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.