Enthalpy Change Example Problem

Enthalpy Change of the Decomposition of Hydrogen Peroxide

Enthalpy is a measure of a system's energy.
Enthalpy is a measure of a system's energy. PM Images, Getty Images

This example problem shows how to find the enthalpy for the decomposition of hydrogen peroxide.

Enthalpy Review

You may wish to review the Laws of Thermochemistry and Endothermic and Exothermic Reactions before you begin. Enthalpy is a thermodynamic property that is the sum of the internal energy that is added to a system and the product of its pressure and volume. It's a measure of the system's capacity to release heat and perform non-mechanical work.

In equations, enthalpy is denoted by the capital letter H, while specific enthalpy is lowercase h. Its units are usually joules, calories, or BTUs.

The change in enthalpy is directly proportional to the amount of reactants and products, so you work this type of problem using the change in enthalpy for the reaction or by calculating it from the heats of formation of the reactants and products and then multiplying this value times the actual quantity (in moles) of material that is present.

Enthalpy Problem

Hydrogen peroxide decomposes according to the following thermochemical reaction:

H2O2(l) → H2O(l) + 1/2 O2(g); ΔH = -98.2 kJ

Calculate the change in enthalpy, ΔH, when 1.00 g of hydrogen peroxide decomposes.


This sort of problem is solved by using a table to look up the change in enthalpy unless it's given to you (as it is here). The thermochemical equation tells us that ΔH ​for the decomposition of 1 mole of H2O2 is -98.2 kJ, so this relationship can be used as a conversion factor.

Once you know the change in enthalpy, you need to know the number of moles of the relevant compound to calculate the answer. Using the Periodic Table to add up the masses of hydrogen and oxygen atoms in hydrogen peroxide, you find the molecular mass of H2O2 is 34.0 (2 x 1 for hydrogen + 2 x 16 for oxygen), which means that 1 mol H2O2 = 34.0 g H2O2.

Using these values:

ΔH = 1.00 g H2O2 x 1 mol H2O2 / 34.0 g H2O2 x -98.2 kJ / 1 mol H2O2

ΔH = -2.89 kJ


The change in enthalpy, ΔH, when 1.00 g of hydrogen peroxide decomposes = -2.89 kJ

It's a good idea to check your work to make sure the conversion factors all cancel out to leave you with an answer in energy units. The most common error made in the calculation is accidentally switching the numerator and denominator of a conversion factor. The other pitfall is significant figures. In this problem, the change in enthalpy and mass of sample both were given using 3 significant figures, so the answer should be reported using the same number of digits.