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### Problem 8.65

Posted: Fri Jan 19, 2018 3:47 pm
Calculate the standard enthalpy of formation of dinitrogen pentoxide from the following data,

2NO(g) + O2(g) --> 2NO2(g) deltaH = -114.1 kJ
4NO2(g) + O2(g) --> 2N2O5(g) delta H = -110.2 kJ

and from the standard enthalpy of formation of nitric oxide, NO (which is 90.25 kJ).

I understand that the equation I'm looking for is N2(g) + 5/2O2(g) --> N2O5(g)
I don't know what I'm missing, but I don't understand how these equations form the one that I'm looking for, since I don't see N2 in either of the first two equations.

### Re: Problem 8.65

Posted: Fri Jan 19, 2018 4:07 pm
I'm sure there are multiple ways to do this, but what I did was:

2NO + O2 -> 2NO2 H = -114.1 kJ

The standard enthalpy of formation for NO is 90.25 kJ and for O2, as a pure element, is 0 so
2(NO2) - (2(90.25 kJ)+ 0 kJ) = -114.1 kJ // solving for NO2 gives you 33.2 kJ

Then:
4NO2 + O2 -> 2N2O5 H=-110.2 kJ
So: 2(N2O5) - (4(33.2 kJ) - 0 kJ) = -110.2 kJ // solving for N2O5 gives you 11.3 kJ

Again, this method probably only really worked because it involved a pure element, but it gets the job done for this scenario