Why Doesn't Nitrous Oxide Decompose at RTP Despite Negative Gibbs Free Energy?

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quietrain
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why doesn't nitrous oxide decompose to nitrogen and oxygen at rtp even though the gibbs free energy is -105 ?

thanks!
 
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Well, I'm not sure about the free energy and all, but nitrous oxide won't turn back into oxygen and nitrogen unless it receives enough energy, such as heat, to break the bonds and separate the molecule. The basic reason is that the nitrogen atoms bond with the oxygen atom and cause the oxygen to get a full electron shell, which is favorable for the oxygen. At the same time the nitrogen gets closer to a full shell as well, which is also favorable to the nitrogen.
 
yea from the chemistry point of view, it is stable, but the question ask me why it is stable even though the gibbs free energy is such a negative number.

since gibbs free energy is the measure of the spontaneity of a reaction, the lower the more readily it happens, it says that nitrous oxide will decompose. but it doesn't happen :X
 
I'm not sure, but maybe its because the free energy is only applicable to work in the system? I'm not very familiar with free energy, sorry.
 
quietrain said:
why doesn't nitrous oxide decompose to nitrogen and oxygen at rtp even though the gibbs free energy is -105 ?

thanks!
A reaction may be thermodynamically favorable but not kinetically. Do you know the difference? Look it up and get back with what you found.
 
hi
negative free energy indicates that the reaction is spontanious but the reaction needs to start an activation energy
 
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oh i see...

so if my temperature is not high enough, then the molecules do not gain enough energy in the first place to kick start the reaction, so at rtp, no reaction occur

gibbs free energy is a measure of spontaneity of the reaction, the tendency of the reaction to continue, but there is a need for a "spark" to start it?

i was looking at this link http://web.tock.com/kalee/chem32/kinT/
and i realize i actually learned before :X

but i don't remember is it physics or chemistry...

it is all about the activation energy the transition states , reactants and products

so thermodynamically, time is not considered, and so it is an indication of how stable the produce is to the reactant, so -Gibbs energy = more stable, so more favorable thermodynamically

but the activation energy , kinetically favorable, is required for the reaction to occur in the first place, and for rtp, the nitrous oxide does not get enough temp to get enough energy so it does not have enough activation energy so not kinetically favorable?

thanks!