THERMOCHEMISTRY
Part III: Standard Enthalpies of Formation DHfo
When determining the enthalpy change associated with a chemical reaction we have several options:
The enthalpy change for a reaction depends on both the pressure and the temperature. As an example let's consider the combustion of methane, CH4:
CH4 + 2 O2 è CO2 + 2 H2O(l) DH = - 863kJ
CH4 + 2 O2 è CO2 + 2 H2O(g) DH = -775 kJ
Even though it is the same chemical reaction, the enthalpy changes are different: In the first reaction water is produced in liquid form, whereas the second reaction produced water vapor. The difference in reaction enthalpy (88 kJ) reflects that energy remains stored in the system if water vapor is formed. This stored enthalpy (44 kJ/mol H2O) is released when the vapor condenses. The differences are relatively small when compared with typical enthalpy changes for a chemical reaction (i.e. 44 kJ/mol versus 800 kJ/mol. In order to be able to compare enthalpy changes for various reactions chemists defined a "standard state" under which chemical reactions take place. The standard state of an element or a compound is the most stable physical state at given temperature and pressure.
The standard conditions are:
Pressure = 1 atm
Temperature = 289 K
Concentration = 1 M
Under these conditions standard enthalpies of formation DHfofor many substances have been tabulated (see appendix L, page A-31 of our text book). The standard enthalpy of formation for elements in its most stable form is generally zero. Tables of standard enthalpies of formation enable us to calculate the enthalpy changes for many chemical reactions.
For any chemical reaction
the change in enthalpy can be calculated from
DH reaction = Hfofinal - Hfoinitial = H(products) - H(reactants)
For any given chemical reaction such as
A + B è C + D
This equation would translate to
where 
is the sum of all
standard enthalpies of the products or reactants, respectively.
Example: Determine the reaction enthalpy of the combustion of methane using the standard enthalpies of formation:
CH4 + 2 O2 è CO2 + 2 H2O(g) DH = -775 kJ
Look up the values for the standard enthalpies of formation in the table, appendix L in our text book:
CH4(g) : -102 kJ/mol
O2(g): 0 kJ/mol
CO2(g): -393 kJ/mol
H2O(g): -242 kJ/mol
then the reaction enthalpy is
Exercise: (Group): You should get together via CHM152 Online mailing list and in a team approach determine the reaction enthalpy for the combustion of the fuels listed below: