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Another Example: (repeated) with the answer to follow:
Calculate ΔH° for this reaction:
2 N2(g) + 5 O2(g) ---> 2 N2O5(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
I'm going to take this one step at a time, but I'm not going to write all three equations each time. First, I want to focus on the second equation, which I will reverse AND multiply through by two:
4 HNO3(l) ---> 2 N2O5(g) + 2 H2O(l) ΔH° = +153.2 kJ
OK, why did I do all that? (1) I had to get the N2O5 on the right hand side AND (2) I had to have it be 2 N2O5. Notice that all 4 components (the three substances and the enthalpy) all got doubled. Did you catch the change from negative to positive in the ΔH°? Good.
You might be wondering why I chose the second equation to work with instead of the first or third. What I did was survey the equations and mentally compare it to the final equation I want to make. In it you will find the N2O5 that guided my choice.
Now, I'm going to choose the third equation to work with. I will NOT flip it, but I will multiply through by four. Why four? First the equation and then the answer:
2 N2(g) + 6 O2(g) + 2 H2(g) ---> 4 HNO3(l) ΔH° = -696.4 kJ
What does this get me? First, I get the 2 N2 I need on the left side of the final answer. Second, I get 4 HNO3 on the right to cancel with the 4 HNO3 on the left in the second equation. Confusing? Go back over it again.
However, I still am not at the final answer. To get there, I need to reverse (sometimes "flip" is the verb used) the first equation and multiply through by two. What I'm going to do is not explain why specifically. I'm now going to show all three equations in their modified status and then continue the discussion.
2 H2O(l) ---> 2 H2(g) + O2(g) ΔH° = +571.6 kJ 4 HNO3(l) ---> 2 N2O5(g) + 2 H2O(l) ΔH° = +153.2 kJ 2 N2(g) + 6 O2(g) + 2 H2(g) ---> 4 HNO3(l) ΔH° = -696.4 kJ
The change to the first equation will allow us to (1) cancel out the water, (2) cancel out the hydrogen and (3) cancel out one of the oxygens leaving the five we need for the answer.
The ΔH° for the reaction as written is +28.4 kJ.
One last note: I don't write kJ/mol in this case because of the two in front of the N2O5. However, I would need to supply the equation along with the 28.4 value. If I divided through by two, I would get the formation reaction for N2O5:
N2(g) + (5/2) O2(g) ---> N2O5(g)
In this case, I would write ΔH°f = 14.2 kJ/mol. The presence of the subscripted "f" indicates that we are dealing with one mole of the target substance.
Another Example: Calculate ΔHf° for this reaction:
6 C(s) + 6 H2(g) + 3 O2(g) ---> C6H12O6(g)
using the following three equations:
C(s) + O2(g) ---> CO2(g) ΔH° = -393.51 kJ H2(g) + (1/2) O2(g) ---> H2O(l) ΔH° = -285.83 kJ C6H12O6(s) + 6 O2(g) ---> 6 CO2(g) + 6 H2O(l) ΔH° = -2803.02 kJ
The answer is -1273.02 kJ/mol.