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

Skills & knowledge

• Calculate 1/T and ln k correctly, using correct significant figures.
• Use units of k to deduce overall reaction order.
• Plot ln k vs 1/T accurately and draw a best-fit straight line.
• Use Arrhenius form ln k = ln A − Eₐ/(RT) to get Eₐ from gradient.
• Deduce products of a thermal decomposition to an alkene + carbonyl.

How marks are won

• Numerical accuracy is essential (mark scheme allows very specific rounding).
• Graph marks are for points (within ½ square) and best-fit line (within 1 square).
• Eₐ method mark: explicitly link gradient = −Eₐ/R .

✅ Correct answers (573 K)

1/T = 1.75 × 10⁻³ K⁻¹

ln k = −6.46

Mark scheme note: allows 0.00175 . Rejects other rounding such as 1.74 × 10⁻³ and rejects “odd” sig figs like 1.7 × 10⁻³ or −6.5 .

📐 Calculations (what you should do)

1. Compute 1/T: 1/573 = 0.001745… K⁻¹
2. Round to match table style: 1.75 × 10⁻³ K⁻¹
3. Compute ln k using k in s⁻¹: k = 15.6 × 10⁻⁴ s⁻¹ = 1.56 × 10⁻³ s⁻¹
4. ln(1.56 × 10⁻³) = −6.46 (to 2 d.p. like the table)

❌ Common errors (why students lose this mark)

• Rounding 1/T incorrectly (e.g. writing 1.74 × 10⁻³). The mark scheme is strict here.
• Using log (base 10) instead of ln (natural log).
• For ln k: forgetting to convert 15.6 × 10⁻⁴ into standard form first (still possible, but more error-prone).

✅ Correct answer

Piece of information: the unit of k in the table heading is s⁻¹ .

Overall order: first order (order 1).

💡 Key knowledge (units of k)

• Zero order: k has units mol dm⁻³ s⁻¹
• First order: k has units s⁻¹
• Second order: k has units dm³ mol⁻¹ s⁻¹

🧠 Exam technique

• Quote the heading carefully: write “unit of k is s⁻¹”, not just “k is s⁻¹”.
• Then state the order clearly: “therefore first order”.
• The mark scheme says marks are independent: you can still gain M2 if your M1 is clear enough, but don’t rely on that.

❌ Common errors

• Saying “because it’s decomposition it’s first order” (not acceptable).
• Writing “unit is k / s⁻¹” incorrectly or just “k” without unit (mark scheme: NOT just k).

💡 What you are plotting

x-axis: 1/T (K⁻¹)
y-axis: ln k

From the table (including your 573 K values), the plotted coordinate pairs are:

• (1.81 × 10⁻³, −7.68)
• (1.78 × 10⁻³, −7.08)
• (1.75 × 10⁻³, −6.46)
• (1.72 × 10⁻³, −5.88)
• (1.69 × 10⁻³, −5.30)

🧠 Examiner-style plotting advice (to secure both marks)

• M1 (points): all points must be within ½ a small square of the correct location.
• M2 (line): draw a best-fit straight line (not dot-to-dot). The line should pass within about 1 square of each point, with some points above and some below.
• Use a sharp pencil/ruler; extend the line well across the graph area to help with gradient later.

❌ Common graph mistakes

• Joining points one-by-one (this is not a best-fit line).
• Plotting T instead of 1/T (will not be linear as expected).
• Plotting k instead of ln k.
• Misreading scientific notation on the x-axis (e.g. mixing up 1.72 × 10⁻³ and 1.72 × 10⁻²).

💡 Key knowledge (Arrhenius linear form)

ln k = ln A − Eₐ/(RT)

Comparing with y = c + mx:

• y is ln k
• x is 1/T
• gradient m = −Eₐ/R

📐 Calculations (step-by-step)

1. Find the gradient using two well-separated points on your best-fit line (not just two data points).
2. Gradient m = Δ(ln k)/Δ(1/T)
3. Expected gradient from mark scheme: m ≈ −19900 (units: K)
4. Use the relationship: m = −Eₐ/R
5. So Eₐ = −m × R
6. Eₐ = −(−19900) × 8.31 = 165000 J mol⁻¹
7. Convert to kJ mol⁻¹: 165000/1000 = 165 kJ mol⁻¹

Accepted range: a value around +161 to +170 kJ mol⁻¹ scores full marks if consistent with your plotted line. Gradient range allowed: approximately −19400 to −20400.

✅ Correct final answer (from mark scheme)

Eₐ = +165 kJ mol⁻¹

Marking logic:
M1: sensible negative gradient from best-fit line
M2: shows/uses gradient = −Eₐ/R
M3: correct Eₐ value in kJ mol⁻¹

❌ Common calculation traps

• Using two table points instead of points on the best-fit line (can shift gradient).
• Forgetting the minus sign (gradient is negative, Eₐ should be positive).
• Using R = 8.31 but leaving Eₐ in J mol⁻¹ and forgetting to divide by 1000.
• Choosing points that are too close together (large percentage uncertainty → wrong gradient).
• If you accidentally get a positive gradient, the mark scheme allows a negative Eₐ only if it follows logically—but aim for the correct negative slope.

✅ Correct structures (any correct representation)

Alkene: CH₂=CHCH₃ (propene)

Carbonyl compound: (CH₃)₂CO or CH₃COCH₃ (propanone)

Mark scheme notes: C=C must be shown for the alkene. Any correct structural form accepted. If you give both correct but the wrong way round → 1 mark.

💡 Key idea to spot quickly

• The decomposition is analogous to splitting an unsaturated alcohol into an alkene + a carbonyl.
• Keep atoms conserved: count carbons on each product side.
• Tertiary alcohol fragments often form a ketone (here: propanone) rather than an aldehyde.

🧠 Exam technique (to avoid losing the 2 marks)

• Write the carbon skeleton first and ensure total C atoms = 6 (same as the reactant).
• For the alkene, explicitly draw/write the double bond: CH₂=CH– must appear.
• For the carbonyl, ensure C=O is present (e.g. CH₃COCH₃ ).

❌ Common errors

• Drawing an alkane (no C=C) → no alkene mark.
• Giving an aldehyde instead of a ketone for this substrate.
• Swapping which product is labelled “alkene” vs “carbonyl” (mark scheme: then only 1 mark even if both structures appear somewhere).

🧠 In the exam, do this

• Q03.1: match the table’s rounding; don’t “over-round”.
• Q03.2: explicitly reference units of k → order.
• Q03.3: plot all 5 points within ½ square; draw a true best-fit line.
• Q03.4: use widely spaced points on the line; show gradient = −Eₐ/R ; convert to kJ mol⁻¹.
• Q03.5: show C=C and C=O clearly; keep labels correct.

✅ Final answers summary

• 573 K: 1/T = 1.75 × 10⁻³ K⁻¹ , ln k = −6.46
• Overall order: 1 (first order) because k has units s⁻¹
• Eₐ ≈ 165 kJ mol⁻¹ (typically 161–170 kJ mol⁻¹)
• Products: CH₂=CHCH₃ and CH₃COCH₃