[Needaask]

When two reactants collide with activation energy, there would be a partial bond breaking and partial bond forming. This is the transition state. However, what happens after the transition state? Would more energy be put in to break the partial bonds from the old reactants and would more energy be given out from the formation of the full bonds? If so, then won't it be possible for the enthalpy graph to continue going up since the old bonds might require more energy than the energy given out from the new bonds formed?

And for decomposition reactions, does concentration affect the rate of reaction? In this case I don't think any collisions would result in reactions because in decomposition, the bonds are broken and formed. So how do I put the collision theory and transition state theory here?

Initially, before learning about the transition state, I thought that the bonds just get broken first then they get formed. But that wouldn't be favorable as the transition state has a lower activation energy. So the collision and transition state theory made sense. But thinking about it in these 2 reactions caused some problems understanding the full picture.

Hope you guys can help

[curiouscat]

When two reactants collide with activation energy, there would be a partial bond breaking and partial bond forming. This is the transition state. However, what happens after the transition state? Would more energy be put in to break the partial bonds from the old reactants and would more energy be given out from the formation of the full bonds? If so, then won't it be possible for the enthalpy graph to continue going up since the old bonds might require more energy than the energy given out from the new bonds formed?

No. If that was the case your TS would occur later.

Overall Energy has to go down after the TS.

[Needaask]

When two reactants collide with activation energy, there would be a partial bond breaking and partial bond forming. This is the transition state. However, what happens after the transition state? Would more energy be put in to break the partial bonds from the old reactants and would more energy be given out from the formation of the full bonds? If so, then won't it be possible for the enthalpy graph to continue going up since the old bonds might require more energy than the energy given out from the new bonds formed?

No. If that was the case your TS would occur later.

Overall Energy has to go down after the TS.

TS is transition state right? Not the entropy times temperature. Just checking.

But why would this be so? Because once the TS is formed, the partial bonds from the reactants would have to continue breaking while the partial bonds to form the new bonds in the products would continue forming. So isn't there a possibility that the energy level goes ups even after the TS?

Or I thought it could be possible that the activation energy provides the energy to continue breaking those partial bonds? So no extra energy is needed to break them while energy would start to be given out during the bond forming of those new partial bonds? But this is just my speculation.

Hope you can help out thanks

[gritch]

TS is transition state right? Not the entropy times temperature. Just checking.

But why would this be so? Because once the TS is formed, the partial bonds from the reactants would have to continue breaking while the partial bonds to form the new bonds in the products would continue forming. So isn't there a possibility that the energy level goes ups even after the TS?

Or I thought it could be possible that the activation energy provides the energy to continue breaking those partial bonds? So no extra energy is needed to break them while energy would start to be given out during the bond forming of those new partial bonds? But this is just my speculation.

Hope you can help out thanks

I think you're under the assumption that breaking bonds always requires energy but that's not the case when we look at transition states. In the formation of a typical transition state we see transfer from a lower energy electron state localized on your reactant molecules to a high energy bonding state between our reactant molecules. The bonds within this type of transition state are higher in energy (if you're familiar with molecular orbital theory, transition state bonding typically involves anti-bonding molecular orbitals) and the relaxation of these high energy bonds occur spontaneous resulting the formation of our products.

A transition state is basically the structure with just enough energy to spontaneously decompose into our reaction products.

[Corribus]

With respect to the previous poster, I'm not sure I follow your reasoning and it seems that you, too, may have some misunderstanding about what a transition state is.

Needaask, we've been down this road before.  The definition of the transition state precludes the scenario you are describing.  A transition state is a saddle point on the N-dimensional potential energy surface.  Mathematically, along the reaction coordinate (which is the lowest energy path between the global minima that correspond to reactants and products) it is a local maximum value of energy (the derivative is equal to zero).  This is the definition of a transition state.  Therefore by definition it is not possible for the energy at any infinitesimal value to the left or right of the transition state along the reaction coordinate to have a higher energetic value.  If it did, it would not be a saddle point on the PES, and therefore it would not qualify as a transition state.  Note that it is possible for there to be other maxima along the reaction coordinate, in the case of a multistep mechanism that features several transition states and quasi-stable intermediates, but the energy of the activated complex still has to decrease before it may increase again to form a higher-energy transition state.

Important: the difference in energy between the reactants and the transition state is NOT necessarily (and not usually) equal to the energy required to break bonds necessary for the reaction to occur.  This is because the activated complex at the transition state is characterized as a structure with both a breaking bond and a forming bond, simultaneously.  If you were to undergo a process with complete breaking of a bond, followed by complete forming of a new bond, the energy required would typically be (far) greater than that characterized by the true transition state, where forming and breaking occur simultaneously.  The former process would not be the lowest energy path, and so MOST reaction events would not occur via this mechanism.  Although, strictly speaking, the probability of it happening is not zero.  The reaction coordinate represents the most probable pathway that a reaction pathway can take.  There are an infinite number of true pathways available, and the probability of them occuring depends on temperature and their relative positions on the PES.  (In a real chemical reaction, not all pathways lead to the intended products, either.  The PES is an idealization.)  Do note, however, that the potential energy is a state function.  The pathway taken only impacts kinetics.  The total energy gained and released by all reaction pathways is the same, as long as the starting and ending points are identical.

Example: Suppose you have a reaction A-B + C --> A + B-C

Now we focus on two possible mechanisms: (1) A-B + C --> A + B + C --> A + B-C and (2) A-B + C --> A--B--C --> A + B-C

In (1), the bond between A-B is completely broken, yielding two very high energy (presumably) isolated atoms that are an infinite distance apart.  Then a new bond between B-C forms.  This process requires (all things being equal, and assuming only population of ground vibrational states) an input activation energy equal to the bond energy of A-B.  This energy is equal to the difference in potential energy between A/B separated by their equilibrium bond distance, and A/B infinitely far apart, which can be determined roughly by any anharmonic oscillator approximation of your choice (Morse potential, say).  Conversely, when B-C forms, a quantity of energy is released equal to the inverse process - bringing B/C infinitely far apart to their equilibrium bond distance.  The difference in bond energies between A-B and B-C would be roughly equivalent to the overall change in energy from reactants to products, which would determine the overall thermodynamic spontaneity of the reaction (endergonic or exergonic, depending on the sign).  Of course, here I am treating enthalpy only and ignoring entropy, but for this simple hypothetical reaction, it's all right I think to paint a conceptual picture.  Note also that I'm neglecting any energy of having A or C sitting around all by themselves.  In a low pressure gas phase reaction, this is probably fine to do for sake of argument, but in solution there will be solvent interactions (enthalpies) to worry about that could not be neglected  Note finally that for this process to happen SOME collision must occur because the energy has to come from somewhere; this could be some spectator gas molecule.

In (2), a different process happens: the bond between A-B begins to break, but at the same time, B-C begins to form.  The transition state is a point somewhere in the middle, where A-B is "half" broken and B-C is "half" formed.  (It won't really probably be "half" and "half", but whatever.)  Note this is not the same as a hypothetical molecule A-B-C!  It should be readily apparent that the energy of this transition state A--B--C MUST be equal to or lower than the energy of A + B + C, all isolated.  To understand why, just take A-B for a moment.  Now start to pull B away.  Gradually the potential energy increases until at large separations some asymptotic limit is reached.  This asymptotic limit is the maximum energy of the system attainable by pulling B away from A.  Likewise, start with B-C and pull B away.  Same thing, same energy.  So at the most, A--B--C has the energy of A + B + C in the limit of the "--" being infinitely long.  Shorter, and there has to be some stabilization, because bonds stabilize.  Of course, if you were to shrink both "--" such that they were on the order of A-B and B-C (A-B-C), you'd probably have a complex with higher energy than A + B + C, but this isn't the transition state.  This is all to say, that A--B--C must be lower energy than A + B + C, else no reaction would proceed through A--B--C.  It wouldn't be along the most probably reaction coordinate, because reactions proceed through the path of least resistance.

Anyway, what you end up with is a transition state energy that is more than either the reactant or the product but less than blasting everything to individual atoms and then reassembling them.  The latter process, by the way, is effectively what we do to determine reaction enthalpies via heats of formation and Hess's law.  Because remember, enthalpy and Gibbs energy are both a state functions.

I realize this is very crude, but here's a figure demonstrating what I mean.  It is two Morse potentials for A-B and B-C, where B-C has a lower energy than A-B.  You can see that if I wanted the reaction to occur, I could completely break A-B (an input energy of "10", and then form B-C, release an energy of "13", for a net energy gain of "3".  However, if I start at the minimum of the blue line and go right by stretching the A-B bond out, and simultaneously form the B-C bond by following the red line also to the right, there is a "crossing" point, where A and C are both partially bonded to B.  Following the reaction in this way is following the reaction coordinate.  A more realistic path is shown in green, which would be akin to both of these processes happening truly spontaneously.  At the maximum of the green line is the transition state.  As I keep following it right, you can see that I continue to "lose" energy from continuing to pull apart the A-B bond.  However this is more than compensated by the formation of B-C, which simultaneously releases potential energy.  Again, very crude figure and totally made up numbers, but maybe it helps convey the idea. 



Please also bear in mind that once the transition state is reached, it is not mandatory that the reaction proceed to generation of products.  The transition state also represents the mid-point in the reverse reaction as well.  All reaction dynamics are governed by statistics.  The state of the system once equilibrium has been reached is determined by the relative energies of the reactants and products, which is why the Gibbs energy change is directly related to the equilibrium constant.  The time it takes to reach equilibrium is related to (among other things) the energy of the transition state, which governs the probability that two colliding molecules at a certain temperature will have enough kinetic energy to form the transition state.  The shape of the potential energy surface also can be used to predict what vibrational modes lead to formation of the activated complex and what kinds of collision trajectories are most likely to bring about a successful reaction.  However these topics are getting rather far afield.

NB: I notice a minor error in the figure.  The "red" line legends should be "B + C to B-C", not "B-C to B + C".

[Needaask]

Hi Corribus thanks for the great post.

Ohh so mathematically, the transition state is the highest energy level. But when I was looking at (2), when A--B--C is formed, won't the A--B part have to stretch and break to form A and B while B--C would shrink and form B-C? In the graph you mentioned that B--C to B-C is greater than A--B to A and B. So won't it also be possible for A--B to form A and B be greater than for B--C to form B-C? In that case won't the energy go up? Even though in a mathematical standpoint it's not supposed to happen. But isn't this still possible?

Thanks

[magician4]

in addition:

Ohh so mathematically, the transition state is the highest energy level.

I think this is a misunderstanding of the situation "what happens when two molecules will collide":

of cause  you could have a greater energy of the total system in transition state, yes... but that would for example be due to kinetic energy excess , when part of the collision energy is not consumed to feed the reach of the transition state, i.e. is not transformed (as it is not needed for this) in "high energy bonds"*⁾
... given that whilst colliding, the molecules were exactly on the minimum pathway at all, which by no means is a given ( au contraire: most of even successfull collisions will happen "offroad", and even more collisions will happen with even higher energies , where nothing will have happened by the end of the day at all, as the molecules didn't do nothing but just crash into each other, and then separate again)


regards

Ingo


*⁾ additionally, in this case the inertia of the colliding molecules would influence the question in which direction (i.e. educts, products or "even higher energy bonds") the system will develop, as it is not free to follow the minimum energy pathway in this situation

[gritch]

With respect to the previous poster, I'm not sure I follow your reasoning and it seems that you, too, may have some misunderstanding about what a transition state is.

I think I've failed to explain myself carefully enough. Your explanation is very heavy on mathematical concept, which is perfectly valid, but I was attempting to explain the concept through chemical means rather than with mathematics which seems to be more in line with the OPs original inquiry. I'll see if I can explain my understanding a bit more thoroughly. I admit I might have confused the concept of intermediate and transition state at my first post but I think it's been worked out now.

The concept of a chemical reaction is for there to be a rearrangement of electron density between several species (the reactants) resulting in the formation of new species (the products). Collisions between reactant molecules with enough energy result in formation of high energy quasi-stable intermediates whose bonds typically involve interactions between reactant molecules' higher energy anti-bonding molecular orbitals.

I found a paper with a nice diagram of the concept: http://link.springer.com/article/10.1007%2Fs11237-011-9188-8
It's a sort of a long read that goes in-dept with IR studies but towards the end (Figure 2) there's a diagram of the intermediate as predicted from DFT calculations of the oxidation of NO into NO₂.

Here is where I will admit I was mistaken, and I believe the OP made the same error as me, there is a slight energy barrier required to transition from this high energy intermediate to the true transition state, in which there is now enough energy to break these intermediate bonds and form the new bonds necessary to form your products. The energy of this high energy intermediate would be represented in your graph as the peak of the A--B--C formation and the energy difference between that and the true transition state is the energy it takes to decompose the intermediate.

I believe most of the confusion comes when we have a very high energy intermediate. Often the difference between the intermediate to TS is so small relative to the energy of the intermediate that the immediate is often taken to be the transition state itself. That is definitely an approximation that the OP and I seem to have been confused with.

I probably didn't need to post again. This is mostly just been for my own benefit rather than the OP's at this point but I'm glad to have learned something new today.

[curiouscat]

Here is where I will admit I was mistaken, and I believe the OP made the same error as me, there is a slight energy barrier required to transition from this high energy intermediate to the true transition state, in which there is now enough energy to break these intermediate bonds and form the new bonds necessary to form your products.

I think you are still mistaken: A Transition State is a transition state and an intermediate is an intermediate.

One is a saddle point and the other isn't.

The energy of this high energy intermediate would be represented in your graph as the peak of the A--B--C formation and the energy difference between that and the true transition state is the energy it takes to decompose the intermediate.

I think that's wrong. The Energy peak is the Transition State. The high energy intermediate cannot be higher than the TS.
[/quote]

[gritch]

Here is where I will admit I was mistaken, and I believe the OP made the same error as me, there is a slight energy barrier required to transition from this high energy intermediate to the true transition state, in which there is now enough energy to break these intermediate bonds and form the new bonds necessary to form your products.

I think you are still mistaken: A Transition State is a transition state and an intermediate is an intermediate.

One is a saddle point and the other isn't.

The energy of this high energy intermediate would be represented in your graph as the peak of the A--B--C formation and the energy difference between that and the true transition state is the energy it takes to decompose the intermediate.

I think that's wrong. The Energy peak is the Transition State. The high energy intermediate cannot be higher than the TS.

I think you've misunderstood me now. I'm not assuming the high energy immediate is higher in energy than the transition state. The fact that I stated that it requires additional energy to decompose into to products implies it's not higher in energy than the TS.

Corribus' graph was a nice representation. The intermediate of this reaction should be shown in the lower A--B--C graph and the transition state is the interception of the other two graphs above it, even when the immediate is formed there is still a barrier of energy up to the transition state to form the products.

[Corribus]

@gritch

You can have quasi-stable intermediates in a reaction mechanism; however as curiouscat has noted they are not transition states.  Intermediates are high-energy minima along the reaction coordinate.  Because they are minima, they can be isolated and studied, in principle, by dropping the temperature sharply in the middle of a reaction.  A--B--C, in my plot above, is NOT an intermediate.  It is a transition state.

Transition states are maxima along the potential energy surface.  There will be a transition state (energy maximum) between every two minima.  So, for a simple reaction with just one reactant and one product, there will be a single transition state.  For a more complex reaction which involves the reactant, the products, and an intermediate, there will be two transition states. 

Transition states are not quasi-stable because they cannot be isolated by removing thermal energy from the system.  They cannot be isolated because there are always other states nearby which have lower energy.

Many gas phase reactions like the one you have linked to exhibit complicated reaction mechanisms that have multiple reasonably stable intermediates.  In my (long distant) past, I published an ab initio study on a different atmospheric chemistry reaction - OH + CO --> H + CO2.  This reaction features a very low energy intermediate, the formyloxyl radical (HOCO).  In fact, the energy of this "intermediate" is lower (if I recall; it's been a while) than either the reactants OR the products.  The reason it is not formed with any abundance is complicated and probably beyond the scope of what needs to be discussed here, but merits at least me pointing out that the potential energy surface is just that - a measure of potential energy only.  A real reaction involves both kinetic and potential energy, so the reaction need not form every intermediate, particularly when there is a lot of kinetic energy around. 

All that said, not all potential energy surfaces or reaction coordinates need include intermediates, as my plot above indicates.  Moreover, with a simple Arrhenius model of kinetics, in most cases intermediates are inconsequential.  The model considers the overall activation energy as the energy difference between the reactants and the highest energy transition state.  At equilibrium, you will have some of the intermediate in your reaction vessel, but still it is not needed to model the rate of appearance of products or disappearance of reactants.

Drawing a potential energy surface with an intermediate is a pinch beyond my capabilities on my home computer.  If it is of interest to you, I can do so fairly easily from work, and post an image tomorrow.

Final note: Invoking bonding or antibonding orbitals isn't really necessary to understand a potential energy surface, and in any case is an over-generalization of reaction mechanisms.  Most surfaces these days are generated by DFT calculations, which don't have any direct or simple relationship to molecular orbital frameworks.

@Needaask

No, the mathematics does not lie.

EDIT: Gritch, based on your more recent post, let me add that you should not take my plot too literally.  The data is made up and is only for illustration.  And it's a 2D abstraction of a true, 3D surface.  The crossing point between the red and blue lines is completely artificial and nothing should be inferred from it.  The green line would be a more realistic depiction of what a reaction coordinate would look like, but again the data is completely fabricated.  Most importantly, there is no significance to the difference between the maximum of the green line and the crossing point of the red and blue lines.  And most certainly, neither one is an intermediate.  I have not build in any intermediates into this hypothetical mechanism.  It's a simple "reactants --> TS --> products" process.