AQA A-Level Chemistry Paper 2, 2018: Question 5

14 marks · Medium difficulty · Practical Techniques & Data Analysis

Use initial rate data to calculate the rate constant k and complete missing values, then complete temperature data and plot an Arrhenius graph to determine the activation energy of the reaction.

Practise this question

Question

Question 05 covers reaction kinetics between bromate(V) ions and bromide ions in acidic conditions with the rate equation rate = k[BrO3-][Br-][H+]^2. Part 05.1 asks to calculate the rate constant k and its units from Experiment 1 in Table 2 to an appropriate number of significant figures (3 marks). Part 05.2 asks to fill in the missing concentrations and rates in Table 2 (3 marks). Part 05.3 provides Table 3 containing temperature T, 1/T, time t, 1/t, and ln(1/t) and asks to complete missing values (2 marks). Part 05.4 provides the rearranged Arrhenius equation involving ln(1/t) and asks the student to plot ln(1/t) against 1/T on the provided grid to determine the activation energy Ea in kJ mol^-1 using R = 8.31 J K^-1 mol^-1 (6 marks).
Question text

05 Bromate(V) ions and bromide ions react in acid conditions according to the equation

BrO −(aq) + 5Br −(aq) + 6H+(aq) → 3Br (aq) + 3H O(l)

32 2

05.1 A series of experiments was carried out at a given temperature. The results were

used to deduce the rate equation for the reaction.

rate = k [BrO −][Br −][H+]2

Table 2 shows an incomplete set of results.

Table 2

+ Initial rate of

Initial [BrO3 ] Initial [Br ] Initial [H ]

Experiment –3 –3 –3 reaction

/ mol dm / mol dm / mol dm –3 –1

/ mol dm s

10.10 0.20 0.30 2.4 × 10–2

20.20 0.30 3.6 × 10–2

30.20 0.40 0.50

40.10 0.10 2.7 × 10–2

Use the data from Experiment 1 to calculate a value for the rate constant, k, at this

temperature and give its units.

Give your answer to an appropriate number of significant figures.

[3 marks]

k Units

05.2 Complete Table 2.

[3 marks]

Space for working 14

05.3 A second series of experiments was carried out to investigate how the rate of the

reaction varies with temperature.

The results were used to obtain a value for the activation energy of the reaction, Ea

Identical amounts of reagents were mixed at different temperatures.

The time taken, t, for a fixed amount of bromine to be formed was measured at

different temperatures.

The results are shown in Table 3.

Table 3

Temperature, T 1 –1 Time, t –1

/ K / s ln

/ K T / s

286 3.50 × 10–3 54 1.85 × 10–2 −3.99

295 3.39 × 10–3 27 3.70 × 10–2

302 15 6.67 × 10–2 −2.71

312 3.21 × 10–3 8 1.25 × 10–1 −2.08

Complete Table 3.

[2 marks]

05.4 The Arrhenius equation can be written as

In this experiment, the rate constant, k, is directly proportional to

t

Therefore

where C1 and C2 are constants.

Use values from Table 3 to plot a graph of ln (y axis) against on the grid.

t T

Use your graph to calculate a value for the activation energy, in kJ mol–1, for this

reaction.

The value of the gas constant, R = 8.31 J K15–1 mol–1

[6 marks]

Activation energy kJ mol−1

Mark scheme

Show the mark scheme Mark scheme for Question 05 gives: 05.1: k = 2.4x10^-2 / (0.10 x 0.20 x 0.30^2) = 13 mol^-3 dm^9 s^-1 (2 sf). 05.2: Experiment 2 [BrO3-] = 0.15, Experiment 3 rate = 0.26 or 0.27, Experiment 4 [H+] = 0.45 or 0.46. 05.3: 1/T = 3.31x10^-3 K^-1, ln(1/t) = -3.30. 05.4: M1 for y-axis scale/labels using over half the grid, M2 points plotted correctly, M3 best-fit straight line, M4 gradient = -6.64 x 10^3 K (range -6500 to -6800), M5 Ea = -gradient x 8.31, M6 Ea = 55.2 kJ mol^-1 (range 54.0 to 56.5 kJ mol^-1).

Question Answers Mark Additional Comments/Guidance

2.4 10 2

k = 2 (= 13.333) Mark is for insertion of numbers into a correctly

0.10 0.20 (0.30) 1

05.1 re-arranged equation.

= 13 (must be 2 sfs) 1

Units mol 3 dm+9 s 1

1 Can be in any order

If k wrong in 5.1 : allow the expected answer OR

Experiment 2 [BrO ] = 0.15 values conseq to their k (allow mix & match)

05.2 3 1

Ex 2 [BrO ] = 2/k

Marked Experiment 3 rate = 0.26 or 0.27 3

with 5.1 + Ext 3 rate = 0.02 × k

Experiment 4 [H ] = 0.45 or 0.46

1 +

Ex 4 [H ] = square root of (2.7/k)

1/T value 3.31(1) × 10 3 or 0.00331(1) Must be 3 sig figs or more

05.3 1

G ln(1/t) value 3.30 or 3.297 1

Not allow 3.29

M1 y axis labelled with values (no units) and plotted points use 1

over half of the axis

M2 points plotted correctly (see graph below) 1

05.4 + - one small square for line of best fit

M3 best fit straight line (minimum 3 points plotted) 1

Can see

05.3 M4 gradient = 6.64 × 103 (K) or -6640 (K) 1 3 3

Range - 6.5 × 10 to - 6.8 × 10 or -6500 to -6800

M5 Ea = M4 × 8.31 1 If gradient outside range then max 4 for M1,M2,M3

1 and M5

M6 = 55.2 kJ mol 1 Range 54.0 - 56.5

– – –

Total 14

How to answer it

Reaction Kinetics & Arrhenius Analysis

AQA A-Level Chemistry • Physical Chemistry • Rates of Reaction

What this question tests

This 14-mark multi-step question assesses core competencies in quantitative chemical kinetics:

  • Rearranging Rate Equations: Calculating the rate constant k and deducing its overall units from multi-reactant experimental data.
  • Data Deductions: Using a determined rate equation and k to find missing concentrations and initial rates across different experiments.
  • Arrhenius Data Processing: Processing raw experimental time/temperature data into reciprocal absolute temperatures (1/T) and logarithmic rate values (ln(1/t)).
  • Graphical Arrhenius Analysis: Setting up axes, accurate coordinate plotting, drawing a line of best fit, finding a negative gradient, and converting that into activation energy (Ea) in kJ mol⁻¹.

Question 05.1: Calculating Rate Constant k and its Units

Data from Experiment 1 • [3 Marks]

📐 Calculations & Step-by-Step Method

Given rate equation: rate = k [BrO₃⁻][Br⁻][H⁺]²

Step 1: Rearrange for k
k = rate / ([BrO₃⁻] × [Br⁻] × [H⁺]²)
Step 2: Substitute values from Exp 1
k = (2.4 × 10⁻²) / (0.10 × 0.20 × (0.30)²)
k = (2.4 × 10⁻²) / (0.10 × 0.20 × 0.090) = (2.4 × 10⁻²) / (1.8 × 10⁻³)
k = 13.333...
Step 3: Apply appropriate significant figures
Input concentrations are given to 2 s.f. (0.10, 0.20, 0.30) and rate is 2.4 × 10⁻² (2 s.f.).
Value of k = 13 (must be 2 s.f.).
Step 4: Determine units of k
Units = (mol dm⁻³ s⁻¹) / ((mol dm⁻³) × (mol dm⁻³) × (mol dm⁻³)²)
Units = (mol dm⁻³ s⁻¹) / (mol⁴ dm⁻¹²)
Units = mol⁻³ dm⁹ s⁻¹

✅ Correct Answer

Value of k: 13 (2 s.f. strictly required)

Units: mol⁻³ dm⁹ s⁻¹ (indices can be in any order, e.g. dm⁹ mol⁻³ s⁻¹ )

❌ Common Errors

  • Forgetting to square [H⁺]: Using 0.30 instead of 0.30² gives k = 4.0 (incorrect).
  • Significant figure penalty: Writing 13.3 or 13.33 loses Mark 2. The question explicitly instructed: "appropriate number of significant figures".
  • Incorrect sign on unit indices: Forgetting that dividing by mol³ gives mol⁻³ and dividing by dm⁻⁹ gives dm⁹.
Mark Breakdown:
• Mark 1: Correct substitution of numbers into rearranged equation: (2.4 × 10⁻²) / (0.10 × 0.20 × 0.30²).
• Mark 2: 13 (must be strictly 2 significant figures).
• Mark 3: mol⁻³ dm⁹ s⁻¹.

Question 05.2: Completing Table 2

Deducing Missing Values • [3 Marks]

Exp Initial [BrO₃⁻] / mol dm⁻³ Initial [Br⁻] / mol dm⁻³ Initial [H⁺] / mol dm⁻³ Initial rate / mol dm⁻³ s⁻¹
1 0.10 0.20 0.30 2.4 × 10⁻²
2 0.15 0.20 0.30 3.6 × 10⁻²
3 0.20 0.40 0.50 0.26 or 0.27
4 0.10 0.10 0.45 (or 0.46) 2.7 × 10⁻²

📐 Step-by-Step Working

Experiment 2: Find [BrO₃⁻]

Compare Exp 1 and 2: [Br⁻] and [H⁺] are constant. Since the reaction is 1st order with respect to BrO₃⁻:

[BrO₃⁻]₂ = 0.10 × (3.6 × 10⁻² / 2.4 × 10⁻²) = 0.10 × 1.5 = 0.15 mol dm⁻³

Experiment 3: Find Rate

rate = k [BrO₃⁻][Br⁻][H⁺]² = (13.333) × (0.20) × (0.40) × (0.50)²
rate = 13.333 × 0.020 = 0.27 mol dm⁻³ s⁻¹ (or 0.26 using k = 13)

Experiment 4: Find [H⁺]

[H⁺]² = rate / (k [BrO₃⁻][Br⁻]) = (2.7 × 10⁻²) / (13.333 × 0.10 × 0.10)
[H⁺]² = (2.7 × 10⁻²) / 0.1333 = 0.2025
[H⁺] = √(0.2025) = 0.45 mol dm⁻³

🧠 Exam Technique & Error Carried Forward (ECF)

  • Two valid methods: You can either use your calculated value of k directly, or use ratio comparisons between experiments (e.g., Exp 1 vs Exp 2).
  • ECF is allowed: If you miscalculated k in 05.1, you can still gain full marks here if your answers are mathematically consequential to your wrong k value:
    • Exp 2 [BrO₃⁻] = 2 / k
    • Exp 3 rate = 0.02 × k
    • Exp 4 [H⁺] = √(2.7 / k)
  • Don't forget the square root: In Exp 4, because the reaction is 2nd order with respect to H⁺, you must take the square root of 0.2025! Leaving the answer as 0.20 is a frequent error.

Question 05.3: Completing Table 3 (Arrhenius Data)

Data Processing for Graph Plotting • [2 Marks]

T / K (1 / T) / K⁻¹ Time, t / s (1 / t) / s⁻¹ ln(1 / t)
286 3.50 × 10⁻³ 54 1.85 × 10⁻² -3.99
295 3.39 × 10⁻³ 27 3.70 × 10⁻² -3.30 (or -3.297)
302 3.31 × 10⁻³ (or 0.00331) 15 6.67 × 10⁻² -2.71
312 3.21 × 10⁻³ 8 1.25 × 10⁻¹ -2.08

📐 Calculations

1. Row 3: Calculate 1/T for T = 302 K:

1 / 302 = 0.00331125... = 3.31 × 10⁻³ K⁻¹ (or 0.00331)

2. Row 2: Calculate ln(1/t) for t = 27 s:

1 / t = 1 / 27 = 0.037037...
ln(1 / 27) = -3.2958... = -3.30 (or -3.297)

❌ Common Errors

  • Significant figures: The 1/T value must be given to at least 3 significant figures to match the table format (3.31 × 10⁻³). Writing 3.3 × 10⁻³ loses the mark.
  • Premature rounding: Calculating ln(0.037) instead of ln(1/27) gives -3.2968, which if rounded wrongly to -3.29 is explicitly NOT allowed by the mark scheme!
  • Sign omission: Forgetting the negative sign on logarithmic values less than 1.

Question 05.4: Arrhenius Graph & Activation Energy

Plotting & Determining Ea • [6 Marks]

💡 Mathematical Foundation

The Arrhenius relationship provided in the question is:

ln(1/t) = (-Ea / R) × (1/T) + C₂

Comparing this to the linear equation y = mx + c:

  • y-axis: ln(1/t)
  • x-axis: (1/T) / K⁻¹
  • Gradient (m): m = -Ea / R
  • Therefore: Ea = -m × R

🧠 Graph Plotting Checklist (M1 - M3)

  • M1: y-axis scale and label:
    • The vertical axis must be labeled ln(1/t) (no units required since ln has no units).
    • Choose a sensible linear scale that spans more than half the grid. For instance, -2.0 at the top down to -4.0 near the bottom (with major lines every 0.4 or 0.2).
  • M2: Accurate plotting:
    • Plot all four points to within ±0.5 small square.
    • Points: (3.50 × 10⁻³, -3.99), (3.39 × 10⁻³, -3.30), (3.31 × 10⁻³, -2.71), (3.21 × 10⁻³, -2.08).
  • M3: Best-fit line:
    • Draw a single straight line using a clear ruler.
    • Ensure a balanced distribution of points above and below the line. Do not force the line through the origin!

📐 Gradient & Ea Calculation (M4 - M6)

Step 1: Calculate the gradient (m)
Pick two points far apart on your line of best fit (e.g. at 1/T = 3.20 × 10⁻³ and 3.50 × 10⁻³):
Δy = -3.99 - (-2.00) = -1.99
Δx = (3.50 × 10⁻³ - 3.20 × 10⁻³) = 0.30 × 10⁻³ K⁻¹
m = Δy / Δx = -1.99 / (0.30 × 10⁻³) = -6.64 × 10³ K
(Accepted range: -6.5 × 10³ to -6.8 × 10³)
Step 2: Multiply by Gas Constant R
Ea (in J mol⁻¹) = -m × R
Ea = -(-6.64 × 10³) × 8.31 = +55 178 J mol⁻¹
Step 3: Convert to kJ mol⁻¹
Divide by 1000:
Ea = 55 178 / 1000 = 55.2 kJ mol⁻¹
(Accepted range: 54.0 to 56.5 kJ mol⁻¹)

❌ Examiner Trap Alert: Where Marks are Lost in Arrhenius Graphs

  • Forgetting 10⁻³ from the x-axis: The x-axis values are multiplied by 10⁻³. If you calculate Δx as 0.30 instead of 0.30 × 10⁻³, your gradient will be off by a factor of 1000!
  • Negative activation energy: Activation energy must be positive. The gradient is negative, so Ea = -gradient × R yields a positive value. Giving -55.2 loses the final mark.
  • Forgetting to divide by 1000: The question asks for Ea in kJ mol⁻¹. Leaving the answer as 55 200 without converting forfeits M6.
  • Small gradient triangles: Examiners require you to use at least half the length of your plotted line to measure Δy and Δx for accurate gradient determination.
Mark Breakdown (6 Marks):
• M1: y-axis properly labelled ln(1/t) and scaled so points occupy >50% of vertical height.
• M2: All points plotted correctly within ±1 small square.
• M3: Straight line of best fit drawn through points.
• M4: Correct gradient calculation in range -6.5 × 10³ to -6.8 × 10³ (or -6500 to -6800).
• M5: Ea = -gradient × 8.31 (method mark).
• M6: Final value: 55.2 kJ mol⁻¹ (range 54.0 to 56.5 kJ mol⁻¹).

Topics

Physical Chemistry · 3.1.5 Kinetics · 3.1.9 Rate Equations

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 2, 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.