Hess's Law
In 1840, Germain Hess at the St Petersburg Academy of Sciences published his Law of Constant Heat Summation — establishing that the total enthalpy change for a reaction is independent of the pathway taken. His key proof was the formation of carbon dioxide: he could not measure the enthalpy of C(s) → CO(g) directly (carbon always partially burns to CO₂), so he combined two measurable equations — C(s) + O₂(g) → CO₂(g) at −393.5 kJ mol⁻¹ and CO(g) + ½O₂(g) → CO₂(g) at −283.0 kJ mol⁻¹ — to derive ΔH(C → CO) = −110.5 kJ mol⁻¹. This is still the accepted value today.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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.
Key Facts
- Hess's Law: ΔH is the same regardless of pathway — enthalpy is a state function
- Reverse equation → ΔH changes sign; scale equation by n → ΔH scales by n
- The NESA prototype: C → CO₂ via CO(g) as intermediate
Concepts
- Why Hess's Law follows from enthalpy being a state function (First Law of Thermodynamics)
- Why some ΔH values can only be found by Hess's Law (reactions unmeasurable directly)
- How intermediate species cancel when stepped equations are added correctly
Skills
- Reverse and scale thermochemical equations, updating ΔH correctly at each step
- Add two or three equations to produce a target equation by cancelling intermediates
- Verify the result matches the target equation before calculating ΔH
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:
- Write down the target equation — this is what you need ΔH for
- Identify which species appear in the target. For each species, find a given equation that contains it
- For each given equation: reverse it (if needed) so the species appears on the correct side; scale it (if needed) to match the coefficient in the target
- Add all modified equations together. Cancel any species that appear on both sides (these are intermediates)
- Confirm the remaining species match the target equation exactly
- Sum all modified ΔH values — the result is ΔH for the target reaction
Hess’s Law: ΔH for a target reaction equals the algebraic sum of ΔH values for a series of steps, regardless of pathway. Rules: reverse an equation → flip sign of ΔH; scale an equation → multiply ΔH by the same factor.
Pause — copy the highlighted definition into your book before moving on.
Quick check: A reaction A → B has ΔH = −80 kJ mol⁻¹. You reverse the equation AND double it. What is the new ΔH?
We just saw the rules for manipulating thermochemical equations — reversing flips ΔH, scaling multiplies it. That raises a question: when is Hess’s Law actually necessary rather than direct measurement? This card answers it → with the formation of CO, a reaction impossible to measure directly.
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:
(2) CO₂(g) → CO(g) + ½O₂(g) ΔH₂ = +283.0 kJ mol⁻¹
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
Carbon monoxide’s ΔHf cannot be measured directly because C + ½O₂ → CO cannot be run to completion without CO₂ forming. Hess’s Law solves this by combining C + O₂ → CO₂ and reversing CO₂ → CO + ½O₂, giving ΔHf(CO) = −110 kJ mol⁻¹.
Add the highlighted point to your notes before the check below.
Explain it: In the NESA example, why must equation (2) be reversed before adding to equation (1)? Explain in your own words using the idea of intermediate species cancellation.
We just saw how a two-step Hess’s Law cycle solved the CO formation problem using measured combustion data. That raises a question: how do you extend this to more complex problems involving three separate combustion equations? This card answers it → with a systematic align-cancel-sum strategy for three-equation cycles.
Most HSC Hess's Law problems use combustion enthalpies — all measurable — to derive the enthalpy of a reaction that cannot be measured directly. Three-equation cycles are the standard exam format.
The same six-step strategy applies, just with three equations to manipulate and more intermediates to cancel. Common example: deriving ΔH of formation for an organic compound from combustion data for the compound and its elements.
Key insight: Combustion enthalpies are always measurable (you can always burn something in a calorimeter). But the enthalpy of formation — the energy change when a compound forms from its elements — often cannot be measured directly. Hess's Law bridges this gap.
Most HSC Hess’s Law problems provide combustion enthalpies for three substances to derive a target ΔH. Strategy: align each equation so desired species appear as products or reactants; reverse and scale to cancel intermediates; sum remaining ΔH values.
Pause — write the highlighted rule into your book.
Fill in the blanks: Complete the Hess's Law strategy: Write the ______ equation first. For each given equation, ______ or ______ it so each species appears on the correct side. Add all equations and ______ species appearing on both sides. Verify the remaining species match the ______ before summing ΔH.
We just saw how three-equation combustion cycles are solved systematically. That raises a question: what are the essential rules to remember for any Hess’s Law problem? This card answers it → with four key ideas that cover every exam question type.
The four key ideas from this lesson — distilled into book notes. Each one connects to a specific exam question type.
Hess's Law Statement
- The total enthalpy change for a reaction is the same regardless of the pathway taken
- Follows from enthalpy being a state function (depends only on initial and final states)
- Consequence of the First Law of Thermodynamics — energy is conserved
Equation Manipulation Rules
- Reverse equation → flip sign of ΔH (× −1)
- Scale equation by n → scale ΔH by n
- Both operations must be applied to the equation AND its ΔH together
NESA Prototype (C → CO₂ via CO)
- Given: C + ½O₂ → CO, ΔH₁ = −110.5 kJ mol⁻¹
- Given: CO₂ → CO + ½O₂, ΔH₂ = +283.0 → reverse → −283.0
- Sum: C + O₂ → CO₂, ΔH = −393.5 kJ mol⁻¹
Problem-Solving Strategy
- Write target equation first
- Reverse and/or scale given equations to get each species on the correct side
- Add equations — cancel intermediates (appear on both sides)
- Verify result = target equation before summing ΔH
Hess’s Law four key rules: (1) ΔH is independent of pathway (state function); (2) reversing an equation flips the sign of ΔH; (3) scaling coefficients scales ΔH proportionally; (4) intermediates must cancel exactly when the target equation is obtained.
Add the highlighted point to your notes before the check below.
Odd one out: Three of the following statements about Hess's Law are correct. Which one is the odd one out (incorrect)?
Worked example · reveal as you go
Three-Equation Hess's Law Cycle. 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
Target needs CH₄ as a product. Equation (3) has CH₄ as a reactant — reverse it:
(3 rev) CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g) ΔH = +890.3 kJ mol⁻¹Target needs C as a reactant. Equation (1) already has C as a reactant — use it unchanged:
(1) C(s) + O₂(g) → CO₂(g) ΔH = −393.5 kJ mol⁻¹Target needs 2 mol H₂ as reactants. Equation (2) has 1 mol H₂ — multiply by 2:
(2 × 2) 2H₂(g) + O₂(g) → 2H₂O(l) ΔH = 2(−285.8) = −571.6 kJ mol⁻¹Adding (3 rev) + (1) + (2 × 2):
CO₂ + 2H₂O + C + O₂ + 2H₂ + O₂ → CH₄ + 2O₂ + CO₂ + 2H₂OCancel: CO₂ (both sides) ✓ | 2H₂O (both sides) ✓ | O₂ + O₂ − 2O₂ = 0 ✓
C(s) + 2H₂(g) → CH₄(g) ✓ Matches targetΔH = +890.3 + (−393.5) + (−571.6)
= 890.3 − 393.5 − 571.6
= −74.8 kJ mol⁻¹
This is ΔH°f[CH₄(g)] — the standard enthalpy of formation of methane. The accepted value is exactly −74.8 kJ mol⁻¹. Hess's Law just derived the enthalpy of formation via combustion data — which is precisely how ΔH°f values are measured experimentally.
Put the Hess's Law calculation method in the correct order.
- Reverse any equations needed, flipping the sign of their ΔH.
- Write down the target equation — the reaction whose ΔH you want to find.
- Add all the adjusted ΔH values to get ΔH for the target reaction.
- Identify which given equations contain the species in the positions you need.
- Multiply equations as needed; scale their ΔH by the same factor.
Formula Reference — This Lesson
Common errors · the 3 traps that cost marks
Forgetting to flip the sign when reversing
Students reverse a thermochemical equation but copy the original ΔH value unchanged. In the NESA example, reversing equation (2) gives ΔH = +283.0 → the correct value after reversal is −283.0 kJ mol⁻¹. Forgetting this sign flip is the single most common error in Hess's Law questions.
Fix: Treat the equation and its ΔH as one inseparable object. Write the reversed equation and immediately write the new ΔH (with flipped sign) on the same line. Never copy-paste the original ΔH after a reversal.
Scaling the equation but not ΔH (or vice versa)
Students multiply all coefficients by 2 but write "ΔH = −285.8 kJ mol⁻¹" instead of "ΔH = −571.6 kJ mol⁻¹". Or they change ΔH but forget to update one coefficient. The equation and ΔH must always be scaled together.
Fix: When scaling, write the new equation on its own line with the new ΔH already calculated. Use brackets: ΔH = 2 × (−285.8) = −571.6. Showing the multiplication step earns method marks even if the arithmetic is wrong.
Skipping the target-verification step
Students sum the ΔH values before checking that the remaining species after cancellation actually match the target equation. If an intermediate species has not cancelled correctly, the ΔH sum is for the wrong reaction — a critical error that yields no marks.
Fix: Always write out the combined equation after cancellation and compare it species-by-species to the target before calculating ΔH. Write "✓ Matches target" as an explicit step — HSC markers look for this verification.
Quick-fire practice · 5 reps +2 XP per reveal
Calculate ΔH for the target reaction using the two equations below. Show each manipulation step clearly before summing ΔH.
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
(2 reversed): CO(g) + ½O₂(g) → CO₂(g) ΔH = −283.0 kJ mol⁻¹ (sign flipped)
Step 2: Add (1) + (2 reversed):
C + ½O₂ + CO + ½O₂ → CO + CO₂
Cancel CO(g) from both sides; ½O₂ + ½O₂ = O₂:
C(s) + O₂(g) → CO₂(g) ✓
Step 3: ΔH = −110.5 + (−283.0) = −393.5 kJ mol⁻¹
A blast furnace engineer knows C(s) + O₂(g) → CO₂(g) has ΔH = −393.5 kJ mol⁻¹ and C(s) + ½O₂(g) → CO(g) has ΔH = −110.5 kJ mol⁻¹. Use Hess's Law to calculate ΔH for the combustion of CO: CO(g) + ½O₂(g) → CO₂(g).
−393.5 = −110.5 + ΔH(CO → CO₂)
ΔH(CO → CO₂) = −393.5 − (−110.5) = −393.5 + 110.5 = −283.0 kJ mol⁻¹
This is ΔH for CO(g) + ½O₂(g) → CO₂(g) — the second-stage combustion of CO. Knowing this lets engineers calculate how much additional heat is produced when CO burns in the upper zone of the blast furnace.
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.
Calculate ΔH for: 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.
C₂H₄: target needs it as reactant; (1) has it as reactant — use (1) unchanged: ΔH = −1411
C₂H₆: target needs it as product; (2) has it as reactant — reverse (2): ΔH = +1560
(2 rev): 2CO₂(g) + 3H₂O(l) → C₂H₆(g) + 7/2 O₂(g) ΔH = +1560
H₂: target needs 1 mol H₂ as reactant; (3) has H₂ as reactant — use (3) unchanged: ΔH = −286
Add (1) + (2 rev) + (3):
C₂H₄ + 3O₂ + 2CO₂ + 3H₂O + H₂ + ½O₂ → 2CO₂ + 2H₂O + C₂H₆ + 7/2 O₂ + H₂O
Cancel: 2CO₂ both sides ✓ | 3H₂O both sides ✓ (2 on right + 1 = 3 ✓) | O₂: 3O₂ + ½O₂ − 7/2 O₂ = 0 ✓
Remaining: C₂H₄(g) + H₂(g) → C₂H₆(g) ✓
ΔH = −1411 + 1560 + (−286) = −137 kJ mol⁻¹
This is the hydrogenation of ethene — exothermic, consistent with adding H₂ across the double bond releasing energy.
A student attempts the hydrogenation of ethene question (Drill 4) but uses equation (2) in its original (unreversed) form. Without calculating, predict what error they will make in the final answer, and explain why the answer they get should immediately signal a mistake.
In terms of the numerical error: they would use ΔH₂ = −1560 instead of +1560. The answer would be −1411 + (−1560) + (−286) = −3257 kJ mol⁻¹ — an absurdly large value for a simple hydrogenation reaction. Any time a Hess's Law answer has an unexpectedly large magnitude, check whether an equation reversal was missed.
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 — and verify using Germain Hess's 1840 carbon dioxide example from the St Petersburg Academy of Sciences:
- Why pathway independence holds: Enthalpy is a state function — like altitude. Whether carbon burns directly to CO₂ or via CO as an intermediate, the total ΔH is identical. Hess showed this experimentally: his two-step pathway gave ΔH(C → CO) = −393.5 − (−283.0) = −110.5 kJ mol⁻¹, and later direct measurements confirmed this value.
- Why this is powerful: Hess could not measure C → CO directly (you always get some CO₂). But he could calculate it from two experiments he could perform — exactly as you can derive any inaccessible ΔH by combining measurable equations.
- The manipulation rules are consequences, not arbitrary: Reversing undoes the energy change (releasing energy going forward means absorbing the same going backward). Scaling doubles the amount of reaction. Both rules follow directly from conservation of energy — Hess understood this in 1840 before the first law of thermodynamics was formally stated.
Pick your answer, then rate your confidence — that tells the system what to drill next.
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.
Wrong: Scaling an equation by ×2 keeps ΔH the same because the same substances are reacting.
Right: ΔH belongs to the equation as written. If you double all coefficients (2× the moles of everything), you must double ΔH. It is a property of the equation, not just the substances.
Wrong: Reversing the equation just means the products and reactants swap — ΔH stays the same.
Right: Reversing flips the sign of ΔH. If the forward reaction is exothermic (ΔH < 0), the reverse is endothermic (ΔH > 0) with the same magnitude. The + must be written explicitly.
Q6. 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
Q7. (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
Q8. 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
Show comprehensive answers ▼
Drill 1 — NESA Prototype
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.
Drill 2 — CO Combustion
−393.5 = −110.5 + ΔH(CO → CO₂) → ΔH(CO → CO₂) = −393.5 + 110.5 = −283.0 kJ mol⁻¹.
Drill 4 — Hydrogenation of Ethene
Target: C₂H₄ + H₂ → C₂H₆. Equation (1) used as-is (ΔH = −1411). Equation (2) reversed (ΔH = +1560). Equation (3) used as-is (ΔH = −286). After cancellation: C₂H₄ + H₂ → C₂H₆ ✓. ΔH = −1411 + 1560 + (−286) = −137 kJ mol⁻¹.
Multiple Choice
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.
Short Answer Model Answers
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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