AQA A-Level Chemistry Paper 1, 2017: Question 5

📐 Calculations (step-by-step)

1. Write the expression with coefficients:
   ΔS = [S°(TiCl₄) + 2S°(CO)] − [S°(TiO₂) + 2S°(C) + 2S°(Cl₂)]
   This expression scores M1 if set up correctly (products − reactants, and coefficients included).
2. Substitute values (all in J K⁻¹ mol⁻¹):
   ΔS = [253 + 2×198] − [50.2 + 2×5.70 + 2×223]
   ΔS = [253 + 396] − [50.2 + 11.4 + 446]
   ΔS = 649 − 507.6 = 141.4 J K⁻¹ mol⁻¹
   Correct value scores M2.

✅ Correct answer

ΔS° = +141.4 J K⁻¹ mol⁻¹

Examiner note: the mark scheme shows ΔS = 141.(4) J K⁻¹ mol⁻¹ (i.e. 141.4). Either is fine if consistent.

🧠 Exam technique

• Always do products − reactants (many students lose M1 by reversing).
• Include every coefficient: especially 2Cl₂ and 2CO.
• Keep units on ΔS as J K⁻¹ mol⁻¹ (don’t convert yet—conversion is usually done in the ΔG step).

❌ Common errors (why marks are lost)

• Forgetting the 2× for CO or Cl₂ (gives a much smaller ΔS).
• Using ΔS = reactants − products (sign error → wrong feasibility later).
• Mixing units: writing kJ K⁻¹ mol⁻¹ here without converting properly.

💡 Key knowledge

• Temperature must be in K: T(K) = T(°C) + 273
• Equation: ΔG = ΔH − TΔS
• Unit match: if ΔH is in kJ mol⁻¹, then TΔS must be in kJ mol⁻¹ too.
• Convert: J → kJ using ×10⁻³

📐 Calculations (step-by-step)

1. Convert temperature:
   T = 989 + 273 = 1262 K
   Using Kelvin appropriately is essential for method marks.
2. Calculate TΔS and convert to kJ mol⁻¹:
   TΔS = 1262 × 141.4 = 178 466.8 J mol⁻¹
   TΔS = 178.467 kJ mol⁻¹ (since ×10⁻³)
3. Substitute into ΔG = ΔH − TΔS:
   ΔG = −60.0 − 178.467 = −238.467 kJ mol⁻¹
   ΔG = −238 kJ mol⁻¹ (3 s.f.)
   Mark scheme shows: ΔG = −60 − (1262 × 141.(4) × 10⁻³) giving −238 kJ mol⁻¹ to 3 s.f.

✅ Correct answer (with sig figs)

ΔG = −238 kJ mol⁻¹ (to 3 significant figures)

M3: uses ΔG = ΔH − TΔS.
M4: correctly substitutes T = 1262 K and ΔS (with correct unit conversion).
M5: correct numerical value with appropriate units/sig figs.

❌ Calculation traps (very common)

• Not converting °C to K: using 989 instead of 1262 massively changes ΔG.
• Unit mismatch: ΔH is kJ mol⁻¹ but ΔS is J K⁻¹ mol⁻¹. You must convert J → kJ (×10⁻³), or convert ΔH to J (×10³). The mark scheme allows either if done consistently.
• Sign mistake: writing ΔG = ΔH + TΔS or accidentally subtracting a negative incorrectly.

Examiner insight: students often got method marks but lost the final mark by forgetting the ×10⁻³ conversion.

✅ Full-mark explanation

The reaction is feasible at 989 °C because ΔG is negative (ΔG < 0).

Mark scheme: “feasible since ΔG is negative/less than zero”. Consequential marking applies: if your ΔG was (incorrectly) positive, you’d be expected to say “not feasible”.

🧠 Exam technique (wording that scores)

• State the sign and the conclusion: “ΔG is negative, therefore feasible/spontaneous.”
• Avoid vague phrases like “it will happen” without linking to ΔG.
• Don’t confuse with rate: feasibility is thermodynamic, not kinetic.

💡 Why ΔG becomes strongly negative here

• ΔS is positive (more gas moles formed overall: 2 mol gas products vs 2 mol gas reactants, plus different species/entropy values).
• At high T, the TΔS term becomes large, making ΔG = ΔH − TΔS more negative when ΔS > 0.

❌ Common errors

• Saying “feasible because ΔH is negative” (not sufficient; ΔG decides feasibility at temperature).
• Claiming “not feasible because it’s endothermic” (it isn’t here, and even if it were, ΔS and T matter).

🧾 What examiners reward

• M1: Correct ΔS setup: ΣS°(products) − ΣS°(reactants) with coefficients.
• M2: ΔS = 141.4 J K⁻¹ mol⁻¹.
• M3: Uses ΔG = ΔH − TΔS.
• M4: Correct substitution including T = 1262 K and correct J↔kJ conversion.
• M5: ΔG = −238 kJ mol⁻¹ (3 s.f., units shown).
• M6: Feasible because ΔG < 0.

✅ Final answers to write on the paper

Gibbs free-energy change: −238 kJ mol⁻¹

Explanation: feasible because ΔG is negative (less than zero).