Unit 5: Thermochemistry
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 H°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 H°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 = H°f(products) - H°f(reactants)
This equation means to take the sum of the H°f of the products, and subtract the sum of the H°f of the reactants. In the reaction above, you find on a table that:
H°f for Fe2O3 = -826 kJ/mol
H°f for Al2O3 = -1676 kJ/mol
H°f for Al = 0
H°f for Fe = 0
(Note that the H°f for Al and Fe are 0, since they are in their standard states. For any element in its standard stae, the H°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.
