Edexcel A-Level Chemistry Paper 2, June 2025: Question 9

📐 Step-by-Step Calculation

1. Calculating Temperature (T):

1/T = 3.29 × 10⁻³ K⁻¹

T = 1 / (3.29 × 10⁻³) = 303.95 K ≈ 304 K

2. Calculating Rate Constant (k):

ln k = -8.4

k = e⁻⁸·⁴ = 2.2487 × 10⁻⁴ ≈ 2.25 × 10⁻⁴ dm³ mol⁻¹ s⁻¹

✅ Correct Answer

❌ Common Errors & Traps

• Significant Figure Penalty: The table provides data to 3 significant figures (e.g., 298, 3.36 × 10⁻³, 1.36 × 10⁻⁴). Writing unrounded numbers (303.95 or 2.2487 × 10⁻⁴) loses a mark.
• Exponential button misuse: Miscalculating e⁻⁸·⁴ as 10⁻⁸·⁴. Remember that natural logarithm (ln) has base e, not 10.

🧠 Exam Technique

Always inspect the other columns in a data table before writing your answer. If all values are given to 3 SF, match that precision exactly.

💡 The Arrhenius Relationship

The logarithmic form of the Arrhenius equation is:

ln k = - (E_a / R) × (1/T) + constant

Comparing to the linear equation y = mx + c :

• y-axis: ln k
• x-axis: 1/T (in K⁻¹)
• Gradient (m): - E_a / R

📐 Gradient & E_a Calculation Steps

1. Choose coordinates on the line of best fit:
   E.g., (x₁, y₁) = (3.03 × 10⁻³, -6.50) and (x₂, y₂) = (3.37 × 10⁻³, -9.00).
2. Calculate Gradient:
   m = Δy / Δx = (-9.00 - (-6.50)) / (3.37 × 10⁻³ - 3.03 × 10⁻³)
   m = -2.50 / (3.4 × 10⁻⁴) = -7353 K
   (Acceptable range: -6953 to -7753 K)
3. Calculate E_a:
   E_a = - gradient × R
   E_a = - (-7353) × 8.31 = +61 103 J mol⁻¹
4. Convert to kJ mol⁻¹:
   E_a = +61 103 / 1000 = +61.1 kJ mol⁻¹
   (Acceptable range: +57.1 to +65.1 kJ mol⁻¹)

✅ Complete Mark Breakdown (5 Marks)

• Mark 1: Axes labelled correctly with quantities and units ( ln k on y-axis, 1/T / K⁻¹ on x-axis). Note: do NOT write 1/t.
• Mark 2: All points plotted correctly within ±½ small square; sensible scale occupying at least 50% of the grid on both axes; straight line of best fit.
• Mark 3: Gradient clearly calculated from a large triangle showing working. Value within -7353 ± 400 .
• Mark 4 (Stand-alone): Negative sign AND unit K (Kelvin) for the gradient.
• Mark 5: Correct calculation of E_a in kJ mol⁻¹ (gradient × 8.31 ÷ 1000), giving a positive value.

❌ Costly Mistakes Identified by Examiners

• Omitting the negative sign or unit on the gradient: The gradient is downward-sloping, so it is strictly negative. The unit is K (from dimensionless / K⁻¹).
• Forgetting to divide by 1000: The question explicitly requests E_a in kJ mol⁻¹. Leaving it in J mol⁻¹ loses the final mark unless clearly labelled.
• Negative E_a: Activation energy is an energy barrier and must ALWAYS be positive! ( E_a = -m × R cancels the negative gradient).

💡 Unit Deduction Method

For any reaction, the rate equation is:

rate = k [concentration]ⁿ

Rearranging for concentration units:

[concentration]ⁿ = rate / k

📐 Step-by-Step Unit Working

Step 1: Divide rate units by rate constant units

(mol dm⁻³ s⁻¹) / (dm³ mol⁻¹ s⁻¹)

Step 2: Simplify powers

• Seconds cancel: s⁻¹ / s⁻¹ = 1
• Moles: mol / mol⁻¹ = mol²
• Volume: dm⁻³ / dm³ = dm⁻⁶

Resulting units = mol² dm⁻⁶

Step 3: Relate to concentration

mol² dm⁻⁶ = (mol dm⁻³)² , which represents [concentration]². Therefore, the reaction is second order overall.

🧠 Diagram Guide: S_N1 Mechanism

The starting material is 2-bromo-2-methylpropane, (CH₃)₃CBr.

Step 1 (Slow / Rate-Determining Step): Heterolytic Fission

• Show dipole on C—Br bond: C^δ+—Br^δ-.
• Draw curly arrow starting directly from the C—Br bond pointing towards the Br atom (or just beyond it).
• Forms a planar tertiary carbocation intermediate: (CH₃)₃C⁺ and a bromide ion: :Br⁻.

Step 2 (Fast Step): Nucleophilic Attack

• Draw a lone pair of electrons on the oxygen of the hydroxide ion: :OH⁻ (with a negative charge).
• Draw curly arrow from the lone pair on O to the positively charged carbon atom (C⁺) of the carbocation.
• Forms the final alcohol: (CH₃)₃COH (2-methylpropan-2-ol).

✅ Mark Scheme Breakdown

Part (b)(i) — 3 Marks:

• Mark 1 (Step 1): Correct dipole on C—Br bond AND curly arrow from C—Br bond onto Br.
• Mark 2 (Intermediate): Correct structural formula of the tertiary carbocation intermediate (CH₃)₃C⁺ and Br⁻ .
• Mark 3 (Step 2): Curly arrow from lone pair on :OH⁻ to the C⁺ atom of the carbocation.

Part (b)(ii) — 1 Mark:

• State that Step 1 is the slow / rate-determining step (RDS) AND that hydroxide ions (:OH⁻) do not appear in Step 1.