In simple terms
A friendly intro before the formal notes — no formulas yet.
The Chemical Detour
Hess's law says the total energy change of a reaction is the same no matter which route you take from reactants to products. So an enthalpy change you cannot measure directly can be found by adding up the steps of an easier, indirect route that starts and ends at the same place.
Think of the altitude difference between the base and the summit of a mountain. Measuring it as a single vertical drop is awkward, but you can hike any winding trail to the top and add up the small ups and downs along the way — the net change in height is identical for every route, because height depends only on where you start and finish. Enthalpy behaves exactly like altitude: it is a state function, so the enthalpy change depends only on the initial and final states, not the path.
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Write the target reaction whose enthalpy change you want — this is the 'direct route'.
- 2
Choose an 'indirect route' that links the same reactants and products through a common set of species, using enthalpy values you already know (formation or combustion).
- 3
Sketch the cycle and mark every arrow with its known ; going with an arrow adds it, going against an arrow subtracts it (reverse the sign).
- 4
Apply Hess's law: the direct route equals the sum of the indirect route, and solve for the unknown .
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Key formulas
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Full topic notes
Formal explanation with the rigour you need for the exam.
Hess's law and why it works
Hess's law states that the total enthalpy change during a reaction is the same regardless of the route by which the reaction is carried out. It is not a separate rule of nature but a direct consequence of the conservation of energy: enthalpy is a state function, so its value depends only on the current state of the system (temperature, pressure, composition), not on the path taken to get there. The enthalpy difference between reactants and products is therefore fixed. If it were not — if one route released more energy than another — you could loop around the cycle and create energy from nothing.
Enthalpy is a state function; depends only on the initial and final states.
Hess's law: the enthalpy change of a reaction is independent of the pathway taken between fixed start and end points.
It lets us calculate enthalpy changes that cannot be measured directly, by routing through species with known enthalpies.
States must match at the start and end (e.g. H₂O(l) is not H₂O(g)) or the enthalpy values will not be consistent.
Constructing enthalpy cycles
An enthalpy cycle is the picture of Hess's law. The target reaction with the unknown is the 'direct route' along the top; the 'indirect route' links the same reactants and products through a common set of species. The two workhorse cycles use standard enthalpies of formation (), where the common species are the constituent elements, and standard enthalpies of combustion (), where the common species are the products of complete combustion. Once the cycle is drawn, the rule is mechanical: travelling WITH an arrow adds its ; travelling AGAINST an arrow subtracts it (reverse the sign).
Formation cycles: the constituent elements in their standard states sit at the bottom; arrows point UP to reactants and products.
Combustion cycles: the complete-combustion products (CO₂(g) and H₂O(l)) sit at the bottom; arrows point DOWN from reactants and products.
Mark every arrow with its known , remembering to multiply by the stoichiometric coefficient of each substance.
Read the relationship by following a complete path from reactants to products: with-arrow adds, against-arrow subtracts.
Standard enthalpies of formation and combustion
The standard enthalpy of formation, , is the enthalpy change when one mole of a compound forms from its constituent elements in their standard states under standard conditions (100 kPa, usually 298 K). A crucial consequence: the of any element in its standard state — O₂(g), Na(s), C(s, graphite) — is exactly zero, because forming an element from itself is no change at all. The standard enthalpy of combustion, , is the enthalpy change when one mole of a substance burns completely in excess oxygen under standard conditions; it is always negative because combustion is exothermic. These two quantities anchor the two standard cycles.
Calculations using enthalpies of formation
In a formation cycle the elements sit at the bottom. The indirect route 'decomposes' the reactants into their elements (the reverse of formation, so those signs flip) and then 'forms' the products from those same elements. Reversing the reactant side is exactly what turns the relationship into products minus reactants.
Calculations using enthalpies of combustion
In a combustion cycle the common species are the combustion products at the bottom. The indirect route combusts the reactants (downwards) and then reverses the combustion of the products to climb back up to them. Reversing the product side is what flips the relationship into reactants minus products — the mirror image of the formation formula.
The single most common mix-up is the order of subtraction. Formation is 'products − reactants'; combustion is 'reactants − products'. If you are ever unsure, sketch the cycle and follow the arrows: going with an arrow is positive, against an arrow is negative. The arrows never lie, and they will always hand you the correct order.
(HL) Born–Haber cycles, lattice enthalpy and enthalpy of solution
For an ionic solid, some enthalpy changes cannot be measured directly at all — you cannot assemble a crystal from a jar of isolated gaseous ions. The lattice enthalpy (here, the enthalpy of forming one mole of the ionic solid from its gaseous ions, an exothermic quantity) is found instead from a Born–Haber cycle: an enthalpy cycle that decomposes the formation of the ionic compound into a chain of measurable steps — atomisation of the elements, ionisation energies to make cations, electron affinities to make anions, and the lattice-formation step. Applying Hess's law around the closed cycle lets you solve for whichever single step is unknown, exactly as in the SL cycles. The enthalpy of solution — the enthalpy change when one mole of solute dissolves to give a very dilute solution — is likewise found from a cycle linking the lattice enthalpy to the hydration enthalpies of the separated ions.
HL only: Born–Haber cycles and lattice enthalpy are assessed at Higher Level; SL students are not examined on them.
A Born–Haber cycle is just Hess's law applied to an ionic compound, with the sum around the closed cycle equal to zero.
Lattice enthalpy is found by rearranging the cycle for the one step that cannot be measured directly.
Enthalpy of solution = lattice enthalpy (bond-breaking, endothermic in the reverse convention) combined with the hydration enthalpies of the gaseous ions (exothermic).
Common mistakes examiners penalise
Not reversing the sign of a reversed step — if you travel a known reaction backwards in a cycle, flip its sign. An exothermic forward step becomes endothermic in reverse.
Getting the subtraction order wrong — formation is Σproducts − Σreactants; combustion is Σreactants − Σproducts. They are mirror images; picking the wrong one negates your answer.
Forgetting that of an element is zero — elements in their standard states (H₂, O₂, Na, C) contribute 0, but they must still appear in the working, not be silently dropped or given a made-up value.
Not multiplying by the stoichiometric coefficients — enthalpy values are per mole, so 3 mol H₂ contributes , not one lot. Skipping the multiplier is a frequent and costly slip.
Mismatched states — H₂O(l) and H₂O(g) have different enthalpies; using the wrong state, or letting reactant and product states differ from the data, breaks the cycle.
Dropping the sign or units on the final answer — an enthalpy change without its sign (and kJ mol⁻¹) is incomplete and loses the answer mark, even when the arithmetic is right.
Model answer — marked the way our engine marks it
This is the showcase for a calculation topic. In Paper 2 the marks are analytic: each is tied to a specific line of working — a method mark (M) or an answer mark (A) — and error-carried-forward (ECF) means a wrong number early on does not cost you every mark that follows, provided the method is written down. Study how each mark below is earned by a specific line, and notice that two of the three marks survive an arithmetic slip.
Where this leads
Energy cycles are the reasoning engine behind much of what follows. Bond-enthalpy calculations are Hess's law with 'break all bonds' as the common intermediate; at HL, Born–Haber and solution cycles extend the same detour logic to ionic compounds; and the sign and size of feed directly into spontaneity and Gibbs energy later in R1. Master the habit — draw the cycle, follow the arrows, reverse the sign when you go backwards, multiply by the coefficients — and every enthalpy problem in the course becomes a variation on a method you already own.
Worked examples
See the formulas applied — reveal one step at a time, like the exam.
Carbon can be oxidised in two steps. Using the cycle data below, calculate for the direct combustion of carbon to carbon dioxide, C(s) + O₂(g) → CO₂(g).
Step 1: C(s) + ½O₂(g) → CO(g), kJ mol⁻¹ Step 2: CO(g) + ½O₂(g) → CO₂(g), kJ mol⁻¹ [3]
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Step 1 — see the cycle. The direct route C(s) + O₂(g) → CO₂(g) can be reached indirectly by going C → CO (Step 1) and then CO → CO₂ (Step 2). Both indirect steps run in the same forward direction as the target, so neither sign is reversed. [M1: correct route — target = Step 1 + Step 2]
Calculate the standard enthalpy change for the hydrogenation of propene to propane: C₃H₆(g) + H₂(g) → C₃H₈(g).
kJ mol⁻¹ kJ mol⁻¹ [3]
- 1
Step 1 — choose the formation formula and list the species. Products: C₃H₈(g). Reactants: C₃H₆(g) and H₂(g). H₂(g) is an element in its standard state, so . [M1: correct method, Σproducts − Σreactants]
Calculate the standard enthalpy of formation of ethanol, C₂H₅OH(l), for the reaction 2C(s, graphite) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l).
kJ mol⁻¹ kJ mol⁻¹ kJ mol⁻¹ [3]
- 1
Step 1 — draw the combustion cycle. The reactants (2C, 3H₂) and the product (C₂H₅OH) all combust downwards to the same species, CO₂(g) and H₂O(l). Following the route — reactants down, then up the reverse of the product's combustion — gives Σ(reactants) − Σ(products). Oxygen does not combust, so it contributes nothing. [M1: correct method with stoichiometric coefficients]
Calculate the standard enthalpy change, , for the reaction 2H₂S(g) + 3O₂(g) → 2SO₂(g) + 2H₂O(l), using standard enthalpies of formation: kJ mol⁻¹ kJ mol⁻¹ kJ mol⁻¹ [3]
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Glossary
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Hess's law
The total enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions (and states) are the same. It follows directly from conservation of energy.
Key takeaways
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- ✓
Enthalpy is a state function; depends only on the initial and final states.
- ✓
Hess's law: the enthalpy change of a reaction is independent of the pathway taken between fixed start and end points.
- ✓
It lets us calculate enthalpy changes that cannot be measured directly, by routing through species with known enthalpies.
- ✓
States must match at the start and end (e.g. H₂O(l) is not H₂O(g)) or the enthalpy values will not be consistent.
Practice — then mark it
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Get a Paper 2 calculation marked: find ΔH_reaction from formation data with full working
Get a Paper 2 calculation marked: find ΔH_reaction from formation data with full working
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