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 1/2 mole of N2 and 1 mol of O2. The reaction is as
follows:

1/2N2 + 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.