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Jesus Rodriguez 1J
Posts: 53
Joined: Sat Jul 22, 2017 3:00 am


Postby Jesus Rodriguez 1J » Sat Jan 20, 2018 7:05 pm

If the reaction we want is N2 + 5/2 O2 --> N205,
then why does the solution stop at 2NO + (3/2)O2 --> N205
I'm wondering about the unequal amount of oxygen

Hazem Nasef 1I
Posts: 51
Joined: Tue Oct 10, 2017 7:13 am

Re: 8.65

Postby Hazem Nasef 1I » Sun Jan 21, 2018 12:14 am

If you notice, in the original equation, it was just N2 reacting with 2.5 moles of O2 ( or 5 moles of O). In this second reaction, instead of N2, it is two moles of NO, which contains two moles of O, and then 1.5 moles of O2 (or 3 moles of O). So both equations essentially have 5 moles of O, but they are spread out differently. So the enthalpy of formation of N2O5 is also equivalent to the enthalpy of the second reaction.

Angel Ni 2K
Posts: 50
Joined: Sat Jul 22, 2017 3:01 am

Re: 8.65

Postby Angel Ni 2K » Sun Jan 21, 2018 1:41 am

The solution stops at 2NO(g) + 3/2O2(g) --> N2O5(g) because you can use the enthalpy of formation of NO(g) to figure out the enthalpy of formation of N2O5(g). For 2NO(g) + 3/2O2(g) --> N2O5(g), deltaH is -169.2 kJ/mol. Thus, 1(deltaHfN2O5) - 2(deltaHfNO) = -169.2 kJ/mol. Using the appendix, 1(deltaHfN2O5) - 2(90.25 kJ/mol) = -169.2 kJ/mol. Therefore, deltaHfN2O5 = 11.3 kJ/mol.

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