AQA GCSE Chemistry Chemistry Paper 2 (Higher), 2023: Question 7
10 marks · Standard Demand difficulty · Short Answer
Analyze the rate of decomposition of hydrogen peroxide catalysed by manganese dioxide, including explaining catalyst action, calculating reaction rate from a tangent, and sketching a concentration change curve.
Practise this questionQuestion
Question text
07 Manganese dioxide catalyses the decomposition of hydrogen peroxide solution.
Oxygen and water are produced.
07.1 Explain how a manganese dioxide catalyst increases the rate of decomposition of
hydrogen peroxide.
[2 marks]
A student investigated the rate of this reaction.
This is the method used.
1. Add 50 cm3 of 2.0 mol/dm3 hydrogen peroxide solution to a conical flask.
2. Add 1.0 g of manganese dioxide to the conical flask.
3. Place the conical flask on a balance and start a timer.
4. Record the total mass lost from the conical flask every 20 seconds for 180 seconds.
07.2 Explain why the mass of the conical flask and contents decreased.
[2 marks]
07.3 Figure 4 shows the results for 50 cm3 of 2.0 mol/dm3 hydrogen peroxide solution and
1.0 g of manganese dioxide.
A tangent to the line has been drawn at 75 seconds.
Figure 4
Determine the rate of reaction when the time was 75 seconds.
Give your answer to 2 significant figures.
[4 marks]
Rate (2 significant figures) = g/s
07.4 The results for 50 cm3 of 2.0 mol/dm3 hydrogen peroxide solution and 1.0 g of
manganese dioxide are shown again on Figure 5.
Figure 5
The student repeated the investigation using 50 cm3 of 1.0 mol/dm3 hydrogen
peroxide solution and 1.0 g of manganese dioxide.
Sketch the expected results for 1.0 mol/dm3 hydrogen peroxide solution on Figure 5.
[2 marks]
Mark scheme
Show the mark scheme
Question 7
AO /
Question Answers Extra information Mark
Spec. Ref.
07.1 (a catalyst) provides a different 1 AO1
pathway for the reaction
4.6.1.3
(which has a) lower activation 4.6.1.4
energy
AO /
Spec. Ref.
07.2 (oxygen is) a gas 1 AO2
4.3.1.3
(which) escaped from the flask 1
4.6.1.1
AO /
Spec. Ref.
07.3 correct value for x step and y 1 AO2
step from tangent
4.6.1.1
value for y step allow correct use of incorrectly 1
(rate =) determined x and/or y step
value for x step
correct calculation of rate 1
answer to 2 significant figures allow an answer correctly 1
rounded to 2 significant figures
from an incorrect calculation
which uses values determined
from the graph.
AO /
Spec. Ref.
07.4 line starting at 0,0 which is less 1 AO2
steep than existing line
4.3.4
4.6.1.2
which becomes level at 0.80 g allow a tolerance of ± ½ a small 1
square
Total Question 7 10
How to answer it
Rates of Reaction & Catalysts: Hydrogen Peroxide Decomposition
What this question tests
This question assesses your understanding of reaction rates, how catalysts function at a molecular level, and how to interpret experimental data. You will need to explain mass loss in open systems, calculate reaction rates using the gradient of a tangent, and sketch a rate curve showing the effect of changing reactant concentration.
Part 07.1: How Catalysts Work
Explain how a manganese dioxide catalyst increases the rate of decomposition of hydrogen peroxide. [2 marks]
Correct Answer
- Provides an alternative reaction pathway [1 mark]
- Which has a lower activation energy [1 mark]
Key Knowledge
A catalyst speeds up a chemical reaction without being used up itself. It does this by offering a different route for the reactants to turn into products, requiring less energy (lower activation energy, Eₐ) to break the initial bonds.
Exam Technique
Do not just write "it speeds up the reaction". The question asks how it increases the rate. You must mention both the alternative pathway and the lower activation energy to secure both marks.
Common Errors
Students often incorrectly state that catalysts "provide energy" or "lower the energy of the reactants". Catalysts only lower the activation energy barrier.
Part 07.2: Explaining Mass Loss
Explain why the mass of the conical flask and contents decreased. [2 marks]
Correct Answer
- Oxygen is produced, which is a gas [1 mark]
- The gas escapes from the flask [1 mark]
Key Knowledge
The reaction equation is:
2H₂O₂(aq) → 2H₂O(l) + O₂(g)
Because the flask is open, the gaseous product (O₂) escapes into the atmosphere, causing the total mass of the flask and its contents to decrease over time.
Common Errors
Simply stating "gas is made" or "a reaction happened" only gets 1 mark. You must explicitly state that the gas escapes from the flask to get the second mark.
Part 07.3: Calculating Rate from a Tangent
Determine the rate of reaction when the time was 75 seconds. Give your answer to 2 significant figures. [4 marks]
Step-by-Step Calculation
- Identify two points on the drawn tangent line:
Look at the ends of the tangent line drawn on Figure 4:
• Point 1 (start of tangent): x₁ = 25 s , y₁ = 1.24 g
• Point 2 (end of tangent): x₂ = 125 s , y₂ = 1.76 g [1 mark for reading values] - Calculate the change in y (Δy) and change in x (Δx):
• Δy = 1.76 - 1.24 = 0.52 g
• Δx = 125 - 25 = 100 s - Calculate the gradient (Rate):
Rate = Δy / Δx
Rate = 0.52 g / 100 s = 0.0052 g/s [1 mark for formula, 1 mark for calculation] - Round to 2 Significant Figures:
0.0052 is already 2 significant figures (leading zeros do not count!).
Rate = 0.0052 g/s [1 mark for 2 sig figs]
Exam Technique: Error Carried Forward (ECF)
Even if you misread the graph coordinates, you can still get 3 out of 4 marks if you show your working clearly: divide your y-step by your x-step, calculate the final value, and round your incorrect final value to 2 significant figures!
Calculation Traps
Do not just divide the y-value at 75 seconds by 75! You must use the gradient of the tangent line, not the curve itself.
Part 07.4: Sketching a Rate Curve
Sketch the expected results for 1.0 mol/dm³ hydrogen peroxide solution on Figure 5. [2 marks]
How to Draw the Curve
- Start at (0,0): The line must begin exactly at the origin.
- Shallower gradient: The curve must be less steep than the original line at all points because the concentration is lower (slower rate of reaction). [1 mark]
- Level off at 0.80 g: The line must become perfectly horizontal exactly at 0.80 g (with a tolerance of ± half a small square). [1 mark]
The Science Behind the Sketch
Why is it less steep?
Lower concentration ( 1.0 mol/dm³ vs 2.0 mol/dm³ ) means fewer reactant particles per unit volume, leading to less frequent successful collisions and a slower rate.
Why does it level off at 0.80 g?
The concentration is exactly halved, while the volume remains the same. This means there are exactly half as many moles of reactant, producing exactly half the amount of oxygen gas ( 1.60 g / 2 = 0.80 g ).
Examiner Tip: Precision Drawing
Use a sharp pencil. Make sure your line does not go above the original curve at any point, and ensure it goes completely flat (horizontal) at exactly 0.80 g (which is 5 small grid squares up from 0.60 g on the y-axis).
Topics
Chemistry · C3: Quantitative Chemistry · C6: The Rate and Extent of Chemical Change
Question and mark scheme from the AQA GCSE Chemistry examination, Chemistry Paper 2 (Higher), 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.