AQA A-Level Chemistry Paper 3, June 2022: Question 4

10 marks · Medium difficulty · Practical Techniques & Data Analysis

Analyze the kinetics of the decomposition of hydrogen peroxide by plotting and interpreting concentration-time and rate-concentration graphs to determine reaction order.

Practise this question

Question

Question 04 shows the catalytic decomposition of hydrogen peroxide. Part 04.1 asks why the reaction is fastest at the start. Part 04.2 provides Figure 1, a concentration versus time curve for H2O2, and asks students to draw a tangent at 0.05 mol dm⁻³ to calculate the gradient. Part 04.3 gives a formula linking [H2O2] at time t to the volume of gas collected and asks to calculate [H2O2] when 20 cm³ of oxygen is collected. Part 04.4 gives a table of rate against [H2O2] to plot on a grid (Figure 2) and draw a line of best fit. Part 04.5 asks to deduce the order of reaction from Figure 2 and justify the answer.
Question text

04 Hydrogen peroxide solution decomposes to form water and oxygen.

2H2O2(aq) → 2H2O(l) + O2(g)

The reaction is catalysed by manganese(IV) oxide.

A student determines the order of this reaction with respect to hydrogen peroxide.

The student uses a continuous monitoring method in the experiment.

The student places hydrogen peroxide solution in a conical flask with the catalyst and

uses a gas syringe to collect the oxygen formed. The student records the volume of

oxygen every 10 seconds for 100 seconds.

04.1 Explain why the reaction is fastest at the start.

[2 marks]

04.2 The graph in Figure 1 shows how the concentration of hydrogen peroxide changes

with time in this experiment.

Figure 1

Tangents to the curve in Figure 1 can be used to determine rates of reaction.

Draw a tangent to the curve when the concentration of hydrogen peroxide solution is

0.05 mol dm–3

Use your tangent to calculate the gradient of the curve at this point.

*14* [2 marks]

16 –3 –1

Gradient mol dm s

04.3 The concentration of hydrogen peroxide solution at time t during the experiment can

be calculated using this expression.

Vmax −Vt

[H O22]t = [H O22]initial

Vmax

[H O ] = concentration of hydrogen peroxide solution at time t / mol dm–3

22 t

[H O ] = concentration of hydrogen peroxide solution at the start / mol dm–3

22 initial

V = total volume of oxygen gas collected during the whole experiment / cm3

max

V = volume of oxygen gas collected at time t / cm3

t

In this experiment,V = 100 cm3

max

Use Figure 1 and the expression to calculate [H O ] when 20 cm3 of oxygen has

22 t

been collected.

[2 marks]

[17H O ] –3

22 t mol dm

Table 5 shows data from a similar experiment.

Table 5

[H O ] / mol dm–3 0.02 0.03 0.05 0.07 0.09

Rate / mol dm–3 s–1 0.00049 0.00073 0.00124 0.00168 0.00219

04.4 Plot the data from Table 5 on the grid in Figure 2.

Draw a line of best fit.

[2 marks]

Figure 2

04.5 Use Figure 2 to determine the order of reaction with respect to H2O2

State how the graph shows this order.

[2 marks]

Order

How the graph shows this order

Mark scheme

Show the mark scheme Mark scheme for Question 04 outlines 10 marks across 5 parts: 04.1 awards marks for highest concentration of particles and more frequent successful collisions; 04.2 awards 1 mark for drawing a valid tangent at 0.05 mol dm⁻³ and 1 mark for gradient between -0.00120 and -0.00155; 04.3 awards 1 mark for reading initial concentration as 0.083 and 1 mark for calculating 0.0664; 04.4 awards 1 mark for plotting points correctly within half a square and 1 mark for a straight line of best fit; 04.5 awards 1 mark for first order and 1 mark for stating that it is a straight line through the origin.

Question Answers Additional Comments/Guidelines Mark

Alternative approach

M1 Higher/est concentration of / more H2O2 / particles / molecules M1 Lower/est concentration of / fewer particles 1

/ reactants / molecules / reactants as time goes on

04.1

M2 More frequent successful collisions M2 Less frequent successful collisions

(look for both ideas even if separated) 1

Ignore ‘chance’ / ‘probability’ (2 x AO1)

M1 Suitable tangent drawn M1 Tangent must be drawn with ruler and touch 1

line at 0.05 mol dm–3 (± 1 square) and not

cross the curve (if white seen between lines

it crosses) 1

04.2 (2 x AO2)

M2 –0.00120 to –0.00155 (mol dm–3 s–1) M2 Ignore units

Allow ecf from unsuitable tangent i.e if M1

not awarded

Ignore sign of gradient

M1 [H O ] = 0.083 mol dm−3 Allow 0.082 – 0.084 1

22 initial

M2 [H O ] = 0.0664 (mol dm−3) Allow 0.0656 – 0.0672 (scores 2/2) 1

04.3 2 2 t

2SF minimum–A-LEVEL CHEMISTRY – – (2 x AO3)

Allow ecf from M1 (M2 = M1 x 0.8)

M1 Points plotted 1 1

M1 allow each point (± square)

18 M2 best fit straight line drawn 1

M2 line should be drawn with a ruler and cover

0.0025 the five points given going within 1 square of (1 x AO2,

each point, no doubles no kinks. The line 1 x AO3)

does not need to be extended to the origin

0.002

Allow reasonable best fit line if points plotted

incorrectly

0.0015

04.4

0.001

0.0005

00.02 0.04 0.06 0.08 0.1

[H O ] / mol dm-3

M1 1st (order) 1

M2 straight line graph through the origin Ignore rate is (directly) proportional to [H2O2] 1

Allow constant gradient line through the origin (2 x AO3)

04.5 Allow use of data from line to show

e.g. x2 conc = x2 rate

Allow if M1 missing

Not if M1 wrong

How to answer it

Kinetics: Decomposition of Hydrogen Peroxide

WHAT THIS QUESTION TESTS

AQA A-Level Chemistry • Physical Chemistry • Rate Equations & Kinetics

  • Applying collision theory to explain rate variations with concentration.
  • Extracting reaction rates by drawing tangents to concentration–time curves and determining gradients.
  • Using mathematical relationships to link gas collection data to concentration remaining ( [H₂O₂]ₜ ).
  • Plotting experimental rate–concentration data with precision and drawing suitable lines of best fit.
  • Deducing reaction order from the shape and features of a rate–concentration graph.

Question 04.1

Collision Theory & Initial Rate • [2 marks]

✅ Correct Answer

  • Mark 1: The concentration of hydrogen peroxide ( H₂O₂ ) / reactant molecules is at its highest.
  • Mark 2: Collisions between reactant particles are at their most frequent (or higher frequency of successful collisions).

🧠 Exam Technique: Rate Explanations

Always state two linked factors when explaining rate in terms of concentration:

  1. Concentration of particles (number of particles per unit volume).
  2. Collision frequency — you must specify rate or frequency of collisions, not just "more collisions".

❌ Common Errors

  • Writing "there are more collisions" without referring to time or frequency (fails M2).
  • Mentioning "chance" or "probability" of collisions rather than frequency.
  • Confusing the effect of concentration with temperature (e.g. claiming particles have "more kinetic energy").
Mark Scheme Breakdown:
M1: Highest concentration of / most H₂O₂ / reactant particles per unit volume [1 mark]
M2: More frequent successful collisions / highest frequency of collisions [1 mark]

Question 04.2

Tangent Construction & Gradient Calculation • [2 marks]

✅ Acceptable Values

Gradient range: −0.00120 to −0.00155 mol dm⁻³ s⁻¹

Note: The mark scheme ignores negative signs and units for this step, though negative is physically correct as [H₂O₂] is decreasing.

📐 Step-by-Step Calculation

  1. Find [H₂O₂] = 0.050 mol dm⁻³ on the y-axis. Follow horizontally to the curve; this corresponds to t ≈ 17–18 s .
  2. Place a ruler touching only that point on the curve (without cutting through it). Draw an extended straight line.
  3. Construct a large right-angled triangle to calculate:
    Gradient = Δ[H₂O₂] / Δt = (y₂ − y₁) / (x₂ − x₁)
  4. Example readings:
    At t = 0 s, y ≈ 0.075 mol dm⁻³
    At t = 50 s, y ≈ 0.010 mol dm⁻³
    Gradient = (0.010 − 0.075) / (50 − 0) = −0.065 / 50 = −0.00130 mol dm⁻³ s⁻¹

🧠 Drawing the Tangent

Examiner rule: Place your ruler so that equal spaces appear between the ruler and the curve on either side of the target point ( 0.05 mol dm⁻³ ). If the tangent cuts across the curve so white space is visible inside the arc, M1 is lost. Make your triangle as large as possible to reduce reading errors.

❌ Common Errors

  • Drawing a chord instead of a tangent (line crosses through the curve).
  • Drawing too short a tangent line, leading to significant truncation errors when reading coordinates.
  • Inverting the gradient formula (e.g. doing Δx / Δy ).
Mark Scheme Breakdown:
M1: Suitable tangent drawn with a ruler touching the line at 0.05 mol dm⁻³ (±1 small square) without crossing the curve [1 mark]
M2: Gradient calculated in range −0.00120 to −0.00155 (allow ECF from incorrect tangent) [1 mark]

Question 04.3

Concentration from Volume Collected • [2 marks]

📐 Step-by-Step Working

  1. Step 1: Read initial concentration ( [H₂O₂]ᵢₙᵢₜᵢₐₗ ) from Figure 1
    At Time = 0 s , the y-intercept is 0.083 mol dm⁻³ (allowed: 0.082 to 0.084).
  2. Step 2: Substitute into the given expression
    Given: Vₘₐₓ = 100 cm³ and Vₜ = 20 cm³
    [H₂O₂]ₜ = [H₂O₂]ᵢₙᵢₜᵢₐₗ × ((100 − 20) / 100)
    [H₂O₂]ₜ = 0.083 × (80 / 100) = 0.083 × 0.80
  3. Step 3: Calculate the final value
    [H₂O₂]ₜ = 0.0664 mol dm⁻³

✅ Correct Answer

[H₂O₂]ₜ = 0.0664 mol dm⁻³

Range allowed: 0.0656 to 0.0672 mol dm⁻³ (minimum 2 significant figures).

Calculation Trap:

Do NOT calculate 20 / 100 = 0.20 and multiply by that. That would give the concentration that has reacted, not the concentration remaining.

Mark Scheme Breakdown:
M1: Correct reading of [H₂O₂]ᵢₙᵢₜᵢₐₗ = 0.083 mol dm⁻³ (0.082 – 0.084) [1 mark]
M2: [H₂O₂]ₜ = 0.0664 mol dm⁻³ (or ECF: M1 × 0.8) [1 mark]

Question 04.4

Plotting Rate against Concentration • [2 marks]

💡 Target Data Coordinates

  • (0.02, 0.00049) → just below the 0.0005 line
  • (0.03, 0.00073) → roughly 1.5 small squares below 0.0010
  • (0.05, 0.00124) → roughly halfway between 0.0010 and 0.0015
  • (0.07, 0.00168) → just below 0.0017
  • (0.09, 0.00219) → just below 0.0022

🧠 Exam Technique: Best Fit Lines

  • Accuracy: Points must be plotted within ±0.5 small grid square. Use sharp pencil crosses ( × ).
  • Ruler: A straight line of best fit must be drawn with a single continuous stroke of a ruler.
  • No multiple lines: "Hairy" lines or sketching by hand without a ruler immediately loses the mark.
  • Coverage: The line must span all five plotted points.
Mark Scheme Breakdown:
M1: All 5 points plotted correctly to within ±0.5 small square [1 mark]
M2: Best-fit straight line drawn with a ruler passing within 1 small square of each point (no kinks, no double lines) [1 mark]

Question 04.5

Deducing Reaction Order • [2 marks]

✅ Correct Answer

  • Order: 1st (First order / 1)
  • How the graph shows this: It is a straight line that passes through the origin (0,0).

❌ Critical Examiner Trap

Writing "rate is directly proportional to concentration" by itself scores 0 for the explanation!

The question asks how the graph shows this. Proportionality is the mathematical conclusion, but the graphical evidence is:
1. Straight line (constant gradient), AND
2. Passes through the origin (0,0).

💡 Alternative Valid Explanations

  • "The line has a constant gradient and goes through the origin."
  • Using paired data points from the line to show direct proportionality: e.g. "When [H₂O₂] doubles from 0.04 to 0.08 mol dm⁻³, the rate doubles from 0.0010 to 0.0020 mol dm⁻³ s⁻¹."

🧠 Quick Summary: Rate vs Concentration Shapes

  • Zero Order: Horizontal line (gradient = 0; rate does not change with [A]).
  • First Order: Straight line passing through the origin (Rate ∝ [A]).
  • Second Order: Upward curve through the origin (Rate ∝ [A]²).
Mark Scheme Breakdown:
M1: 1st / first order [1 mark]
M2: Straight line graph through the origin / constant gradient through (0,0) [1 mark]
Note: M2 is not awarded if M1 is incorrect.

Topics

Physical Chemistry · 3.1.5 Kinetics · 3.1.9 Rate Equations

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.