Edexcel A-Level Chemistry Paper 3, June 2025: Question 4

✅ Model Answer

Description: Rinse the pipette with (deionised/distilled water, followed by) the 2.00 mol dm⁻³ sulfuric acid solution. [1 Mark]

Justification: To avoid diluting the acid solution (so its concentration remains precisely 2.00 mol dm⁻³) / to ensure the droplets do not react with the reagents. [1 Mark]

🧠 Exam Technique & Dependency

• Dependent Marks: Mark 2 is strictly dependent on scoring Mark 1. If you suggest rinsing with water only, you cannot get either mark!
• Always specify the exact identity and concentration of the solution being transferred when stating what to rinse with.

❌ Common Student Errors

• Rinsing only with water: Leaving water droplets inside dilutes the acid, decreasing its concentration and spoiling the kinetic rate measurement.
• Drying with paper towels: Paper can leave fibres that block pipette tips or contaminate the liquid.

✅ Correct Answers

(i) Why quenching is necessary [2 Marks]:

• To stop (or freeze) the reaction at that specific time. [1 Mark]
• So that there is no further change in concentration while the titration is carried out. [1 Mark]

(ii) Reagent and Equation [2 Marks]:

Compound: Sodium hydrogencarbonate / NaHCO₃ (or KHCO₃ / Na₂CO₃ / K₂CO₃) [1 Mark]

Equation: [1 Mark]

NaHCO₃ + H⁺ → Na⁺ + CO₂ + H₂O

or: 2NaHCO₃ + H₂SO₄ → Na₂SO₄ + 2CO₂ + 2H₂O

or ionic: HCO₃⁻ + H⁺ → CO₂ + H₂O

💡 Key Knowledge: How Quenching Works

• The iodination of propanone is acid-catalysed by H⁺ ions.
• Adding a weak base (like NaHCO₃) removes the H⁺ catalyst by converting it into CO₂ and H₂O. Without the acid catalyst, the rate becomes negligibly slow.
• Sodium hydroxide is generally avoided because strongly alkaline conditions cause iodine to disproportionate ( I₂ + 2OH⁻ → I⁻ + IO⁻ + H₂O ).

❌ Examiner Pitfalls to Avoid

• Equilibrium Trap: Do not say "to freeze the position of equilibrium". This is not a reversible equilibrium reaction; it is a one-way kinetic process! The mark scheme specifically states: "Ignore references to stopping/freezing an equilibrium".
• Vague accuracy statements: Stating just "to make the titration accurate" scores 0. You must state that concentration would otherwise keep changing.

✅ Model Answer (Any 2 points)

• Risk vs Hazard: The overall risk depends on both the hazard of the substance and the level of exposure / quantity formed. [1 Mark]
• Low yield/quantity: Only a very small amount of iodopropanone is formed in the experiment (due to the very low initial concentration/moles of I₂). [1 Mark]
• Remains in solution: The iodopropanone remains dissolved in aqueous solution / has a high boiling point / is not volatile. [1 Mark]

🧠 Exam Technique: Hazard vs Risk

Examiners look for the clear distinction:

• Hazard: The intrinsic potential to cause harm (e.g. corrosive, lachrymator).
• Risk: The likelihood of harm occurring under actual working conditions (controlled by low quantities and high dilution).

📐 Step-by-Step Calculation

Step 1: Calculate moles of I₂ initially added

Moles = volume (dm³) × concentration

n(I₂) = (50.0 / 1000) × 0.0200 = 1.00 × 10⁻³ mol

Step 2: Calculate moles of propanone initially added

n(CH₃COCH₃) = (25.0 / 1000) × 2.00 = 0.0500 mol

Step 3: Compare moles and deduce constant concentration

Compare: 0.0500 mol >> 1.00 × 10⁻³ mol (a 50-fold excess!).

Therefore, propanone is in great excess, meaning its concentration remains effectively constant (negligible change).

Step 4: Conclude for the rate

Since [propanone] and [H⁺] remain virtually unchanged, any change in rate is solely due to the changing concentration of iodine.

❌ Common Errors

• Forgetting cm³ to dm³: Forgetting to divide by 1000 leads to incorrect powers of 10.
• Qualitative statements: Writing "propanone is in excess" without doing both mole calculations scores 0 out of 4. The question command word is "Show, by calculating...".
• Missing the deduction: Simply finding the two mole numbers without explicitly stating that 0.0500 >> 1.00 × 10⁻³ misses Mark 3 and 4.

(i) Plotting the Graph [2 Marks]

• Axes & Scale: x-axis = Time / min (scale 0 to 14), y-axis = Titre / cm³ (scale 0 to 20 or 8 to 18). Must occupy more than 50% of the grid. [1 Mark]
• Plotting & Best Fit: Points plotted within ± half a small square:
  (2, 17.00), (4, 15.60), (6, 14.00), (8, 12.50), (10, 11.00), (12, 9.40)
• Draw a single straight line of best fit through the points with a ruler. [1 Mark]

(ii) Reason [1 Mark]

Why is it unnecessary to calculate [I₂]?

The titre (volume of Na₂S₂O₃) is directly proportional to the concentration of iodine.

(iii) Order Deduction & Justification [2 Marks]

Order with respect to iodine: 0 (Zero order) [1 Mark]

Justification: The graph is a straight line / has a constant gradient (rate does not change as concentration falls). [1 Mark]

💡 Distinguishing Kinetics Curves

• Zero Order (Conc vs Time): Straight line with a negative gradient. Constant rate ( rate = k[A]⁰ = k ).
• First Order (Conc vs Time): Downward curve with a constant half-life ( t₁/₂ = ln 2 / k ).
• Second Order (Conc vs Time): Steep initial drop with successive half-lives doubling.

❌ Common Examiner Criticisms

• Connecting dots: Students often join data points dot-to-dot rather than drawing a single smooth straight line of best fit.
• Confusing Rate vs Conc graphs: In a Rate vs [Conc] graph, zero order is a horizontal flat line. In a [Conc] vs Time graph, zero order is a straight line with a constant negative downward gradient. Do not confuse the two!
• M2 Dependency: Stating "constant gradient" without deducing zero order (or vice versa) loses marks.