Hess's Law

Part of the Energy, Rates and Equilibrium section of Coddy's Chemistry journey. Lesson 8 of 35.

Hess's law: the same start and end give the same $\Delta H$ , whatever the route.

C → CO: $-110\ \mathrm{kJ}$ ; CO → CO₂: $-283\ \mathrm{kJ}$ . So C → CO₂: $-393\ \mathrm{kJ}$ .

Add the steps. A step used backwards changes sign; a step used twice counts twice.

Challenge

Sulfur burns in two steps: S + O₂ → SO₂, $\Delta H_1 = -297\ \mathrm{kJ}$ , then SO₂ + ½O₂ → SO₃, $\Delta H_2 = -99\ \mathrm{kJ}$ . Find $H$ for S + 1½O₂ → SO₃.

Try it yourself

Solve for $H$

Test yourself

This lesson includes a short quiz. Start the lesson to answer it and track your progress.

All lessons in Energy, Rates and Equilibrium

1Energy in Reactions

Exothermic and Endothermic Energy Diagrams Enthalpy Change Activation Energy Recap - Hand Warmer

4Rates of Reaction

Collision Theory Concentration and Rate Temperature and Rate Surface Area Catalysts Recap - Speed Trials

7Practice Problems

Recap - Methanol Plant Recap - Butane Lighter Recap - Limestone Kiln

2Bond Energies

Breaking and Making Bonds Delta H from Bond EnergiesHess's LawEnergy and Moles Recap - Fuel Value

5Equilibrium

Reversible Reactions Dynamic Equilibrium The Equilibrium Constant Writing K Solids and Liquids in K Recap - K Builder

3Changes of State

Heating Curves Melting and Boiling Energy of State Changes Recap - Ice to Steam

6Changing an Equilibrium

Le Chatelier's Principle Changing Concentration Changing Pressure Changing Temperature The Haber Process Recap - Factory Settings