OCR A-Level Chemistry Unified chemistry (03), June 2017: Question 3

✅ Correct Answer

• Rate equation: rate = k[H₂O₂][I⁻]
• Rate constant (k): 0.02 (or 0.020)
• Units: dm³ mol⁻¹ s⁻¹

💡 Key Knowledge

• Comparing experiments 1 and 2: doubling [I⁻] doubles the rate, showing first order with respect to I⁻.
• Comparing experiments 1 and 3: doubling [H₂O₂] doubles the rate, showing first order with respect to H₂O₂.
• Experiment 4 shows doubling [H⁺] has no effect on rate, meaning it is zero order with respect to H⁺.

🧠 Exam Technique

• Always include square brackets around concentrations in rate equations. State symbols are ignored, but format correctly.
• Rearrange your rate equation for k before substituting values to avoid algebraic slips.

❌ Common Errors

• Including [H⁺] in the rate equation (zero-order species should be omitted or written to the power of 0).
• Forgetting to factor in the scale factor given in the table header ( ×10⁻⁶ ) when calculating k .
• Incorrectly deriving or cancelling units for k .

✅ Correct Answer

• Plot a graph of ln k (y-axis) against 1/T (x-axis).
• Measure the gradient of the line.
• Calculate activation energy using Eₐ = (-)R × gradient (where R = 8.314 J K⁻¹ mol⁻¹).

💡 Key Knowledge

">
• Based on the Arrhenius equation: ln k = (-Eₐ / R)(1/T) + ln A
• The gradient of this straight-line graph equals -Eₐ / R .

🧠 Exam Technique

• Explicitly state what goes on each axis so the examiner has zero ambiguity.
• Remember to account for the negative sign: gradients for Arrhenius plots are negative, so multiplying by -R yields a positive value for Eₐ .

❌ Common Errors

• Failing to include units for gas constant R or confusing temperature scales (T must be in Kelvin).
• Forgetting the negative sign when converting gradient to activation energy.

✅ Correct Answer

• Reaction 1 (with MnO₂): H₂O₂ + MnO₂ + 2H⁺ → O₂ + Mn²⁺ + 2H₂O
• Reaction 2 (with Mn²⁺): H₂O₂ + Mn²⁺ → MnO₂ + 2H⁺
• Conclusion: MnO₂ is consumed in the first step and regenerated in the second step, fulfilling the definition of a catalyst.

💡 Key Knowledge

• A catalyst provides an alternative reaction pathway with a lower activation energy.
• Electrode data analysis: The more positive E value system oxidises the less positive system (or apply principles of redox feasibility using electrode potentials).

🧠 Exam Technique

• To score full marks, you must link the electrode potential values to the feasibility of both half-reactions and explicitly state that MnO₂ is reformed at the end.
• Ensure all equations are balanced for mass and charge.

❌ Common Errors

• Listing half-equations without showing how they combine into overall catalytic steps.
• Omitting state symbols or failing to show the regeneration of MnO₂ .

✅ Correct Answer

Displayed or skeletal formula clearly showing the COOOH functional group:

CH₃-C(=O)-O-OH

💡 Key Knowledge

• Peroxycarboxylic acids contain an extra oxygen atom within the functional group compared to standard carboxylic acids ( -C(=O)-O-OH instead of -C(=O)-OH ).

🧠 Exam Technique

• The question specifies that the COOOH functional group must be clearly displayed. Avoid condensing this part into CO₃H .

❌ Common Errors

• Drawing a standard carboxylic acid ( CH₃COOH ) by missing out the bridging peroxy oxygen atom.

✅ Correct Answer

• Kc expression: Kc = [CH₃COOOH][H₂O] / ([H₂O₂][CH₃COOH])
• Equilibrium concentration of product: 0.0925 mol dm⁻³
• Final equilibrium amount: 0.023 (or 0.0231) mol

💡 Key Knowledge

• Water is included in heterogeneous equilibria expressions when present as a liquid if its concentration is considered constant, but here examine the stoichiometry: equation is H₂O₂ + CH₃COOH ⇌ CH₃COOOH + H₂O (1:1 ratio for products and reactants, so [H₂O] = [CH₃COOOH] at equilibrium).
• Concentration = moles / volume in dm³.

🧠 Exam Technique

• Read carefully whether the question asks for concentration or amount in moles. Here, the final step requires converting concentration back to moles using the given volume ( 250 cm³ ).

❌ Common Errors

• Forgetting to square terms or missing out water/reactants in the Kc expression setup.
• Forgetting to convert cm³ to dm³ by dividing by 1000 during the final moles conversion step.