Edexcel A-Level Chemistry Paper 2, June 2022: Question 9

✅ Correct Answer

Rate = k[CH₃CHO]²

💡 Key Knowledge

The overall order of the reaction is equal to the sum of the individual reactant orders in the rate equation. Since the reaction is second order with respect to ethanal and second order overall, ethanal is the only reactant term raised to the power of 2.

❌ Common Errors

Writing rounded brackets instead of square brackets [ ] , or including "Rate =" twice. Ensure you use square brackets for concentrations.

✅ Correct Answer

dm³ mol⁻¹ s⁻¹ (or mol⁻¹ dm³ s⁻¹ )

📐 Derivation Steps

1. Rearrange the rate equation for k: k = Rate / [CH₃CHO]²
2. Substitute units: k = (mol dm⁻³ s⁻¹) / (mol dm⁻³)².
3. Simplify: k = (mol dm⁻³ s⁻¹) / (mol² dm⁻⁶) = dm³ mol⁻¹ s⁻¹ .

❌ Common Errors

Failing to square the concentration term in the denominator. Always derive the units dynamically from your specific rate equation rather than memorising them blindly.

✅ Correct Answer

0 s to 420 s: 8.6 × 10⁻⁴ to 9.0 × 10⁻⁴ mol dm⁻³ s⁻¹
420 s to 1260 s: 2.0 × 10⁻⁴ to 2.1 × 10⁻⁴ mol dm⁻³ s⁻¹

📐 Step-by-Step Calculation

1. Interval 1 (0 to 420 s): Change in concentration / change in time = (0.72 - 0.36) / (420 - 0) = 8.57 × 10⁻⁴ . Round to 2 significant figures: 8.6 × 10⁻⁴ mol dm⁻³ s⁻¹ .
2. Interval 2 (420 to 1260 s): (0.36 - 0.18) / (1260 - 420) = 0.18 / 840 = 2.14 × 10⁻⁴ . Round to 2 significant figures: 2.1 × 10⁻⁴ mol dm⁻³ s⁻¹ .

🧠 Exam Technique

Pay close attention to the instruction: "Give your answers to an appropriate number of significant figures." The source data is given to 2 significant figures, so your final answers must match.

✅ Correct Answer

Not zero order: The rate is not constant as concentration changes.
Not first order: Successive half-lives are not constant (or rate change is not directly proportional to concentration change; concentration halves from 0.72 to 0.36 in 420 s, but doesn't halve again in the next equal time interval).

💡 Key Knowledge

• Zero order: Rate remains constant; concentration-time graph is a straight line.
• First order: Constant half-life regardless of concentration.
• Second order: Half-life increases as concentration decreases.

🧠 Exam Technique

You must address both statements (why it's not zero AND why it's not first order) to secure all available marks. Explicitly reference your calculated values from part (c) to back up your reasoning.

💡 Completing the Table

Missing 1/T value (at T = 790 K): 1 / 790 = 1.27 × 10⁻³ K⁻¹
Missing ln k value (at k = 0.343): ln(0.343) = -1.07

📐 Calculating Activation Energy (Eₐ)

1. Plotting: Plot ln k (y-axis) against 1/T (x-axis). Ensure points occupy more than half the grid. Draw a line of best fit.
2. Gradient: Determine the gradient from your line. Examiner expected value: gradient = -21250 K (units are K since y is dimensionless and x is K⁻¹).
3. Relate to Arrhenius: From ln k = -(Eₐ / R) × (1/T) + constant , the gradient equals -Eₐ / R .
4. Calculate Eₐ: Eₐ = - (gradient × R) = -(-21250 × 8.31) = 176587 J mol⁻¹ = 177 kJ mol⁻¹ .

❌ Common Errors & Traps

• Forgetting to divide by 1000 when converting J mol⁻¹ into kJ mol⁻¹.
• Omitting the negative sign on the gradient or failing to give a positive final activation energy.
• Missing units for the gradient (K) or activation energy (kJ mol⁻¹).