AQA A-Level Chemistry Paper 1, 2018: Question 3

📐 Step-by-Step Calculation

1. Sum of products' entropies:
   ΣS(products) = (4 × 211) + (6 × 189) = 844 + 1134 = 1978 J K⁻¹ mol⁻¹
2. Sum of reactants' entropies:
   ΣS(reactants) = (4 × 193) + (5 × 205) = 772 + 1025 = 1797 J K⁻¹ mol⁻¹
3. Calculate ΔS:
   ΔS = ΣS(products) − ΣS(reactants)
   ΔS = 1978 − 1797 = +181 J K⁻¹ mol⁻¹

✅ Mark Scheme Breakdown

• Mark 1 (Method): Correctly multiplying by molar coefficients:
  [(4 × 211) + (6 × 189)] - [(4 × 193) + (5 × 205)] or showing 1978 - 1797 .
• Mark 2 (Answer): 181 (or +181) J K⁻¹ mol⁻¹.

🧠 Exam Technique: Common Sense Check

Look at the moles of gas: 9 moles of gas react to form 10 moles of gas. An increase in the number of gaseous molecules creates greater disorder, so ΔS must be a positive value. If your final value is negative, re-check your subtraction order!

❌ Common Errors

• Reactants minus Products: Subtracting backwards gives -181 , losing the final mark.
• Ignoring balancing numbers: Just adding table values without multiplying by 4, 5, and 6.

📐 Step-by-Step Calculation

1. Convert temperature to Kelvin:
   T = 600 + 273 = 873 K
2. Convert ΔS to kJ K⁻¹ mol⁻¹:
   ΔS = 181 ÷ 1000 = 0.181 kJ K⁻¹ mol⁻¹ (or convert ΔH to J).
3. Substitute into Gibbs equation:
   ΔG = −905 − (873 × 0.181)
   ΔG = −905 − 158.013 = −1063 kJ mol⁻¹ (allow −1060)

✅ Mark Scheme Breakdown

• Mark 1 (Method): Correct substitution with T in K and matching units:
  ΔG = -905 - [(600 + 273) × (181 × 10⁻³)]
• Mark 2 (Answer): −1063 kJ mol⁻¹ (or −1060).
• Note: If using given dummy value ΔS = 211 J K⁻¹ mol⁻¹:
  ΔG = -905 - (873 × 0.211) = −1089 kJ mol⁻¹.

❌ Common Trap: Unit Mismatch

The #1 error here is failing to divide ΔS by 1000! ΔH is in kJ mol⁻¹ while ΔS is in J K⁻¹ mol⁻¹ . You cannot add or subtract them directly without matching units.

🧠 Error-Carried-Forward (ECF)

If your answer to 03.1 was wrong, you still get full marks here via ECF as long as you use: -905 - (873 × your_ΔS × 10⁻³) .

✅ Correct Answer

• Effect on ΔG: ΔG becomes more negative (or less positive). [1 mark]
• Explanation: Because entropy change (ΔS) is positive (or: the term TΔS gets larger / −TΔS becomes more negative). [1 mark]

❌ Precise Terminology Warning

Examiner Warning: The mark scheme explicitly states:

"Ignore increase / decrease / larger / smaller ΔG"

When numbers are negative, "decreases" is ambiguous (does it mean more negative or closer to zero?). You must state "more negative" or "less positive".

💡 The 3 Key Sequential Stages

1. Stage 1 — Adsorption: Reactants (NH₃ and O₂) form bonds and are adsorbed onto the platinum surface / active sites.
2. Stage 2 — Reaction: Bonds in reactants are weakened/broken and new bonds form on the surface.
3. Stage 3 — Desorption: Products (NO and H₂O) break away / are desorbed from the surface, freeing active sites.

✅ Mark Scheme Breakdown (1 mark per stage)

• Mark 1: Reactant(s) adsorbed onto the (platinum) surface OR platinum provides active sites.
• Mark 2: Reaction on the surface OR bond breaking / weakening / bond making occurs on the surface.
• Mark 3: Desorption of the product(s) (or products leave the surface).

❌ Critical Spelling Trap

Adsorption vs Absorption:
Ensure you write a d sorption (substance adhering to a solid surface), NOT absorption (taking inside like a sponge). Confusing these words can cost you the mark.

🧠 Exam Technique

Whenever AQA asks for the "stages" of heterogeneous catalysis, structure your answer chronologically using the key terms: Adsorption → Surface reaction/bond weakening → Desorption. This guarantees all 3 marks.

✅ 03.5: Change in Oxidation State [1 mark]

Answer: Changes from −3 to +2  OR  (+) 5

✅ 03.6: Equation for Formation of N₂O [1 mark]

Answer:

2NH₃ + 2O₂ → N₂O + 3H₂O

Multiples are allowed (e.g. 4NH₃ + 4O₂ → 2N₂O + 6H₂O). State symbols are not required.

❌ Common Errors in 03.5

• Forgetting signs: Writing "3 to 2" scores 0. Signs ( - and + ) are essential for oxidation states.
• Inverting signs: Incorrectly identifying H as −1 (hydride) to get N as +3.

🧠 Quick Balancing Tip for 03.6

Write out skeleton: NH₃ + O₂ → N₂O + H₂O .
Notice 2 N's in N₂O, so start with 2NH₃ . That gives 6 H's, requiring 3H₂O . Total oxygens on RHS = 1 (in N₂O) + 3 (in H₂O) = 4 O atoms = 2O₂ . Clean, simple, integer ratio!