The big idea: The enthalpy change of a reaction, ΔH, depends only on the starting materials and the finishing materials — not on the path taken to get there.
This is Hess's law: if a reaction can happen by two different routes, the total enthalpy change is the same for both.
Think of climbing a hill: your change in height depends only on where you start and where you finish, not on which path you walk up.
Because ΔH depends only on the start and end states, the direct route (top) equals the indirect route ΔH₁ + ΔH₂. This is Hess's law.
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
Define: state function: A state function is a property that depends only on the current state of the system, not on how it got there.
Enthalpy is a state function, so ΔH is path-independent. That is exactly why we can add up the steps of a longer route to find the ΔH of a reaction we cannot measure directly.
Free preview
This is the free notes preview
You're reading the free notes. Aimnova Pro unlocks the full study experience — and you can try it with your first topic free to keep:
- FlashcardsLock in vocabulary and key terms with spaced repetition.
- Practice questionsAnswer exam-style questions and get instant AI marking.
- Mock exams & past-paper vaultSit full mocks and see exactly how examiners award marks.
- Personalised study planA daily plan built around your exam date and weak areas.
Some enthalpy changes are very hard to measure directly (the reaction is too slow, or side reactions occur). Hess's law lets us find them indirectly: build an energy cycle (a triangle of arrows) and travel by an alternative route whose ΔH values we already know.
- enthalpy change of the direct route (often the unknown)
- enthalpy changes of each step of the indirect (alternative) route
The golden rule of cycles: Follow the arrows around the cycle.
- Going with an arrow: add its ΔH. - Going against an arrow (the reverse reaction): subtract its ΔH (the sign flips).
The direct route (the one you want) must equal the indirect route. Set them equal and solve for the unknown.
A very common indirect route goes down to the elements (using standard enthalpies of formation) and back up to the products. Written as one equation, that cycle becomes the data-booklet relationship:
- standard enthalpy change of the reaction (kJ mol⁻¹)
- standard enthalpy of formation of each product (kJ mol⁻¹)
- standard enthalpy of formation of each reactant (kJ mol⁻¹)
Learn what examiners really want
See exactly what to write to score full marks. Our AI shows you model answers and the key phrases examiners look for.
Worked example — adding two known reactions
Find ΔH for C(s) + ½O2(g) → CO(g), given:
(1) C(s) + O2(g) → CO2(g), ΔH₁ = −394 kJ mol⁻¹
(2) CO(g) + ½O2(g) → CO2(g), ΔH₂ = −283 kJ mol⁻¹
Solution
- Build the route. The target makes CO; reaction (1) makes CO2 directly, reaction (2) makes CO2 from CO. So go forward by (1), then reverse (2) to get back to CO.
- Formula first — the direct route equals 'do (1), then reverse (2)':
- Substitute (reversing (2) flips its sign):
- Work it out:
Final answer
ΔH = −111 kJ mol⁻¹ (exothermic). Reversing reaction (2) flips its sign, so you subtract it.
Worked example — using enthalpies of formation
Calculate ΔH for CH4(g) + 2O2(g) → CO2(g) + 2H2O(l). Use ΔHf: CH4 = −75, CO2 = −394, H2O(l) = −286 kJ mol⁻¹. (ΔHf of O2 = 0, an element.)
Solution
- Formula first — products minus reactants:
- Products: one CO2 and two H2O:
- Reactants: one CH4 (O2 is an element, ΔHf = 0):
- Subtract:
Final answer
ΔH = −891 kJ mol⁻¹ (exothermic — the combustion of methane).
How this is tested: Hess's law is a staple Paper 2 determine/calculate question (2–3 marks): you are given two or three reactions (or a table of ΔHf values) and must construct a cycle to find an ΔH that cannot be measured directly.
On Paper 1A it appears as a quick 'which combination of the given ΔH values gives the target ΔH?' multiple-choice item.
How the marks fall: (1) Sketch the cycle (or line up the equations) so the arrows point the right way. (2) Reverse any reaction you use backwards (flip its sign) and multiply if you need more than one mole. (3) Add them up — watch every sign.
IB-style question — enthalpy of formation of propane
Determine the standard enthalpy of formation of propane, 3C(s) + 4H2(g) → C3H8(g), using these combustion data:
(1) C(s) + O2(g) → CO2(g), ΔHc = −394 kJ mol⁻¹
(2) H2(g) + ½O2(g) → H2O(l), ΔHc = −286 kJ mol⁻¹
(3) C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l), ΔHc = −2220 kJ mol⁻¹. [3]
Solution
- Build the cycle. Combust the elements (down via the combustion products), then come back up by reversing the combustion of propane. So: 3 × (1) + 4 × (2) − (3).
- Burn the elements — three carbons and four hydrogens:
- Reverse the combustion of propane (flip its sign):
- Add the two parts of the route:
Final answer
ΔHf(propane) = −106 kJ mol⁻¹. Reversing reaction (3) flips its sign, and each combustion is multiplied by how many moles the formation needs.
Watch the multipliers: The single most common error is forgetting to multiply a step by the number of moles in the target equation.
Forming propane needs 3 carbons and 4 hydrogens — so multiply reaction (1) by 3 and reaction (2) by 4 before adding. Miss this and the whole answer is wrong.