Edexcel A-Level Chemistry Paper 2, June 2024: Question 3
18 marks · Hard difficulty · Calculations
Investigate the decomposition of hydrogen peroxide including its dot-and-cross diagram, redox/disproportionation analysis, kinetics via experimental data, rate equation and constant calculations, gas testing, colour changes, Arrhenius graph gradient and activation energy determination, and collision factor calculation.
Practise this questionQuestion
Question text
3 This question is about hydrogen peroxide, H2O2 .
(a) Draw a dot‑and‑cross diagram of a molecule of hydrogen peroxide.
(1)
(b) Hydrogen peroxide decomposes to form water and oxygen.
2H2O2(aq) → 2H2O(l) + O2(g)
Explain, using oxidation numbers, why the decomposition of hydrogen peroxide
is classified as a disproportionation reaction.
(3)
(c) The decomposition of hydrogen peroxide is catalysed by iodide ions, I–(aq).
The kinetics of this reaction were investigated using different concentrations of
hydrogen peroxide and iodide ions.
The results are shown in the table.
Experiment [H O2(aq)] / mol dm–3 [I–(aq)] / mol dm–3 Rate / mol dm–3 s–1
1 0.100 0.0500 8.90 × 10–7
2 0.400 0.0500 3.56 × 10–6 7
3 *P76896A0732*0.2000.100 3.56×10–6
(i) Deduce the order of reaction with respect to hydrogen peroxide and to
iodide ions.
(2)
Order with respect to hydrogen peroxide …
Order with respect to iodide ions …
(ii) Write the rate equation for the reaction using your answer to (c)(i).
(1)
(iii) Calculate the rate constant, k, using data from Experiment 3.
Include units in your answer.
(2)
(d) The breakdown of hydrogen peroxide with iodide ions as a catalyst is the basis of
the demonstration ‘Elephant’s Toothpaste’.
The presence of a detergent results in a rapid eruption of foam as the oxygen gas
is released.
(i) Describe the test and the positive result that confirms the gas produced
is oxygen.
(1)
… *P76896A0832*
(ii) The foam produced often has a slight yellow/brown colour.
Explain what causes this colour, which is not caused by the detergent.
(2)
(e) The effect of temperature on the rate of the decomposition of hydrogen peroxide
without a catalyst was also investigated.
A graph of lnk against 1/temperature (1/T) was plotted.
1/T / K–1
0.0030 0.0032 0.0034 0.0036
–6
–7
–8
lnk
–9
–10 *P76896A0932*
–11
(i) Determine the gradient of the graph.
Include units in your answer. You must show your working on the graph.
(3)
(ii) Calculate the activation energy, E , of the reaction, in kJ mol–1, using your
a
answer to (e)(i) and the Arrhenius equation shown.
Ea 1
lnk = – × + constant
R T
[Gas constant (R) = 8.31 J mol–1 K–1]
(1)
*P76896A01032*
(f ) Another way to write the Arrhenius equation is shown.
k = Ae–Ea/RT
The constant A is often called the collision factor as it is linked to the orientation
of the particles colliding in a reaction.
The decomposition of hydrogen peroxide is also catalysed by aluminium.
Calculate a value for A at 370K, for the catalysed decomposition of
hydrogen peroxide with an activation energy, E , of 5.02 × 104 J mol–1 and a
a
numerical value for the rate constant of 1.60 × 10–3.
Units are not required.
(2)
(Total for Question 3 = 18 marks)
Mark scheme
Show the mark scheme
Question
Answer Additional Guidance Mark
Number
3(a) (1)
Ignore omission of circles
Ignore line to show covalent bonds
Allow reversal of dots and crosses
Allow all dots or all crosses
Question
Answer Additional Guidance Mark
Number
3(b) An explanation that makes reference to the following points: Oxidation numbers may be shown above / (3)
below species in equation
• oxidation number of oxygen has changed from –1 (in
H2O2) (1) Allow Roman numerals to show oxidation
numbers
• to –2 in water, so reduced (1)
• and 0 in (elemental) oxygen, so is oxidised (1)
Allow 1 rescue mark for correct oxidation
numbers of oxygen in both products if references
to oxidation and reduction incorrect or omitted
or not linked to a specific reaction in M2 and M3
Allow 1 rescue mark for correct oxidation
numbers and references to oxidation and
reduction if products omitted in M2 and M3
Question
Answer Additional Guidance Mark
Number
3(c)(i) An answer that makes reference to the following points: (2)
• (order with respect to hydrogen peroxide =) 1 / first (1)
• (order with respect to iodide ions =) 1 / first (1)
Question
Answer Additional Guidance Mark
Number
3(c)(ii) An answer that makes reference to the following point: Rate equation must be consistent on orders from (i) (1)
• rate / r = k[H O ][I–] Allow use of correct names
Allow TE from 3(c)(i)
Do not award round brackets
Do not award omission of rate / r
Question
Answer Additional Guidance Mark
Number
3(c)(iii) Example of calculation (2)
Calculation must be consistent on rate equation from
(ii)
• calculation of k (1) k = rate ÷ [H O ][I–]
3.56 × 10–6 ÷ (0.200 × 0.100) = 1.78 × 10–4
(1) dm3 mol–1 s–1
• units of k
Allow dm3/mol/s
Allow dm3 mol– s–
Allow units in any order
Ignore SF except 1 SF
Allow TE from incorrect rate equation in 3(c)(ii)
Allow calculation based on data from experiments 1
or 2
NOTE – with TE on incorrect order, calculation and
units require checking as different experiments give
different values
Question
Answer Additional Guidance Mark
Number
3(d)(i) An answer that makes reference to the following points: (1)
• put glowing splint (into bubbles of gas / presence of Ignore ‘a splint that has (just) been blown out
gas) relights’
and
splint relights Do not award ‘relights a burnt splint’
Do not award squeaky pop
Question
Answer Additional Guidance Mark
Number
3(d)(ii) An explanation that makes reference to the following points: (2)
• (colour caused by aqueous solution of) iodine (1) Accept (colour caused by) aqueous (solution of) I –
Ignore iodide ions are yellow in solution
Do not award iodine ions
• (formed as excess hydrogen) peroxide oxidises iodide (1) Allow oxygen oxidises / converts / reacts with the
(ions) / converts iodide (ions) / reacts with iodide iodide (ions) / iodide reduces (hydrogen) peroxide
(ions)
Allow correct equation
2H+ + H O + 2I– → I + 2H O
22 2 2
Ignore incomplete equations (as considered part of
the working out process)
Question
Answer Additional Guidance Mark
Number
3(e)(i) Example of calculation (3)
• triangle or similar shown on graph to indicate Allow circling of points / marking of points on line
changes in x and y (1) For example
–11– - 6 = –5
0.003588 – 0.003038 = 0.00055
• calculation of gradient (1) –5 ÷ 0.00055 = – 9090.9
Allow gradient in range −8500 to – 9500, consistent
with correctly determined y and x values , though
gradient must be checked for calculator errors
Allow TE on incorrectly determined y and x values
from M1
Allow negative gradient even if values used in Δy / Δx
would give a positive value
Ignore SF
Ignore rounding errors in final value of gradient
• units of gradient (1) K
No TE for units based on inverted axes
Question
Answer Additional Guidance Mark
Number
3(e)(ii) Example of calculation (1)
• calculation of Ea –9090.9 × 8.31 = – Ea
E = (+) 75.545 (kJ mol–1)
a
Allow (+) 75545 J mol–1
Ignore SF except 1 SF
Allow TE from (e)(i)
Do not award negative activation energies.
Note : if answer is in J mol–1 units must be given
Question
Answer Additional Guidance Mark
Number
3(f) Example of calculation (2)
• expression for, using numbers, or calculation of (1) e–50200 ÷ (8.31 × 370) / 8.1164 × 10–8
e−Ea/RT
(1) 1.60 × 10–3 ÷ 8.1164 × 10–8 = 1.9713 × 104
• calculation of A
Ignore SF
Ignore units
TE from M1 to M2
Correct answer with no working scores 2
Note – use of rounded values of e−Ea/RT is acceptable
and final answer may need checking as it will make a
difference to the numerical value
M1 can be subsumed in M2
(Total for Question 3 = 18 marks)
How to answer it
Hydrogen Peroxide: Bonding, Kinetics & Arrhenius Chemistry
What this question tests
This comprehensive Edexcel A-Level question evaluates bonding representation (dot-and-cross diagrams), redox mechanisms (identifying disproportionation via oxidation numbers), experimental rate kinetics (deducing orders, rate equations, and calculating rate constants with units), practical observations (gas testing and iodine coloration), and graphical/algebraic manipulation of the Arrhenius equation (calculating gradients, activation energy Eₐ , and the collision factor A ).
Dot-and-Cross Diagram of Hydrogen Peroxide
✅ Correct Answer
Draw a single covalent bond between the two oxygen atoms ( O-O ), with each oxygen bonded to a hydrogen atom ( H-O-O-H ). Show shared electron pairs between H and O, and between O and O. Include two non-bonding lone electron pairs on each oxygen atom.
💡 Key Knowledge
Hydrogen peroxide has an open-book geometry with a single covalent O-O peroxide bond. Each oxygen atom has an octet achieved by sharing electrons with H and O, leaving two unshared lone pairs.
❌ Common Errors
Failing to include the lone pairs on the oxygen atoms, or incorrectly drawing a double bond between the oxygen atoms which would give oxygen ten valence electrons.
Explaining Disproportionation via Oxidation Numbers
✅ Correct Answer
In H₂O₂ , the oxidation number of oxygen is -1 . In water ( H₂O ), oxygen's oxidation number changes to -2 (it is reduced). In oxygen gas ( O₂ ), oxygen's oxidation number becomes 0 (it is oxidized). Since the same element ( O ) is simultaneously oxidized and reduced, it is a disproportionation reaction.
🧠 Exam Technique
To secure all 3 marks, you must explicitly state:
1. The initial oxidation number ( -1 in H₂O₂ ).
2. The species and final oxidation number where reduction occurs ( -2 in H₂O ).
3. The species and final oxidation number where oxidation occurs ( 0 in O₂ ).
Rate Orders, Rate Equations, and Rate Constants
(i) Deduce the order of reaction
✅ Correct Answers
- Order with respect to hydrogen peroxide: 1 (first order)
- Order with respect to iodide ions: 1 (first order)
📐 Working Out
Comparing Exp 1 and 2: [H₂O₂] quadruples (0.100 to 0.400) while [I⁻] is constant. The rate quadruples ( 8.90×10⁻⁷ to 3.56×10⁻⁶ ), meaning it is first order with respect to [H₂O₂] . Comparing Exp 2 and 3: doubling [I⁻] while keeping [H₂O₂] constant doubles the rate, making it first order with respect to [I⁻] .
(ii) Write the rate equation
✅ Correct Answer
Rate = k[H₂O₂][I⁻]
(iii) Calculate the rate constant, k
📐 Step-by-Step Calculation (Experiment 3)
- Rearrange: k = Rate / ([H₂O₂][I⁻])
- Substitute values: 3.56×10⁻⁶ / (0.200 × 0.100)
- Evaluate: 1.78×10⁻⁴
- Units: dm³ mol⁻¹ s⁻¹
❌ Common Errors
Forgetting or messing up units for a second-overall-order rate constant. Ensure your units cancel out correctly: (mol dm⁻³ s⁻¹) / (mol dm⁻³ × mol dm⁻³) = dm³ mol⁻¹ s⁻¹ .
Gas Tests and Colour Explanations
(i) Gas test for oxygen
✅ Correct Answer
Add a glowing splint into the gas / bubbles. Result: The splint relights.
(ii) Yellow/brown foam colour explanation
✅ Correct Answer
The colour is caused by the formation of aqueous iodine ( I₂(aq) or I₃⁻ ), produced because excess hydrogen peroxide oxidises iodide ions.
💡 Key Knowledge
Iodide ions ( I⁻ ) act as a homogeneous catalyst. Intermediate steps form iodine, which imparts a yellow-brown hue to the mixture before it reacts further or dilutes in the foam.
Gradient Determination and Activation Energy ( Eₐ )
(i) Determine the gradient of the graph
📐 Working Out
- Draw a large triangle on the line of best fit showing clear coordinate selections.
- Use coordinates from your triangle: e.g., Δy = -11 - (-6) = -5 and Δx = 0.00358 - 0.003038 = 0.00055 .
- Calculate gradient: -5 / 0.00055 = -9090.9 (Acceptable range: -8500 to -9500 ).
- Units: K (or no units since it's dimensionless on inverted axes).
❌ Common Errors
Drawing triangles that are too small (less than half the line length) introduces significant reading inaccuracies. Always pick points far apart on your constructed triangle.
(ii) Calculate the activation energy, Eₐ
📐 Calculation Steps
From the Arrhenius equation provided: ln k = -(Eₐ / R) × (1 / T) + constant , the gradient equals -Eₐ / R .
- Set up equation: -9090.9 = -Eₐ / 8.31
- Solve for Eₐ : 9090.9 × 8.31 = 75545 J mol⁻¹
- Convert to kJ mol⁻¹ : 75.5 kJ mol⁻¹
🧠 Exam Technique
Watch out for units! R is given in J mol⁻¹ K⁻¹ , so your initial calculation for Eₐ will be in joules ( J mol⁻¹ ). You must divide by 1000 to get kJ mol⁻¹ unless specified otherwise.
Calculating the Collision Factor (A)
📐 Step-by-Step Calculation
- Equation: k = A × e^(-Eₐ / RT) rearranged gives A = k / e^(-Eₐ / RT)
- Calculate exponent term (-Eₐ / RT):
-5.02×10⁴ / (8.31 × 370) = -16.326...
Therefore, e^(-16.326) = 8.1164 × 10⁻⁸ - Substitute into rearranged A expression:
A = (1.60 × 10⁻³) / (8.1164 × 10⁻⁸) - Final Answer: 19713 or 1.97 × 10⁴
💡 Key Knowledge
The pre-exponential factor A (or collision factor) relates to the frequency of collisions with the correct spatial orientation. Because units were explicitly stated as not required, you only need to provide the numerical value.
Topics
Physical Chemistry · Inorganic Chemistry · Organic Chemistry · Topic 2: Bonding and Structure · Topic 3: Redox I · Topic 16: Kinetics II · Topic 4: Inorganic Chemistry and the Periodic Table
Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 2, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.