---
title: "Hess's Law and Enthalpy Cycles"
description: "The enthalpy change of a reaction does not depend on the route taken. Build a cycle out of enthalpies of formation or of combustion and you get a value for a reaction nobody can measure directly."
canonical: https://lightmysky.com/learn/science/hesss-law-and-enthalpy-cycles-mt_530V9czLOb
source: https://lightmysky.com/learn/science/hesss-law-and-enthalpy-cycles-mt_530V9czLOb.md
retrieved: 2026-09-12
---

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# Hess's Law and Enthalpy Cycles

The enthalpy change of a reaction does not depend on the route taken. Build a cycle out of enthalpies of formation or of combustion and you get a value for a reaction nobody can measure directly.

Subject: Science · Area: Chemistry · Ages 16 to 17
Page: https://lightmysky.com/learn/science/hesss-law-and-enthalpy-cycles-mt_530V9czLOb

## Ready when they can

- States Hess's law and explains why it follows from conservation of energy
- Draws a cycle with the arrows pointing the right way for formation data and for combustion data
- Calculates an enthalpy change from standard enthalpies of formation of reactants and products
- Explains why the enthalpy of formation of an element in its standard state is zero

## Lesson: Adding detours to find the heat

Hess's law says the total enthalpy change of a reaction is the same no matter which route you take. Enthalpy is a state function, so only the start and finish matter, not the path between them. That is why made-up steps still give the real answer. It matters because some reactions, like the thermal decomposition of calcium carbonate, cannot be measured cleanly in a beaker.

With formation data, draw the target across the top and point every arrow down from the elements to their compounds, since formation builds compounds from elements. Then take products minus reactants. The formation enthalpy of any element in its standard state is zero, which fixes the starting line.

**Example.** Decompose calcium carbonate: CaCO3 breaks into CaO plus CO2. Formation values are minus 1207 for CaCO3, minus 635 for CaO and minus 394 for CO2, all in kilojoules per mole. Products sum to minus 1029, reactants to minus 1207, so products minus reactants gives plus 178 kilojoules per mole.

**Tip.** With combustion data the arrows point up instead, since everything burns to shared products. The classic slip is one backwards arrow, which flips a sign. Check every direction before adding anything up.

**Recap.** Draw the cycle, point the arrows right, and add the known steps to reach the missing value.

## Practice

8 questions on this page, each with its working shown.

## Needs first

- [Enthalpy Change, Standard Conditions and Calorimetry](https://lightmysky.com/learn/science/enthalpy-change-standard-conditions-and-calorimetry-mt_GPgwgyqyvN)
- [Bond Energies and the Overall Energy Change](https://lightmysky.com/learn/science/bond-energies-and-the-overall-energy-change-mt_VI0_1ogIOy)

## Opens up

- [Benzene: Delocalisation and Electrophilic Substitution](https://lightmysky.com/learn/science/benzene-delocalisation-and-electrophilic-substitution-mt_jFqF7cZ0Ij)
- [Free Energy and Whether a Reaction Is Feasible](https://lightmysky.com/learn/science/free-energy-and-whether-a-reaction-is-feasible-mt_o1kZgBYWoM)
- [Lattice Enthalpy and Born-Haber Cycles](https://lightmysky.com/learn/science/lattice-enthalpy-and-born-haber-cycles-mt_pUku4iUjYa)