Chemistry • Year 11 • Module 4 • Lesson 9
Hess’s Law Applied: Photosynthesis & Respiration
Synthesise, evaluate, and justify multi-step reasoning about biological energy systems using Hess’s Law at Band 5–6 level.
Question 1 — Evaluating biofuel carbon neutrality using Hess’s Law
8 marks
Scenario. The Queensland government is evaluating two carbon-sequestration strategies to offset industrial emissions in the Mackay region.
Strategy P — Sugarcane biofuel (CSR Ltd, Mackay QLD). Sugarcane is grown, fermented to bioethanol, and burned as fuel. CO₂ captured during photosynthesis is released on combustion. Supporters claim this is “carbon neutral” because the CO₂ is recently fixed, not ancient.
Strategy Q — Native forest conservation (wet sclerophyll eucalypt, Mackay hinterland). Existing forest is preserved, continuously removing CO₂ from the atmosphere via net photosynthesis and locking carbon into biomass (wood). Estimated net uptake: 7.2 t CO₂-eq ha⁻¹ yr⁻¹.
Supporting data.
| Parameter | Strategy P — Biofuel | Strategy Q — Forest |
|---|---|---|
| Net CO₂ change after 1 year (per ha) | ~0 t (CO₂ in = CO₂ out) | −7.2 t (net removal) |
| ΔH(overall pathway) | ≈0 kJ (cycle closes) | +2803 kJ per mol glucose stored |
| Carbon residence time | <1 growing season | Decades–centuries in wood |
| Land use (ha per tonne CO₂ offset) | High (ethanol yield ~6 t CO₂-eq ha⁻¹ yr⁻¹ displaced) | Moderate (7.2 t CO₂-eq ha⁻¹ yr⁻¹ removed) |
Sources: hypothetical data based on Mackey et al. (2013); IEA biofuel LCA estimates; CSR Ltd operational data.
Extended response prompt.
Using Hess’s Law and the data provided, evaluate which strategy more effectively reduces atmospheric CO₂ in the short term (1–5 years) and in the long term (50+ years). In your response you must:
- Explain the Hess’s Law basis of the biofuel carbon-neutrality claim (refer to net ΔH of the photosynthesis–combustion cycle).
- Compare the two strategies on at least two criteria from the table (net CO₂ change, ΔH, carbon residence time, or land use).
- Identify one limitation of using Hess’s Law alone to evaluate environmental impact.
- Reach an evidence-based judgement about which strategy is preferable for long-term atmospheric CO₂ reduction.
Question 2 — Evaluating breath analysis as a metabolic diagnostic tool
7 marks
Scenario. Paramedics at a road crash in Sydney detect elevated acetone (CH₃COCH₃) in a patient’s expired breath. The hospital biochemist explains that the patient’s metabolism has shifted from complete aerobic respiration of glucose to incomplete oxidation of fatty acids, producing ketone bodies (including acetone) as intermediates.
Relevant thermochemical data:
| Reaction | ΔH (kJ mol⁻¹) |
|---|---|
| Complete respiration: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O | −2803 |
| Incomplete: C₆H₁₂O₆ + 3O₂ → 3CH₃COCH₃ + 3H₂O | −960 (estimated) |
| Complete combustion of acetone: CH₃COCH₃ + 4O₂ → 3CO₂ + 3H₂O | −1790 |
Source: hypothetical data based on published combustion enthalpies; ΔH values are approximate.
Extended response prompt.
Using Hess’s Law and the data above, evaluate the statement: “A patient producing acetone in their breath is releasing less energy per mole of glucose than a patient undergoing complete aerobic respiration.”
In your response you must:
- Use Hess’s Law to verify the statement by calculating the energy released via the incomplete pathway (glucose → acetone) and comparing it with complete respiration.
- Explain what “the incomplete pathway does not reach the same final state” means in terms of enthalpy and Hess’s Law.
- Assess one implication of this for the patient’s cells (e.g. ATP production, body temperature maintenance).
Q1 — Marking guidelines (8 marks)
Mark 1 (Hess’s Law basis of carbon neutrality). The biofuel cycle: photosynthesis (ΔH = +2803) + fermentation (ΔH = −68) + 2×combustion (ΔH = −2732) sums to approximately zero kJ mol⁻¹. This is because the initial state (CO₂ + H₂O in atmosphere) and final state (CO₂ + H₂O released on combustion) are identical — enthalpy is a state function, so net ΔH ≈ 0. This is the Hess’s Law basis of the claim: no net addition of carbon to the atmosphere if the cycle is truly closed.
Mark 2 (Criterion comparison — net CO₂ change). Strategy P results in net CO₂ change of ~0 t ha⁻¹ yr⁻¹ in a closed cycle; Strategy Q actively removes 7.2 t CO₂-eq ha⁻¹ yr⁻¹ from the atmosphere. For the same land area, Strategy Q removes more CO₂ than Strategy P offsets.
Mark 3 (Criterion comparison — carbon residence time). Strategy P has a residence time of less than one season (CO₂ fixed in summer is released in the same year on combustion). Strategy Q locks carbon into wood for decades to centuries. For long-term atmospheric reduction, Strategy Q provides durable sequestration; Strategy P’s benefits disappear immediately on combustion.
Mark 4 (Limitation of Hess’s Law alone). Hess’s Law describes enthalpy changes only — it cannot account for the timing of CO₂ emissions vs. uptake, infrastructure emissions from biofuel production (tractors, processing plant), soil carbon disturbance from land clearing, or the ecological value of forests beyond carbon storage. An evaluation based on ΔH alone may find Strategy P “neutral” when in practice it is not if lifecycle emissions are included.
Mark 5 (Short-term judgement). In the short term (1–5 years), Strategy Q is superior: it actively removes CO₂ from the atmosphere (net negative flux), whereas Strategy P only prevents adding new CO₂ (net zero flux). [1 mark]
Mark 6 (Long-term judgement). Over 50+ years, Strategy Q continues to build carbon stores in biomass; Strategy P must continuously grow and burn crops to maintain its offset and cannot accumulate a carbon debt. Evidence-based judgement: Strategy Q is the superior long-term CO₂ reduction tool on both the thermodynamic data (7.2 t ha⁻¹ yr⁻¹ vs 0) and carbon residence time criteria. [1 mark]
Marks 7–8 (Quality of reasoning). Award 1 mark for coherent structure linking Hess’s Law logic to environmental criteria; 1 mark for an explicit, evidence-anchored conclusion that goes beyond restating the data.
Q2 — Marking guidelines (7 marks)
Mark 1 (Hess’s Law calculation — incomplete pathway). ΔH(glucose → 3 acetone) = −960 kJ mol⁻¹. This is the energy released when glucose is converted to acetone. It is less than 2803 kJ mol⁻¹. The statement is therefore confirmed: less energy is released per mole of glucose via the incomplete pathway. [1 mark for correct identification of −960 kJ mol⁻¹ as the relevant value.]
Mark 2 (Hess’s Law verification by two-step route). Hess’s Law allows us to calculate ΔH for glucose → CO₂ + H₂O via the acetone intermediate: Step 1 (glucose → 3 acetone): ΔH = −960 kJ mol⁻¹; Step 2 (3 acetone → CO₂ + H₂O): ΔH = 3(−1790) = −5370 kJ mol⁻¹. Total via both steps: ΔH = −960 + (−5370) = −6330 kJ mol⁻¹. This exceeds the accepted −2803 kJ mol⁻¹ for complete respiration, indicating the estimated ΔH values given are illustrative/approximate rather than consistent — accept any student working that correctly applies the two-step Hess’s Law addition. The key conceptual point is that stopping at acetone (not completing combustion) means 960 < 2803 kJ is released. [1 mark for showing the two-step addition regardless of the discrepancy in approximate values.]
Mark 3 (Final-state argument). Hess’s Law states ΔH depends only on the initial and final states. Complete respiration ends at CO₂ + H₂O (the lowest enthalpy state for these elements). The incomplete pathway ends at acetone, which is at a higher enthalpy than CO₂ + H₂O because acetone still contains chemical energy that could be released by further oxidation. Therefore the enthalpy drop from glucose to acetone is smaller than the drop from glucose to CO₂ + H₂O, confirming less energy is released. [1 mark.]
Mark 4 (Implication for ATP production). Less energy released per mole of glucose means fewer ATP molecules can be produced via the incomplete pathway. If ATP production falls, cells cannot sustain high-demand activities (muscle contraction, ion pumping, biosynthesis). The patient’s cells are therefore energy-deficient despite metabolising glucose. [1 mark.]
Mark 5 (Implication for body temperature). Complete respiration releases ~58% of its 2803 kJ mol⁻¹ as heat (about 1644 kJ mol⁻¹). With the incomplete pathway releasing only ~960 kJ mol⁻¹ as heat, less body heat is produced per mole of substrate. The patient may have difficulty maintaining core temperature. [1 mark for either ATP or temperature implication, with physiological reasoning; 1 additional mark if both are addressed coherently.]
Mark 6 (Overall evaluation quality). Award 1 mark for a student who clearly connects the thermodynamic conclusion (less energy released) to a clinical implication (cells energy-deficient), demonstrating synthesis rather than mere calculation.