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Example #3 (repeated) with the answer to follow:
Calculate the enthapy for the following reaction:
N2(g) + 2O2(g) ---> 2NO2(g) ΔH° = ??? kJ
Using the following two equations:
N2(g) + O2(g) ---> 2NO(g) ΔH° = +180 kJ 2NO2(g) ---> 2NO(g) + O2(g) ΔH° = +112 kJ
In order to solve this, we must reverse at least one equation and it turns out that the second one will require reversal. Here are both with the reversal to the second:
N2(g) + O2(g) ---> 2NO(g) ΔH° = +180 kJ 2NO(g) + O2(g) ---> 2NO2(g) ΔH° = -112 kJ
Notice that I have also changed the sign on the enthalpy from positive to negative.
Next, we add the two equations together and eliminate identical items. We also add the two enthalpies together.
N2(g) + 2O2(g) ---> 2NO2(g) ΔH° = +68 kJ
The answer has been obtained: +68 kJ/mol.
Two points to make:
(1) When you write the enthalpy with the equation, you do not need to add the "per mole" portion of kJ/mol. However, when the enthalpy is discussed in a paragraph associated with the equation, the "per mole" portion should be included.
(2) I have decided to write the positive sign on every enthalpy that is positive. This is not standard practice and many texts do not do it. I have decided to write it so as to make things as clear as I can. If you think it's overkill, please don't email me with your opinion. Thanks.
Another Example: Calculate ΔH° for this reaction:
2N2(g) + 5O2(g) ---> 2N2O5(g)
using the following three equations:
H2(g) + (1/2) O2(g) ---> H2O(l) ΔH° = -285.8 kJ N2O5(g) + H2O(l) ---> 2HNO3(l) ΔH° = -76.6 kJ (1/2) N2(g) + (3/2) O2(g) + (1/2) H2(g) ---> HNO3(l) ΔH° = -174.1 kJ
This example shows something new not discussed yet. I hope it is obvious that one of the equations with the nitric acid (HNO3) will have to be reversed. In addition (this is the new part), you will need to multiply through an equation by a particular factor. (In fact, in this equation more than one factor will be needed!!!) The reason for this: to make substances not in the final answer (like the HNO3) cancel out, there have to be an EQUAL number of them on each side when you add the three equations together.
When you multiply through by the factor, MAKE sure to multiply every component on the reactant side and the product side AS WELL AS . . .
the enthalpy value!!!!!!
That's right. Multiply the enthalpy value times the factor and use that new value in the calculation. I'll do the reason in the solution file to the problem.