OCR A-Level Chemistry AS Depth in chemistry (02), June 2024: Question 4
13 marks · Medium difficulty · Structured Questions
Explain the effect of a catalyst using a Boltzmann distribution, analyze gas volume data to determine reaction rate via a tangent, calculate reactant concentration from gas volume, and suggest control variables.
Practise this questionQuestion
Question text
4 Aqueous hydrogen peroxide, H2O2(aq), gradually decomposes to produce water and oxygen.
2H O (aq) 2H O(l) + O (g) ΔH = –196 kJ mol–1 Equation 4.1
22 2 2
The rate of decomposition of H2O2 can be increased by adding a small amount of
manganese(IV) oxide, MnO2, which acts as a catalyst.
(a) Explain, using a Boltzmann distribution model, why the rate of a reaction increases in the
presence of a catalyst.
You are provided with the axes below, which you should label.
… [4]
(b) A student investigates the rate of decomposition of H2O2, on addition of MnO2 catalyst, using a
gas syringe.
100 cm3 gas syringe
The student obtains the results shown in graph 4.1.
Graph 4.1
Volume
3 50
/ cm
0 20 40 60 80 100 120 140 160 180 200
Time / s
(i) On graph 4.1, draw a best-fit smooth curve of the results and circle the anomalous result. [2]
(ii) Use your graph to determine the rate of reaction, in cm3 s–1, at 50 s.
Show your working below and on the graph.
(iii) The student uses 50.0 cm3 of H O in the experiment. Equation 4.1 shows the reaction that
takes place.
2H2O2(aq) 2H2O(l) + O2(g) Equation 4.1
Calculate the concentration of H O , in mol dm–3, required to produce 90 cm3 of O (g) at RTP.
22 2
rate = … cm3 s–1 [2]
concentration = … mol dm–3 [3]
(c) A student plans to compare the rate of decomposition of H2O2 using different metal oxides as the
catalyst.
Suggest two variables which should be kept constant.
1 …
2 …
[2]
Mark scheme
Show the mark scheme
Question Answer Marks Guidance
4 (a) 4 ANNOTATE ANSWER WITH TICKS AND CROSSES
(Number of)
molecules
Energy Ecatalyst Ea
Correct drawing of Boltzmann distribution DO NOT ALLOW two curves
Curve starts within one small square of origin Confusion with effect of temperature
AND
not touching the x axis at high energy ✓ IGNORE a slight inflexion on the curve if less than one small
square
Axes labels
y: (number of) molecules/particles
AND DO NOT ALLOW ‘atoms’ as y-axis label
x: (kinetic) energy ✓ DO NOT ALLOW ‘enthalpy’ for x-axis label
Catalyst and activation energy
Catalyst provides a lower activation energy
OR
Ec shown below Ea on Boltzmann distribution ✓
Particles with E > Ea
More molecules/particles/collisions have energy above IF y axis labelled as ‘atoms’
activation energy (with catalyst) ALLOW ECF for atoms (instead of molecules/particles)
OR
more molecules have enough energy to react IGNORE (more) successful collisions
OR IGNORE response implying ‘more collisions’
greater area under curve above activation (confusion with effect of greater temperature)
energy ✓
(b) (i) Line 17 2
Smooth curve using all points ALLOW flexibility around point at 120 s
EXCEPT point at 100 s. ✓ Graph should be seen to level off on or very near to 90 cm3
Anomaly
Point at 100 s circled ✓
(ii) Tangent on graph 2 DO NOT ALLOW interpolation (taking a direct reading from
drawn at = 50 s (± 10 s) ✓ graph),
Answer must be derived from taking a gradient
Calculation of rate
= gradient (y/x) of tangent drawn ALLOW ECF from incorrectly drawn tangent or a straight line
of best fit
= 0.67 ± 0.2 cm3 s–1✓
(iii) FIRST CHECK ANSWER ON THE ANSWER LINE 3 ALLOW ECF
If answer = 0.15 (mol dm-3) award 3 marks
COMMON ERRORS
n(O2) = 90/24000 OR 0.09/24 OR 0.00375 (mol) ✓ For 2 marks:
0.075 missing x 2
n(H O ) = 2 0.00375 OR 0.0075 (mol) ✓ 150 missing a cm3 to dm3 conversion
c(H O ) = 0.0075 1000/50.0 = 0.15 mol dm–3 ✓ ---------------------------------------------------------------------------
ALLOW use of ideal gas equation using sensible p and T
for first mark. e.g.
from 100 kPa and 293 K
pV
n =
RT
pV (100 103) (90 10–6)
→ n = = = 0.00369… (mol)
RT 8.314 293
Examples of ‘sensible’ p and T:
p = 100 kPa, 101 kPa, 101,325 Pa
T = 273 – 298 K
(c) ANY two ✓✓ 2
• Amount of catalyst/metal oxide DO NOT ALLOW concentration/volume of catalyst/metal oxide
(allow same mass OR same moles)
• Temperature
• Volume of H2O2
• Concentration of H2O2
• Moles/amount of H2O2
• Pressure
• Surface area of catalyst
How to answer it
Decomposition of Hydrogen Peroxide: Rates, Catalysis & Calculations
Core AS Physical Chemistry & Practical Skills:
- Boltzmann Distribution: Accurate sketching of energy distribution curves, labelling axes correctly, and using the distribution to explain how a catalyst accelerates reaction rates.
- Graphical Analysis: Plotting smooth lines of best fit, identifying experimental anomalies, and measuring instantaneous rates via tangents.
- Stoichiometry & Gas Volumes: Combining molar gas volume ( 24.0 dm³ mol⁻¹ at RTP) with mole ratios and solution concentration formulas.
- Experimental Design: Selecting valid control variables when comparing solid heterogeneous catalysts.
Part (a) — Boltzmann Distribution Model & Catalysis
Sketching the distribution and explaining the effect of a catalyst [4 Marks]
✅ Mark Scheme Breakdown (4 Marks)
- M1 (Axes Labels): y-axis: (number of) molecules or particles; x-axis: (kinetic) energy.
- M2 (Curve Shape): Starts within one small square of the origin (0,0) , asymmetric curve with a peak to the left, approaches but never touches the x-axis at high energy.
- M3 (Catalyst Action): Catalyst provides an alternative pathway with a lower activation energy (or label an activation energy Ec or Ecat distinctly to the left of uncatalysed Ea ).
- M4 (Molecules with E ≥ Ea): A greater proportion/number of molecules have energy equal to or greater than the activation energy (or significantly greater area under the curve to the right of Ec ).
💡 Diagram Description for Full Marks
How to draw the Boltzmann curve:
- Start strictly at or within 1 small grid square of (0,0) .
- Rise steeply to a rounded maximum, then tail off gently to the right.
- Keep the high-energy tail above the x-axis (asymptotic; never touching or crossing it).
- Do NOT draw a second shifted curve! (A second curve is only used when temperature changes).
- Draw two vertical lines marking activation energies: uncatalysed Ea further to the right, and catalysed Ecat to the left of Ea .
❌ Common Student Errors
- Drawing a second curve: Students frequently confuse the effect of a catalyst with a temperature increase and shift the peak. Only one distribution curve should be drawn.
- Incorrect axis labels: Labelling the x-axis as "enthalpy" or "progress of reaction", or labelling the y-axis as "atoms" or "rate".
- Vague collision claims: Writing "more collisions occur per second". A catalyst increases the proportion of effective/successful collisions, not the overall collision frequency.
🧠 Exam Technique Tip
Always write: "A catalyst provides an alternative reaction pathway with lower activation energy. Therefore, a greater proportion of molecules have energy ≥ Ea, leading to more frequent successful collisions."
Part (b)(i) & (b)(ii) — Graph Work & Rate Determination
Line of best fit, identifying anomalies, and calculating initial/instantaneous rate
✅ Correct Responses
(b)(i) [2 Marks]:
- Curve (1 mark): A smooth single curve starting at (0,0) passing through all points except the anomalous point, levelling off smoothly at or very close to 90 cm³ .
- Anomaly (1 mark): Clearly circle the plotted point at t = 100 s (volume = 72 cm³).
(b)(ii) [2 Marks]:
- Tangent (1 mark): A straight ruler line drawn tangent to the curve precisely at t = 50 s (acceptable between 40 s and 60 s).
- Calculation (1 mark): Rate = Gradient of tangent ( Δy / Δx ).
Allowed range: 0.67 ± 0.20 cm³ s⁻¹ (i.e. 0.47 – 0.87 cm³ s⁻¹ ).
📐 How to Calculate the Gradient at t = 50 s
- Place a clear ruler at the curve at exactly t = 50 s so the ruler forms a balanced tangent (equal angles between ruler and curve on both sides).
- Draw the tangent line extended as long as possible across the grid to minimise reading error.
- Pick two easily readable coordinates on the tangent line, e.g.:
(x₁, y₁) = (0 s, 22 cm³)
(x₂, y₂) = (90 s, 82 cm³) - Calculate the slope:
Gradient = (82 - 22) / (90 - 0) = 60 / 90 = 0.67 cm³ s⁻¹
❌ Common Errors in Rate Graphs
- Direct reading (Interpolation): Reading the y-value at 50 s ( ~56 cm³ ) and dividing by 50 s ( 56 / 50 = 1.12 ). The mark scheme states: DO NOT ALLOW interpolation. Answer MUST be derived from a tangent gradient.
- Forcing curve through anomaly: Wobbly lines that kink downwards to hit the 100 s point lose the curve mark.
- Small tangents: Drawing tiny tangents leads to severe rounding and reading errors outside the ±0.20 window.
🧠 Exam Technique
Always draw a large right-angled triangle on your tangent line. Show Δy and Δx working directly on the question page to secure method marks even if your arithmetic slips.
Part (b)(iii) — Quantitative Stoichiometry
Calculating original concentration of H₂O₂ from gas volume produced [3 Marks]
📐 Step-by-Step Calculation
Reaction: 2H₂O₂(aq) → 2H₂O(l) + O₂(g)
Data provided: Volume of O₂ formed = 90 cm³ ; Volume of H₂O₂ solution used = 50.0 cm³ .
- Step 1: Calculate moles of O₂(g) produced
At RTP, 1 mole of gas occupies 24.0 dm³ = 24 000 cm³ .
n(O₂) = 90 / 24 000 = 0.00375 mol (or 3.75 × 10⁻³ mol )
[Award 1 Mark] - Step 2: Determine moles of H₂O₂ reacted
From Equation 4.1, the stoichiometric ratio is 2 mol H₂O₂ : 1 mol O₂ .
n(H₂O₂) = 2 × 0.00375 = 0.0075 mol (or 7.50 × 10⁻³ mol )
[Award 1 Mark] - Step 3: Calculate concentration of H₂O₂
Convert volume of solution to dm³: 50.0 cm³ = 50.0 / 1000 = 0.050 dm³ .
c(H₂O₂) = moles / volume = 0.0075 / 0.050 = 0.15 mol dm⁻³
[Award 1 Mark]
✅ Final Answer
0.15 mol dm⁻³
Full 3 marks awarded directly if final answer 0.15 is seen on the answer line.
❌ Common Calculation Traps
- Missing the 2:1 ratio: Getting 0.075 mol dm⁻³ because of forgetting to multiply moles of O₂ by 2. (Scores 2/3 marks).
- Unit conversion error: Failing to convert 50 cm³ into dm³ , yielding 150 mol dm⁻³ . (Scores 2/3 marks).
Part (c) — Fair Testing & Control Variables
Comparing different metal oxide catalysts [2 Marks]
✅ Accepted Control Variables (Any Two)
- Amount of catalyst / metal oxide (allow: same mass OR same moles)
- Surface area / particle size of the catalyst (e.g. all powdered)
- Temperature of the mixture
- Concentration of H₂O₂
- Volume of H₂O₂
- Pressure
❌ Rejected Answers (Examiner Guidance)
- DO NOT ALLOW: "Concentration of catalyst" or "Volume of catalyst". Manganese(IV) oxide is a solid heterogeneous catalyst, so referring to its concentration or solution volume is chemically invalid.
- Avoid vague answers such as "amount of H₂O₂" — be precise and specify volume or concentration.
Topics
Module 1: Development of practical skills in chemistry · Module 2: Foundations in chemistry · Module 3: Periodic table and energy · Practical Activity Groups · 1.1 Practical skills assessed in a written examination · 2.1 Atoms and reactions · 3.2 Physical chemistry · PAG 9: Rates of reaction – continuous monitoring method
Question and mark scheme from the OCR A-Level Chemistry examination, AS Depth in chemistry (02), June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.