AQA A-Level Chemistry Paper 3, June 2018: Question 1
19 marks · Medium difficulty · Practical Techniques & Data Analysis
Determine the rate equation, rate constant, and reaction orders using concentration-time graphs, tangents, and a clock reaction method.
Practise this questionQuestion
![The question consists of eight parts (01.1 to 01.8) focusing on kinetics. It presents the redox reaction of hydrogen peroxide with iodide ions in acidic solution, simplifying the rate equation using large excesses of reactants. Figure 1 shows a linear decrease in [H+] over time from 0.50 mol dm⁻³ at 0 s to 0.00 mol dm⁻³ at approximately 420 s. Table 1 and Figure 2 show data points for a curved concentration-time graph requiring plotting, a curve of best fit, and a tangent drawn at [H+] = 0.35 mol dm⁻³. Part 01.8 is an extended 6-mark question describing a clock reaction where reagent X delays the appearance of dark blue product E, asking for experimental details to find the order of reaction with respect to reactant A.](/.netlify/images?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fexamquestions-bf477.firebasestorage.app%2Fo%2Fquestions%252FAQA.Q18.L3.01.SRC.aqa-alevel-3%25257C2018%25257CJUN%25257Cuntiered-1790797988511.png%3Falt%3Dmedia&w=1000)
Question text
01 Iodide ions are oxidised to iodine by hydrogen peroxide in acidic conditions.
H O (aq) + 2H+(aq) + 2I–(aq) → I (aq) + 2H O(l)
22 2 2
The rate equation for this reaction can be written as
rate = k [H O ]a[I–]b[H+]c
In an experiment to determine the order with respect to H+(aq), a reaction mixture is
made containing H+(aq) with a concentration of 0.500 mol dm–3
A large excess of both H O and I– is used in this reaction mixture so that the rate
equation can be simplified to
rate = k [H+]c
01.1 Explain why the use of a large excess of H O and I– means that the rate of reaction at
a fixed temperature depends only on the concentration of H+(aq).
[2 marks]
01.2 Samples of the reaction mixture are removed at timed intervals and titrated with alkali
to determine the concentration of H+(aq).
State and explain what must be done to each sample before it is titrated with alkali.
[2 marks]
01.3 A graph of the results is shown in Figure 1.
Figure 1
Explain how the graph shows that the order with respect to H+(aq) is zero.
[2 marks]
01.4 Use the graph in Figure 1 to calculate the value of k1
Give the units of k1
[3 marks]
k1
Units
01.5 A second reaction mixture is made at the same temperature. The initial
concentrations of H+(aq) and I–(aq) in this mixture are both 0.500 mol dm–3
*03* There is a large excess of H2O2
In this reaction mixture, the rate depends only on the concentration of I–(aq).
The results are shown in Table 1.
Table 1
Time/s 0 100 200 400 600 800 1000 1200
[H+] / mol dm–3 0.50 0.44 0.39 0.31 0.24 0.19 0.15 0.12
Plot these results on the grid in Figure 2. The first three points have been plotted.
[1 mark]
Figure 2
01.6 Draw a line of best fit on the grid in Figure 2.
[1 mark]
01.7 Calculate the rate of reaction when [H+] = 0.35 mol dm–3
Show your working using a suitable construction on the graph in Figure 2.
[2 marks]
Rate mol dm–3 s–1
01.8 A general equation for a reaction is shown.
A(aq) + B(aq) + C(aq) → D(aq) + E(aq)
In aqueous solution, A, B, C and D are all colourless but E is dark blue.
A reagent (X) is available that reacts rapidly with E. This means that, if a small
amount of X is included in the initial reaction mixture, it will react with any E produced
until all of the X has been used up.
Explain, giving brief experimental details, how you could use a series of experiments
to determine the order of this reaction with respect to A. In each experiment you
should obtain a measure of the initial rate of reaction.
[6 marks]
Mark scheme
Show the mark scheme
Question Answers Additional Comments/Guidelines Mark
H O and/or I– concentration change is negligible / Only the concentration of H+ changes.
H O and/or I– concentration (effectively) constant
01.1 2
so have a constant/no effect on the rate / so is zero order (w.r.t. H O Ignore references to H+ is limiting reagent / rds /
and I–) / a and b are zero –
H2O2/I not in rate equation
Do not allow reference to catalyst.
Stop the reaction / quench
By dilution / cooling / adding a reagent to react with H O /I- Allow valid suggestions about how to stop the
reaction.
Do not allow reaction with acid/alkali / neutralisation
01.2 2
for M2
Do not penalise other named reagents.
Ignore references to measuring volume and adding
indicator
M1: constant gradient Allow constant rate / rate = k
OR Ignore reference to straight line
change/decrease in concentration is proportional to time Not increase in concentration / concentration is
inversely proportional / concentration (on its own) is
proportional
01.3 2
M2: as [H+] changes/decreases
M2 dependent on correct M1
Allow rate v concentration graph would give
horizontal straight line owtte
Allow so [H+] has no effect on the rate–– –
evidence of attempt at calculation of gradient via ∆y/∆x allow construction lines on graph
k = 0.0012 / 1.2 x 10–3
1 At least 2 sf (0.00118 – 0.00122)
Correct answer scores 2/2
No ecf from incorrect or inverted numbers in M1
01.4 3
k1 = – 0.0012 scores 1/2
Additional processing of data such as including [H+]
loses M2
units = mol dm–3 s–1
M3 mark independently
5 points correctly plotted Allow ±half a small square for each point
01.5 1
– – –
Smooth curve only within one small square of all
12 points (ecf on 01.5)
Not a series of straight lines between points
01.6 1
M1 for a tangent / triangle / other suitable working
Allow ECF for both M1 and M2 following on from
straight line drawn in 01.6, but must show suitable
construction on graph for M1
01.7 2
+ –3 Ignore negative signs
M1: Tangent to curve drawn at [H3O ] = 0.35 mol dm –4 –4
Allow value in range 3.70 x 10 - 4.50 x 10
e.g. 0.18/440
–4 –3 –1 At least 2sf
M2: Rate = 4.09 x 10 (mol dm s )
ecf from any straight line for correctly calculated
gradient
– – –
This question is marked using levels of response. Refer to the Mark Indicative Chemistry content Method 1
Scheme Instructions for examiners for guidance on how to mark
Stage 1 Preparation
this question
1a Measure (suitable/known volumes of) some reagents
Level 3 All stages are covered and the explanation of (ignore quoted values for volume)
each stage is correct and virtually complete. 1b Measure (known amount of) X / use a colorimeter
5-6 marks 1c into separate container(s) – (allow up to two reagents
Answer is coherent and shows progression and X measured together into one container); reference
through all three stages. to A, B or C added last. NOT if X added last.
Stage 2 Procedure
A clear explanation of how the order is
2a Start clock/timer at the point of mixing (don’t allow if only
determined from the results is needed to show
2 reagents mixed)
coherence.
(allow even if X not added or added last)l
Level 2 All stages are covered (NB ‘covered’ means min 2
2b Time recorded for appearance of blue colour/specific
from stage 2) but the explanation of each stage
reading on colorimeter/disappearing cross
3-4 marks may be incomplete or may contain inaccuracies
2c Use of same concentration of B and C / same total
01.8 OR two stages covered and the explanations are 6
volume / same volume/amount of X
generally correct and virtually complete
2d Same temperature/use water bath
2e Repeat with different concentrations of A (can be implied
Answer is coherent and shows some progression
through different volumes of A and same total volume)
through all three stages. Some steps in each
stage may be out of order and incomplete Stage 3 Use of Results
Level 1 Two stages are covered but the explanation of 3a 1/time taken is a measure of the rate
each stage may be incomplete or may contain 3b plot of 1/time against volumes/concentrations of A or plot
1-2 marks inaccuracies log(1/time) vs log(volume or concentration of A)
OR only one stage is covered but the explanation 3c description of interpreting order from shape of 1/time vs
is generally correct and virtually complete volume or concentration graph / gradient of log plot gives
order / allow interpretation of time vs concentration graph
Answer shows some progression between two / ratio between change in concentration and change in
stages rate (e.g, 2x[A] = 2 x rate so 1st order)
Level 0
Insufficient correct Chemistry to warrant a mark
0 marks
Total 19
– – –
Indicative Chemistry content – Alternative Method Using Colorimetry and repeated Continuous Monitoring
Stage 1 Preparation
1a Measure (suitable/known volumes of) A, B and C (ignore quoted values for volume)
1b Use of colorimeter
1c into separate container(s) – (allow up to two reagents measured together into one container) – ignore use of X
Stage 2 Procedure
2a Start clock/timer at the point of mixing
2b Take series of colorimeter readings at regular time intervals
2c Use of same concentration of B and C / same total volume / (same volume/amount of X)
2d Same temperature
2e Repeat with different concentrations of A (can be implied through different volumes of A and same total volume)
Stage 3 Use of Results
3a Plot absorbance vs time and measure/calculate gradient at time=0
3b plot of gradient against volumes/concentrations of A or plot log(1/time) vs log(volume or concentration of A)
3c description of interpreting order from shape of 1/time vs volume or concentration graph / gradient of log plot gives order
How to answer it
Kinetics: Orders of Reaction, Graphical Analysis & Clock Reactions
This multi-part physical chemistry question assesses core competencies across AQA A-Level Chemistry Topic 3.1.9 (Rate Equations):
- Pseudo-order kinetics: Using large excesses of reagents to keep their concentrations virtually constant.
- Sampling and quenching: Experimental methods for sampling, stopping a reaction, and titrimetric analysis.
- Graphical orders: Interpreting concentration–time graphs for zero and non-zero orders.
- Tangent & gradient calculations: Finding rate constants ( k ) and instantaneous rates from tangent slopes, alongside correct dimensional unit analysis.
- Experimental design (6-marker): Planning an iodine clock or colorimetric initial-rate investigation to determine the order of reaction with respect to a single reagent.
Pseudo-Order Reaction Conditions
Explaining the effect of large excesses of reactants
✅ Model Answer
- The concentration of H₂O₂ and I⁻ is effectively constant (change is negligible). [1 mark]
- So H₂O₂ and I⁻ have no effect on the rate (they behave as zero order, so a = 0 and b = 0 ). [1 mark]
💡 Key Knowledge
When a reactant is in large excess, its concentration changes by such an infinitesimal percentage throughout the reaction that it is treated as a constant.
The constant terms merge into the rate constant:
k₁ = k[H₂O₂]ᵃ[I⁻]ᵇ .
❌ Common Misconceptions & Examiner Traps
Do not say "H₂O₂ and I⁻ are catalysts" or "they are not in the rate-determining step". They are reactants. Never simply write "their amounts don't change" without explicitly referring to concentration.
Quenching Before Titration
Continuous sampling technique
✅ Model Answer
- Action: Stop the reaction / quench the reaction. [1 mark]
- Method: By rapidly cooling (in an ice bath), or by dilution with a large volume of cold water, or by adding a chemical to consume H₂O₂ / I⁻. [1 mark]
🧠 Exam Technique
If you take a sample to titrate acid with alkali, the titration takes several minutes. Without quenching, the reaction continues during titration, altering [H⁺] and giving an incorrect concentration at time t .
❌ Dangerous Error
Never suggest quenching by adding an acid or an alkali! Since the titration itself measures [H⁺] using alkali, adding an acid or base would completely invalidate the titration measurement.
Zero Order Concentration–Time Graph
Interpreting the straight line in Figure 1
✅ Model Answer
- The graph has a constant gradient (or the decrease in concentration is directly proportional to time). [1 mark]
- Therefore, the rate remains constant as [H⁺] changes (rate is independent of [H⁺]). [1 mark]
💡 Distinguishing Top Answers
Merely stating "it is a straight line" is not enough for mark 1. You must state that the gradient / slope is constant, meaning the rate does not change as the concentration falls.
Calculating the Rate Constant k₁ & Units
Determining the gradient from Figure 1
📐 Step-by-Step Calculation
k₁ = (0.50 - 0.00) / (415 - 0) = 0.00120 mol dm⁻³ s⁻¹ (Acceptable range: 0.00118 to 0.00122 or 1.2 × 10⁻³). [2 marks]
units of k₁ = rate / [H⁺]⁰ = mol dm⁻³ s⁻¹ . [1 mark]
❌ Common Traps
- Sign error: Stating k₁ = -0.0012 loses 1 mark. Rate constants must always be positive values.
- Over-processing: Do not try to multiply or divide by [H⁺] after obtaining the gradient.
Plotting Data & Drawing Line of Best Fit
Figure 2: Non-zero order curve
✅ 01.5 Plotting Points (1 mark)
Plot the remaining 5 data points accurately to within ±0.5 small grid square:
- (400, 0.31)
- (600, 0.24)
- (800, 0.19)
- (1000, 0.15)
- (1200, 0.12)
✅ 01.6 Line of Best Fit (1 mark)
Draw a single smooth curve passing within one small square of all plotted points.
Do not join points with straight "dot-to-dot" lines or draw double/feathered lines.
Calculating Reaction Rate via Tangent Construction
Determining instantaneous rate at [H⁺] = 0.35 mol dm⁻³
📐 Tangent Construction & Calculation
Rate = |Δy / Δx| = 0.18 / 440 = 4.09 × 10⁻⁴ mol dm⁻³ s⁻¹ .
Allowable range: 3.70 × 10⁻⁴ to 4.50 × 10⁻⁴ mol dm⁻³ s⁻¹ (at least 2 sig figs). [1 mark]
🧠 Exam Technique: Tangents
Make your tangent triangle as large as possible. If examiners cannot see your working or construction lines on Figure 2, you automatically forfeit the first mark even if your final answer is within range!
Extended Practical Design: Clock Reaction / Initial Rates
A(aq) + B(aq) + C(aq) → D(aq) + E(aq)
💡 The Principle of an Iodine-style "Clock" Reaction
Product E immediately reacts with reagent X and is removed. When X is completely used up, excess E suddenly accumulates, turning the mixture dark blue. Since a constant, known amount of X is used each time, 1 / time (1/t) serves as a direct measure of initial rate.
📋 Full Mark Scheme Breakdown (Structured in 3 Stages)
Stage 1: Preparation
- 1a. Measure known volumes of reagents A, B, and C using a burette/pipette.
- 1b. Measure a constant, known amount (volume) of reagent X.
- 1c. Place solutions into separate containers before mixing (e.g. mix A, B, and X in one flask, then add C to start).
Stage 2: Procedure & Control Variables
- 2a. Start the stopwatch immediately at the moment of mixing.
- 2b. Stop the timer when the dark blue colour appears.
- 2c. Keep the concentrations and volumes of B and C constant, and maintain a constant total volume (by adding deionised water). Keep the amount of X constant.
- 2d. Keep temperature constant (using a water bath).
- 2e. Repeat the experiment with different volumes (concentrations) of A.
Stage 3: Analysis & Determining the Order
- 3a. 1 / time is proportional to initial rate.
- 3b. Plot a graph of 1 / time against [A] (or volume of A), or plot log(1/time) against log[A] .
- 3c. Deduce the order from graph shape:
- Horizontal flat line → Zero order (rate independent of [A])
- Straight line through the origin → First order (rate directly proportional to [A])
- Upward curve → Second order (or slope of log-log plot gives the order)
🧠 Level 3 Requirements (5–6 Marks)
To reach Level 3, your answer must cover all 3 stages logically:
- Stage 1: Apparatus, known volumes of A, B, C, X.
- Stage 2: Timing to colour change, varying [A] while keeping B, C, X, and total volume constant.
- Stage 3: Clear mathematical link ( Rate ∝ 1/t ) and graph interpretation.
❌ Common Errors in 6-Mark Questions
- Forgetting to state that total volume must be kept constant using water.
- Varying more than one reactant concentration at a time.
- Failing to explain how the graph shape translates into 0, 1st, or 2nd order.
Topics
Physical Chemistry · Required Practicals · 3.1.9 Rate Equations · Required Practical 7: Measuring the rate of reaction by an initial rate method · Required Practical 3: Investigation of how the rate of a reaction changes
Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.