# How to Calculate Delta H F

By Robert Schrader; Updated April 24, 2017

In a chemical reaction, both reactants and the products they form have what are called "heats of formation." Expressed by the symbol "ΔHf" (delta HF), heats of formation are an important part of understanding energy transfer during chemical reactions. In order to calculate ΔHf for any product or reactant, you must have on hand the total amount of heat the reaction produces (ΔH), as well as the ΔHf value for all the other reactants and/or products, all of which your chemistry problem will provide you.

Arrange your given ΔHf and ΔH values according to the following equation: ΔH = ΔHf (products) - ΔHf (reactants). For example, imagine that you want to know ΔHf for acetylene, C2H2, for the reaction C2H2 (g) + 5/2O2 (g) --> 2CO2 (g) + H2O (g), the combustion of acetylene, whose ΔH is -1,256 kJ/mol. You also know that ΔHf for CO2 is -394 kJ/mol and ΔHf for H2O is -242 kJ/mol. Elemental reactants and products such as oxygen gas have no "heat of formation" by definition--they exist is their form naturally. Knowing this, you can express the following: ΔH = ΔHf (products) - ΔHf (reactants), or -1,256 = (2_-394 + -242) - ΔHfC2H2, which you can rearrange as follows: ΔHfC2H2 = -1,256 - (2_-394 + -242). Notice that you must multiply ΔHfCO2 by two because of the "2" coefficient in front of it in the reaction equation.

Solve your equation for ΔHf, in the case of the example ΔHfC2H2. ΔHfC2H2 = -1,256 - (2*-394 + -242) = -1,256 - (-788 + -242) = -1,256 - (-1,030) = -1,256 + 1030 = -226 kJ/mol.

Adjust your ΔHf value's "sign" depending on whether it's for a product or a reactant. Product ΔHf values will always be negative, while those for reactants are always positive. As C2H2 is a reactant, its ΔHf is positive. Therefore, ΔHfC2H2 = 226 kJ/mol.

#### Tip

ΔHf and ΔH values are always given in kilojoules per moles, where a "kilojoule" is the international unit of heat or energy and a "mole" is a unit which describes a very large number of molecules of a compound.