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Topic: Enthalpy , Heat of formation  (Read 18074 times)

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Offline Aznhmonglor

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Enthalpy , Heat of formation
« on: September 17, 2009, 08:40:30 PM »
Determine  :delta:H, q, w, and  :delta:E at 298 K and 1 atm for the complete reaction of 8.610 g of N2O4.

N2O4(g) + 2N2H4(l)  3N2(g) + 4H2O(l).

The heat of formation of liquid hydrazine (N2H4) is 50.63 kJ/mol. Use tabulated data for other heats of formation.

The tolerance on each question is only 0.03 kJ, so express all answers to 0.01 kJ.

Wow I have been stuck on this problem for an hour now. I can't get the  :delta:H right for some reason. I looked up the values and N2O4's heat of formation is 9.66, N2H4 as stated in problem is 50.63, N2 is 0 and H20 is -285.83.

Now to get  :delta:H I did products heat of formation minus the reactants and I still can't get it.

Heres what it looks like:

(3(0) + 4(-285.83)) - (9.66 + 2(50.63)) = -1254.24 kJ

It is wrong. What am I doing wrong?

Offline UG

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Re: Enthalpy , Heat of formation
« Reply #1 on: September 18, 2009, 01:14:13 AM »
Dunno what's wrong  :-X
Your working looks right. Have you got all of the states correct in your equation?

Offline Aznhmonglor

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Re: Enthalpy , Heat of formation
« Reply #2 on: September 18, 2009, 08:09:34 AM »
Dunno what's wrong  :-X
Your working looks right. Have you got all of the states correct in your equation?

Yes all of the states are right. But for some reason the answer is just wrong. =/

Offline sjb

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Re: Enthalpy , Heat of formation
« Reply #3 on: September 18, 2009, 09:33:14 AM »
8.610g of dinitrogen tetroxide is not 1 mol of dinitrogen tetroxide?

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