Steel production requires carbon to reduce iron ore — and chemists needed to know the exact ΔH for C(s) + O₂(g) → CO₂(g) via the intermediate CO(g). The problem: you can't burn carbon to CO without also producing CO₂. The solution: Hess's Law. Add two measurable equations together, cancel the intermediate, and the answer falls out — no impossible experiment required.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
Imagine you are trying to work out how much it costs to drive from Sydney to Melbourne. You could go directly, or you could go via Canberra. The total distance — and therefore the petrol cost — depends only on where you start and where you end, not on the route you took.
Now imagine the same rule applies to enthalpy. Germain Hess discovered in 1840 that chemical energy follows exactly this pattern.
Before this lesson: Why would the total enthalpy change for a reaction be the same regardless of the pathway? Think about what enthalpy actually measures, and whether there is any energy that could "hide" in the intermediate steps. Write your prediction before the lesson explains it.
Type your initial response below — you will revisit this at the end of the lesson.
Write your initial response in your book. You will revisit it at the end of the lesson.
📚 Core Content
To apply Hess's Law, you need to reverse and scale thermochemical equations — and both operations change ΔH in predictable, non-negotiable ways.
The two rules — both must be applied every time:
| Operation | Effect on equation | Effect on ΔH | Example |
|---|---|---|---|
| Reverse | Reactants ↔ Products swap sides | ΔH × (−1) — sign flips | A → B (ΔH = −50) becomes B → A (ΔH = +50) |
| Scale by n | All coefficients × n | ΔH × n — magnitude scales | A → B (ΔH = −50) doubled: 2A → 2B (ΔH = −100) |
| Reverse then scale | Both operations applied | ΔH × (−1) × n | A → B (ΔH = −50), reversed and doubled: 2B → 2A (ΔH = +100) |
Strategy for solving Hess's Law problems:
The NSW Chemistry syllabus explicitly references this example. Learn it as a prototype — every Hess's Law problem follows the same logic.
Target equation: C(s) + O₂(g) → CO₂(g) ΔH = ?
Given equations:
Step-by-step solution:
Step 1 — Identify the intermediate. CO(g) and ½O₂(g) appear in both given equations but not in the target. They are the intermediates that must cancel.
Step 2 — In equation (1): CO(g) is a product. In equation (2): CO(g) is also a product. To cancel CO, we need it on different sides — so reverse equation (2).
Equation (2) reversed:
(2 rev) CO(g) + ½O₂(g) → CO₂(g) ΔH = −283.0 kJ mol⁻¹ (sign flipped)
Step 3 — Add equation (1) and reversed equation (2):
C(s) + ½O₂(g) + CO(g) + ½O₂(g) → CO(g) + CO₂(g)
Cancel CO(g) from both sides; ½O₂ + ½O₂ = O₂:
C(s) + O₂(g) → CO₂(g) ✓ Matches target
Step 4 — Sum ΔH values:
ΔH = −110.5 + (−283.0) = −393.5 kJ mol⁻¹
🔬 Worked Examples
Calculate ΔH for the reaction: C(s) + 2H₂(g) → CH₄(g)
Given: (1) C(s) + O₂(g) → CO₂(g) ΔH₁ = −393.5 kJ mol⁻¹
(2) H₂(g) + ½O₂(g) → H₂O(l) ΔH₂ = −285.8 kJ mol⁻¹
(3) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH₃ = −890.3 kJ mol⁻¹
Calculate ΔH for: N₂(g) + O₂(g) → 2NO(g)
Given: (1) N₂(g) + 2O₂(g) → 2NO₂(g) ΔH₁ = +68 kJ mol⁻¹
(2) NO(g) + ½O₂(g) → NO₂(g) ΔH₂ = −57 kJ mol⁻¹
🔬 Activities
a Given the two equations below, calculate ΔH for the target reaction.
Target: C(s) + O₂(g) → CO₂(g)
Given: (1) C(s) + ½O₂(g) → CO(g) ΔH₁ = −110.5 kJ mol⁻¹
(2) CO₂(g) → CO(g) + ½O₂(g) ΔH₂ = +283.0 kJ mol⁻¹
Show each manipulation step clearly before summing ΔH.
b In a blast furnace, the actual reaction of carbon with limited oxygen produces CO(g), not CO₂(g):
C(s) + ½O₂(g) → CO(g) ΔH = −110.5 kJ mol⁻¹.
A furnace engineer knows that C(s) + O₂(g) → CO₂(g) has ΔH = −393.5 kJ mol⁻¹. Use Hess's Law to calculate ΔH for the subsequent combustion CO(g) + ½O₂(g) → CO₂(g).
c Explain, using the concept of enthalpy as a state function, why Hess's Law gives the same answer whether C burns directly to CO₂ or via CO as an intermediate.
Type your working and responses below:
Answer in your workbook — show every manipulation step.
a Calculate ΔH for the reaction: C₂H₄(g) + H₂(g) → C₂H₆(g)
Given: (1) C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(l) ΔH₁ = −1411 kJ mol⁻¹
(2) C₂H₆(g) + 7/2 O₂(g) → 2CO₂(g) + 3H₂O(l) ΔH₂ = −1560 kJ mol⁻¹
(3) H₂(g) + ½O₂(g) → H₂O(l) ΔH₃ = −286 kJ mol⁻¹
Show all steps: identify manipulations needed for each equation, cancel intermediates, verify target, then calculate ΔH.
b A student attempts the same question but uses equation (2) in its original (unreversed) form. Without calculating, predict what error they will make in the final answer, and explain why.
Type your full working below:
Answer in your workbook.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
Wrong: In Hess's Law, you can change the chemical formula of a substance to make the equations add up.
Right: You can only reverse equations and scale coefficients — you cannot change chemical formulas. If the target equation requires CO₂, you must find an equation that produces CO₂, not modify a CO equation. The substances themselves must remain unchanged.
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
6. Calculate ΔH for the reaction: S(s) + 3/2 O₂(g) → SO₃(g)
Given: (1) S(s) + O₂(g) → SO₂(g) ΔH₁ = −297 kJ mol⁻¹
(2) 2SO₂(g) + O₂(g) → 2SO₃(g) ΔH₂ = −198 kJ mol⁻¹
Show all manipulation steps clearly.
4 MARKS
Type your full working below:
Show full working in your workbook.
7. (a) Explain why it is not possible to directly measure the standard enthalpy of formation of carbon monoxide, CO(g), in a laboratory calorimetry experiment. (2 marks)
(b) Describe how Hess's Law can be used to calculate ΔH°f[CO(g)] from measurable combustion data. Name the two equations you would use. (3 marks)
5 MARKS
Type your full answer:
Answer in your workbook.
8. A steel manufacturer needs to calculate the total heat energy released when 500 kg of carbon (as coke) reacts completely to form CO₂ in a blast furnace. The relevant thermochemical data is:
C(s) + ½O₂(g) → CO(g) ΔH₁ = −110.5 kJ mol⁻¹
CO(g) + ½O₂(g) → CO₂(g) ΔH₂ = −283.0 kJ mol⁻¹
(a) Use Hess's Law to write the thermochemical equation for C(s) + O₂(g) → CO₂(g) and state ΔH. (2 marks)
(b) Calculate the total heat energy released (in MJ) when 500 kg of carbon reacts to form CO₂. (Molar mass of C = 12.01 g mol⁻¹.) (3 marks)
(c) In practice, not all carbon burns fully to CO₂ — some stops at CO. Explain whether the actual heat released would be more or less than your calculated value, and why. (2 marks)
7 MARKS
Type your answer below:
Answer in your workbook.
Go back to your Think First response about why the total enthalpy change would be the same regardless of pathway. Now you can answer precisely:
Type your reflection below:
Write your reflection in your book.
(a) Intermediate = CO(g). Equation (2) has CO as product — reverse (2): CO(g) + ½O₂(g) → CO₂(g), ΔH = −283.0 kJ mol⁻¹. Add (1) + (2 rev): C + ½O₂ + CO + ½O₂ → CO + CO₂. Cancel CO; combine ½O₂ + ½O₂ = O₂: C(s) + O₂(g) → CO₂(g) ✓. ΔH = −110.5 + (−283.0) = −393.5 kJ mol⁻¹.
(b) −393.5 = −110.5 + ΔH(CO → CO₂) → ΔH(CO → CO₂) = −393.5 + 110.5 = −283.0 kJ mol⁻¹. This is the enthalpy of combustion of CO to CO₂.
(c) Enthalpy is a state function — it depends only on the initial state (C + O₂) and final state (CO₂), not the route between them. Whether the reaction proceeds directly or via CO as an intermediate, the system starts and ends at the same energy levels. Energy is conserved (First Law), so the total ΔH must be identical via both pathways.
(a) Target: C₂H₄ + H₂ → C₂H₆. Equation (1) used as-is (ΔH = −1411). Equation (2) reversed: 2CO₂ + 3H₂O → C₂H₆ + 7/2 O₂ (ΔH = +1560). Equation (3) used as-is (ΔH = −286). After cancellation: C₂H₄ + H₂ → C₂H₆ ✓. ΔH = −1411 + 1560 + (−286) = −137 kJ mol⁻¹. Exothermic hydrogenation.
(b) Using (2) unreversed leaves C₂H₆ as a reactant. The equations don't cancel to give the target. Numerically, ΔH = −1411 + (−1560) + (−286) = −3257 kJ mol⁻¹ — an implausibly large value that immediately flags the error.
1. C — Hess's Law: ΔH is path-independent. It depends only on initial and final states, not on whether the reaction occurs in one step or multiple steps.
2. B — Reverse: ΔH = +120 kJ mol⁻¹. Then double: ΔH = +240 kJ mol⁻¹. Both operations applied to ΔH in sequence.
3. B — Intermediates appear as products in one stepped equation and reactants in another. They cancel when equations are added, leaving only the target equation species.
4. A — Target: W + Y → Z. Equation (1) as-is (X + Y → Z, ΔH = −200). Equation (2) as-is (W → X + 2Y, ΔH = +80). Add: W + X + 2Y → X + 3Y + Z → cancel X (both sides) and one Y: W + Y → Z ✓. ΔH = −200 + 80 = −120 kJ mol⁻¹.
5. D — Target: 2CO + O₂ → 2CO₂. Use: C + O₂ → CO₂ (×2): ΔH = 2(−393.5) = −787; and C + ½O₂ → CO (×2, reversed): 2CO → 2C + O₂, ΔH = +2(110.5) = +221. Add: 2CO + O₂ → 2CO₂ ✓. ΔH = −787 + 221 = −566 kJ mol⁻¹. Alternatively: ΔH(CO → CO₂) = −283.0 × 2 = −566.
Q6 (4 marks): Target: S + 3/2 O₂ → SO₃. Equation (1) used as-is: S + O₂ → SO₂, ΔH = −297 [1]. Equation (2) scaled by ½: SO₂ + ½O₂ → SO₃, ΔH = −99 [1]. Add: S + O₂ + SO₂ + ½O₂ → SO₂ + SO₃. Cancel SO₂; O₂ + ½O₂ = 3/2 O₂: S + 3/2 O₂ → SO₃ ✓ [1]. ΔH = −297 + (−99) = −396 kJ mol⁻¹ [1].
Q7 (5 marks):
(a) Burning carbon in limited oxygen always produces a mixture of CO and CO₂, not pure CO [1]. There is no experimental setup that cleanly isolates C + ½O₂ → CO without CO₂ also forming, making direct calorimetric measurement of ΔH°f[CO(g)] impossible [1].
(b) Two measurable equations needed: (i) C(s) + O₂(g) → CO₂(g), ΔH = −393.5 kJ mol⁻¹ (combustion of carbon) [1]; (ii) CO(g) + ½O₂(g) → CO₂(g), ΔH = −283.0 kJ mol⁻¹ (combustion of CO) [1]. Reverse equation (ii): CO₂(g) → CO(g) + ½O₂(g), ΔH = +283.0. Add to (i): CO₂ cancels; result: C(s) + ½O₂(g) → CO(g), ΔH = −393.5 + 283.0 = −110.5 kJ mol⁻¹ [1].
Q8 (7 marks):
(a) Add equations (1) and (2): CO(g) cancels (product in 1, reactant in 2); ½O₂ + ½O₂ = O₂. Result: C(s) + O₂(g) → CO₂(g) [1]. ΔH = −110.5 + (−283.0) = −393.5 kJ mol⁻¹ [1].
(b) n(C) = 500,000 ÷ 12.01 = 41,632 mol [1]. Q = 41,632 × 393.5 = 16,382,192 kJ ÷ 1000 = ≈ 16,400 MJ [1]. Answer in range 16,360–16,420 MJ accepted [1].
(c) If some carbon stops at CO, only ΔH₁ = −110.5 kJ mol⁻¹ is released per mole of C (compared to −393.5 for full combustion). The second stage (CO → CO₂, ΔH = −283.0) is not reached for those moles. Therefore the actual heat released would be less than the calculated value [1], by approximately −283.0 × n(moles stopping at CO) [1].
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