AQA A-Level Chemistry Paper 1, 2021: Question 4

✅ Model Answer by Stages (Level 3: 5–6 Marks)

Stage 1: Definition & Basic Role

• A heterogeneous catalyst is in a different phase/physical state from the reactants (solid Fe vs. gaseous N₂/H₂).
• It speeds up the rate of reaction by providing an alternative reaction pathway with a lower activation energy, remaining chemically unchanged at the end.

Stage 2: Mechanism of Action

• Adsorption: N₂ and H₂ molecules adsorb onto the active sites on the iron surface.
• Reaction: Covalent bonds within N₂ and H₂ weaken; reactant molecules react on the surface to form NH₃.
• Desorption: NH₃ molecules desorb (leave) from the catalyst surface.

Stage 3: Efficiency & Lifetime Factors

• Efficiency: Increased by providing a large surface area (e.g., using a finely divided powder, porous pellets, or a honeycomb/mesh support).
• Lifetime: Reduced by catalyst poisoning (e.g., sulfur impurities in natural gas feedstocks adsorb irreversibly to active sites, blocking them).

💡 Key Knowledge

• Adsorption vs. Absorption: Adsorption is strictly a surface phenomenon. Never write "absorption".
• Active Sites: Specific regions on the transition metal surface where reactant molecules bond using vacant d-orbitals.
• Poisoning: Impurities bond more strongly to active sites than reactants do, which is irreversible and lowers the catalyst's useful lifespan.

🧠 Exam Technique: Level of Response Strategy

To reach Level 3 (5–6 marks), you must cover all three stages coherently:

• Stage 1: Definition of heterogeneous + lowers Eₐ.
• Stage 2: Adsorption → bond weakening/reaction → desorption.
• Stage 3: Increasing surface area + sulfur poisoning blocking active sites.

❌ Common Errors

• Discussing Le Chatelier's principle, equilibrium yield, or temperature/pressure tradeoffs (the mark scheme explicitly instructs: "Ignore references to temperature and pressure").
• Forgetting the final step of the mechanism: desorption. Without desorption, active sites cannot be regenerated.

🧠 Exam Technique

The equations can be written in either order because Fe³⁺ can also act as the initial catalyst. Notice that opposite charges attract (Fe²⁺ + negative ion, Fe³⁺ + negative ion), which eliminates the high activation energy barrier caused by anion-anion repulsion.

❌ Common Errors

• Writing equations that are not balanced for charge: ensure 2 Fe²⁺ balance S₂O₈²⁻.
• Stating "iodine repels peroxodisulfate" instead of mentioning that both ions are negative.

✅ Correct Answer

Zinc has only one oxidation state (+2) in its compounds / Zn²⁺ cannot be oxidised to Zn³⁺ / Zinc does not possess variable oxidation states. [1 mark]

❌ Common Error

Writing merely that "zinc has a full d-subshell". While true ([Ar] 3d¹⁰), the mark scheme specifically notes: "Ignore has a full d shell". The question asks why it cannot catalyse this redox cycle, which requires alternating between oxidation states.

Step 1: Calculate moles of Iron (Fe)
Amount of Fe = mass / Aᵣ = 0.998 g / 55.8 g mol⁻¹ = 0.017885 mol [M1]

Step 2: Calculate moles of Hydrochloric Acid (HCl)
Amount of HCl = concentration × volume = 1.00 mol dm⁻³ × (30.0 / 1000 dm³) = 0.0300 mol [M2]

Step 3: Determine the Limiting Reagent
From stoichiometry, 1 mole of Fe requires 2 moles of HCl.
0.017885 mol Fe would require 2 × 0.017885 = 0.03577 mol HCl.
Since we only have 0.0300 mol of HCl, HCl is the limiting reagent (Fe is in excess). [M3]

Step 4: Calculate moles of H₂ produced
Ratio is 2 HCl : 1 H₂
Moles of H₂ = moles of HCl / 2 = 0.0300 / 2 = 0.0150 mol [M4]

Step 5: Convert Units for the Ideal Gas Equation (pV = nRT)
Temperature, T = 30 °C + 273 = 303 K
Pressure, P = 100 kPa = 100,000 Pa (or 1.00 × 10⁵ Pa) [M5]

Step 6: Calculate Volume in m³
V = nRT / P = (0.0150 mol × 8.31 J K⁻¹ mol⁻¹ × 303 K) / 100,000 Pa
V = 37.76895 / 100,000 = 3.78 × 10⁻⁴ m³ (3 significant figures) [M6]

❌ Calculation Pitfalls

• Wrong Limiting Reagent: Assuming Fe is limiting because 0.998 is smaller than 30.0, or failing to multiply by the 1:2 ratio.
• Unit Conversions: Forgetting that P must be in Pa (multiply kPa by 1000) and T must be in Kelvin (+ 273).
• Significant Figures: The question strictly demands 3 sig figs. Writing 3.8 × 10⁻⁴ or 0.0003777 loses M6.

🧠 Exam Technique

Always show explicit calculations for both reagents when determining the limiting reactant. If you make an arithmetic error in M2, error-carried-forward (ECF) allows you to score M4 and M6 provided your working is clearly shown.

✅ 04.5: Precipitate A (Fe²⁺ + Carbonate)

• Identity: FeCO₃ (or iron(II) carbonate) [1 mark]
• Colour: Green (allow white) [1 mark]

Note: Fe²⁺ is insufficiently acidic to liberate CO₂ gas from carbonate ions, so a direct precipitation reaction occurs.

✅ 04.6: Precipitate B (Fe³⁺ + Carbonate)

• Formula: Fe(H₂O)₃(OH)₃ [1 mark]
• Colour: Brown [1 mark]
• Ionic Equation:
  2 [Fe(H₂O)₆]³⁺ + 3 CO₃²⁻ → 2 Fe(H₂O)₃(OH)₃ + 3 H₂O + 3 CO₂ [1 mark]

💡 Key Knowledge: Why Do Fe²⁺ and Fe³⁺ React Differently with CO₃²⁻?

• Fe²⁺ (+2 ion): Lower charge density; less hydrolysed. CO₃²⁻ simply displaces water to form a precipitate of FeCO₃.
• Fe³⁺ (+3 ion): High charge density; acidic enough to react in an acid-base reaction with CO₃²⁻. CO₃²⁻ acts as a base and removes protons, forming CO₂ bubbles and the neutral hydroxide precipitate Fe(H₂O)₃(OH)₃. Fe₂(CO₃)₃ does NOT form!

❌ Common Errors

• Writing "Fe₂(CO₃)₃" as Precipitate B — this compound does not exist in aqueous solution.
• Omitting CO₂ or H₂O from the ionic equation in 04.6.
• Confusing colours: Fe(OH)₂ is green (turns brown on oxidation in air), whereas Fe(OH)₃ is brown.

🧠 Exam Technique: Structuring Acidity Answers

Always use the standard three-step causation chain for aqueous ion acidity:

1. Compare ionic size and charge (charge density) of the metal cation.
2. State that it polarises the attached H₂O ligands more.
3. State that this weakens the O–H bond, releasing more H⁺ ions into the aqueous solution.

❌ Common Misconceptions

• Penalised Phrasing: Writing that "Fe³⁺ releases 3 H⁺ ions" directly. Fe³⁺ itself does not release protons; the coordinated water ligands undergo deprotonation:
  [Fe(H₂O)₆]³⁺ + H₂O ⇌ [Fe(H₂O)₅(OH)]²⁺ + H₃O⁺
• Comparing the complex ions instead of the bare metal ions (e.g. stating "[Fe(H₂O)₆]³⁺ has a smaller radius"). The mark scheme penalises referring to the complex ion instead of the metal ion in M1/M2.