Chemistry • Year 11 • Module 4 • Lesson 9
Hess’s Law Applied: Photosynthesis & Respiration
Apply Hess’s Law to biological energy systems, interpret real data, and reason about ATP coupling and carbon sequestration.
1. Interpret ATP coupling data
A student investigated four different biosynthesis reactions in human muscle cells and the number of ATP molecules coupled to each reaction. The table below shows the ΔH of each uncoupled biosynthesis reaction and the number of ATP hydrolysis events used. 8 marks
| Reaction | ΔH of biosynthesis (kJ mol⁻¹) | Moles of ATP coupled | ΔH of combined reaction (kJ mol⁻¹) |
|---|---|---|---|
| A — Alanine synthesis | +42 | 2 | — |
| B — Muscle glycogen synthesis | +68 | 3 | — |
| C — Membrane lipid synthesis | +55 | 2 | — |
| D — Protein subunit assembly | +83 | 3 | — |
Note: ΔH(ATP hydrolysis) = −30.5 kJ mol⁻¹ per mole. Source: hypothetical data based on published biochemical free-energy values.
1.1 Using Hess’s Law, calculate the combined ΔH for Reactions A and B. Show your working. 3 marks
1.2 Reactions C and D both use 3 moles of ATP, yet Reaction D’s biosynthesis ΔH is larger. Without calculating, predict whether Reaction D will have a more negative combined ΔH than Reaction C. Justify your prediction. 2 marks
1.3 Evaluate the claim: “The more ATP molecules coupled, the more enthalpy-favourable the combined reaction will be.” Use one example from the table to support your evaluation. 3 marks
2. Interpret data graph — Australian forest carbon sequestration
The graph below shows estimated annual net carbon uptake (tonnes of CO₂-equivalent per hectare per year) for three Australian forest types. Net uptake represents the excess of photosynthesis over respiration — carbon that is stored as biomass rather than returned to the atmosphere. 9 marks
2.1 Describe the trend shown in Figure 2.1. Include reference to the highest and lowest values. 2 marks
2.2 Using Hess’s Law and the equation for photosynthesis (ΔH = +2803 kJ mol⁻¹), explain what it means for a forest ecosystem to have a “positive net CO₂ uptake” in terms of the relative rates of photosynthesis and respiration. 3 marks
2.3 Calculate the energy stored (in kJ) per hectare per year in the wet sclerophyll eucalypt forest, assuming that all net carbon uptake is in the form of glucose and that the molar mass of CO₂ = 44.01 g mol⁻¹. Sequestration rate = 7.2 t CO₂-eq ha⁻¹ yr⁻¹; 6 mol CO₂ fixed per mol glucose; ΔH(photosynthesis) = +2803 kJ mol⁻¹. 4 marks
3. Cause-and-effect chain — Australian sugarcane biofuel
CSR Ltd processes sugarcane grown in the Mackay region of Queensland into bioethanol. The process can be summarised as a multi-step Hess’s Law chain from photosynthesis to combustion of bioethanol. Use the chain below to trace the energy transformations. 6 marks
Given data: ΔH(photosynthesis, glucose) = +2803 kJ mol⁻¹; ΔH(fermentation: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂) = −68 kJ mol⁻¹; ΔH(combustion of ethanol) = −1366 kJ mol⁻¹ per mole C₂H₅OH.
3.1 Write the overall Hess’s Law sum for the complete pathway: sunlight absorbed (photosynthesis) → fermentation → combustion of 2 mol ethanol. Calculate the net ΔH for the process starting from CO₂ + H₂O through to the combustion products CO₂ + H₂O. 3 marks
3.2 The net ΔH you calculated should be close to zero (or zero). Explain what this result means for the carbon neutrality claim of biofuels, using Hess’s Law as your conceptual framework. 3 marks
4. Predict and justify — incomplete respiration in breath analysis
Breath analysis devices used by paramedics and emergency physicians can detect acetone (CH₃COCH₃) in expired air. Elevated acetone indicates incomplete carbohydrate metabolism — the body is oxidising fat instead of glucose, producing ketone bodies as intermediates.
The standard complete respiration route: C₆H₁₂O₆ → 6CO₂ + 6H₂O ΔH = −2803 kJ mol⁻¹
An incomplete “partial oxidation” route produces intermediate organic acids with a less negative ΔH. 4 marks
4.1 Using Hess’s Law, predict whether the energy released per mole of glucose via an incomplete pathway would be greater than, equal to, or less than 2803 kJ mol⁻¹. Justify your prediction with reference to enthalpy as a state function and the concept of pathway independence. 4 marks
Q1 — ATP coupling data
1.1 Reaction A: ΔH(combined) = +42 + 2(−30.5) = +42 − 61 = −19 kJ mol⁻¹. Reaction B: ΔH(combined) = +68 + 3(−30.5) = +68 − 91.5 = −23.5 kJ mol⁻¹. [1 mark per correct calculation; 1 mark for showing working.]
1.2 Reaction D will have a less negative (closer to zero or potentially still positive) combined ΔH than Reaction C, not more negative. Both use 3 mol ATP (−91.5 kJ total from hydrolysis), but Reaction D has a larger positive biosynthesis term (+83 vs +55). ΔH(C, combined) = 55 − 91.5 = −36.5; ΔH(D, combined) = 83 − 91.5 = −8.5 kJ mol⁻¹. Reaction C is more enthalpy-favourable despite Reaction D using the same ATP. [1 mark for correct prediction with valid reasoning.]
1.3 The claim is partially true but incomplete. Adding more ATP does increase the total exothermic contribution from hydrolysis, making the combined ΔH more negative. However, the net result also depends on the size of the endothermic biosynthesis term. For example, Reaction A (2 ATP, +42) gives combined ΔH = −19 kJ mol⁻¹, while Reaction B (3 ATP, +68) gives −23.5 kJ mol⁻¹ — a larger endothermic reaction with one more ATP coupling ends up less negative per unit ATP than it might appear. The claim holds directionally but must always be evaluated against the size of ΔH(biosynthesis). [1 mark for partially correct claim; 1 mark for valid example from table; 1 mark for nuanced evaluation.]
Q2 — Australian forest carbon sequestration graph
2.1 Net carbon sequestration is highest in the wet sclerophyll eucalypt forest (7.2 t CO₂-eq ha⁻¹ yr⁻¹) and lowest in the Great Barrier Reef seagrass meadow (1.8 t CO₂-eq ha⁻¹ yr⁻¹). All three ecosystem types show positive net carbon uptake, but uptake differs by a factor of four between the highest and lowest values.
2.2 Positive net CO₂ uptake means the rate of photosynthesis exceeds the rate of respiration. From a Hess’s Law perspective, more CO₂ is being converted to glucose (endothermic step, ΔH = +2803 kJ per mole) than is being released by the reverse reaction (respiration, ΔH = −2803 kJ per mole). The net result is a build-up of glucose — stored solar energy — in the form of wood and biomass. [1 mark rate comparison; 1 mark linking to photosynthesis > respiration; 1 mark biomass/stored energy consequence.]
2.3 Step 1: Convert 7.2 t CO₂ to moles. 7.2 t = 7.2 × 10⁶ g; moles CO₂ = 7 200 000 ÷ 44.01 = 163 599 mol CO₂ (approx. 1.636 × 10⁵ mol). Step 2: Moles of glucose (6 mol CO₂ per mol glucose) = 163 599 ÷ 6 = 27 267 mol glucose. Step 3: Energy stored = 27 267 × 2803 = 7.64 × 10⁷ kJ ha⁻¹ yr⁻¹ (approx. 76 400 GJ). [1 mark for correct mass-to-moles conversion; 1 mark for dividing by 6; 1 mark for multiplying by 2803; 1 mark for final answer with correct units.]
Q3 — Sugarcane biofuel Hess’s Law chain
3.1 Write each step:
(1) Photosynthesis: 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂ ΔH = +2803 kJ mol⁻¹
(2) Fermentation: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂ ΔH = −68 kJ mol⁻¹
(3) Combustion ×2: 2C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O ΔH = 2(−1366) = −2732 kJ mol⁻¹
Net ΔH = +2803 + (−68) + (−2732) = +3 kJ mol⁻¹ ≈ 0 (the small difference reflects rounding in the given values; conceptually the cycle sums to zero).
3.2 The near-zero net ΔH reflects the fact that the carbon pathway is cyclic — CO₂ fixed in photosynthesis is returned to the atmosphere on combustion of the bioethanol derived from that same glucose. By Hess’s Law, the total enthalpy change depends only on initial and final states (CO₂ + H₂O in both cases), not on the pathway. This is the thermodynamic basis of the carbon-neutrality claim: if the CO₂ released by burning bioethanol was recently captured from the atmosphere (not released from ancient fossil stores), the net atmospheric CO₂ change is approximately zero. [1 mark for identifying cyclic pathway; 1 mark for Hess’s Law state-function argument; 1 mark for correctly applying this to the carbon-neutrality claim.]
Q4 — Incomplete respiration
4.1 Via an incomplete pathway, the energy released would be less than 2803 kJ mol⁻¹. Reasoning: Hess’s Law states that ΔH depends only on the initial and final states of the system, not the pathway. Complete respiration takes glucose all the way to CO₂ + H₂O, releasing 2803 kJ mol⁻¹. An incomplete pathway stops at an intermediate (e.g. lactate, acetone) that still contains chemical energy — the final state is at a higher enthalpy than CO₂ + H₂O. Therefore the enthalpy drop from glucose to intermediate is smaller than the drop from glucose to CO₂ + H₂O, so less energy is released. [1 mark for “less than”; 1 mark for state-function reasoning; 1 mark for identifying intermediate as higher-enthalpy final state; 1 mark for correct direction of argument.]