4D.23 7th Ed.

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JulieAljamal1E
Posts: 71
Joined: Fri Sep 28, 2018 12:24 am

4D.23 7th Ed.

Postby JulieAljamal1E » Tue Jan 29, 2019 7:55 pm

The question asks to calculate the standard enthalpy formation of dinitrogen pentoxide from the info:
2NO+O2–>2NO2 delta H= -114.1kJ
4NO2+O2->2N2O5 delta H= -110.2 kJ
I’m confused on how to do this. The solutions show to add the first reaction with half of the second, but why are you allowed to multiply the second equation by 1/2. Then after it adds the two up it says the enthalpy of the reaction equals the enthalpy of formation of N2O5 minus twice the enthalpy formation of NO, how is that so? Can you step by step explain this problem, thank you!

Sarah Kiamanesh 1D
Posts: 30
Joined: Fri Sep 28, 2018 12:22 am

Re: 4D.23 7th Ed.

Postby Sarah Kiamanesh 1D » Wed Jan 30, 2019 10:17 am

For this problem, we want to cancel out NO2 using Hess's Law because that's the only component for which we don't know the standard enthalpy of formation for. To cancel this, we multiply the second equation by 1/2 and add the two equations. Remember to adjust the standard enthalpies of the reactions as you manipulate the equations. Adding the equations results in
2NO + 3/2O2 --> N2O5
Delta H = -114.1 + (-55.1) = -169.2KJ
Now, we use Delta Hrxn = (nHf(products))-(nHf(reactants))
-169.2KJ = (1)Delta Hf(N2O5) - (3/2)Delta Hf(O2) - (2)Delta Hf(NO)
-169.2KJ = Delta Hf(N2O5) - 0 - 2(90.25)
Delta Hf(N2O5) = 11.3KJ


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