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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.
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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