## Section 4: Enthalpy of Formation

The H for a reaction can also be determined by using the standard enthalpies of formation of the compounds involved in the reaction. The standard enthalpy of formation is symbolized f. The standard enthalpy of formation of a compound means the change in enthalpy that goes with the formation of one mole of the compound from its elements, with all substances in their standard states at 25°C. For example, it takes 34 kJ to make 1 mole of NO2 from ½ mole of N2 and 1 mol of O2. The reaction is as follows:

½N2 + O2 --> NO2       f = 34 kJ/mol

So, the standard enthalpy of formation of NO2 is 34 kJ. In your textbook, there is probably a table that has the standard enthalpies of formation for most common compounds.

You can use these tables to determine the H of some reactions. For example, the reaction between aluminum and iron (III) oxide.

2Al + Fe2O3 --> Al2O3 + 2Fe       H = ???

In order to find the H, you use this equation:

H = f(products) - f(reactants)

This equation means to take the sum of the f of the products, and subtract the sum of the f of the reactants. In the reaction above, you find on a table that:

f for Fe2O3 = -826 kJ/mol
f for Al2O3 = -1676 kJ/mol
f for Al = 0
f for Fe = 0

(Note that the f for Al and Fe are 0, since they are in their standard states. For any element in its standard stae, the f will be 0.)

Using the equation:

H = (-1676 kJ/mol) - (-826 kJ/mol)
H = -850 kJ/mol

So, the H for the reaction above is -850 kJ/mol, a highly exothermic reaction.