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Topic: thermochemistry  (Read 4892 times)

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Offline Moneyking

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thermochemistry
« on: December 11, 2012, 10:53:58 PM »
Assuming the delta H and delta S do not change with temperature, determine the temperature above which you can theoretically prepare carbon dioxide CO2(g) direct from its elements:

C(s, graphite) + O2 ---> CO2(g)


SO FAR WHAT I GOT:

delta H = (-393.5) - (0+0)
delta S = (213.7) - (5.7 + 205.1)
=2.9/1000
t=(-393.5)/(0.0029)
=-135689.7

IS this correct?

Offline curiouscat

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Re: thermochemistry
« Reply #1 on: December 11, 2012, 11:39:27 PM »
Units?

Offline Moneyking

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Re: thermochemistry
« Reply #2 on: December 12, 2012, 12:22:53 AM »

Offline curiouscat

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Re: thermochemistry
« Reply #3 on: December 12, 2012, 12:43:45 AM »
Your data and calcs. seem ok. Yet the answer worries my intuition.

Yes, graphite is very stable to oxidation but don't think it's that stable.

Maybe the assumptions cause a glitch.

PS. Online estimates for pure graphite combustion Temps. vary but 4000°C was one estimate.
« Last Edit: December 12, 2012, 12:54:35 AM by curiouscat »

Offline Moneyking

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Re: thermochemistry
« Reply #4 on: December 12, 2012, 10:05:10 PM »
Your data and calcs. seem ok. Yet the answer worries my intuition.

Yes, graphite is very stable to oxidation but don't think it's that stable.

Maybe the assumptions cause a glitch.

PS. Online estimates for pure graphite combustion Temps. vary but 4000°C was one estimate.
Ye, I am confused about the answer. the answer is VERY large!

Offline curiouscat

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Re: thermochemistry
« Reply #5 on: December 13, 2012, 12:38:14 AM »
On more thought, I think  we have a blunder somewhere though I can't find it.

Offline curiouscat

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Re: thermochemistry
« Reply #6 on: December 13, 2012, 12:46:07 AM »
I think I have a hunch. Note you are getting a -ive T.

So the equation has no solution. i.e. ΔG is never zero; always negative. So this conversion might be spontaneous at all temperatures.  Ughh. Hate myself for being careless with signs.  >:(

Wish someone else can verify this!

PS. Thermodynamically spontaneous can still be kinetically improbable. Interesting way to ask the question; almost a trick.

Offline Moneyking

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Re: thermochemistry
« Reply #7 on: December 13, 2012, 02:29:45 PM »
On more thought, I think  we have a blunder somewhere though I can't find it.
I am just wondering, but if I didn't divide by 1000, then the answer would be positive. in degrees though. but i am sure  you have to divide by 1000 because delta S is joules, and delta H is in kilo-joules.

Offline curiouscat

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Re: thermochemistry
« Reply #8 on: December 13, 2012, 02:35:30 PM »

I am just wondering, but if I didn't divide by 1000, then the answer would be positive.

No it wont.

Offline Moneyking

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Re: thermochemistry
« Reply #9 on: December 13, 2012, 04:15:22 PM »

I am just wondering, but if I didn't divide by 1000, then the answer would be positive.

No it wont.
ye my mad, i was adding 273 instead of substracting 273.

Offline curiouscat

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Re: thermochemistry
« Reply #10 on: December 13, 2012, 04:17:18 PM »

ye my mad, i was adding 273 instead of substracting 273.

Still not convinced that it is always spontaneous?

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