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.
Step 1: Set Up the Equation
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) + (5/2)O2 (g) --> 2CO2 (g) + H2O (g), the combustion of acetylene, the ΔH of which is -1,256 kJ/mol.
You know that the ΔHf of CO2 is -394 kJ/mol and the ΔHf of 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 all of this, you can write the following: ΔH = ΔHf (products) - ΔHf (reactants), or
-1,256 = (2 × (-394) + (-242)) - ΔHf(C2H2),
which you can rearrange as follows:
ΔHf(C2H2) = [2 ×(-394) + (-242)] +1,256.
Note that you must multiply the ΔHf of CO2 by two because of the "2" coefficient in front of it in the reaction equation.
Step 2: Solve the Equation
Solve your equation for ΔHf. In the case of the example ΔHf(C2H2),
ΔHf(C2H2) = [2 ×(-394) + (-242)] - (-1,256).
= (-1,030) + 1,256 = 226 kJ/mol.
Step 3: Validate the Sign
Adjust your ΔHf value's sign depending on whether it is 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, ΔHf(C2H2) = 226 kJ/mol.
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.