AQA GCSE Chemistry Chemistry Paper 2 (Higher), June 2025: Question 8

13 marks · High Demand difficulty · Extended Answer

Investigate the rate of decomposition of hydrogen peroxide with different catalysts, including reaction profile analysis, tangent calculation of rate at 60 seconds in standard form, and effect of temperature.

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Question

Question text

08 Hydrogen peroxide decomposes into water and oxygen.

The equation for the reaction is:

2 H2O2(aq) → 2 H2O(I) + O2(g)

A student investigated the effect of catalyst A and of catalyst B on the rate of

this reaction.

Figure 6 shows the apparatus.

Figure 6

This is the method used.

1. Put the conical flask on the balance.

2. Add 50 cm3 of 1.5 mol/dm3 hydrogen peroxide solution to the conical flask.

3. Add 1.00 g of catalyst A to the conical flask.

4. Start a timer.

5. Record the loss in mass of the conical flask and contents every 30 seconds

for 3 minutes.

6. Repeat steps 1 to 5 using catalyst B.

08.1 Explain why the conical flask and contents lost mass.

[2 marks]

08.2 Figure 7 shows the reaction profile for the decomposition of hydrogen peroxide using:

• catalyst A

*22* • catalyst B.

Figure 7

Explain why catalyst A and catalyst B give different rates of reaction for the

decomposition of hydrogen peroxide.

[3 marks]

08.3 Figure 8 shows the results for catalyst A.

Figure 8

Determine the rate of the reaction when the time was 60 seconds.

Give your answer in standard form.

[5 marks]

Rate (standard form) =25 g/s

08.4 Explain how the rate of decomposition of hydrogen peroxide depends on the

temperature of the solution.

[3 marks]

Mark scheme

Show the mark scheme

Question 8

AO /

Question Answers Extra information Mark

Spec. Ref.

08.1 allow oxygen for gas AO2

4.3.1.3

a gas was produced 1 4.6.1.4

(and the gas) escaped from the 1

flask

AO /

Spec. Ref.

08.2 catalysts provide a different 1 AO3

pathway for the reaction 4.6.1.3

4.6.1.4

(and) the activation energy allow converse 1

using B is lower

(so) the rate of reaction is higher allow converse 1

using B

allow (so) the reaction using B

has more particles with an

energy exceeding the activation

energy

AO /

Spec. Ref.

08.3 tangent at 60 s 1 AO2

4.6.1.1

value for x step and y step from allow a tolerance of ± ½ a small 1

tangent square

allow correct use of an incorrect

tangent

value for y step allow correct use of incorrectly 1

(rate =) determined x and / or y step

value for x step

from a drawn tangent

correct calculation of rate 1

answer in standard form allow an answer correctly given 1

in standard form from an

incorrect calculation which uses

values determined from the

graph

AO /

Spec. Ref.

08.4 allow converse AO1

4.6.1.2

4.6.1.3

the rate (of decomposition / 1

reaction) increases at higher

temperatures

(because) particles have more allow (because) particles move 1

energy faster

(so) collide more frequently allow (and) a greater proportion 1

collide with sufficient energy to

exceed the activation energy

Total Question 8 13

How to answer it

Rates of Reaction: Decomposition of Hydrogen Peroxide

📋 What this question tests

This 13-mark question examines key aspects of Rate and Extent of Chemical Change (AQA Topic 6):

Apparatus and mass change: Linking mass loss to gas escaping an open system.
Catalysts and reaction profiles: Explaining catalytic pathways, activation energy differences, and reaction rates.
Mathematical skills (graph work): Constructing a tangent at a specified time (t = 60 s), calculating the gradient, and reporting the rate in standard form.
Collision theory: Describing and explaining the effect of temperature on rate using particle kinetic energy and collision frequency.

Question 08.1 · 2 Marks

Explaining Mass Loss During the Reaction

Explain why the conical flask and contents lost mass.

✅ Model Answer

A gas (or oxygen) was produced. [1 mark]
The gas escaped from the flask. [1 mark]

Examiner note: You must state both that a gas forms AND that it escapes into the air. Stating only "oxygen was formed" earns 1 mark max.

❌ Common Errors

Missing the escape: Writing only "gas was given off" without mentioning it left/escaped the open flask.
Violating conservation of mass: Writing "matter was destroyed" or "liquid evaporated".
Vague references: Stating "hydrogen peroxide was used up" without linking it to the gas produced.

Question 08.2 · 3 Marks

Comparing Catalysts Using Reaction Profiles

Explain why catalyst A and catalyst B give different rates of reaction.

✅ Model Answer

Catalysts provide an alternative reaction pathway. [1 mark]
The activation energy using catalyst B is lower (or using catalyst A is higher). [1 mark]
Therefore, the rate of reaction is higher with catalyst B (or lower with catalyst A) / more particles have energy greater than or equal to the activation energy. [1 mark]

💡 Key Knowledge

Activation Energy (Eₐ): The minimum amount of energy that colliding particles must possess to react.
On Figure 7, the curve for Catalyst B has a lower peak from the reactant line than Catalyst A.
Lower peak = lower activation energy = faster reaction because a higher fraction of collisions succeed.

🧠 Exam Technique

Make sure to explicitly refer to both catalysts or clearly compare them. A common pitfall is giving the generic definition of a catalyst without using the comparative graph data to say which catalyst has the lower activation energy and which produces the faster rate.

Question 08.3 · 5 Marks

Determining Rate of Reaction from a Tangent

Determine the rate of the reaction when the time was 60 seconds. Give your answer in standard form.

📐 Step-by-Step Calculation

Step 1: Draw a tangent at t = 60 s [1 mark]
Place a ruler exactly touching the curve at t = 60 s. Ensure the angle balances the curve evenly on either side. Extend the line across the grid to make reading points easy.
Step 2: Read values for the triangle (Δy and Δx) [1 mark]
For a typical accurate tangent touching the curve at (60 s, 1.04 g):
Let the tangent extend between:
- Point 1: (0 s, 0.44 g)
- Point 2: (120 s, 1.28 g)
Change in y (mass lost) = 1.28 − 0.44 = 0.84 g
Change in x (time) = 120 − 0 = 120 s
Step 3: Calculate the gradient (Rate = Δy / Δx) [2 marks]
Rate = 0.84 g ÷ 120 s = 0.0070 g/s
Mark for substitution [1 mark] + Mark for calculated value [1 mark]
Step 4: Convert into standard form [1 mark]
0.0070 = 7.0 × 10⁻³ g/s (Acceptable range typically 6.0 × 10⁻³ to 8.5 × 10⁻³ depending on individual tangent line)

🧠 How Examiners Award These 5 Marks

Mark 1: A straight tangent drawn at exactly 60 s.
Mark 2: Correct coordinates read from the drawn tangent (tolerance ± ½ small square).
Mark 3: Correct formula set up: (y₂ − y₁) ÷ (x₂ − x₁).
Mark 4: Correct arithmetic calculation of the rate.
Mark 5: Final answer converted into correct standard form: A × 10ⁿ where 1 ≤ A < 10.

❌ Common Errors on Tangent Questions

Reading directly from the curve: Dividing 1.04 g by 60 s gives the average rate from 0 to 60 s, NOT the rate at 60 s. You score 0 for the gradient marks if no tangent is drawn!
Tiny tangent triangles: Drawing a 1 cm triangle leads to huge reading errors. Make your tangent line long!
Forgetting standard form: Writing 0.007 and stopping loses the final mark.

Question 08.4 · 3 Marks

Collision Theory: Temperature and Rate

Explain how the rate of decomposition of hydrogen peroxide depends on the temperature of the solution.

✅ Model Answer

The rate of reaction increases as temperature increases. [1 mark]
(Because) particles gain kinetic energy / move faster. [1 mark]
Therefore, particles collide more frequently (or a greater proportion of collisions have energy ≥ activation energy). [1 mark]

💡 Collision Theory Checklist

For full marks on any temperature question, always state:

Trend: Higher temperature = higher rate.
Energy: Particles have more kinetic energy (move faster).
Collisions: Collisions happen more frequently (per unit time) AND more collisions are successful (energy ≥ Eₐ).

❌ Examiner Warning: Key Word Traps

Saying "there are more collisions" loses the mark. You must say more frequent collisions or more collisions per second / per unit time.
Forgetting to state the initial direction: state clearly that rate increases as temperature rises.

Topics

Chemistry · C3: Quantitative Chemistry · C5: Energy Changes · C6: The Rate and Extent of Chemical Change

Question and mark scheme from the AQA GCSE Chemistry examination, Chemistry Paper 2 (Higher), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.