AQA A-Level Chemistry Paper 1, 2021: Question 7
9 marks · Medium difficulty · State/Explain/Numerical
Explain differences in standard entropy, state the temperature for zero entropy, and calculate the minimum temperature for reaction feasibility using enthalpy and entropy changes.
Practise this questionQuestion
Question text
07 This question is about thermodynamics.
Consider the reaction shown.
2 Al2O3(s) + 3 C(s) → 4 Al(s) + 3 CO2(g)
Table 7 shows some thermodynamic data.
Table 7
Substance Al2O3(s) Al(s) C(s) CO2(g)
Δ Hϴ/ kJ mol–1 –1669 0 0 –394
f
Sϴ/ J K–1 mol–1 51 28 6 214
07.1 Explain why the standard entropy value for carbon dioxide is greater than that for
carbon.
[1 mark]
07 2 State the temperature at which the standard entropy of aluminium is 0 J K–1 mol–1
.
[1 mark]
07.3 Use the equation and the data in Table 7 to calculate the minimum temperature,
in K, at which this reaction becomes feasible.
[7 marks]
Minimum temperature K
Mark scheme
Show the mark scheme
Question Answers Additional comments/Guidelines Mark
CO2 / gas is more disordered (than solid) Allow answers based on carbon
07.1 1
Ignore CO2 is a gas and C is a solid
0 K Units essential
07.2 1
Allow absolute zero OR –273 oC
M1 ΔH = (3 ͯ – 394) –(–1669 x 2)
M1 correct expression 1
M2 = 2156 (kJ mol–1)
M2 if –2156 seen allow 1 mark out M1 and M2
M3 ΔS = (28 x 4 + 214 x 3) – ( 51 x 2 + 6 x 3) M3 correct expression
M4 = 634 (J K–1mol –1) M4 if – 634 allow 1 mark from M4 and M4
07.3
M5 ΔG= ΔH – T ΔS or ΔH =T ΔS or T = ΔH ÷ΔS M5 expression or rearranged expression or with 1
numbers
M6 ΔS = 0.634 kJ K–1mol –1 1
M6 ΔS = M4 ÷ 1000
M7 T = 2156 = 3400 to 3401 (K) 1
0.634 M7 = M2 ÷ M6 but must be a positive answer
How to answer it
Thermodynamics & Feasibility: Reduction of Al₂O₃
This question assesses mastery of A-Level Thermodynamics (AQA 3.1.8):
- Qualitative understanding of entropy (S) and states of matter (disorder).
- The Third Law of Thermodynamics: conditions under which standard entropy is zero.
- Calculation of reaction enthalpy change (ΔH) from standard enthalpies of formation (ΔfH⦵).
- Calculation of reaction entropy change (ΔS) from absolute standard entropies (S⦵).
- Applying the Gibbs free energy relationship (ΔG = ΔH − TΔS) to determine the feasibility threshold temperature (T = ΔH / ΔS) including unit conversion.
Reaction Reference
2 Al₂O₃(s) + 3 C(s) → 4 Al(s) + 3 CO₂(g)
| Substance | Al₂O₃(s) | Al(s) | C(s) | CO₂(g) |
|---|---|---|---|---|
| ΔfH⦵ / kJ mol⁻¹ | −1669 | 0 | 0 | −394 |
| S⦵ / J K⁻¹ mol⁻¹ | 51 | 28 | 6 | 214 |
Part (a): Comparing Entropy of CO₂(g) and C(s)
Explain why the standard entropy value for carbon dioxide is greater than that for carbon.
✅ Correct Answer
CO₂ is a gas, so it is more disordered than solid carbon (or carbon solid is more ordered than CO₂ gas).
🧠 Exam Technique
- Entropy is a direct measure of disorder / randomness of particles.
- Always link the difference in entropy directly to the degree of disorder between the two physical states.
❌ Common Errors & Examiner Pitfalls
- Simply stating states of matter: Writing only "CO₂ is a gas and C is a solid" gets 0 marks. The mark scheme explicitly states: "Ignore CO₂ is a gas and C is a solid". You must use the word disorder or random.
- Confusing mass or complexity with disorder: Do not just state "CO₂ has more atoms" or "CO₂ is a bigger molecule"; standard entropy values are primarily driven by the physical state.
• [1 mark]: CO₂ / gas is more disordered (than solid). Allow answers framed in terms of carbon (e.g. solid carbon has a more ordered regular arrangement).
Part (b): Temperature Where Entropy is Zero
State the temperature at which the standard entropy of aluminium is 0 J K⁻¹ mol⁻¹.
✅ Correct Answer
0 K (or −273 °C, or "absolute zero")
💡 Key Knowledge (Third Law)
At absolute zero (0 K), a perfect crystal lattice possesses minimal thermal motion and only one possible microstate, hence standard entropy is defined as exactly 0 J K⁻¹ mol⁻¹.
❌ Common Errors
- Missing units: Writing just 0 scores 0 marks. The mark scheme specifies: "Units essential" unless written as "absolute zero".
- Confusing temperature scales: Stating 0 °C (273 K) is incorrect. A solid still has considerable vibrational entropy at 0 °C!
• [1 mark]: 0 K (units essential). Also accept "absolute zero" or "−273 °C".
Part (c): Minimum Feasible Temperature Calculation
Use the equation and data in Table 7 to calculate the minimum temperature, in K, at which this reaction becomes feasible.
📐 Step-by-Step Calculation
Formula: ΔH = Σ ΔfH(products) − Σ ΔfH(reactants)
ΔH = [4(0) + 3(−394)] − [2(−1669) + 3(0)]
ΔH = (−1182) − (−3338) = +2156 kJ mol⁻¹
✓ M1 awarded for the correct expression. M2 awarded for +2156 kJ mol⁻¹.
Formula: ΔS = Σ S(products) − Σ S(reactants)
ΔS = [4(28) + 3(214)] − [2(51) + 3(6)]
ΔS = (112 + 642) − (102 + 18) = 754 − 120 = +634 J K⁻¹ mol⁻¹
✓ M3 awarded for correct expression. M4 awarded for +634 J K⁻¹ mol⁻¹.
A reaction is feasible when ΔG ≤ 0.
ΔG = ΔH − TΔS = 0 ⟹ T = ΔH / ΔS
ΔS is in J K⁻¹ mol⁻¹, but ΔH is in kJ mol⁻¹. Convert ΔS to kJ K⁻¹ mol⁻¹:
ΔS = 634 ÷ 1000 = 0.634 kJ K⁻¹ mol⁻¹
(Alternatively, convert ΔH to J: 2156 × 1000 = 2 156 000 J mol⁻¹)
T = 2156 / 0.634 = 3400.6 K
Final Answer: 3400 K to 3401 K
❌ Calculation Traps & Common Errors
- Forgetting Stoichiometric Coefficients: Multiplying by the big balancing numbers in the balanced equation (2 for Al₂O₃, 3 for C, 4 for Al, 3 for CO₂) is essential. Forgetting the '4' for Al or the '3' for C ruins M1/M3.
- The "Double Negative" Sign Error: Subtracting a negative enthalpy of formation: −1182 − (−3338) = +2156 . Getting −2156 loses M2.
- Unit Mismatch: Dividing 2156 by 634 directly gives 3.4 K — an impossible answer! You must divide ΔS by 1000 first (or multiply ΔH by 1000).
- Negative Temperature: Absolute temperature in Kelvin cannot be negative. If your answer is negative, you made a sign error earlier in ΔH or ΔS.
🧠 Examiner Commentary
- High Mark Yield: 7 marks make up almost 10% of a typical paper! Lay out each line clearly so examiners can award method marks even if an arithmetic slip occurs.
- Error Carried Forward (ECF): If you miscalculate ΔH or ΔS, you can still gain marks M5, M6, and M7 if you clearly show the relationship T = ΔH / ΔS and do the correct unit conversion.
- Significant Figures: 3400 K (3 s.f.) or 3401 K (4 s.f.) are both accepted by the mark scheme.
• M1: Expression for ΔH: (3 × −394) − (−1669 × 2)
• M2: ΔH = +2156 kJ mol⁻¹ (allow 1 mark out of M1 & M2 if −2156 seen)
• M3: Expression for ΔS: (28 × 4 + 214 × 3) − (51 × 2 + 6 × 3)
• M4: ΔS = +634 J K⁻¹ mol⁻¹ (allow 1 mark out of M3 & M4 if −634 seen)
• M5: Rearrangement: ΔG = ΔH − TΔS or T = ΔH / ΔS
• M6: Conversion: ΔS = 0.634 kJ K⁻¹ mol⁻¹ (M4 ÷ 1000)
• M7: T = 2156 / 0.634 = 3400 to 3401 (K) (must be positive)
Topics
Physical Chemistry · 3.1.8 Thermodynamics
Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.