1. Hess's law
Hess's law: the enthalpy change for a reaction is independent of the route taken, provided the initial and final states (substances, amounts and physical states) are the same. It follows from conservation of energy.
This lets us find ΔH for reactions we cannot measure directly, such as forming methane from carbon and hydrogen, by going round a cycle of reactions we can measure.
2. Rules for manipulating equations
- Reverse an equation → reverse the sign of ΔH.
- Multiply the coefficients → multiply ΔH by the same factor.
- Add equations → add their ΔH values. Species that appear on both sides cancel.
3. Using formation data
ν is each species' coefficient. Formation arrows run from the common elements to the reactants and to the products; subtract the reactant formation total from the product formation total.
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Formation cycle for the combustion of ethanol
Labels to include:
- Top left: C₂H₅OH(l) + 3O₂(g)
- Top right: 2CO₂(g) + 3H₂O(l)
- Top arrow left → right: ΔrH (unknown)
- Bottom centre: 2C(s) + 3H₂(g) + 3½O₂(g) (elements in standard states)
- Arrow bottom → top left: ΔfH(C₂H₅OH) (O₂ contributes 0)
- Arrow bottom → top right: 2ΔfH(CO₂) + 3ΔfH(H₂O)
Both upward arrows start from the same elements. Going the direct way (elements → products) equals going via the reactants, so ΔHr = (product formation arrow) − (reactant formation arrow).
4. Using combustion data
Note the order is the other way round. This works because both reactants and products burn to the same final combustion products (CO2(g) and H2O(l)), so the cycle arrows point down to them.
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Combustion cycle for the formation of ethane
Labels to include:
- Top left: 2C(s) + 3H₂(g) (+ 3½O₂(g))
- Top right: C₂H₆(g) (+ 3½O₂(g))
- Top arrow left → right: ΔfH(C₂H₆) (unknown)
- Bottom centre: 2CO₂(g) + 3H₂O(l)
- Arrow top left → bottom: 2ΔcH(C) + 3ΔcH(H₂)
- Arrow top right → bottom: ΔcH(C₂H₆)
Both routes end at the same combustion products. Direct route = route via the products, so ΔH = (reactant combustion arrow) − (product combustion arrow).
5. Zero-value traps
| Quantity | Zero? | Why |
|---|---|---|
| ΔfH of O2(g) | Yes | It is an element in its standard state |
| Bond enthalpy of O=O in O2 | No | Breaking the O=O bond needs energy (about +498 kJ mol⁻¹) |
| ΔfH of H2(g) | Yes | Element in its standard state |
| ΔcH of H2(g) | No | Hydrogen burns to water (about −286 kJ mol⁻¹) |
| ΔcH of O2(g) or CO2(g) | Not used | Neither burns further in oxygen; they have no combustion value in cycles |
6. Finding an unknown formation enthalpy
7. Hydrated salt cycle
The enthalpy change of hydration of a salt, e.g. CuSO4(s) + 5H2O(l) → CuSO4·5H2O(s), is difficult to measure reliably by direct mixing, because the hydration (and any dissolving) and the heat transfer are hard to control. An indirect dissolution cycle gives a controlled route instead: dissolve the anhydrous salt and the hydrated salt separately in water to make the same final solution.
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Hydration cycle for copper(II) sulfate
Labels to include:
- Top left: CuSO₄(s) + 5H₂O(l)
- Top right: CuSO₄·5H₂O(s)
- Top arrow left → right: ΔH(hydration) (unknown)
- Bottom centre: CuSO₄(aq) (same final solution and dilution)
- Arrow top left → bottom: ΔH(solution, anhydrous) (+ water)
- Arrow top right → bottom: ΔH(solution, hydrate) (+ water)
Both salts end in the same solution. ΔH(hydration) = ΔH(solution, anhydrous) − ΔH(solution, hydrate).
8. Indirect route: NaHCO₃ decomposition
This needs strong heating, so its temperature change cannot be measured in a cup. Instead, measure two reactions with hydrochloric acid that both end in the same products:
| Reaction | ΔH / kJ mol⁻¹ | |
|---|---|---|
| 1 | NaHCO3(s) + HCl(aq) → NaCl(aq) + CO2(g) + H2O(l) | ΔH₁ = +28.0 |
| 2 | Na2CO3(s) + 2HCl(aq) → 2NaCl(aq) + CO2(g) + H2O(l) | ΔH₂ = −35.0 |
Algebra
Double reaction 1: 2NaHCO3 + 2HCl → 2NaCl + 2CO2 + 2H2O, ΔH = 2ΔH1.
Reverse reaction 2: 2NaCl + CO2 + H2O → Na2CO3 + 2HCl, ΔH = −ΔH2.
Add: 2HCl, 2NaCl, one CO2 and one H2O cancel, leaving the target equation.
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Hess cycle for NaHCO₃ decomposition
Labels to include:
- Top left: 2NaHCO₃(s) (+ 2HCl(aq))
- Top right: Na₂CO₃(s) + CO₂(g) + H₂O(l) (+ 2HCl(aq))
- Top arrow: ΔH (target)
- Bottom: 2NaCl(aq) + 2CO₂(g) + 2H₂O(l)
- Left arrow down: 2ΔH₁
- Right arrow down: ΔH₂
The same 2HCl is added on both sides so the two acid reactions reach one common final state. The algebra above is complete without the diagram.
Quick checks
These are Finesse practice questions. The step-by-step answers are indicative worked solutions, not official AQA mark allocations.
Q1. Use ΔfH values (kJ mol−1: CH4(g) −75, CO2(g) −394, H2O(l) −286) to find ΔcH of methane.Show answer
Step 1: CH4 + 2O2 → CO2 + 2H2O(l).
Step 2: products = −394 + 2(−286) = −966; reactants = −75 + 0 = −75.
Step 3: ΔH = −966 − (−75) = −891 kJ mol−1.
Q2. Find ΔfH of propane, C3H8(g), using ΔcH (kJ mol−1): C(s) −394, H2(g) −286, C3H8(g) −2220.Show answer
Step 1: 3C(s) + 4H2(g) → C3H8(g).
Step 2: reactants = 3(−394) + 4(−286) = −2326.
Step 3: ΔH = −2326 − (−2220) = −106 kJ mol−1.
Q3. 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g), ΔH = −905 kJ mol−1. ΔfH: NO(g) +90, H2O(g) −242. Find ΔfH of NH3(g).Show answer
Step 1: −905 = [4(+90) + 6(−242)] − [4x + 0] = −1092 − 4x.
Step 2: 4x = −1092 + 905 = −187.
Step 3: x = −187 ÷ 4 = −46.8 kJ mol−1.
Q4. ΔHsol(MgSO4) = −91.2 kJ mol−1; ΔHsol(MgSO4·7H2O) = +16.1 kJ mol−1, both to the same final solution. Find ΔH for MgSO4(s) + 7H2O(l) → MgSO4·7H2O(s).Show answer
Step 1: ΔHhydration = ΔHsol(anhydrous) − ΔHsol(hydrate).
Step 2: −91.2 − (+16.1) = −107.3 kJ mol−1.
Q5. A student writes ΔcH(H2) = 0 in a cycle “because H2 is an element”. Correct this.Show answer
Elements in their standard states have ΔfH = 0, not ΔcH = 0.
Hydrogen burns to form water with a large negative ΔH (about −286 kJ mol−1), so its combustion value must be used.
Sources
Sources and examiner guidance (reviewed 1 October 2026)
- Chemrevise — AQA 1.4 Energetics revision guide (N. Goalby) — Checklist for Hess cycles, formation/combustion routes and hydration cycles.
- AQA 7405 specification — 3.1.4 Energetics (incl. Required Practical 2) — 3.1.4.3 Applications of Hess's law.
- AQA 7404/2 mark scheme, November 2020 series (footer reads June 2020) — Q01.2: signs, coefficients, dividing the unknown by 3.
- AQA 7404/2 mark scheme, June 2019 — Q03.3: combustion cycle with coefficients and reversal.
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