Enthalpy Diagrams
The enthalpy relationships involved in adding thermochemical equations are most easily visualized by means of an enthalpy diagram, such as the one below. Horizontal lines in such a diagram correspond to different absolute values of enthalpy, H. A horizontal line drawn higher in the diagram represents a larger value of H. Changes in enthalpy, H, are represented by the vertical distances between the lines. Take another look at the diagram below. It shows all three of the CO2 reactions already discussed. The total decrease in energy, however, is the same regardless of which path is taken, so the total energy evolved in the two-step path has to be the same as in the one-step reaction.
 

 
Law of Heat Summation (Hess's Law) For any reaction that can be written in steps, the standard heat of reaction is the same as the sum of the standard heats of reactions for the steps.
 
One of the most useful applications of Hess's law is the calculation of the value of ΔHo for a reaction whose ΔHo is unknown or cannot be measured. Hess's law says that we can add thermochemical equations, including their values of ΔHo, to obtain some desired thermochemical equation and its ΔHo.
 
Example Problem:
Consider the following thermochemical equations:

                    C(s) + ½O2(g) ------> CO(g)        ΔHo = -10.5 kJ
                CO(g) + ½O2(g) ------> CO2(g)       ΔHo = -283.0 kJ
 
Use them to find the Ho in kilojoules for the reaction.

                                 C(s) + O2(g) ------> CO2(g)
 
Solution
This is a particularly simple problem, but it illustrates a few important points about all such problems. Let's add the two given thermochemical equations:
 
             C(s) + ½O2(g) ------> CO(g)             ΔHo = -10.5 kJ
         CO(g) + ½O2(g) ------> CO2(g)            ΔHo = -283.0 kJ

        C(s) + ½O2(g) + ½O2(g) + CO(g) -----> CO(g) + CO2(g)
 
The resulting equations can be simplified. Cancel the CO(g) because it appears on both sides. You can do this as long as you have the same chemical in the same physical state. Add the two oxygen terms together. This gives us the target equation.
 
                   C(s) + ½O2(g) ------> CO(g)           ΔHo = -10.5 kJ
                CO(g) + ½O2(g) ------> CO2(g)         ΔHo = -283.0 kJ
                     C(s) + O2 (g) ------> CO2(g)          ΔHo = -393.5 kJ
 
Example Problem #2
Carbon monoxide is often used in metallurgy to remove oxygen from metal oxides and thereby give the free metal. The thermochemical equation for the reaction of CO with iron(III) oxide, Fe2O3, is
 
              Fe2O3(s) + 3 CO(g) ------> 2 Fe(s) + 3 CO2 (g)          ΔHo = -26.74 kJ
 
Use this equation and the equation for the combustion of CO

                          CO(g) + ½O2(g) ------> CO2(g)       ΔHo = -283.0 kJ
 
to calculate the value of Ho for the reaction

                               2 Fe(s) + 1½O2 (g) -----> Fe2O3
 
Solution
Combine the equations in such a way that we can add them to the final target equation. Then we add the corresponding ΔHo's to obtain the ΔHo of the target equation.
 
Step 1    The target equation must have 2 Fe on the left, but the first equation above has 2 Fe on the right. To move it left, reverse the entire equation and remember to reverse the sign of Ho. When the equation is flipped over the Fe2O3 also falls into the correct position.
Step 2  There must be 1½O2 of the left, and we must be able to cancel three CO and three CO2 when the equations are added. Multiply the second equation by 3 and we get the necessary coefficients. Multiply the Ho values for the second equation by 3 as well. The adjusted equations are:

2 Fe(s) + 3 CO2(g) ----> Fe2O3(s) + 3 CO(g)          ΔHo = +26.74 kJ
3 CO(g) + 1½O2(g) ----> 3 CO2(g)   ΔHo = 3(-283.0 kJ) = -849.0 kJ
 2 Fe(s) + 1½O2 (g) -----> Fe2O3                                     ΔHo = -822.26 kJ