OCR A-Level Chemistry AS Depth in chemistry (02), June 2019: Question 4

13 marks · Medium difficulty · Structured Questions

Calculate the volume of hydrogen produced from the reaction of zinc with hydrochloric acid, plot experimental rate data on a grid, sketch a new rate curve, and explain rate factors using collision theory and catalysis.

Practise this question

Question

A multi-part question about the reaction between zinc and hydrochloric acid. Part (a) asks to calculate the volume of hydrogen produced at RTP from 50.0 cm3 of 0.100 mol dm-3 HCl and 0.200 g of zinc. Part (b) includes a diagram of a conical flask connected to a 100 cm3 gas syringe, a table of results (Table 4.1) measuring volume of gas against time, and a blank grid (Fig. 4.1) for labeling axes, plotting points, identifying an anomaly, and drawing a best-fit curve. Part (c) asks to sketch a new curve on Fig. 4.1 for a different concentration of acid. Part (d) asks to explain in terms of collision theory why the rate of reaction decreases over time. Part (e) asks to explain how a copper(II) sulfate catalyst increases the rate of reaction and why it is difficult to classify as homogeneous or heterogeneous in this reaction.
Question text

4 Zinc reacts with hydrochloric acid, HCl(aq), as shown in the following equation.

Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

A student investigates the rate of this reaction.

(a) The student plans to react 50.0 cm3 of 0.100 mol dm–3 HCl with 0.200 g of zinc (an excess).

Calculate the volume, in cm3, of hydrogen that should be produced at RTP.

volume = … cm3 [3]

(b) The student uses the apparatus in the diagram.

100 cm3 gas syringe

The student’s method is outlined below:

• Pour 50.0 cm3 of 0.100 mol dm–3 HCl into the conical flask.

• Add 0.200 g of zinc (an excess), and quickly attach the delivery tube and gas syringe.

• Measure the volume of gas collected every 20 seconds until the reaction stops.

The student obtains the results shown in Table 4.1.

Time / s 0 20 40 60 80 100 120 160 200

Volume of gas / cm3 0 16 27 37 39 50 53 58 58

Table 4.1

(i) On the graph paper in Fig. 4.1, label the x axis and plot the results in Table 4.1. [1]

(ii) Circle any anomalous results present in the graph you have drawn in Fig. 4.1. [1]

(iii) Draw a best-fit smooth curve on the graph you have drawn in11 Fig. 4.1. [1]

volume

3 40

/cm

Fig. 4.1

(c) The student repeats the experiment using:

• zinc with the same mass (0.200 g) and same surface area

• the same temperature and pressure

• 40.0 cm3 of 0.125 mol dm–3 HCl, instead of 50.0 cm3 of 0.100 mol dm–3 HCl.

On your graph in Fig. 4.1 sketch the curve you would expect in this experiment. [2]

(d) The graph shows that rate of reaction decreases over time.

Explain why, in terms of collision theory.

… [2]

(e) (i) The rate of the reaction between zinc and hydrochloric acid can be increased using a

solution of copper(II) sulfate as a catalyst.

Explain how a catalyst increases the rate of reaction.

… [2]

(ii) Why is it difficult to classify the solution of copper(II) sulfate as a homogeneous or

heterogeneous catalyst in this reaction?

… [1]

Mark scheme

Show the mark scheme The mark scheme provides numerical calculation steps for part (a) yielding 60 cm3, marking points for labeling the x-axis, plotting points, identifying the anomaly at 80 s, and drawing a smooth curve for part (b). Part (c) awards marks for a steeper initial slope, earlier leveling, and same final volume of 58 cm3. Part (d) requires mentioning decreasing acid concentration and fewer frequent collisions per second. Part (e) awards marks for lower activation energy providing an alternative route and a greater proportion of molecules having energy equal to or greater than the activation energy, and for part (ii) notes that reactants have different physical states.

AO

Question Answer Marks Guidance

element

4 (a) FIRST CHECK THE ANSWER ON ANSWER LINE 3 AO2.6 ALLOW 120 cm3 for 2 marks (no ÷ 2)

If answer = 60 cm3 award 3 marks ×3 ALLOW 240 cm3 for 2 marks (× 2 not ÷ 2)

---------------------------------------------------------------------------

50.0 –3

n(HCl) = × 0.100 = 5.00 × 10 (mol)

1000 IGNORE absence of trailing zeroes,

–3 e.g. for 0.100, ALLOW 0.1

5.00 × 10 –3

n(H ) = = 2.50 × 10 (mol)

Volume = 2.5(0) × 10–3 × 24.0 × 1000 = 60(.0) cm3 ALLOW ECF from n(HCl)

ALLOW ECF from n(HCl) and/or n(H2)

(b) (i) Use of graph paper 1 AO2.4 ALLOW Time (s) OR Time in s

linear numerical scale chosen for x axis ×1 ALLOW seconds OR sec OR secs

AND

Time / s added as label Tolerance ± 1 small square

AND

ALL points plotted correctly Point at 0,0 NOT required

ALLOW up to 3 plotting errors

(ii) Anomaly 1 AO2.4 ALLOW one more anomalous point NOT on the

point at 80 s circled ×1 curve drawn in (iii)

(iii) Line 1 AO3.1

smooth curve using all points

EXCEPT point at 80 s

(c) Initial slope is steeper 2 AO2.8

AND curve levels off at an earlier time ×2

3 Tolerance ± 1 small square

Same volume of gas produced (58 cm )

17 AO

element

4 (d) Rate 2 AO1.1 IGNORE amount of acid decreases, response

(Acid) concentration decreases. ×2 must imply a volume and NOT area,

e.g. fewer particles/molecules/ions

in same space /volume

Collisions

Fewer collisions per second

OR less frequent collisions ‘fewer collisions’ alone is not sufficient (no rate)

(e) (i) Catalyst lowers the activation energy (by providing an 2 AO1.2

alternative route) ×2

A greater proportion of molecules have more energy ALLOW ‘more’ for ‘greater proportion’

greater than/equal to activation energy

ALLOW more molecules have sufficient energy to

react

IGNORE (more) successful collisions

(ii) Reactants have different physical states 1 AO2.1 ALLOW idea that copper(II) sulfate solution is

homogeneous in relation to the acid, but

heterogeneous in relation to the zinc

Total 13

How to answer it

Rates of Reaction Study Guide

OCR AS Level Chemistry • Kinetics

What this question tests

This multi-part question covers core kinetics concepts: calculating gas volumes from stoichiometry and limiting reagents, graphical data processing (plotting, identifying anomalies, drawing curves), predicting kinetic trends from changing reactant volumes/concentrations, explaining collision theory qualitatively, and defining catalysts (including homogenous vs heterogeneous systems).

Part (a): Gas Volume Calculation

Calculate the volume of hydrogen produced at RTP. [3 marks]

📐 Step-by-Step Calculation

  1. Find moles of HCl:
    n(HCl) = (50.0 / 1000) × 0.100 = 5.00 × 10⁻³ mol
  2. Use stoichiometry (1:2 ratio):
    From Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g), 2 moles of HCl produce 1 mole of H₂.
    n(H₂) = (5.00 × 10⁻³ ) / 2 = 2.50 × 10⁻³ mol
  3. Calculate gas volume at RTP:
    Volume = moles × molar gas volume (24.0 dm³ mol⁻¹)
    Volume = 2.50 × 10⁻³ × 24.0 = 0.060 dm³ = 60.0 cm³

❌ Common Calculation Traps

  • Stoichiometry errors: Forgetting to divide by 2 for the 2:1 mole ratio between HCl and H₂. This often leads to 120 cm³ or 240 cm³.
  • Unit mix-ups: Forgetting to convert cm³ to dm³ when using concentrations or molar volumes (molar volume is 24.0 dm³ mol⁻¹, so multiplying by 1000 is required to get cm³).
Mark Scheme Breakdown: 1 mark for correct moles of HCl • 1 mark for correct moles of H₂ (accounting for stoichiometry) • 1 mark for final correct volume (60.0 cm³). Note: If final answer is 60 cm³, award all 3 marks immediately.

Part (b): Graphical Analysis

(i) Labelling & Plotting • (ii) Identifying Anomalies • (iii) Curve of Best Fit [3 marks total]

(i) Plotting & Axis Label: X-axis must be clearly labelled as Time / s (or equivalent with units) with a linear numerical scale. All points accurately plotted to within ±1 small square. [1 mark]
(ii) Anomaly: The point at 80 s (volume 39 cm³) must be identified/circled as it deviates from the smooth trend. [1 mark]
(iii) Curve: A smooth curve of best fit drawn through all points except the anomalous 80 s point. [1 mark]

🧠 Exam Technique for Graphs

  • Never force an anomaly onto your line of best fit—ignore it when drawing the curve.
  • Ensure smooth lines; do not use a ruler to connect-the-dots for kinetic curves.
  • Always check that axis labels include both the quantity and the slash separator followed by units (e.g., Time / s ).

Part (c): Sketching a Modified Experiment Curve

Sketch the curve for 40.0 cm³ of 0.125 mol dm⁻³ HCl instead of 50.0 cm³ of 0.100 mol dm⁻³ HCl. [2 marks]

💡 Determining the New Parameters

  • Calculate new moles of HCl: (40.0 / 1000) × 0.125 = 5.00 × 10⁻³ mol .
  • Since the moles of acid are identical to the original experiment, the maximum volume of gas produced remains the same ( 58 cm³ ).
  • Concentration is higher (0.125 vs 0.100 mol dm⁻³), meaning the initial rate of reaction is faster.

✅ What Examiners Look For

  • Initial slope: Must be steeper than the original curve (faster initial rate due to higher concentration). [Mark 1]
  • Levelling off: Must level off earlier, finishing at the exact same final height (58 cm³) because the total moles of limiting reagent are unchanged. [Mark 2]

Part (d): Collision Theory Explanation

Explain why the rate of reaction decreases over time. [2 marks]

✅ Full-Mark Answer

As the reaction proceeds, reactant particles are consumed, so the concentration of acid decreases. This results in fewer collisions per second (or less frequent successful collisions) between particles.

❌ Common Student Pitfalls

  • Vague wording: Saying "there are fewer particles" without specifying that concentration decreases or referencing the volume/space.
  • Incomplete statements: Stating "fewer collisions" without qualifying it with a time unit (must be "per second" or "less frequent").
  • Amount vs Concentration: Avoid stating "amount of acid decreases" unless strictly tied to volume/space context; OCR explicitly looks for concentration decreases.

Part (e): Catalysts and Classification

(i) How a catalyst increases rate • (ii) Why CuSO₄ solution is hard to classify [3 marks total]

💡 (i) Action of a Catalyst [2 marks]

  • Provides an alternative reaction pathway with a lower activation energy. [1 mark]
  • A greater proportion of molecules have energy greater than or equal to the new, lower activation energy. [1 mark]

💡 (ii) Catalyst Classification [1 mark]

Answer: The reactants have different physical states (Zinc is a solid, hydrochloric acid and copper(II) sulfate are in aqueous solution).

Examiner Guidance: CuSO₄ solution is homogenous relative to the aqueous acid, but heterogeneous relative to the solid zinc. This hybrid state makes strict classification difficult.

Topics

Module 2: Foundations in chemistry · Module 3: Periodic table and energy · Practical Activity Groups · Module 5: Physical chemistry and transition elements · PAG 9: Rates of reaction – continuous monitoring method · 5.3 Transition elements · 2.1 Atoms and reactions · 3.2 Physical chemistry

Question and mark scheme from the OCR A-Level Chemistry examination, AS Depth in chemistry (02), June 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.