from the count, while the right has two gas molecules from $\mathrm{NH_3}+\mathrm{HCl}$ . A right shift uses $\mathrm{NH_4Cl}$ and forms more $\mathrm{NH_3}$ and $\mathrm{HCl}$ .	5
Total Question 3			5
04.1	There are three gas molecules on the left and two on the right. Increasing pressure shifts equilibrium to the right. The number of C particles increases by $16$ . An increase of $16$ C particles represents $8$ reaction changes as written. The new number of A particles is $32$ . The new number of B particles is $54$ .	The gaseous coefficients total $1+2=3$ on the left and $2$ on the right, so increased pressure favours the right. The C increase is $46-30=16$ particles. Each reaction change produces two C particles, so the extent is $16/2=8$ . This consumes $8$ A particles and $2\times8=16$ B particles, leaving $40-8=32$ A and $70-16=54$ B.	6
Total Question 4			6
05.1	The unchanged $64\%$ yield shows that pressure does not shift the equilibrium position. The left side has three gas molecules. Therefore $a=3$ , giving equal gas-molecule totals on both sides. Higher pressure reduces the time taken to reach equilibrium, so it increases the reaction rates. The statement is false overall: pressure does not affect the equilibrium position here, but it does affect how quickly equilibrium is reached.	A pressure change leaves equilibrium position unchanged when the gaseous coefficients have equal totals. The left-hand total is $2+1=3$ , so the data imply $a=3$ . The fall in time from $90$ to $42$ to $28$ minutes shows faster opposing reactions at higher pressure, even though their eventual equal-rate composition remains at $64\%$ Z.	5
Total Question 5			5