Standard Heats of Formation and Hess's Law
The preceding sections have lead to this point.
 
        ΔHo = [Ho products] - [Ho reactants]
 
What goes into the products and reactants brackets are the standard heats of formation. We'll begin by the defining the term standard state. Any substance in its most stable physical form (gas, liquid, or solid) at 25oC and under a pressure of 1 atm is said to be in its standard state. The element oxygen for example, is in its standard state when its exists as a molecule of O2 -not as atoms and not as molecules of O3. The element carbon is in its standard state when it exits as graphite - not as diamond - at 25oC. Diamond is also a form of carbon, but it is actually slightly less stable than graphite.
 
The quantities that we'll use from now on to compute values for Ho are called the standard enthalpies of formation of standard heats of formation. The standard heats of formation of a compound Hfois the amount of heat absorbed or evolved when one mole of the compound is formed from its elements in their standard states. Thus, the thermochemical equation for the formation of one mole of liquid water from oxygen and hydrogen in their standard states is
 
                 H2(g) + ½O2(g) ------> H2O(l)             Hfo = -285.8 kJ/mol
 
The standard enthalpy change for this reaction, that is, the enthalpy change at 25oC and 1 atm, is called the standard heat of formation of liquid water. This is a point that often causes confusion among students. Each of the following equations, for example involves the formation of CO(g)
 
                                 C(s) + O2(g) ----> CO2(g)
                           CO(g) + ½O2(g) ----> CO2(g)
                           2 C(s) + 2 O2(g) ----> 2 CO2(g)
 
However, only the first involves just the elements as reactants and the formation of just one mole of CO2. In the second equation, one of the reactants is a compound, carbon monoxide, and in the third, two moles of CO2 are formed. Only the first equation, therefore, has a standard enthalpy change that we identify as Hfo.

Your databook has tables of standard heats of formation for a variety of substances. They are listed alphabetically and in some cases not logically. There are also standard heats of formation values for the organic chemicals, too.
 

***************** IMPORTANT **********************
The Hfo for any element in its standard state is zero. (0 kJ/mol)
 
This makes sense if you think about it. There would be no enthalpy change if you "form an element in its standard state from itself."
 
Using Hess's Law with Standard Heats of Formation
The Ho for any reaction must be the difference between the total enthalpies of formation of the products and those of the reactants. Any reaction can be generalized by the equation

                          aA + bB + ..... -----> nN + mM + .....
 

where a, b, n, m, etc, are the coefficients of substances A, B, N, M, etc. The value of Ho can be found using Hess's law equation.
 
                    Δ Ho = [nHfo(N) + mHfo(M) + ...] - [aHfo(A) + bHfo(B) + ...]
 
Putting this into a more useful form we get:
 
                    Δ Ho = [sum of the Hfo of products]-[sum of the Hfo of reactants]
 
Example Problem
Some chefs keep baking soda, NaHCO3, handy to put out grease fires. When thrown on the fire, baking soda partly smothers the fire, and the heat decomposes it to give CO2, which further smothers the flame. The equation for the decomposition of NaHCO3 is

                            2 NaHCO3(s) -------> Na2CO3(s) + H2O(g) + CO2(g)
 

Calculate the ΔHo for this reaction in kilojoules and kilocalories.
 
Solution
              ΔHo = sum of products - sum of reactants
                    = [ Na2CO3(s) + H2O(g) + CO2(g) ] - [ (2)NaHCO3(s)]
 
Look up the values in the databook tables for each substance.
Make sure the physical states are identical.
 
                   = [-1130.7 -241.8 -393.5 kJ/mol ]-[(2)(-950.8 kJ/mol)]
                   = -1766 kJ/mol - (-1901.6 kJ/mol)
                   = +135.6 kJ/mol

Under standard conditions, the reaction is endothermic by 135.6 kJ/mol.
To calculate kilocalories, remember the conversion factor of

                       1 cal = 4.184 J or 1 kcal = 4.184 kJ

In kilocalories the reaction is endothermic by 135.6 kJ/mol
                                                                             4.184 kJ/cal

which is 32.41 kcal/mole.
 

Example Problem #2
What is the ΔHo in kilojoules for the combustion of 1 mol of ethanol, C2H5OH(l), to form gaseous carbon dioxide and gaseous water?
 
Solution
First write and balance the combustion equation.

                        C2H5OH(l) + 3 O2(g) -----> 2 CO2(g) + 3 H2O(g)

Hess's law for this equation is:

               ΔHo = [(2)CO2(g) + (3)H2O(g)] - [C2H5OH(l) + (3)O2(g)]
                     = [ (2)-393.5 + (3)-241.8 kJ/mol] - [ -277.1 + (3)0 kJ/mol]
                     = [-787 -725.4 kJ/mol ] - [ -277.1 kJ/mol ]
                     = -1512.4 + 277.1 kJ/mol
                     = -1235.3 kJ/mol

The reaction for the combustion of ethanol is exothermic by 1235.3 kJ/mol.