Hess's Law
Enthalpy doesn't care how you got there — so you can build any reaction's ΔH out of steps you already know.
⏱️ About 18 min
You can't burn carbon and stop cleanly at carbon monoxide — some always races on to carbon dioxide. So how does anyone know the ΔH of that half-finished reaction? You never measure it. You assemble it from two reactions you <em>can</em> measure. That trick is Hess's law.
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The big idea: Enthalpy is a state function, so ΔH depends only on the starting and ending states — not the path between them. That means you can add, reverse, and scale known reactions like algebra to reach a target reaction's ΔH.
Why the path doesn't matter
Recall from the first lesson that enthalpy is a state function: it depends only on the current state, like altitude on a mountain. Whether you drive up the gentle road or scramble straight up the cliff, the summit is the same height above where you started.
Hess's law is exactly this idea applied to reactions: the total enthalpy change of a reaction is the same whether it happens in one step or many. So if a target reaction can be written as a sum of other reactions, its ΔH is the sum of their ΔH values.
\[ \Delta H_{\text{rxn}} = \sum \Delta H_{\text{steps}} \]
The enthalpy change of an overall reaction equals the sum of the enthalpy changes of the steps you build it from.
🔑 Two moves you're allowed to make
Reverse a reaction → flip the sign of ΔH. If A → B has ΔH = −100 kJ, then B → A has ΔH = +100 kJ (running it backwards absorbs what it released). Scale a reaction → scale ΔH by the same factor. Double every coefficient and you double ΔH; halve them and you halve ΔH. ΔH is an amount of energy, so it scales with the amount of reaction.
✨ The strategy
Look at your target equation. For each substance, decide which given reaction contains it and whether you must flip that reaction (to move the substance to the correct side) and/or scale it (to get the right coefficient). Apply the same operation to its ΔH. Then add everything up — species that appear on both sides cancel, and the ΔH values add to your answer.
📝 Worked example: Find ΔH for S(s) + 3⁄2 O₂(g) → SO₃(g), given: (1) S(s) + O₂(g) → SO₂(g), ΔH₁ = −296.8 kJ; (2) 2 SO₂(g) + O₂(g) → 2 SO₃(g), ΔH₂ = −197.8 kJ.
- The target needs 1 S and 1 SO₃. Reaction (1) already gives S → SO₂ with the right direction and amount — keep it as is: ΔH = −296.8 kJ.
- Reaction (2) makes 2 SO₃; we only want 1, so halve it: SO₂ + 1⁄2 O₂ → SO₃, ΔH = −197.8 ÷ 2 = −98.9 kJ.
- Add the two steps. The SO₂ produced in (1) is consumed in the halved (2), so it cancels.
- ΔH = (−296.8) + (−98.9) = −395.7 kJ.
✓ ΔH = −395.7 kJ for S(s) + 3⁄2 O₂(g) → SO₃(g).
✏️ Practice: Find ΔH (in kJ) for C(s) + 1⁄2 O₂(g) → CO(g), given: (1) C(s) + O₂(g) → CO₂(g), ΔH₁ = −393.5 kJ; (2) CO(g) + 1⁄2 O₂(g) → CO₂(g), ΔH₂ = −283.0 kJ.
kJ
Solution
- Target has CO as a product, but in (2) CO is a reactant — so reverse (2): CO₂ → CO + 1⁄2 O₂, ΔH = +283.0 kJ.
- Keep (1) as is: C + O₂ → CO₂, ΔH = −393.5 kJ.
- Add them: the CO₂ cancels (product of (1), reactant of the reversed (2)), leaving C + 1⁄2 O₂ → CO.
- ΔH = (−393.5) + (+283.0) = −110.5 kJ.
✏️ Practice: Find ΔH (in kJ) for N₂(g) + 2 O₂(g) → 2 NO₂(g), given: (1) N₂(g) + O₂(g) → 2 NO(g), ΔH₁ = +180.5 kJ; (2) 2 NO(g) + O₂(g) → 2 NO₂(g), ΔH₂ = −114.1 kJ.
kJ
Solution
- Both reactions are already pointed the right way and correctly scaled — just add them.
- The 2 NO produced in (1) is consumed in (2), so NO cancels.
- Overall: N₂ + 2 O₂ → 2 NO₂.
- ΔH = (+180.5) + (−114.1) = +66.4 kJ — endothermic overall.
Check your understanding
1. The reaction A → B has ΔH = −250 kJ. What is ΔH for B → A?
Reversing a reaction flips the sign of ΔH. If the forward reaction releases 250 kJ, the reverse absorbs 250 kJ: ΔH = +250 kJ.
2. Hess's law works because enthalpy is a…
Enthalpy is a state function — it depends only on the initial and final states, not the route. So the total ΔH is the same however you split the reaction into steps.
3. If X → Y has ΔH = −40 kJ, what is ΔH for 3X → 3Y?
Scaling a reaction scales its ΔH by the same factor. Tripling the amounts triples the energy: 3 × (−40) = −120 kJ.
✅ Key takeaways
- Hess's law: because enthalpy is a state function, ΔH is path-independent.
- A target reaction's ΔH = the sum of the ΔH values of the steps that build it.
- Reverse a reaction → flip the sign of ΔH.
- Scale a reaction → multiply ΔH by the same factor.
- Manipulate the given reactions so species cancel, then add the ΔH values.
➡️ Hess's law tells us ΔH comes from a difference in stored energy — but where is that energy actually stored? In the chemical bonds themselves. Next we estimate ΔH directly from the bonds broken and formed.
Want to test yourself on this?
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