Chemistry • Year 11 • Module 4 • Lesson 8
Hess's Law
Synthesise and evaluate — two extended-response questions requiring sustained multi-criteria reasoning, evidence-based judgement, and real-world connections.
Question 1 — Evaluating energy efficiency in industrial SO₃ production
8 marks
Scenario. A chemical plant produces SO₃ for sulfuric acid manufacture (Contact Process). The plant manager proposes two routes:
Route A (one-step): S(s) + ¾O₂(g) → SO₃(g)
Route B (two-step): Step 1: S(s) + O₂(g) → SO₂(g); Step 2: SO₂(g) + ½O₂(g) → SO₃(g)
The thermochemical data available are:
| Equation | ΔH (kJ mol⁻¹) |
|---|---|
| S(s) + O₂(g) → SO₂(g) | −297 |
| 2SO₂(g) + O₂(g) → 2SO₃(g) | −198 |
| S(s) + ¾O₂(g) → SO₃(g) | −396 |
The plant’s energy recovery system can only recapture heat released in the first step of a multi-step route, not from intermediate-step reactions. Assume all reactions go to completion.
Question. Evaluate whether the plant manager’s choice of route affects the total enthalpy change for the production of SO₃, and assess which route is more useful for the plant’s energy recovery system.
In your response you must:
- State and explain Hess’s Law as it applies to both routes (ΔH of Route A vs total ΔH of Route B).
- Calculate ΔH for Step 2 of Route B using the data provided; show your working.
- Evaluate on at least two criteria (total energy released; energy available in Step 1 alone) which route is more practical for the energy recovery system described.
- Identify one limitation of assuming the reactions “go to completion” in an industrial context.
Question 2 — Methane formation and the limits of Hess’s Law in practice
8 marks
Scenario. The standard enthalpy of formation of methane, CH₄(g), cannot be measured directly because the reaction C(s) + 2H₂(g) → CH₄(g) does not proceed at measurable rate under standard conditions. A researcher uses Hess’s Law with three combustion enthalpies:
| Equation | ΔH (kJ mol⁻¹) |
|---|---|
| (1) C(s) + O₂(g) → CO₂(g) | −393.5 |
| (2) H₂(g) + ½O₂(g) → H₂O(l) | −285.8 |
| (3) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) | −890.3 |
The researcher's lab notebook shows the following working — which contains one deliberate error:
Step 2: Use (1): C + O₂ → CO₂ ΔH = −393.5
Step 3: Use (2) × 2: 2H₂ + O₂ → 2H₂O ΔH = −285.8
Step 4: Sum: ΔH = 890.3 + (−393.5) + (−285.8) = +210.8 kJ mol⁻¹
The accepted value is ΔHₑf[CH₄(g)] = −74.8 kJ mol⁻¹.
Question. Identify and explain the error in the researcher’s working, then provide a complete corrected calculation. Evaluate the reliability of Hess’s Law as a method for determining enthalpy values that cannot be measured directly.
In your response you must:
- Identify the specific step where the error occurs and state precisely what rule was violated.
- Provide the corrected calculation, showing all steps and the correct ΔH.
- Verify your answer matches the accepted value and explain what this confirms about Hess’s Law.
- Evaluate two reasons why Hess’s Law results from combustion data may still carry small uncertainties in practice, even when the algebra is correct.
Q1 — Industrial SO₃ production (8 marks)
1 — Hess’s Law applied: Hess’s Law states that ΔH for a reaction is independent of the pathway taken; enthalpy is a state function. Route A converts S + ¾O₂ → SO₃ in one step (ΔH = −396 kJ mol⁻¹). Route B covers the same conversion in two steps. By Hess’s Law, ΔH(total, B) = ΔH(Step 1) + ΔH(Step 2) = ΔH(Route A) = −396 kJ mol⁻¹. [2 marks: 1 for stating Hess’s Law correctly; 1 for correctly equating total ΔH across both routes]
2 — Calculate ΔH(Step 2): ΔH(Step 2) = total ΔH − ΔH(Step 1) = −396 − (−297) = −99 kJ mol⁻¹. (Equivalently: ½ × (−198) = −99 kJ mol⁻¹ from the given equation scaled by ½.) [1 mark for correct value with working]
3 — Evaluate routes on two criteria:
Total energy released: Both routes release −396 kJ mol⁻¹ total — identical by Hess’s Law. Neither is superior on this criterion. [1 mark]
Energy available in Step 1 for recovery: Route B Step 1 releases −297 kJ mol⁻¹ in a single measurable combustion event (S + O₂ → SO₂) — a larger, easily recoverable heat release. Route A releases all energy in one step too (−396 kJ mol⁻¹). However, Route B’s Step 1 is the most exothermic stage; Step 2 releases only −99 kJ mol⁻¹ and is catalysis-dependent (requires V₂O₅ catalyst at 450°C in practice). If the energy recovery system captures only Step 1 heat, Route B captures −297 kJ mol⁻¹ from Step 1 (76% of total) while Route A captures the full −396 kJ mol⁻¹ in its single step. On balance, Route A is simpler for recovery (one heat exchange point), but Route B may be industrially preferred because SO₂ is an intermediate that can be stored and converted separately under optimised conditions. [2 marks: 1 per criterion evaluated with reference to data]
4 — Limitation: In practice, reactions rarely go to 100% completion — SO₂/SO₃ equilibrium (the Contact Process) typically achieves ~98% conversion at best. Assuming complete conversion overestimates the heat released and the yield of SO₃; actual industrial performance requires correction for equilibrium conversion. [1 mark]
Marking note: award 1 additional quality mark for a response that integrates all four requirements into a coherent, structured argument rather than listing isolated points.
Q2 — Methane formation and error analysis (8 marks)
1 — Identify error: The error is in Step 3. Equation (2) was correctly scaled by 2 (giving 2H₂ + O₂ → 2H₂O), but ΔH was NOT scaled: the student used −285.8 instead of 2 × (−285.8) = −571.6 kJ mol⁻¹. This violates the manipulation rule: any operation applied to the equation must also be applied to ΔH. [2 marks: 1 for identifying Step 3 as the error; 1 for naming the violated rule precisely]
2 — Corrected calculation:
Reverse (3): CO₂ + 2H₂O → CH₄ + 2O₂ ΔH = +890.3 [1]
Use (1) as written: C + O₂ → CO₂ ΔH = −393.5 [1]
Scale (2) × 2: 2H₂ + O₂ → 2H₂O ΔH = 2(−285.8) = −571.6 [1]
Add all: cancel CO₂, 2H₂O, 2O₂ ✓. Net: C + 2H₂ → CH₄.
ΔH = +890.3 + (−393.5) + (−571.6) = +890.3 − 965.1 = −74.8 kJ mol⁻¹ [1]
3 — Verify and confirm Hess’s Law: The corrected answer (−74.8 kJ mol⁻¹) matches the accepted value exactly. This confirms that Hess’s Law gives a reliable, reproducible ΔH for the target reaction regardless of whether it can be measured directly, as long as the manipulation rules are applied correctly and the given ΔH values are accurate. [1 mark]
4 — Two practical uncertainties:
(a) Calorimeter heat loss: not all heat produced in the combustion is captured in the calorimeter solution — some escapes to the surroundings. This makes the measured ΔH slightly less negative than the true value, introducing systematic error into each component equation. [½ mark]
(b) Purity of samples: if the carbon sample contains impurities (e.g. moisture, non-carbon ash), the mass used in calculations overestimates the number of moles of C combusted, leading to an underestimated magnitude of ΔH per mole. Impure CH₄ similarly introduces error into combustion enthalpy (3). [½ mark]
Accept any two of: heat loss, impure samples, incomplete combustion (CO in products), uncertainty in mass measurement, phase of water product assumed to be liquid. [1 mark total for two valid points]