AQA A-Level Chemistry Paper 3, 2024: Question 15

Rate from a gas-volume–time graph (tangent method)

AQA A-Level Chemistry • Graph skills • 1 mark

What this question tests

Skills & knowledge

Interpreting a volume of gas vs time graph for Mg + 2HCl → MgCl₂ + H₂.
Finding the instantaneous rate at a specific time using a tangent.
Calculating a gradient with correct units: cm³ s⁻¹ .

Examiner focus

Choosing two clear points on the tangent (not on the curve markers).
Using a large triangle to reduce reading error.
Matching to the closest multiple-choice option.

Mark scheme outcome: Q15 = A (AO2), value 0.4 cm³ s⁻¹.

Part (a)

Instantaneous rate at 70 s

Find the gradient of the tangent at t = 70 s

✅ Correct answer (1 mark)

A = 0.4 cm³ s⁻¹

This is the closest value to the gradient of the tangent to the curve at 70 s.

💡 Key knowledge

Rate from a graph = gradient = Δ(volume) / Δ(time).
At a single time (e.g. 70 s), you need the instantaneous rate → draw a tangent.
Units: volume is in cm³ , time is in s ⇒ rate is cm³ s⁻¹ .

📐 Calculations (how you’d get ≈ 0.4)

Locate 70 s on the x-axis.
Draw a tangent that just touches the curve at 70 s (same local slope).
Pick two far-apart points on the tangent (not the original plotted crosses), e.g. points roughly around:
Example read-off (will vary slightly): (50 s, 56 cm³) and (90 s, 72 cm³)
Compute gradient:
rate = ΔV/Δt = (72 − 56) / (90 − 50) = 16/40 = 0.40 cm³ s⁻¹
Select the closest option: 0.4.

Your exact two points may differ slightly depending on your tangent, but the gradient should be closest to 0.4.

🧠 Exam technique (how to secure the mark)

Do not join two data points around 60 s and 80 s and call it the rate at 70 s; that gives an average rate, not instantaneous.
Make your tangent and your gradient triangle as large as possible to reduce uncertainty.
When the curve is flattening, expect a smaller rate than earlier times (sanity check).

❌ Common errors (what loses the mark)

Using a chord between two plotted points instead of a tangent at 70 s.
Choosing points on the curve crosses rather than on the tangent line you drew.
Using a tiny triangle (e.g. 5 s wide), causing large reading errors → wrong option.
Unit slips: forgetting it is cm³ s⁻¹ (not cm³ or s⁻¹ alone).
Mistaking the y-value at 70 s (volume) for the rate.

Mark breakdown: 1 mark (AO2). Awarded for selecting the correct closest rate value: A (0.4).
Examiner insight: Full-mark responses typically show a clear tangent and a large, well-chosen gradient triangle. Many incorrect answers come from calculating an average gradient between neighbouring points (often giving values nearer 0.9 or 1.1 earlier in the reaction).

Quick self-check

Does your answer make sense?

Reasonableness checks

At early times the curve is steep → rate is larger.
By ~70 s the curve is starting to flatten → rate should be moderate and trending down.
Near the plateau (~140–180 s), the rate should approach 0.

Multiple-choice strategy

If your gradient is around 0.35–0.45, pick 0.4.
If you got ~1.1 or ~2.5 at 70 s, you’ve probably used too early a part of the curve or calculated the wrong gradient.