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Topic: Thermodynamically favorable but kinetically difficult?  (Read 2244 times)

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joomlab

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Thermodynamically favorable but kinetically difficult?
« on: November 02, 2013, 02:07:52 PM »
How can a reaction be thermodynamically favorable but also kinetically difficult?

For example, N2 + 3 H2  ::equil:: 2 NH3

It says in my text that it is kinetically difficult because intermediates rapidly decay...

I guess, my main questions are how is N2 + 3 H2  ::equil:: 2 NH3 thermodynamically favorable? What does thermodynamically favorable mean exactly?


Offline curiouscat

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Re: Thermodynamically favorable but kinetically difficult?
« Reply #1 on: November 02, 2013, 02:31:07 PM »
It could have a very high barrier. Super high activation energy.

Thermodynamically favorable doesn't consider the location of the activated transition state, at all.

Offline magician4

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Re: Thermodynamically favorable but kinetically difficult?
« Reply #2 on: November 02, 2013, 03:07:44 PM »
in addition:
Quote
It says in my text that it is kinetically difficult because intermediates rapidly decay...
I would prefer to look at this from a different point of view: it requires a lot of energy (i.e., in general, high temperatures) to break for example the triple bond of nitrogen, gaining two nitrogen atoms ( as one aspect of the total events).
hence, this doesn't happen very often, therefore the overall speed ( i.e molecules transformed per second) of this is low

Quote
I guess, my main questions are how is N2 + 3 H2  ::equil:: 2 NH3 thermodynamically favorable? What does thermodynamically favorable mean exactly?
in your example: from two nitrogen atoms and 6 hydrogen atoms, you will gain more energy if you formed two ammonia molecules than the what you would gain from forming a nitrogen molecule and three hydrogen molecules

regards

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