AQA A-Level Chemistry Paper 2, 2019: Question 4

📐 Step-by-Step Calculation

Initial rate is calculated as: Rate = Δ[P] / Δt

1. Given (Experiment 1):
   Rate = (1.00 × 10⁻² − 0.92 × 10⁻²) / 5.0 = 0.08 × 10⁻² / 5.0 = 1.6 × 10⁻⁴ mol dm⁻³ s⁻¹
2. Experiment 2:
   Δ[P] = (2.00 × 10⁻² − 1.84 × 10⁻²) = 0.16 × 10⁻² mol dm⁻³
   Rate = (0.16 × 10⁻²) / 5.0 = 3.2 × 10⁻⁴ mol dm⁻³ s⁻¹
3. Experiment 3:
   Δ[P] = (0.50 × 10⁻² − 0.34 × 10⁻²) = 0.16 × 10⁻² mol dm⁻³
   Rate = (0.16 × 10⁻²) / 5.0 = 3.2 × 10⁻⁴ mol dm⁻³ s⁻¹

✅ Correct Answer

❌ Common Errors

• Forgetting to divide by the time interval (Δt = 5.0 s), giving 0.16 × 10⁻² instead of 3.2 × 10⁻⁴.
• Misreading standard index notation on calculators (e.g. entering 0.0032 instead of 0.00032).

🧠 Exam Technique

Always verify the formula by testing it on the row that was already filled in for you (Experiment 1). If your calculation matches 1.6 × 10⁻⁴, your method is guaranteed correct.

💡 Key Knowledge

• Zero order ([A]⁰): Changing [A] has no effect on rate.
• First order ([A]¹): Rate is directly proportional to [A] (doubling [A] doubles rate).
• Second order ([A]²): Rate is proportional to [A]² (doubling [A] quadruples rate).

📐 Logic & Deduction

Order with respect to P:

• Compare Experiment 1 and Experiment 2 where [Q] is constant (1.25 × 10⁻² mol dm⁻³).
• [P] doubles from 1.00 × 10⁻² to 2.00 × 10⁻² (×2).
• Rate doubles from 1.6 × 10⁻⁴ to 3.2 × 10⁻⁴ (×2).
• Therefore, order with respect to P = 1.

Order with respect to Q:

• Compare Experiment 1 and Experiment 3:
• [P] halves (×0.5). Because P is 1st order, this factor alone would halve the rate to (1.6 × 10⁻⁴ × 0.5) = 0.8 × 10⁻⁴.
• The observed rate in Expt 3 is 3.2 × 10⁻⁴, which is 4 times faster than 0.8 × 10⁻⁴ (3.2 × 10⁻⁴ / 0.8 × 10⁻⁴ = 4).
• At the same time, [Q] doubles from 1.25 × 10⁻² to 2.50 × 10⁻² (×2).
• Doubling [Q] causes a 4-fold increase in rate (×2² = 4).
• Therefore, order with respect to Q = 2.

✅ Correct Answer

Order with respect to P: 1

Order with respect to Q: 2

❌ Common Errors & Misconceptions

• Failing to account for P when finding Q: Students notice that the rate in Expt 2 and Expt 3 is identical (3.2 × 10⁻⁴) and jump to the false conclusion that Q is zero order, completely overlooking that [P] dropped fourfold from Expt 2 to Expt 3!
• Always isolate the effect of one reagent at a time when both change simultaneously.

📐 Step-by-Step Calculation

1. Write the rate equation:
   Rate = k[R]²[S]²
2. Rearrange to solve for k [Mark 1]:
   k = Rate / ([R]² × [S]²)
3. Substitute numerical values:
   Rate = 1.20 × 10⁻³
   [R] = 1.00 × 10⁻² ⇒ [R]² = (1.00 × 10⁻²)² = 1.00 × 10⁻⁴
   [S] = 2.45 × 10⁻² ⇒ [S]² = (2.45 × 10⁻²)² = 6.0025 × 10⁻⁴
   Denominator = (1.00 × 10⁻⁴) × (6.0025 × 10⁻⁴) = 6.0025 × 10⁻⁸
4. Calculate k [Mark 2]:
   k = (1.20 × 10⁻³) / (6.0025 × 10⁻⁸) = 19992
   In standard form: 2.00 × 10⁴ (or 1.99 × 10⁴ to 3 s.f.)

🧠 Unit Derivation Step-by-Step [Mark 3]

Substitute concentration units into rearranged rate equation:

Units of k = (mol dm⁻³ s⁻¹) / ((mol dm⁻³)² × (mol dm⁻³)²)

• Denominator power = (mol dm⁻³)⁴ = mol⁴ dm⁻¹²
• Cancel one (mol dm⁻³):
  = s⁻¹ / (mol dm⁻³)³ = s⁻¹ / (mol³ dm⁻⁹)
• Invert denominator powers to bring to numerator:
• mol³ ⇒ mol⁻³
• dm⁻⁹ ⇒ dm⁹
• Resulting units: mol⁻³ dm⁹ s⁻¹

✅ Final Marks Breakdown

• k: 2.00 × 10⁴ (or 19992 / 1.99 × 10⁴)
• Units: mol⁻³ dm⁹ s⁻¹

❌ Common Calculation Traps

• Forgetting to square the concentrations: The prompt explicitly states second order w.r.t both R and S. Omitting the ² powers gives k = 4.90, which loses M2 and leads to incorrect units.
• Sign errors in indices: Writing mol³ dm⁻⁹ s⁻¹ instead of mol⁻³ dm⁹ s⁻¹. Remember that dividing by a negative index makes it positive!
• Bracket errors on calculator: Not enclosing the denominator in brackets, resulting in: Rate / [R]² × [S]².