Chemistry • Year 11 • Module 4 • Lesson 8
Hess's Law
Apply the manipulation rules to real data, interpret enthalpy graphs, trace cause-and-effect chains, and evaluate a perturbation scenario.
1. Structured Hess’s Law calculation chain — SO₃ formation and acid rain
Sulfur trioxide (SO₃) is formed in industrial and atmospheric chemistry and is a key step in the formation of sulfuric acid (acid rain). The table below shows a structured worked framework for calculating ΔH for the target reaction. Complete every shaded cell. 8 marks
Target reaction: S(s) + ¾O₂(g) → SO₃(g) ΔH = ?
Given equations:
(1) S(s) + O₂(g) → SO₂(g) ΔH₁ = −297 kJ mol⁻¹
(2) 2SO₂(g) + O₂(g) → 2SO₃(g) ΔH₂ = −198 kJ mol⁻¹
| Step | Equation used | Manipulation needed | Modified equation | Modified ΔH (kJ mol⁻¹) |
|---|---|---|---|---|
| 1 | (1) | None — use as written | S(s) + O₂(g) → SO₂(g) | |
| 2 | (2) | |||
| 3 | Add modified equations. Cancel intermediates. | |||
| 4 | Verify target matches. State ΔH for target. | |||
1.1 What species acts as the intermediate in this calculation? Explain how you know it is an intermediate and not a species in the target reaction. 2 marks
2. Interpret graph — cumulative ΔH for carbon oxidation at Port Kembla
The graph below models the cumulative enthalpy released per mole of carbon as it oxidises in three scenarios in a blast furnace: (i) all carbon burns directly to CO₂; (ii) carbon burns in two stages (C→CO then CO→CO₂); (iii) carbon stops at CO (incomplete combustion). All ΔH values are based on NESA prototype data. 7 marks
2.1 State the total cumulative ΔH for each scenario at “Stage 2 complete”. 2 marks
2.2 Explain why scenarios (i) and (ii) reach the same final cumulative ΔH. Refer to Hess’s Law in your answer. 2 marks
2.3 A Port Kembla furnace engineer estimates that 15% of the carbon in the furnace charge only reaches CO before being drawn off. Explain whether the actual heat produced would be more or less than the value predicted assuming complete combustion to CO₂, and calculate the approximate percentage shortfall. 3 marks
3. Cause-and-effect chain — applying manipulation rules
Each “cause box” describes an action taken during a Hess’s Law calculation. Fill each empty “effect box” with the consequence of that action. 6 marks
| Cause (action taken) | Effect (consequence for ΔH or equation) |
|---|---|
| Equation A: X → Y ΔH = −80 kJ mol⁻¹ is reversed. | |
| Equation B: 2P → Q + R ΔH = +120 kJ mol⁻¹ is scaled by ½. | |
| Equation C: M + N → L ΔH = −50 kJ mol⁻¹ is reversed, then doubled. | |
| When two modified equations are added together, species T appears as a product in one equation and as a reactant in the other in equal amounts. | |
| After adding all equations, the species remaining do not match the target equation (an extra O₂ appears). | |
| Equation D has ΔH = −200 kJ mol⁻¹. A student scales the equation by 3 but forgets to scale ΔH. |
4. Case study — Hess’s Law and the acid rain chemistry of SO₃
Read the scenario, then answer the question below. 5 marks
Context. Nitrogen oxides and sulfur oxides released from industrial sites and vehicle exhausts react in the atmosphere to form acid rain. A critical step is the oxidation of SO₂ to SO₃, which then reacts with water to form H₂SO₄. The enthalpy change for the conversion SO₂(g) + ½O₂(g) → SO₃(g) cannot be measured directly in most school laboratories because the reaction is slow without a catalyst. However, two measurable enthalpies are available:
(1) S(s) + O₂(g) → SO₂(g) ΔH₁ = −297 kJ mol⁻¹
(2) S(s) + ¾O₂(g) → SO₃(g) ΔH₂ = −396 kJ mol⁻¹
(Both measurable directly from combustion calorimetry.)
4.1 Use Hess’s Law to calculate ΔH for: SO₂(g) + ½O₂(g) → SO₃(g). Show all manipulation steps. Then explain why this enthalpy value is chemically significant in the context of acid rain formation. 5 marks
5. Compare & contrast — direct measurement vs Hess’s Law approach
Complete the table comparing the two approaches to determining ΔH. 6 marks
| Feature | Direct calorimetric measurement | Hess’s Law (indirect) approach |
|---|---|---|
| What is measured directly? | ||
| When is it the only option? | When reaction can be performed cleanly in a calorimeter | |
| NESA example where this applies | C + ½O₂ → CO (cannot isolate CO product) | |
| Required mathematical operations | q = mcΔT; ΔH = −q/n | |
| Accuracy limitation | ||
| Real-world Australian use |
Q1 — Structured Hess’s Law chain (SO₃ from S)
Step 1: Equation (1) used as written. Modified ΔH = −297 kJ mol⁻¹.
Step 2: Equation (2) needs SO₂ to appear as a reactant (so it cancels with the SO₂ produced in Step 1), and only 1 mol SO₃ as product. Equation (2) must be scaled by ½: SO₂(g) + ½O₂(g) → SO₃(g). Modified ΔH = ½ × (−198) = −99 kJ mol⁻¹.
Step 3 — combined equation: S + O₂ + SO₂ + ½O₂ → SO₂ + SO₃. Cancel SO₂ (both sides); O₂ + ½O₂ = ¾O₂. Result: S(s) + ¾O₂(g) → SO₃(g).
Step 4 — verify and state ΔH: Matches target. ΔH = −297 + (−99) = −396 kJ mol⁻¹.
1.1 Intermediate = SO₂(g). It appears as a product in equation (1) and as a reactant in the scaled equation (2), so it cancels when the equations are added. SO₂ does not appear in the target equation, confirming it is an intermediate.
Q2 — Graph interpretation
2.1 Scenario (i): −393.5 kJ per mol C at Stage 2 complete. Scenario (ii): −393.5 kJ per mol C. Scenario (iii): −110.5 kJ per mol C (flat after Stage 1).
2.2 Hess’s Law: enthalpy is a state function, so ΔH depends only on the initial state (C + O₂) and the final state (CO₂), not on the route. Whether C burns directly or via CO as an intermediate, the initial and final states are identical, so the total cumulative ΔH must be equal (−393.5 kJ mol⁻¹).
2.3 If 15% of carbon stops at CO: those moles release only −110.5 kJ each; the remaining 85% release −393.5 kJ each. Expected average per mol C = 0.85 × 393.5 + 0.15 × 110.5 = 334.5 + 16.6 = 351.1 kJ. Predicted (complete) = 393.5 kJ. Shortfall = (393.5 − 351.1)/393.5 × 100 ≈ 10.8% less heat than predicted, so actual heat released is less than the complete-combustion prediction.
Q3 — Cause-and-effect chain
Row 1: New equation is Y → X; ΔH = +80 kJ mol⁻¹ (sign flipped).
Row 2: New equation is P → ½Q + ½R; ΔH = +60 kJ mol⁻¹ (scaled by ½).
Row 3: Reversed: L → M + N, ΔH = +50. Then doubled: 2L → 2M + 2N, ΔH = +100 kJ mol⁻¹.
Row 4: Species T cancels — it does not appear in the combined (net) equation.
Row 5: The manipulation is incorrect — an equation must be checked and adjusted (reversed or scaled differently) so the extra O₂ cancels. Do not proceed to sum ΔH until the target equation is matched exactly.
Row 6: The calculated ΔH is wrong — it would be −200 instead of the correct −600 kJ mol⁻¹. The student must multiply ΔH by 3 as well: ΔH(correct) = 3 × (−200) = −600 kJ mol⁻¹.
Q4 — Acid rain case study (5 marks)
Calculation: Target: SO₂(g) + ½O₂(g) → SO₃(g).
Reverse (1): SO₂(g) → S(s) + O₂(g); ΔH = +297 kJ mol⁻¹ [1].
Add reversed (1) + (2): SO₂ + S + ¾O₂ → S + O₂ + SO₃. Cancel S(s) and subtract O₂ from both sides (¾O₂ − O₂ = −¼O₂... re-check: reversed (1) has O₂ as product; (2) has ¾O₂ as reactant; net O₂ side: ¾O₂ [reactant] − O₂ [product in rev(1)] = ½O₂ reactant) [1]. Remaining: SO₂(g) + ½O₂(g) → SO₃(g) ✓ [1].
ΔH = +297 + (−396) = −99 kJ mol⁻¹ [1].
This is exothermic (−99 kJ mol⁻¹), meaning the conversion of SO₂ to SO₃ in the atmosphere releases energy, making it thermodynamically favourable. SO₃ then reacts rapidly with atmospheric moisture to form H₂SO₄ (sulfuric acid), the main contributor to acid rain — a process that is also highly exothermic and spontaneous [1].
Q5 — Compare and contrast table
Direct: What is measured? Temperature change (ΔT) in a calorimeter; converted to q = mcΔT and then ΔH = −q/n. Hess’s Law: What is measured? ΔH values of a set of simpler, measurable reactions that can be algebraically combined to give the target.
Direct: When the only option? When the reaction can be cleanly isolated in a calorimeter and no significant side reactions occur. Hess’s Law: When the only option? When the target reaction cannot be measured directly — e.g. reactions that produce a mixture of products, extremely fast reactions, reactions that require impractical conditions.
NESA example: Direct — combustion of C to CO₂; combustion of CO to CO₂. Hess’s Law — formation of CO from C (cannot isolate).
Hess’s Law operations: Reverse equations (flip sign of ΔH), scale equations (multiply ΔH by n), then add and cancel intermediates.
Accuracy limitation: Direct — heat loss to surroundings, incomplete combustion. Hess’s Law — accumulated uncertainty in each component ΔH measurement; errors compound.
Real-world Australian use: Direct — lab measurement of combustion fuels. Hess’s Law — Port Kembla blast furnace energy calculations; SO₃/acid rain atmospheric chemistry modelling.