Edexcel A-Level Chemistry Paper 2, November 2020: Question 3
6 marks · Medium difficulty · Short Open Response
Explain whether the reaction of bromate ions with bromide ions is disproportionation, determine its overall order, and calculate the maximum volume of oxygen produced from the thermal decomposition of potassium bromate.
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Question text
3 This question is about the compound potassium bromate, KBrO3.
(a) These bromate ions react with bromide ions in acidic solution.
BrO−(aq) + 5Br−(aq) + 6H+(aq) → 3Br (aq) + 3H O(l)
32 2
(i) Explain, in terms of oxidation numbers, whether or not this is a
disproportionation reaction.
(2)
(ii) What is the overall order of this reaction?
(1)
A 3
B 6
C 12
D cannot tell from this information
(b) Potassium bromate decomposes on heating.
2KBrO3 → 2KBr + 3O2
Calculate the maximum volume of oxygen, in dm3, measured at room temperature
and pressure (r.t.p.), that could be produced from the complete decomposition
of 5.20 g of potassium bromate.
[Molar volume of gas at r.t.p. = 24.0 dm3 mol−1]
(3)
*P62669A0832*
(Total for Question 3 = 6 marks)
Mark scheme
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How to answer it
Reactions of Potassium Bromate (KBrO₃)
Assigning oxidation numbers to identify redox processes, understanding the definition of a disproportionation reaction, recognizing kinetic principles (stoichiometry vs. rate equations), and performing multi-step stoichiometry calculations involving gas molar volumes at r.t.p.
Part (a)(i): Disproportionation Analysis
Explain, in terms of oxidation numbers, whether or not this is a disproportionation reaction. (2 marks)
✅ Correct Answer
Not a disproportionation reaction.
- Oxidation number of Br in BrO₃⁻ is +5 .
- Oxidation number of Br in Br⁻ is -1 .
- Oxidation number of Br in Br₂ is 0 .
- Reason: Two different bromine-containing species ( BrO₃⁻ and Br⁻ ) are reacting to form a single oxidation state ( 0 in Br₂ ), rather than a single species being simultaneously oxidized and reduced. (This is the reverse process: comproportionation).
💡 Key Knowledge
A disproportionation reaction occurs when a single element in a single chemical species is simultaneously oxidized and reduced. Here, you have two different starting reactants containing bromine, making this a comproportionation reaction instead.
❌ Common Errors
- Failing to state clearly whether it is or isn't disproportionation at the beginning of the explanation.
- Inversely assigning oxidation states (e.g., mixing up +5 and -1 ).
- Modifying or discussing oxidation numbers of hydrogen ( +1 ) or oxygen ( -2 ), which do not earn credit here.
🧠 Exam Technique
Always structure your answer in two clear parts to secure both marks: (1) State the exact oxidation numbers for all bromine species involved, and (2) explicitly link those numbers back to the definition of disproportionation.
Part (a)(ii): Overall Order of Reaction
What is the overall order of this reaction? Multiple choice. (1 mark)
✅ Correct Answer
D — cannot tell from this information
💡 Key Knowledge
Stoichiometric coefficients in a balanced equation (like 1, 5, and 6) never automatically give the reaction order. Reaction orders can only be determined experimentally.
❌ Common Errors
- Choosing B (6) or C (12) by mistakenly adding up the stoichiometric coefficients from the equation. This is a classic trap for students confusing balanced equation moles with rate-determining steps or overall orders.
🧠 Exam Technique
If you see a balanced equation and are asked for the rate equation or overall order without experimental concentration-time or initial rate data, the answer is always that it cannot be determined.
Part (b): Gas Volume Calculation
Calculate the maximum volume of oxygen, in dm³, at r.t.p. from the complete decomposition of 5.20 g of potassium bromate. (3 marks)
📐 Step-by-Step Calculation
- Calculate the molar mass of KBrO₃:
Molar Mass = 39.1 + 79.9 + (3 × 16.0) = 167.0 g mol⁻¹ - Calculate the moles of KBrO₃:
Moles = Mass / Molar Mass = 5.20 / 167.0 = 0.031138 mol - Use the stoichiometric ratio to find moles of O₂:
Balanced equation: 2KBrO₃ → 2KBr + 3O₂
Ratio is 2 mol KBrO₃ : 3 mol O₂
Moles of O₂ = 0.031138 × (3 / 2) = 0.046707 mol - Calculate the volume of O₂ at r.t.p.:
Volume = Moles × Molar Volume (24.0 dm³ mol⁻¹) = 0.046707 × 24.0 = 1.12 dm³ (or 1.12096 dm³)
❌ Common Calculation Traps
- Inverted mole ratio: Multiplying by (2/3) instead of (3/2).
- Atomic mass errors: Forgetting to multiply oxygen's mass by 3 in the molar mass calculation.
- Significant Figures: Answers rounding to 2 s.f. (1.1) or using unrounded intermediate values correctly. Examiners accept 2 to 4 significant figures, but avoid 1 s.f.
🧠 Exam Technique & Error Carried Forward (TE)
Always show your working clearly line-by-line. If you make a minor arithmetic error in step 1, examiners use Error Carried Forward (TE), meaning subsequent correct steps using your incorrect value will still earn the remaining marks!
Topics
Physical Chemistry · Inorganic Chemistry · Topic 3: Redox I · Topic 5: Formulae, Equations and Amounts of Substance · 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, November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.