AQA A-Level Chemistry Paper 3, 2024: Question 2

✅ Correct answers (what to write)

• Instead of leaving the flask open: to prevent acid/solution/liquid escaping (e.g. splashing/spraying/spitting out).
• Instead of inserting a bung: to allow CO₂ gas to escape.

💡 Key knowledge

• Cotton wool acts like a splash guard but is not gas-tight.
• Continuous monitoring relies on CO₂ escaping so the balance reading decreases.

🧠 Exam technique

• Answer both prompts separately: “instead of leaving open…” and “instead of inserting a bung…”.
• Use the words acid/solution for the first and CO₂ gas for the second to match the mark scheme.

❌ Common errors (examiner insight)

• Writing “to stop evaporation” or “to stop spillage” on its own is typically ignored unless it clearly links to acid/solution escaping.
• For the bung: “pressure would build up” is ignored by the mark scheme—state the key point: CO₂ must escape.
• Do not mention the wrong gas (e.g. “oxygen escapes”). The scheme warns against wrongly identified gases.

✅ Correct answer

Using a large excess of CaCO₃ means the amount/mass (and therefore surface area) of CaCO₃ stays approximately constant, so the only changing factor affecting rate is the changing concentration of HCl.

💡 Key knowledge

• For a solid reactant, the “concentration” idea is replaced by surface area available.
• If CaCO₃ is in large excess, it does not become limiting during the time window.

❌ Common errors (examiner insight)

• “The concentration stays constant” (without saying what) is usually not credited.
• “HCl is the limiting reagent” on its own is not the mark point here unless linked to CaCO₃ surface area staying constant.
• “Volume stays constant” is ignored in the mark scheme for this mark.

💡 Key knowledge (what the symbols mean)

• mₜ = mass of CO₂ produced in time t .
• m_total = total mass of CO₂ produced when reaction finishes.
• So m_total − mₜ = mass of CO₂ still to be produced after time t .

✅ Full-mark explanation (2 clear points)

1. The mass of CO₂ produced by time t ( mₜ ) is proportional to the amount (and therefore concentration) of HCl that has reacted by time t .
2. The total CO₂ produced ( m_total ) is proportional to the total amount (concentration) of HCl present initially (HCl “added”). Therefore m_total − mₜ is proportional to the amount (concentration) of HCl remaining at time t .

🧠 Exam technique

• Use the word proportional (or “equal to/represents” as allowed) and explicitly mention HCl reacted vs HCl remaining.
• It helps to phrase it as: “CO₂ made ↔ HCl used up”, then “CO₂ still to be made ↔ HCl still to react”.

❌ Common errors (examiner insight)

• Only saying “as HCl decreases, rate decreases” doesn’t earn these marks—you must link the mass terms to amount of HCl.
• Missing the “total initially present” step often loses the second mark.

✅ What a full-mark graph must show

• Axes labelled with quantities and units:
  • x-axis: (m_total − mₜ) / g
  • y-axis: Rate of reaction / g s⁻¹
• All five points plotted accurately (within about half a small square).
• A suitable straight line of best fit (should be within one square of all points, allowing one slight outlier).

📐 Plotting data (the five coordinate pairs)

Plot rate on y and (m_total − mₜ) on x:

• (0.340, 23.0 × 10⁻⁴)
• (0.280, 19.0 × 10⁻⁴)
• (0.225, 15.7 × 10⁻⁴)
• (0.170, 11.5 × 10⁻⁴)
• (0.100, 6.67 × 10⁻⁴)

🧠 Exam technique (how to secure M1–M3)

1. Scales (M1): choose scales so the plotted points take up >50% of each axis. (Don’t squash into one corner.)
2. Plot (M2): use sharp crosses (×), consistent size.
3. Line (M3): draw a single straight best-fit line (not dot-to-dot). Use a ruler.

❌ Common errors (directly from mark scheme guidance)

• Reversing axes can lose the scale mark (though you may still gain plotting/line marks).
• Forgetting units on axes loses credit for good labelling.
• Drawing a curve (because the original graph was curved) loses the straight-line mark here.

✅ Correct statement

The graph is a straight line through the origin, showing rate is directly proportional to (m_total − m_t), and since (m_total − m_t) ∝ [HCl], then Rate ∝ [HCl] so Rate = k[HCl].

❌ Common errors

• Only saying “straight line” (without “through the origin”) is not enough for the mark.
• Only saying “directly proportional” may be ignored if you don’t tie it to the graph feature (origin).

✅ Any two acceptable answers

• Volume of CO₂ gas produced (e.g. gas syringe) over time
• pH over time
• Concentration of HCl/acid/H⁺ over time (e.g. titration samples or probe)
• Conductivity over time

🧠 Exam technique

• State an actual measurable quantity: “volume of CO₂”, not just “volume”.
• Avoid “time taken for CaCO₃ to disappear” because CaCO₃ is in excess (and it’s explicitly not credited).

❌ Common errors (examiner insight)

• “Temperature” is not an alternative variable here (and is rejected by the mark scheme).
• “Concentration” without saying of what is not credited.
• “Amount of CO₂” is too vague; use volume of CO₂ (or pressure with appropriate apparatus, but stick to mark scheme-listed answers for safety).

💡 The logic chain the question builds

• Mass loss measures CO₂ produced.
• (m_total − mₜ) measures CO₂ still to be produced.
• CO₂ still to be produced ∝ HCl still to react ∝ [HCl] at time t.
• If rate vs (m_total − mₜ) is linear through origin, then rate ∝ [HCl] → Rate = k[HCl].

🧠 What distinguishes top responses

• They explicitly use proportionality and connect each step to the chemistry (stoichiometry + limiting reagent).
• Graphs are easy to read: large scale, labelled units, best-fit straight line (not join-the-dots).
• They answer exactly what is asked (e.g. origin + straight line, not extra unrelated discussion).