January 27, 2021, 11:49:07 AM
Forum Rules: Read This Before Posting


Topic: Question on collision theory  (Read 8272 times)

0 Members and 1 Guest are viewing this topic.

Offline Needaask

  • Full Member
  • ****
  • Posts: 186
  • Mole Snacks: +6/-16
Question on collision theory
« on: April 10, 2013, 05:58:04 AM »
Hi :)

I have been thinking for a while about this ever since the redox question. For example in this reaction ZnO(s)+C(s) :rarrow: Zn(s)+CO(g), so in this case for the reaction to occur we would say that the zinc oxide and carbon would have to collide with enough energy so that the bonds would be broken.

However, after reading up on Hess's Law I thought that I could split this reaction into 2 parts 1) Bond breaking and 2) Bond forming

So for 1. Bond breaking we have:
ZnO(s) :rarrow: Zn2+ (g) +O2- (g)
C(s) :rarrow: C(g)
O2-  :rarrow: O(g) +2e

And for 2. Bond forming we have:
Zn2+ (g) +2e :rarrow: Zn(g)
C(g)+O(g)  :rarrow: CO(g)
Zn(g)  :rarrow: Zn(s)

So to associate this with the whole energy level diagrams I drew this: http://postimg.org/image/viwl757t1/ so I was thinking that I don't really need any collisions here? Since in those individual bond breaking/bond forming reactions there are no collisions taking place and instead, only stuff breaking into smaller parts.

So I'm not too sure how the collision theory works anymore..

Thanks for the help :)

Offline Rutherford

  • Sr. Member
  • *****
  • Posts: 1868
  • Mole Snacks: +60/-29
  • Gender: Male
Re: Question on collision theory
« Reply #1 on: April 10, 2013, 07:41:30 AM »
How do you think that the bonds got broken?

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Question on collision theory
« Reply #2 on: April 10, 2013, 11:29:39 AM »
The energy for these must come from somewhere. Either it is heat, or collisions. It doesn't matter whether the bonds you break or form are intermolecular (state change), between nuclei and electrons (ionization/affinity) or covalent/ionic, the energy must come from collisions or heat.

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3026
  • Mole Snacks: +458/-22
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Question on collision theory
« Reply #3 on: April 10, 2013, 09:06:52 PM »
Hess's law works because enthalpy is a state function: the enthalpic change of a reaction does not depend on how you get from reactants to products.  However Hess's law does not tell you what the path of minimum energy is, so it can tell you nothing about kinetics! 

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 Needaask

  • Full Member
  • ****
  • Posts: 186
  • Mole Snacks: +6/-16
Re: Question on collision theory
« Reply #4 on: April 11, 2013, 08:04:07 AM »
The energy for these must come from somewhere. Either it is heat, or collisions. It doesn't matter whether the bonds you break or form are intermolecular (state change), between nuclei and electrons (ionization/affinity) or covalent/ionic, the energy must come from collisions or heat.

Hi thanks for the response :)
But in the bond breaking part of the reaction (the 1. Part) how would there be collisions in ionization/electron affinity or intermolecular bond breaking? Because in these cases I'm thinking that the bonds just break as there isn't anything to collide with.


Hess's law works because enthalpy is a state function: the enthalpic change of a reaction does not depend on how you get from reactants to products.  However Hess's law does not tell you what the path of minimum energy is, so it can tell you nothing about kinetics!

Hi Corribus :)

According to the collision theory the bonds are broken once they collide at the peak of activation energy and they also must have the right orientation. But why can't those bonds just break after enough heat is supplied. Then when we stop heating it and let it cool down, then new bonds are formed? Because even though Hess's Law doesn't tell us what goes on exactly (just learned that haha :) ) it made me wonder why I even require that collision to happen between the reactants.

Because in decomposition we don't need any collision between them (is this true?) and they just break their bonds when enough heat is supplied for activation energy, then when I decrease the temperature (or do I even need to decrease the temperature for the new bonds to form here?) the new bonds are being formed. So I was thinking - since decomposition doesn't require any collision why not for those A+B reactions too?

Thanks for the help ;)

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3026
  • Mole Snacks: +458/-22
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Question on collision theory
« Reply #5 on: April 11, 2013, 10:36:08 AM »
Will a rock break apart on its own?  What if you hit it with a hammer?

There's nothing to say that one bond can't break, and then the separated products going on to react to form products.  It may even statistically happen.  Statistics aren't on your side, though, because this won't be the lowest energy (highest probability) path, which usually goes through some "halfway point" transition state, where bond breaking and forming happens in concerted fashion.

That said, ANY bond breaking requires energy in some form.  In most cases this is heat, which is just molecules colliding with each other.  No bond will spontaneously break. (But see footnote.)  If a diatomic molecule is all alone somewhere in the dark, there will be no energy to break the bond.  Now a bond could break due to collision with a spectator molecule, which could give rise to the mechanism you are talking about, but completely ripping apart a bond is a much more energetically unfavorable process than proceding through a transition state, where bond breaking is "pushed along" by simultaneous bond formation. 

Note that energy for bond breaking doesn't have to come from heat.  It can also come from light, which is why you can get photolysis reactions without collisions.

Footnote: What I said isn't technically true.  Bond can break without having enough energy to do so because of quantum tunneling.  But the probabiliy of this happening is so low in all cases except hydrogen that it's practically negligible.
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 Needaask

  • Full Member
  • ****
  • Posts: 186
  • Mole Snacks: +6/-16
Re: Question on collision theory
« Reply #6 on: April 11, 2013, 01:39:27 PM »
Will a rock break apart on its own?  What if you hit it with a hammer?

There's nothing to say that one bond can't break, and then the separated products going on to react to form products.  It may even statistically happen.  Statistics aren't on your side, though, because this won't be the lowest energy (highest probability) path, which usually goes through some "halfway point" transition state, where bond breaking and forming happens in concerted fashion.

That said, ANY bond breaking requires energy in some form.  In most cases this is heat, which is just molecules colliding with each other.  No bond will spontaneously break. (But see footnote.)  If a diatomic molecule is all alone somewhere in the dark, there will be no energy to break the bond.  Now a bond could break due to collision with a spectator molecule, which could give rise to the mechanism you are talking about, but completely ripping apart a bond is a much more energetically unfavorable process than proceding through a transition state, where bond breaking is "pushed along" by simultaneous bond formation. 

Note that energy for bond breaking doesn't have to come from heat.  It can also come from light, which is why you can get photolysis reactions without collisions.

Footnote: What I said isn't technically true.  Bond can break without having enough energy to do so because of quantum tunneling.  But the probabiliy of this happening is so low in all cases except hydrogen that it's practically negligible.

Hi :) Great reply! I just have a  questions about this.

1) So technically the mechanism I talked about where 1. all the bonds in the 2 or more reactants are broken just by heating them up separately and letting them recombine by themselves is possible but it isn't as favorable as 2. where the reactants collide against each other and form a transition state?

2) But why would that be less favorable than the mechanism I thought should/could occur? Because in decomposition reactions there are no collisions but only bond breaking.

3) Also, I'm having a little trouble understanding the transition state. In my textbook it says that initially it  is bond breaking (the endothermic part of the reaction or the reactants to peak of Ea part) then after that its bond forming (the exothermic part of the reaction or the peak of Ea to products). However, it also stated that the bonds are broken only when the reactants have activation energy.

So how should we explain the energy graph - are the bonds being overcame bit by bit during the rise to Ea and are the bonds formed bit by bit during the drop in energy? Because initially I never had a concrete idea of what was going on throughout the reaction and how the transition state fits in. Hope you can clarify things here :)

4) Lastly, how would we relate these to decomposition reactions? Because in this case I'm not sure if again the bonds are slowly overcame during the rise to Ea and slowly forming back to the products. Again I'm not sure where the transition state fits in because now its just bond breaking..

Thanks so much for the help :) i greatly appreciate it :)

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3026
  • Mole Snacks: +458/-22
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Question on collision theory
« Reply #7 on: April 11, 2013, 10:13:42 PM »
First, when you heat a sample up, that is pretty much defined as increasing collisions. You can't have heat without collisions.  I mean, let's suppose you have a completely isolated molecule.  What does that even mean to heat it up?  It's nonsense.  That why it's very cold in space. ;) 

The only time you can have "no collisions" is at absolute zero, where molecules simple aren't moving.  At nonzero temperatures, there are collisions happening all the time.  It then really becomes a question of which collisions have the energy (and geometry) to bring about a chemical change.  And for a large ensemble of molecules, that comes down to probability.  The macroscale observation will be the one with the highest probability of occurance under the conditions the system is subjected to.

The transition state is a saddle point on a potential energy surface. It's precise geometry is not something you can generally define.  It depends totally on the reaction.  Sometimes it looks more like the reactants.  Sometimes more like the products.  Sometimes it's something 50% in between. 
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 Needaask

  • Full Member
  • ****
  • Posts: 186
  • Mole Snacks: +6/-16
Re: Question on collision theory
« Reply #8 on: April 12, 2013, 01:41:07 AM »
First, when you heat a sample up, that is pretty much defined as increasing collisions. You can't have heat without collisions.  I mean, let's suppose you have a completely isolated molecule.  What does that even mean to heat it up?  It's nonsense.  That why it's very cold in space. ;) 

The only time you can have "no collisions" is at absolute zero, where molecules simple aren't moving.  At nonzero temperatures, there are collisions happening all the time.  It then really becomes a question of which collisions have the energy (and geometry) to bring about a chemical change.  And for a large ensemble of molecules, that comes down to probability.  The macroscale observation will be the one with the highest probability of occurance under the conditions the system is subjected to.

The transition state is a saddle point on a potential energy surface. It's precise geometry is not something you can generally define.  It depends totally on the reaction.  Sometimes it looks more like the reactants.  Sometimes more like the products.  Sometimes it's something 50% in between.

Oh that makes sense :) but what about the probability of having just the bonds being broken without any collision?

(For this question I accept that the bonds break only by collision)
So only when they have Ea, then the bonds gets broken during the collision. But when we draw the energy profile diagram from the reactants, the energy level would increase to the peak of Ea and the drop down to the energy level of the products.

But since the bond only breaks when they collide at the peak of the Ea, I don't quite understand what's going on during the reaction. I was taught that the reactant to Ea peak was bond breaking while the Ea peak to the products was bond forming. But now I'm not so sure about it. Could you enlighten me here?

Thanks so much :)

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3026
  • Mole Snacks: +458/-22
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Question on collision theory
« Reply #9 on: April 12, 2013, 09:48:10 AM »
It's usually a more fluid process than "bond breaks then new bond forms".

If you have A-B + C  :rarrow: A + B-C, the most obvious pathway for the reaction will probably go through a transition state that looks like A --- B --- C, where the bond between A and B is breaking at the same time the bond between B and C is forming.  The transition state is a transient species that doesn't persist for any finite period of time, it's just the midway point between reactant and product. 

In this case C is the colliding molecule.  Some mechanisms go through a collision with a spectator gas, which brings about a quasi-stable intermediate, which then undergoes another collision to bring about the products.  But its very situational, as I said previously.  The only way to know how a reaction actually proceeds (or what a transition state actually looks like) is experimentation or a good ab initio calculation.
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 Needaask

  • Full Member
  • ****
  • Posts: 186
  • Mole Snacks: +6/-16
Re: Question on collision theory
« Reply #10 on: April 14, 2013, 06:32:34 AM »
It's usually a more fluid process than "bond breaks then new bond forms".

If you have A-B + C  :rarrow: A + B-C, the most obvious pathway for the reaction will probably go through a transition state that looks like A --- B --- C, where the bond between A and B is breaking at the same time the bond between B and C is forming.  The transition state is a transient species that doesn't persist for any finite period of time, it's just the midway point between reactant and product. 

In this case C is the colliding molecule.  Some mechanisms go through a collision with a spectator gas, which brings about a quasi-stable intermediate, which then undergoes another collision to bring about the products.  But its very situational, as I said previously.  The only way to know how a reaction actually proceeds (or what a transition state actually looks like) is experimentation or a good ab initio calculation.

Hi :) sorry for the late reply

Ohh so can I explain it like this: when the reactants collide with greater or equal to activation energy, a transition state will form.

But what does the spike to activation energy and drop to products mean in terms of that explanation? Because we always explained that energy is absorbed in the rise to activation energy where bonds are broken and energy is given out during the bond forming process.

But now that what actually occurs is that when the reactants collide with activation energy the a bond is partially broken just as it is partially formed so I'm a bit lost now..

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3026
  • Mole Snacks: +458/-22
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Question on collision theory
« Reply #11 on: April 14, 2013, 09:19:11 AM »
Ohh so can I explain it like this: when the reactants collide with greater or equal to activation energy, a transition state will have a higher probability of forming.
With the modification I made in bold, yes.  Not every collision having the requisite energy will yield a reaction.  Consider A-B-C + D  :rarrow: A + B-C-D.

If D comes "from the right" to form the transition state A----B-C----D, and they have enough energy, then the reaction will proceed.  If D comes "from the left", the transition state A----B-C-----D can't form even if the reactants have enough energy, because the geometries aren't right.

Quote
But now that what actually occurs is that when the reactants collide with activation energy the a bond is partially broken just as it is partially formed so I'm a bit lost now..

Well that's exactly what would happen.  In A + B-C  :rarrow: A-B + C, as A approaches B-C, there is an attraction between A and B that grows gradually as the distance between them shrinks.  Simultaneously, the bond between B and C gradually weakens because electron density is shifted away from it.  The transition state is the halfway point.

While "billiard ball" collisions is a helpful way to visualize things, and indeed it's a way reactions are often modelled, that's just an approximation.  It doesn't quite happen like that in reality.
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 Needaask

  • Full Member
  • ****
  • Posts: 186
  • Mole Snacks: +6/-16
Re: Question on collision theory
« Reply #12 on: April 17, 2013, 05:18:19 AM »
Ohh so can I explain it like this: when the reactants collide with greater or equal to activation energy, a transition state will have a higher probability of forming.
With the modification I made in bold, yes.  Not every collision having the requisite energy will yield a reaction.  Consider A-B-C + D  :rarrow: A + B-C-D.

If D comes "from the right" to form the transition state A----B-C----D, and they have enough energy, then the reaction will proceed.  If D comes "from the left", the transition state A----B-C-----D can't form even if the reactants have enough energy, because the geometries aren't right.

Quote
But now that what actually occurs is that when the reactants collide with activation energy the a bond is partially broken just as it is partially formed so I'm a bit lost now..

Well that's exactly what would happen.  In A + B-C  :rarrow: A-B + C, as A approaches B-C, there is an attraction between A and B that grows gradually as the distance between them shrinks.  Simultaneously, the bond between B and C gradually weakens because electron density is shifted away from it.  The transition state is the halfway point.

While "billiard ball" collisions is a helpful way to visualize things, and indeed it's a way reactions are often modelled, that's just an approximation.  It doesn't quite happen like that in reality.

Hi thanks for the great reply :) and sorry for the late reply (just started my tertiary education on applied chemistry wish me luck :) )

Ohh that clarifies a lot of doubts. So when we draw that spike to activation energy, that only shows that the energy level increases as its temperature increases? The once it reaches as specific temperature and bumps in the right orientation then a the transition state is formed? I have a few more questions regarding the aftermath of that collision.

After which would it be correct to say that as the new bond is formed the old bond starts weaken and break off? Because at the transition state a bond is partially formed as another bond is partially broken? If so wouldn't more energy be required to break that old bond?

But what about other reactions that only involve one reactant such as decomposition? Because I'm not sure about how the reaction would take place now that there aren't any collision..

Also what about ionization, electron affinity and melting/boiling reactions?

For instance in ionization Li :rarrow: Li+ +e so in this case again I don't really know if there should be an activation energy or not..

Similarly for Li+ +e :rarrow: Li I'm not sure if the ion and electron would have to collide with an activation energy.

Lastly, for stuff like NaCl(s) :rarrow: Na+ (g)+Cl- (g) I'm not sure if there is an activation energy spike and vice versa.

Thanks so much for the help and I'm so sorry for the late reply. Was quite busy integrating into the polytechnic.

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3026
  • Mole Snacks: +458/-22
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Question on collision theory
« Reply #13 on: April 17, 2013, 09:50:49 AM »
After which would it be correct to say that as the new bond is formed the old bond starts weaken and break off? Because at the transition state a bond is partially formed as another bond is partially broken? If so wouldn't more energy be required to break that old bond?
Well I'm generalizing of course but I'd say the old bond starts to weaken as soon as the new bond starts to form.  This happens, by the way, even in the case of collisions without enough energy to cause a reaction to go.  Electron densities are always in flux - bonds are always vibrating, getting longer and shorter and longer and shorter.  When other atoms and molecules are in close proximity, this can induce dipole moment changes and other such effects that temporarily change molecular structure and chemical properties. 

To really understand chemistry, you have to realize that all equilibria are dynamic and what we observe experimentally are only ensemble averages.

Quote
But what about other reactions that only involve one reactant such as decomposition? Because I'm not sure about how the reaction would take place now that there aren't any collision..

Some decompositions truly have no thermodynamic activation energy, or the activation energy is so small as to be effectively zero.  (Although, this does not mean they are necessarily kinetically fast.)  However, even in the case of a spontaneous decomposition, there usually is some activation energy because bond breaking is not the only consideration.  In condensed phase, for example, formation and breaking of intermolecular interactions is also a factor, as is solvent reorganization (enthalpy and entropy) and so forth.  While these effects are generally small compared to breaking and formation of true covalent bonds, they are often large enough to impose some kind of thermodynamic barrier to reaction.  Such would be the case of some of the ionization and dissolution processes you listed.
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 Needaask

  • Full Member
  • ****
  • Posts: 186
  • Mole Snacks: +6/-16
Re: Question on collision theory
« Reply #14 on: April 17, 2013, 09:06:37 PM »
After which would it be correct to say that as the new bond is formed the old bond starts weaken and break off? Because at the transition state a bond is partially formed as another bond is partially broken? If so wouldn't more energy be required to break that old bond?
Well I'm generalizing of course but I'd say the old bond starts to weaken as soon as the new bond starts to form.  This happens, by the way, even in the case of collisions without enough energy to cause a reaction to go.  Electron densities are always in flux - bonds are always vibrating, getting longer and shorter and longer and shorter.  When other atoms and molecules are in close proximity, this can induce dipole moment changes and other such effects that temporarily change molecular structure and chemical properties. 

To really understand chemistry, you have to realize that all equilibria are dynamic and what we observe experimentally are only ensemble averages.

Quote
But what about other reactions that only involve one reactant such as decomposition? Because I'm not sure about how the reaction would take place now that there aren't any collision..

Some decompositions truly have no thermodynamic activation energy, or the activation energy is so small as to be effectively zero.  (Although, this does not mean they are necessarily kinetically fast.)  However, even in the case of a spontaneous decomposition, there usually is some activation energy because bond breaking is not the only consideration.  In condensed phase, for example, formation and breaking of intermolecular interactions is also a factor, as is solvent reorganization (enthalpy and entropy) and so forth.  While these effects are generally small compared to breaking and formation of true covalent bonds, they are often large enough to impose some kind of thermodynamic barrier to reaction.  Such would be the case of some of the ionization and dissolution processes you listed.

But actually why would we need to reach activation energy since the old bonds start breaking only when the new bonds start to form so I don't quite get why the energy would be given out. Because its only after we provided energy for activation energy, then the bond starts to break/form simultaneously so that breaking process (while the new bond forms give out energy) absorb energy just as it gives out energy. So would that mean even after providing so much energy to reach Ea more energy is required to break those bonds again? Because in this case it seems that the energy required for Ea is useless because the actual bond is only broken after reaching Ea.

As for the decomposition reaction, shouldn't there still be an activation energy since I have to break those intramolecular bonds? But now its a bit confusing because usually a new bond is formed while a bond is broken at the transition state. While in this case I think there's still both bond breaking and bond forming within the molecule but I'm not entirely sure what's the mechanics of that reaction.

Like in more those 2 or more reactants reaction, we would tell that when they have equal or greater activation energy they would collide with the right orientation to form a transition state where an old bond is broken simultaneously as a new bond is formed. But for one reactant stuff it seems very different to me.

Thanks so much for the help and so sorry if I'm not getting it :) I'm really curious about this because I found out my textbook was oversimplifying things..

Sponsored Links