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Offline Corribus

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Re: Question on collision theory
« Reply #15 on: April 19, 2013, 02:46:56 PM »
There is an activation energy because forming the transition state is (usually) energetically unfavorable.  It's like trying to roll a ball over a hill.  If you don't push it hard enough, it's just going to roll right back down in your direction.  The activation energy is the difference in free energy (enthalpy/entropy) between the reactants and the transition state; it is related to the amount of energy required to break required bonds, but I don't think that's necessarily the best way to look at it - although for general chemistry or high school chemistry level, it's sufficient. 

Plus - and I can't stress this enough - these energies we speak of represent statistical averages.  Heat isn't really something that's absorbed and emitted by single molecules that are reacting.  It's a representation of the average amount of kinetic energy in a system.  When heat is released from an exothermic reaction, it's because the average potential energy of the molecules in the system after the reaction is less than that of molecules before the reaction, and the difference has been converted into kinetic energy (vibrations, rotations, translations).  It's not because each reacted molecule gives out a little squirt of "heat".  In other words, energy is more a state of the system than a true "reactant" or "product".  At least, that's the way I like to view it.

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As for the decomposition reaction, shouldn't there still be an activation energy since I have to break those intramolecular bonds?
Yes, there usually is.  Sometimes it's so small that it takes almost nothing to overcome (as in shock sensitive explosives like nitrogen triiodide or fulminated mercury), and in these cases the "transition state" is difficult to precisely define.  Also, I'll remind you that activation energy is NOT just enthalpy.  It's entropy as well. A lot of times a good portion of the energy required to make a reaction go is not just a matter of breaking and forming bonds.  It's the entropic change required for arranging a very specific structural transition state form.  If the transition state of a decomposition requires entropically unfavorable rearrangement of solvent molecules, for example, this could contribute to a reasonably large activation energy even if the amount of energy required to break bonds and interactions is relatively small.  At least in principle; can't think of a good example off the top of my head.

Just keep in mind that EVERY chemical will decompose if enough heat energy is put into the system. 
« Last Edit: April 19, 2013, 03:55:20 PM by Corribus »
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Offline Needaask

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Re: Question on collision theory
« Reply #16 on: April 21, 2013, 03:56:03 AM »
There is an activation energy because forming the transition state is (usually) energetically unfavorable.  It's like trying to roll a ball over a hill.  If you don't push it hard enough, it's just going to roll right back down in your direction.  The activation energy is the difference in free energy (enthalpy/entropy) between the reactants and the transition state; it is related to the amount of energy required to break required bonds, but I don't think that's necessarily the best way to look at it - although for general chemistry or high school chemistry level, it's sufficient. 

Plus - and I can't stress this enough - these energies we speak of represent statistical averages.  Heat isn't really something that's absorbed and emitted by single molecules that are reacting.  It's a representation of the average amount of kinetic energy in a system.  When heat is released from an exothermic reaction, it's because the average potential energy of the molecules in the system after the reaction is less than that of molecules before the reaction, and the difference has been converted into kinetic energy (vibrations, rotations, translations).  It's not because each reacted molecule gives out a little squirt of "heat".  In other words, energy is more a state of the system than a true "reactant" or "product".  At least, that's the way I like to view it.

Quote
As for the decomposition reaction, shouldn't there still be an activation energy since I have to break those intramolecular bonds?
Yes, there usually is.  Sometimes it's so small that it takes almost nothing to overcome (as in shock sensitive explosives like nitrogen triiodide or fulminated mercury), and in these cases the "transition state" is difficult to precisely define.  Also, I'll remind you that activation energy is NOT just enthalpy.  It's entropy as well. A lot of times a good portion of the energy required to make a reaction go is not just a matter of breaking and forming bonds.  It's the entropic change required for arranging a very specific structural transition state form.  If the transition state of a decomposition requires entropically unfavorable rearrangement of solvent molecules, for example, this could contribute to a reasonably large activation energy even if the amount of energy required to break bonds and interactions is relatively small.  At least in principle; can't think of a good example off the top of my head.

Just keep in mind that EVERY chemical will decompose if enough heat energy is put into the system.

Hi thanks so much for the replies :)

I watched Khanacademy and the Central Science video on the transition state along with this http://www.rpi.edu/dept/chem-eng/Biotech-Environ/Projects00/enzkin/transition.htm and I have some queriesThey focused on the transition state and what's going on is that they collision in the correct orientation then an intermediate is formed - bond partially broken and formed at the same time.

But after that transition state is formed, what happens after that? Will the transition state take in more energy to break the partial bond or does it already have enough to do so?

Thanks so much :)

Offline curiouscat

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Re: Question on collision theory
« Reply #17 on: April 21, 2013, 05:44:13 AM »
But after that transition state is formed, what happens after that? Will the transition state take in more energy to break the partial bond or does it already have enough to do so?



The TS is the highest energy point. No more energy needs to be taken in. Net energy is actually being released after the TS.

Offline Needaask

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Re: Question on collision theory
« Reply #18 on: April 21, 2013, 06:06:11 AM »
But after that transition state is formed, what happens after that? Will the transition state take in more energy to break the partial bond or does it already have enough to do so?



The TS is the highest energy point. No more energy needs to be taken in. Net energy is actually being released after the TS.

Ohh! So at that point enough energy is required to break the bonds so only energy is given out when the new bonds form?

Offline Needaask

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Re: Question on collision theory
« Reply #19 on: April 21, 2013, 09:40:54 AM »
But after that transition state is formed, what happens after that? Will the transition state take in more energy to break the partial bond or does it already have enough to do so?



The TS is the highest energy point. No more energy needs to be taken in. Net energy is actually being released after the TS.

Also, in the spike to the transition state, is the energy absorbed or does the temperature of it just increase? Then when the energy level drops from the transition stage does it give out even more energy causing the temperature to rise even more?

Thanks so much for the help everyone :)

Offline Corribus

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Re: Question on collision theory
« Reply #20 on: April 21, 2013, 09:49:21 AM »
Think of it this way: if you have a simple transformation A-B + C  :rarrow: A + B-C.

At the transition state A---B---C, the A-B bond is half broken and the B-C bond is half formed.  It's true that as the reaction proceeds from this point, the A-B bond is still "being broken" and maybe additional energy is required.  However the extra amount of energy required to break it the rest of the way is more than compensated by the energy gains by the formation of B-C at that point.  After the transition state, it's all energy gain.  The transition state is the point where the energy requirement for breaking A-B is exactly balanced by the energy gain for forming B-C.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline curiouscat

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Re: Question on collision theory
« Reply #21 on: April 21, 2013, 10:07:59 AM »

Also, in the spike to the transition state, is the energy absorbed or does the temperature of it just increase?


What is your definition of temperature? Can a single TS molecule even have something called temperature?


Offline curiouscat

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Re: Question on collision theory
« Reply #22 on: April 21, 2013, 10:12:17 AM »
After the transition state, it's all energy gain.  The transition state is the point where the energy requirement for breaking A-B is exactly balanced by the energy gain for forming B-C.

In addition I'll point out that bonds themselves are entirely abstractions. They are all constructs in our mind. There is no particular point where you can say that "Aha! The A--B bond is completely broken now!"

I'll reiterate @Corribus's point about statistical averages and system properties.

Your questions make me think you are taking the bond, TS etc. analogies a tad too far. You really should read up a bit more about the statistical nature of molecular properties.

Offline Needaask

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Re: Question on collision theory
« Reply #23 on: April 22, 2013, 09:04:27 AM »
Think of it this way: if you have a simple transformation A-B + C  :rarrow: A + B-C.

At the transition state A---B---C, the A-B bond is half broken and the B-C bond is half formed.  It's true that as the reaction proceeds from this point, the A-B bond is still "being broken" and maybe additional energy is required.  However the extra amount of energy required to break it the rest of the way is more than compensated by the energy gains by the formation of B-C at that point.  After the transition state, it's all energy gain.  The transition state is the point where the energy requirement for breaking A-B is exactly balanced by the energy gain for forming B-C.

Hi thanks for the great response.

But what if the AB bond is stronger than the BC bond then wouldn't the energy still continue to get absorbed causing the energy level to go up? And actually what would the energy from the activation energy be used for?

Thanks for the help :)

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