Edexcel A-Level Chemistry AS Paper 2, November 2020: Question 6

13 marks · Medium difficulty · Practical Techniques and Data Analysis

Investigate the decomposition of aqueous hydrogen peroxide catalysed by manganese(IV) oxide by measuring oxygen gas volume, analysing rate data graphically, and describing a method to recover and test the heterogeneous catalyst.

Practise this question

Question

An exam question about the catalytic decomposition of hydrogen peroxide, showing a conical flask setup, a table of time versus oxygen volume, a rate graph with a curve of best fit up to 150 seconds, and several parts asking to draw a gas collection apparatus, calculate initial and tangent rates, sketch a higher-temperature rate curve, explain it using collision theory, and outline a method to recover and check the catalyst.
Question text

6 Aqueous hydrogen peroxide decomposes according to the following equation.

2H2O2(aq) → 2H2O(l) + O2(g)

The decomposition is catalysed by manganese(IV) oxide.

This can be investigated by measuring the volume of oxygen produced at various

times as the reaction proceeds. Part of the apparatus used in the experiment is shown.

The manganese(IV) oxide is placed in a small glass container, which is then tipped

over to start the reaction. A stop clock is started at the same time.

(a) Complete the diagram to show how the gas can be collected and its volume measured,

labelling the apparatus used.

(2)

aqueous

hydrogen peroxide

manganese(IV) oxide

(b) An experiment was carried out using 0.25g of manganese(IV) oxide granules and

50 cm3 of aqueous hydrogen peroxide of concentration 0.16 mol dm−3. The results

are shown in the table and plotted on a graph.

Time/s 0.0 20.0 30.0 50.0 60.0 80.0 100 120 150

Volume of O / cm3 0 51 68 85 88 91 92 92 92

*P62307A01928*

Volume

3 60

/cm

0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

Time/s

(i) The rate of reaction may be assumed to be approximately constant up to the

first volume measurement (20.0s in this experiment).

Use this approximation to calculate the initial rate of this reaction, giving the

units with your answer.

(1)

(ii) Draw a tangent at 40s on the graph on Page 20 and use it to calculate the rate

of reaction at this time.

(2)

*P62307A02028*

(iii) The experiment was repeated on a different day when the laboratory was

20°C warmer. The volume of oxygen was recorded for the same total time

of 150 s.

Draw the line that you would expect to obtain in this experiment. Assume the

pressure in the laboratory is the same. No calculation is required.

(2)

Volume

3 60

/cm

0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

Time/s

(iv)Explain, using collision theory, any differences between the line you have*P62307A02128*

drawn and the original line of best fit.

(2)

(c) Catalysts are not used up during a reaction. Manganese(IV) oxide acts as a

heterogeneous catalyst.

Describe in outline a method to show that the manganese(IV) oxide is not used up

in the decomposition of hydrogen peroxide and that it still functions as a catalyst.

(4)

(Total for Question 6 = 13 marks)

Mark scheme

Show the mark scheme The mark scheme for the hydrogen peroxide decomposition question, detailing expected answers and acceptable margins for gas collection diagrams, rate calculations from tangents and gradients, sketch curve shapes for temperature increases, collision theory explanations involving proportion of particles exceeding activation energy, and catalyst recovery steps including filtration, washing, drying, and reweighing.

How to answer it

Decomposition of Hydrogen Peroxide Rates & Catalysis

What this question tests

This question assesses core kinetics and practical chemistry skills: designing apparatus for gas collection, calculating rates from initial approximations and tangents, interpreting rate-concentration-temperature graphs using collision theory, and planning an experimental procedure to prove a heterogeneous catalyst is unconsumed.

Part (a)

Gas Collection Apparatus

✅ Correct Answer

Complete the diagram showing either:

  • Method 1: Delivery tube connected to an inverted measuring cylinder submerged in a water trough (scaled measuring cylinder must be labeled or shown clearly).
  • Method 2: Delivery tube connected directly to a gas syringe with a distinct, separate plunger.

❌ Common Errors

  • Leaving significant air gaps in joints or running delivery tubes through the side walls of water troughs.
  • Failing to use scaled glassware for the water collection method.
Total: 2 marks (1 mark for collection method, 1 mark for correct scaled glassware/distinct syringe).
Part (b)(i)

Initial Rate Calculation

📐 Step-by-Step Calculation

Using the first data point (20.0 s, 51 cm³ of O₂):

  1. Formula: Rate = Change in Volume / Change in Time
  2. Calculation: 51 / 20 = 2.55 cm³ s⁻¹ (or 2.5 cm³ s⁻¹ if using 50 cm³)
  3. Units: cm³ s⁻¹ (or cm³/s)

❌ Common Errors

  • Omitting or incorrectly formatting the unit (e.g., writing cm⁻³ s⁻¹ instead of cm³ s⁻¹).
  • Using points further down the curve instead of the first specified measurement at 20.0 seconds.
Total: 1 mark (calculation + correct units). Note: Ignore significant figures except 1 SF.
Part (b)(ii)

Calculating Rate at a Specific Time via Tangent

🧠 Exam Technique

  • Place a transparent ruler at exactly t = 40 s on the curve so equal amounts of curve show on either side of the touch point.
  • Choose large points far apart on your drawn tangent line to minimize reading errors when calculating the gradient (Δy / Δx).

✅ Expected Range

Correctly calculated gradients falling within the acceptable examiner range of 0.600 to 0.950 cm³ s⁻¹ score full marks.

Total: 2 marks (1 mark for drawing a suitable tangent, 1 mark for calculating the gradient).
Part (b)(iii) & (iv)

Temperature Effects & Collision Theory

💡 Key Knowledge (Graphs & Theory)
  • New Line (iii): Must rise more steeply (steeper initial gradient) and finish at a slightly higher final volume plateau (due to gas expansion at higher temperature), but staying below 100 cm³.
  • Collision Theory (iv): Higher temperature means particles have greater kinetic energy. This leads to a greater proportion of particles having energy greater than or equal to the activation energy (E ≥ Eₐ), increasing successful collision frequency.

❌ Common Errors

  • Drawing the new temperature curve starting at a different origin or crossing the original curve.
  • Just saying "more collisions happen" without qualifying that a *greater proportion* exceed activation energy.
Total: 4 marks combined ((iii) = 2 marks, (iv) = 2 marks).
Part (c)

Proving Catalyst Recovery

🧠 Experimental Outline

  1. Filter: Separate the solid manganese(IV) oxide from the mixture using filtration after the reaction stops.
  2. Wash & Dry: Rinse the recovered solid with distilled water and dry it (e.g., in an oven or desiccator).
  3. Weigh: Weigh the dry solid; it should equal the initial mass ( 0.25 g ).
  4. Reuse: Add the recovered solid to a fresh batch of hydrogen peroxide to confirm it still catalyses the reaction at the same rate.

❌ Common Errors

  • Attempting to measure the *volume* of the catalyst instead of mass.
  • Forgetting the crucial re-testing step to prove it "still functions as a catalyst."
Total: 4 marks (1 mark for filtration, 1 mark for washing/drying, 1 mark for re-weighing to 0.25 g, 1 mark for re-testing functionality).

Topics

Physical Chemistry · Core Practicals · Topic 9: Kinetics I

Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 2, November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.