OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2025: Question 18

✅ Model Answer

Platinum has a different physical state / phase to the reactants.

(Alternatively: Platinum is a solid, whereas the reactants are gases.)

❌ Common Traps & Examiner Notes

• Mentioning products: Do NOT refer to products. Catalysts interact with the reactants. Mentioning "different phase to reactants and products" is often ignored, but saying only "different phase to products" scores 0.
• Vague descriptions: Simply saying "it is in a different state" without specifying "to the reactants" can lose the mark.

🖌 How to Draw the Boltzmann Distribution

• Origin: The curve must start at the origin (within the first tiny grid square at 0,0).
• Asymmetry: The peak should be skewed to the left, followed by a long tail to the right.
• High-Energy Tail: Must never touch or cross the x-axis at high energy (it is asymptotic). Must not become completely horizontal for more than 1 large grid square.
• Labelling Activation Energies: Draw two vertical lines on the energy axis:
  • Uncatalysed: E_a further to the right.
  • Catalysed: E_cat (or E_c ) to the left of E_a .
• Shading: Shade the area under the curve to the right of E_cat to show the larger fraction of molecules.

✅ Written Explanation

• The platinum catalyst provides an alternative reaction pathway with a lower activation energy ( E_cat < E_a ).
• A greater proportion / more of the molecules have energy equal to or greater than the activation energy ( E ≥ E_a ).

❌ Mark Traps to Avoid

• Do NOT draw multiple curves: Temperature is constant. Drawing a second flattened curve scores 0 for the graph.
• "Molecules have more energy": ❌ INCORRECT. Catalysts do not give molecules more energy (only temperature does that). Catalysts lower the energy requirement threshold!
• "More collisions": Insufficient. Collision frequency does not change significantly; only the fraction of successful collisions increases.

🧠 Examiner Tip

Always use the word proportion or fraction: "A greater proportion of molecules have energy ≥ E_a." This guarantees the mark and prevents ambiguous phrasing.

✅ Accepted Answers (Any One)

• Lower operating temperatures can be used to achieve the same reaction rate/yield.
• Less energy demand / less electricity or fuel needed to heat the reaction.
• Less fossil fuels burnt, resulting in a reduction in CO₂ emissions.
• The catalyst is not consumed and can be reused, reducing waste.

🧠 Exam Strategy

Link the chemistry directly to environmental impact: Lower temperature required → less energy needed → fewer fossil fuels burnt → reduced CO₂ / greenhouse gas emissions.

Step 1: Determine order with respect to O₂
Compare Experiment 1 and Experiment 2 where [NO] is held constant at 0.040 mol dm⁻³ :

• [O₂] increases from 0.035 to 0.070 mol dm⁻³ (× 2)
• Rate increases from 7.45 × 10⁻⁴ to 1.49 × 10⁻³ mol dm⁻³ s⁻¹ (× 2)
• Since doubling concentration doubles the rate: Order with respect to O₂ = 1 (first order).

Step 2: Determine order with respect to NO
Compare Experiment 1 and Experiment 3 where [O₂] is held constant at 0.035 mol dm⁻³ :

• [NO] increases from 0.040 to 0.060 mol dm⁻³ : 0.060 / 0.040 = 1.5 (× 1.5)
• Rate increases from 7.45 × 10⁻⁴ to 1.68 × 10⁻³ mol dm⁻³ s⁻¹ : 1.68 × 10⁻³ / 7.45 × 10⁻⁴ = 2.25 (× 2.25)
• Notice that (1.5)² = 2.25 : Order with respect to NO = 2 (second order).

Step 3: Construct the Rate Equation

Rate = k[NO]²[O₂]

Step 4: Calculate the value of k (using Experiment 1)
Rearrange: k = Rate / ([NO]²[O₂])
k = (7.45 × 10⁻⁴) / ((0.040)² × 0.035)
k = (7.45 × 10⁻⁴) / (0.0016 × 0.035) = (7.45 × 10⁻⁴) / (5.60 × 10⁻⁵) = 13.3035... ≈ 13.3 (or 13)

Step 5: Derive the Units for k
Units of k = (mol dm⁻³ s⁻¹) / ((mol dm⁻³)² × (mol dm⁻³)) = (mol dm⁻³ s⁻¹) / (mol³ dm⁻⁹)
= mol¹⁻³ dm⁻³⁻(⁻⁹) s⁻¹ = dm⁶ mol⁻² s⁻¹

✅ Summary of Answers

• Order wrt O₂: 1
• Order wrt NO: 2
• Rate equation: Rate = k[NO]²[O₂]
• Value of k: 13.3 (accepts 13 to calculator precision)
• Units of k: dm⁶ mol⁻² s⁻¹

❌ Calculation Pitfalls

• Forgetting to square [NO]: Neglecting the exponent in the denominator when calculating k gives 532 instead of 13.3.
• Unit inversions: Always check signs carefully: dm⁶ mol⁻² s⁻¹ (dm has positive index, mol has negative index).

Step 1: Best Fit Line Parameters
Drawing a straight line through the 4 valid points gives coordinates:

• At 1/T = 4.0 × 10⁻⁴ K⁻¹ , ln k ≈ 15.6 (range allowed: 15.3 – 15.7)
• At 1/T = 7.0 × 10⁻⁴ K⁻¹ , ln k ≈ 6.6 (range allowed: 6.3 – 6.8)

Step 2: Calculate the Gradient
Gradient = Δy / Δx = (6.6 - 15.6) / ((7.0 - 4.0) × 10⁻⁴) = -9.0 / (3.0 × 10⁻⁴) = -30,000 K
Mark scheme accepts gradients in the range: -28,300 to -31,300

Step 3: Relate Gradient to Activation Energy (E_a)
From the Arrhenius equation: ln k = -E_a / (R × T) + ln A → Gradient = -E_a / R
Therefore:
E_a = -Gradient × R
E_a = -(-30,000) × 8.314 = +249,420 J mol⁻¹

Step 4: Convert to kJ mol⁻¹ and round to 2 Significant Figures

• Convert J to kJ: 249,420 / 1000 = 249.42 kJ mol⁻¹
• Round to 2 significant figures (explicit question instruction): 250 kJ mol⁻¹

Acceptable range: 240 to 260 kJ mol⁻¹

❌ Fatal Calculation Errors

• Missing the 10⁻⁴ scale factor: The x-axis is labelled × 10⁻⁴ . Forgetting this yields a gradient of -3.0 instead of -30,000.
• Forgetting to divide by 1000: The gas constant R = 8.314 J K⁻¹ mol⁻¹ gives E_a in J mol⁻¹. You must divide by 1000 to get kJ mol⁻¹.
• Negative E_a: Activation energy is always positive. An answer of -250 scores 0 for that mark.
• Wrong Significant Figures: Giving 249 or 249.4 loses the final mark. 2 sig figs are explicitly required!

✅ Final Expected Values

• Straight line of best fit correctly excluding anomaly
• Gradient: -28,300 to -31,300
• E_a: 250 kJ mol⁻¹ (to 2 s.f.)