OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2017: Question 19

10 marks · Hard difficulty · Structured Questions

Calculate the total volume of oxygen produced, suggest an alternative experimental method, determine the initial rate, order of reaction, and rate constant for the decomposition of hydrogen peroxide using a concentration-time graph.

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Question

Question text

19 Aqueous solutions of hydrogen peroxide, H2O2(aq), decompose as in the equation below.

2H2O2(aq) 2H2O(l) + O2(g)

A student investigates the decomposition of H2O2(aq) by measuring the volume of oxygen gas

produced over time. All gas volumes are measured at room temperature and pressure.

The student uses 25.0 cm3 of 2.30 mol dm–3 H O .

From the results, the student determines the concentration of H2O2(aq) at each time. The student

then plots a concentration–time graph.

2.5

2.0

1.5

[H2O2(aq)]

/ mol dm–3

1.0

0.5

0.0

0 500 1000 1500 2000 2500 3000

time/s

(a) Determine the total volume of oxygen, measured at room temperature and pressure, that the

student should be prepared to collect in this investigation.

Suggest apparatus that would allow this gas volume to be collected, indicating clearly the

scale of working.

… [3]

(b) Suggest a different experimental method that would allow the rate of this reaction to be

followed over time.

… [1]

(c)* Determine the initial rate of reaction, the order with respect to H2O2, and the rate constant.

Your answer must show full working on the graph and on the lines below.

… [6]

Mark scheme

Show the mark scheme

Question Answer Marks Guidance

19 (a) 25.0 3

n(H2O2) = 2.30 1000 OR = 0.0575 (mol)

0.0575 3 ALLOW 0.69(0) dm3

vol O2 = 24000 = 690 cm nd st

22 mark subsumes 1 mark

Collect in 1000 cm3/1 dm3 measuring cylinder

ALLOW 1000 cm3/1 dm3 syringe

Needs a name of actual apparatus, not just ‘container’

‘measuring cylinder’ without volume is insufficient

DO NOT ALLOW burette

For other possible apparatus, contact Team Leader

ALLOW volumes from 700–1000 cm3 but should be

realistic apparatus, e.g. 700, 750, 800, 850, 900, 950.

(b) Measure mass (loss) 1 ALLOW weight for mass

ALLOW take samples and titrate (remaining H2O2)

(c)* Please refer to the marking instructions on page 5 of 6 Indicative scientific points may include:

mark scheme for guidance on marking this question.

Initial rate

Level 3 (5–6 marks)

Tangent shown on graph as line at t = 0 s

A comprehensive conclusion using quantitative data –3

from the graph to correctly determine initial rate Gradient determined in range: 1.5 – 2.0 × 10

AND half lives/gradient with 1st order conclusion for 2.3 –3

e.g. = 1.77 × 10

H O AND determination of k. 1300

22 –3 –

initial rate as gradient value with units: mol dm s

There is a well-developed line of reasoning which is 1,

clear and logically structured. For other methods contact TL

Clear working for initial rate, half life/gradient and

order and k. Evidence for 1st order 2 methods

Units mostly correct throughout. 1st order clearly linked to half-life OR 2 gradients:

1. Half life

Level 2 (3–4 marks) Half life shown on graph

Attempts to describe all three scientific points but

Half life range 800–1000 s

explanations may be incomplete.

Two ‘constant’ half lives ±50 s

OR Explains two scientific points thoroughly with few

2. Two gradients two rates

omissions.

2 tangents shown on graph at c and c/2

There is a line of reasoning with some structure and Gradient at c/2 is half gradient at c

supported by some evidence. The scientific points are e.g. c = 2.3 mol dm–3, gradient = 1.6 × 10–3

supported by evidence from the graph. AND c = 1.15 mol dm–3, gradient = 0.8 × 10–3

Level 1 (1–2 marks) For chosen method, conclusion: H2O2 is 1st order

Reaches a simple conclusion using at least one piece

of quantitative data from the graph. Determination of k 2 methods

Attempts to calculate initial rate OR half life. k clearly linked to rate OR half-life:

rate 1.6 10–3

There is an attempt at a logical structure with a k = e.g. k = = 7 10–4

reasoned conclusion from the evidence. [H2O2] 2.3

s–1

0 marks No response worthy of credit. ln2 0.693 –4 –1

OR k = e.g. k = 950 = 7.3 10 s

t1/2

Total 10

How to answer it

Kinetics: Decomposition of Hydrogen Peroxide

What this question tests

This multi-part investigation tests your ability to link stoichiometry with molar gas volumes, choose appropriate laboratory apparatus for measuring gas evolution, suggest alternative kinetic monitoring methods, and interpret concentration-time graphs to determine initial rates, reaction orders, and rate constants (level-of-response extended calculation).

Part (a) — Stoichiometry & Gas Collection

Calculating Gas Volumes and Selecting Apparatus

📐 Step-by-Step Calculation

Moles of H₂O₂: n = c × V = 2.30 × (25.0 / 1000) = 0.0575 mol
Moles of O₂ produced: Use the stoichiometric ratio from 2H₂O₂ → 2H₂O + O₂ . Ratio is 2:1, so n(O₂) = 0.0575 / 2 = 0.02875 mol
Volume of O₂ at rtp: Multiply by molar gas volume ( 24000 cm³ mol⁻¹ ): 0.02875 × 24000 = 690 cm³

✅ Correct Answer & Apparatus

Total Volume: 690 cm³ (Accept 690 cm³ to 1000 cm³ depending on rounding, with 690 cm³ being exact).
Apparatus: 1000 cm³ (or 1 dm³ ) measuring cylinder OR gas syringe.

❌ Common Errors

Forgetting to divide by 2 for the reaction stoichiometry ratio.
Failing to name a specific, realistic volume capacity (e.g., just writing "syringe" or "measuring cylinder" without a capacity score 0 for the apparatus mark).
Suggesting a burette (which only holds up to 50 cm³ and is completely inappropriate for 690 cm³ of gas).

🧠 Exam Technique

The mark scheme notes that the 2nd calculation mark subsumes the 1st. Always state your molar calculation clearly before applying the molar gas volume constant ( 24000 cm³ mol⁻¹ ).

Part (b) — Alternative Methods

Suggesting Alternative Kinetic Tracking Methods

✅ Acceptable Answers

Measure mass loss over time (since oxygen gas escapes the open reaction vessel).
Take titration samples (quench aliquots at set times and titrate remaining H₂O₂ with acidified manganate(VII)).

💡 Key Knowledge

When a gas is produced and escapes into the atmosphere, tracking the total mass of the flask and contents on a 2-decimal place balance is a standard, highly accurate alternative to gas collection.

Part (c)* — Level-of-Response Graph Analysis

Initial Rate, Reaction Order, and Rate Constant Calculation

🧠 Level 3 (5–6 Marks) Requirements

Tangent drawn accurately at t = 0 s .
Initial rate calculated from the gradient with correct units ( mol dm⁻³ s⁻¹ ).
Order with respect to H₂O₂ deduced clearly using either successive half-lives or comparing gradients at different concentrations.
Rate constant ( k ) calculated accurately with correct units.

📐 Working Out Step-by-Step

Initial Rate: Draw a sharp tangent at t = 0 where concentration is 2.30 mol dm⁻³ . Calculating the gradient gives roughly 1.77 × 10⁻³ mol dm⁻³ s⁻¹ (Acceptable range: 1.5 – 2.0 × 10⁻³ ).
Order: Measuring half-lives ( t₁/₂ ) from 2.30 to 1.15 , and 1.15 to 0.575 shows a constant half-life of approximately 950 s . Constant half-life proves the reaction is first order with respect to H₂O₂.
Rate Constant ( k ): Using k = rate / [H₂O₂] or k = ln(2) / t₁/₂ yields k ≈ 7.3 × 10⁻⁴ s⁻¹ .

❌ Examiner Warnings & Common Pitfalls

Poor Tangents: Drawing a tiny triangle for the tangent gradient calculation severely impacts accuracy. Construct your triangle over a large span of the line.
Missing Units: Omitting or misstating units for initial rate ( mol dm⁻³ s⁻¹ ) or the rate constant ( s⁻¹ for 1st order) prevents access to top-tier marks.

Topics

Module 5: Physical chemistry and transition elements · Practical Activity Groups · 5.1 Rates, equilibrium and pH · PAG 9: Rates of reaction – continuous monitoring method

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.