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Topic: Internal Energy Q  (Read 5003 times)

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Offline _cheers

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Internal Energy Q
« on: April 08, 2007, 07:21:07 PM »
If I have a thermochemical equation as such:

2NaN3(S) ----> 2 Na (s) + 3N2 (g)  (delta) H = 42.7 kJ and heat of formation for NaN3 is - 21.35 kJ/mol

any hints on how to find (delta) E (not/knot?)

and, where can I input symbols? like delta, etc??

thanks
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Offline Yggdrasil

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Re: Internal Energy Q
« Reply #1 on: April 08, 2007, 07:50:15 PM »
How is enthalpy related to internal energy?

Offline _cheers

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Re: Internal Energy Q
« Reply #2 on: April 08, 2007, 08:17:51 PM »
 ok, I'm trying to grasp this concept..but does most of E occur as heat transfer, so it would be close to H...but somehow include the moles of gas produced? wouldnt some energy be lost to surroundings as gas is produced? 

i dunno...i have some equations to work with, but I cant make any of them fit because I dont seem to have enough data. I know I need to think more than that, but I'm so frustrated!

Somehow, I think I use the H value, and something to do with the 2 moles of N2 produced, disregarding the Na(s) produced....close? do I make sense?
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allanf

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Re: Internal Energy Q
« Reply #3 on: April 08, 2007, 08:40:26 PM »
Notice also that the reaction you provide is just the decomposition reaction for NaN3, the reverse of the formation reaction.  That is to say the reaction is described equally well either by the (delta)H or the heat of formation, having both is just redundant.

Offline Yggdrasil

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Re: Internal Energy Q
« Reply #4 on: April 08, 2007, 08:45:53 PM »
Remember that enthalpy is defined in the following way:

H = U + PV

In other words, enthalpy takes into account both the change in internal energy of the reactants, but it also includes the PV work done by the system during the course of the reaction.

In a more relevant form, you get the following equation:

(Delta)H = (Delta)U + Delta(PV)

If pressure is constant during your reaction then this becomes:

(Delta)H = (Delta)U + P(Delta)V

You can calculate (Delta)V by assuming that the volumes occupied by the solids is negligible to the volumes occupied by a gas.

Offline _cheers

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Re: Internal Energy Q
« Reply #5 on: April 08, 2007, 08:57:56 PM »
hmmmm....ok I see where your going...its just understanding that concept, I didnt think of it that way, but it makes sense.

thanks!
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