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Author: Subject: Does the temperature drop when a molecule splits in half?
SplendidAcylation
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[*] posted on 18-5-2024 at 09:18
Does the temperature drop when a molecule splits in half?


Hi,

I have recently been trying to get a better understanding of thermodynamics and kinetics, as such, I have a long list of problems that remain unsolved (by me).

One particularly troublesome problem is the pursuit of an explanation of how entropy manifests itself on a molecular scale.

What happens when atoms/molecules/ions collide that allows the reaction to proceed further (or less far) than would be expected if we solely looked at the enthalpy changes?

Evidently, when entropy increases, the rate of the backwards reaction is slower than we would expect if we considered only the enthalpy changes, but why?


Although many sleepless nights considering these questions have yielded no useful results, many new questions have emerged, the most troubling of which is as follows:


When a particle splits into two (or more) pieces, does the temperature decrease (or increase in the case of the opposite process)?


Consider the following imaginary reaction, where I have used real molecules do make it a little easier to imagine (but the reaction is totally made up):

Br-Br + Cl-Cl <----> 2Br + 2Cl

Here we have started with two particles (molecules) and ended up with four particles (atoms/radicals).

In this imaginary reaction, the enthalpy change is zero, so no energy is released or absorbed:


Screenshot_2024-05-17_09-54-01.png - 6kB

Because the kinetic energy of the products will be the same as that of the reactants, and we have twice as many particles at the product side, this must mean the kinetic energy is spread out between the larger number of particles, and therefore the average kinetic energy of each particle is lower.
Since temperature is a measure of average kinetic energy, the temperature must decrease.

This seems highly odd to me, is it correct or have I made a mistake somewhere?


Furthermore, if this reaction is carried out at constant-temperature (with the alleged cooling effect nullified by returning the reaction mixture to room temperature), would the equilibrium constant be 1 for this reaction, since the activation energy for both forward and reverse reactions is the same?

I suspect the answer is "no" because the reaction is accompanied by an increase in entropy.
This brings me to the question that has been keeping me awake; How does this increase of entropy manifest itself?


On the product side, there are twice as many particles, so collisions would be twice as frequent, however only half of those collisions would be between reacting species, so the greater collision frequency would cancel out...

I of course understand the thermodynamic concept of entropy (about as well as any mere mortal who is hopeless in mathematics could) however I don't understand how it happens... Entropy doesn't just magically increase, there must be something about this reaction that makes the reverse reaction slower than expected, and I can't figure out what it is.


P.S. someone else has asked the same question regarding the temperature drop:
https://physics.stackexchange.com/questions/506252/do-partic...




Thanks in advance!
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[*] posted on 19-5-2024 at 11:23


I want to recommend a book in place of trying to address the issue, so you can see by yourself "where the floorboard creaks". Thermodynamics was the most confusing subject I was subjected to during collegeNote.

"Chemical Thermodynamics", by Maxwell Len McGlashan (https://archive.org/details/chemicalthermody0000mcgl), seems decent enough.

Note: to make things worse, they used H. B. Callen's "Thermodynamics", which is not itself a paradigm of clarity. Old farts... Thermodynamics is a pain in the neck and other anatomical structures. No wonder the great minds behind statistical physics ended up self-deceasing themselves.




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[*] posted on 19-5-2024 at 17:45


If you have Br2 --> 2 Br, the enthalpy change is NOT zero, as it requires energy to break that bond. So, yes, its temperature will decrease.

Picture a hot molecule of bromine. It's vibrating hard, with a lot of vibrational energy, with the two atoms coming in and out rapidly. At some point, the spring (bond) breaks and the two atoms float away from each other. Since the bond broke when the two atoms were at their maximum distance (and didn't have a lot of kinetic energy), they're now moving away from each other fairly slowly. So they have no vibrational energy any more, and no rotational energy (spinning a sphere isn't like spinning a molecule), and low translational (is that the right word?) energy, they are now at much lower total kinetic energy. They're cold.

[Edited on 20-5-2024 by DraconicAcid]




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[*] posted on 19-5-2024 at 19:30


Quote:
Does the temperature drop when a molecule splits in half?


Yes, energy is required to break bonds. I'm currently "slowly" reading this book,
Nature Of The Chemical Bond, I think also suggested by bnull in another thread. Very interesting, thanks!

Even for very unstable molecules, e.g., nitrogen trichloride, a reasonable amount of energy still needs to be put in to break the bonds (N-Cl) (which comes from the environment e.g., uv light, heat). In this case it is the shear stability of the triple nitrogen bond, and thus energy released when this bond forms, which favours the reaction to proceed. Once a small amount of energy is put into the system, the energy released quickly decomposes the rest of the nitrogen trichloride, leading to the explosion observed.

2 Br --> Br2 is favoured at room temperature, but, at very high temperatures, the reverse reaction will be preferred. Somewhere in the middle (but at still a relatively high temperature) the equilibrium constant will be 1.

Energy is related to entropy, if a lot of energy has to be put in to break bonds, this will lower the entropy of the surrounding environment (temperature will decrease). When we produce Br2 at room temperature, the decrease in the number of particles is more than made up for by the energy released when the molecule forms, and thus increase in the entropy of the environment.

This is a good video on entropy, The Most Misunderstood Concept in Physics.
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SplendidAcylation
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[*] posted on 16-6-2024 at 03:38


Thanks for the replies everyone!

I must apologize profusely for the appallingly late response, unfortunately life has been getting in the way of chemistry lately...

@DraconicAcid,

Yes I am aware that the enthalpy change for splitting a bromine molecule into bromine atoms is positive, I was just using it as a fictional example, sorry, in retrospect that was a bit silly!

Thanks for the explanation about the bonds, though, I didn't know it happened like this, that's really cool.


Basically what I was referring to was a theoretical reaction that results in a greater number of particles, but which has an enthalpy change of zero;

A <--> 2B

If we start off with 100 particles of A, each with 1 joule of energy (stupid, I know!).

The particles split (the mechanism by which this happens is unimportant I guess, you can imagine that two particles of A collide, forming four particles of B).


Since the enthalpy change is zero, no chemical energy is released or absorbed by the particles, so the energy is conserved, therefore if we end up with 200 particles of B, we still have a total of 100 joules of energy in the system, which means each particle of B will now have an average energy of 0.5 joules.


Since temperature is a measure of the average kinetic energy of the particles, we could therefore conclude that the temperature would drop.
Is this correct?



In a more realistic example, we could have a pure sample of A at a low temperature, such that the rate of conversion to B is almost zero.

We could then allow the temperature to rise, which would result in the forward reaction taking place, as the reaction approached equilibrium.

We would then see that the temperature would drop (or it wouldn't rise as much as it should) due to this "spreading out" of energy among a larger number of particles, in spite of no enthalpy change.

This seems strange to me, but what do I know :P




As an aside, one other thing; Larger molecules are said to have a higher entropy than smaller ones.

We could imagine a reaction where a smaller molecule rearranges into a longer one, thus increasing entropy.

This would make the reaction more favourable due to the increase in entropy, even though we have the same number of particles on each side of the equation.

I can't quite figure out why this would make a reaction more favourable.

Is it because a larger molecule is less likely to collide in the correct orientation to result in a reaction?

Imagine the extreme scenario of a bare H+ cation, it could collide in any orientation and react...

I believe this is called "steric factor" and when I search for "steric factor" and "entropy" together I get absolutely no results, which indicates I am completely on the wrong track.

Another problem with this theory is that this would indicate that the thermodynamics depend upon the reaction mechanism (i.e. the way the particles must collide to react) and of course this is nonsense.


Tl;Dr, why is a larger molecule favoured by entropy even though the same number of particles exists at each side of the equation?
What happens at the molecular level to make such a reaction favoured?

[Edited on 16-6-2024 by SplendidAcylation]
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[*] posted on 16-6-2024 at 16:36


I have not read all of this. Just enough to give you a good example that you can experiment with.

NO2 exists in equilibrium with N2O4. The enthalpy change (in that direction) is indeed negative.

NO2 is pretty easy to produce and is of course brown. N2O4 is colourless. So, you can collect a vial of the gas and drop it into ice bath and hot water baths and observe the colour change. (You can also play with the equilibrium using a syringe and changing the pressure. This is more difficult to observe since changing the volume intensified the colour of a gas. But still good to do.)
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[*] posted on 17-6-2024 at 13:49


Quote: Originally posted by SplendidAcylation  

As an aside, one other thing; Larger molecules are said to have a higher entropy than smaller ones.

We could imagine a reaction where a smaller molecule rearranges into a longer one, thus increasing entropy.


How does a molecule rearrange to a bigger one? If you just rearrange the atoms, the molecule stays the same size.




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[*] posted on 17-6-2024 at 15:43


Quote: Originally posted by DraconicAcid  
Quote: Originally posted by SplendidAcylation  

As an aside, one other thing; Larger molecules are said to have a higher entropy than smaller ones.

We could imagine a reaction where a smaller molecule rearranges into a longer one, thus increasing entropy.


How does a molecule rearrange to a bigger one? If you just rearrange the atoms, the molecule stays the same size.


I think the reference is to isomerism. So, dimethyl propane versus n-pentane.
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[*] posted on 17-6-2024 at 16:00


I don't think those will have very different entropies.



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[*] posted on 17-6-2024 at 16:14


I agree. The differences will be marginal. It is just how I interpreted the statement.
Interesting thought experiment though.

I can see that dimethyl propane is more constrained in its movement than n-pentane. It stands to reason that the longer molecule will have greater entropy because it can twist and bend into a greater variety of shapes.

[Edited on 18-6-2024 by j_sum1]
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[*] posted on 17-6-2024 at 19:46


What about something like a carbon bonded to 4 benzene's (around it) or cyclobutane, each corner bonded to a benzene.

When all those benzenes are all split off from the central atom or molecule, does the temperature of the benzene ring actually change?
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[*] posted on 18-6-2024 at 16:17


Here is the experiment I referred to earlier.
https://www.youtube.com/shorts/rO8nIOH4as8

Although if you avoid the low temperatures -- don't use dry ice, but keep it above 2 degrees, you avoid any condensation or solidification or formation of N2O3. Then you can observe the equilibrium between the two gases, which is what you were interested in.
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[*] posted on 7-7-2024 at 03:05


Quote: Originally posted by j_sum1  
I have not read all of this. Just enough to give you a good example that you can experiment with.

NO2 exists in equilibrium with N2O4. The enthalpy change (in that direction) is indeed negative.

NO2 is pretty easy to produce and is of course brown. N2O4 is colourless. So, you can collect a vial of the gas and drop it into ice bath and hot water baths and observe the colour change. (You can also play with the equilibrium using a syringe and changing the pressure. This is more difficult to observe since changing the volume intensified the colour of a gas. But still good to do.)


Thanks for the suggestion! I recall seeing this reaction pictured in a book that I read as a kid, but had since forgotten.

As for the video, it's quite impressive, I didn't realize the colour change would be so dramatic.

Must try this!



Quote: Originally posted by DraconicAcid  
Quote: Originally posted by SplendidAcylation  

As an aside, one other thing; Larger molecules are said to have a higher entropy than smaller ones.

We could imagine a reaction where a smaller molecule rearranges into a longer one, thus increasing entropy.


How does a molecule rearrange to a bigger one? If you just rearrange the atoms, the molecule stays the same size.


Quote: Originally posted by j_sum1  
Quote: Originally posted by DraconicAcid  
Quote: Originally posted by SplendidAcylation  

As an aside, one other thing; Larger molecules are said to have a higher entropy than smaller ones.

We could imagine a reaction where a smaller molecule rearranges into a longer one, thus increasing entropy.


How does a molecule rearrange to a bigger one? If you just rearrange the atoms, the molecule stays the same size.


I think the reference is to isomerism. So, dimethyl propane versus n-pentane.



Yes. The specific example I was imagining was propene undergoing cyclization to cyclopropane, but alkane isomerism also works!




Quote: Originally posted by j_sum1  
I agree. The differences will be marginal. It is just how I interpreted the statement.
Interesting thought experiment though.

I can see that dimethyl propane is more constrained in its movement than n-pentane. It stands to reason that the longer molecule will have greater entropy because it can twist and bend into a greater variety of shapes.

[Edited on 18-6-2024 by j_sum1]





So based on the idea that it can twist into a greater variety of shapes, would it be correct to say that this makes the reaction more favourable because the more bendy molecule is less likely to collide in the correct orientation in order for the reverse reaction to happen?
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[*] posted on 11-7-2024 at 13:48


Quote: Originally posted by j_sum1  
I have not read all of this. Just enough to give you a good example that you can experiment with.

NO2 exists in equilibrium with N2O4. The enthalpy change (in that direction) is indeed negative.

NO2 is pretty easy to produce and is of course brown. N2O4 is colourless. So, you can collect a vial of the gas and drop it into ice bath and hot water baths and observe the colour change. (You can also play with the equilibrium using a syringe and changing the pressure. This is more difficult to observe since changing the volume intensified the colour of a gas. But still good to do.)

You could also play with the equilibrium by changing the concentration at constant pressure, by mixing N2O4 with a different, unreactive gas. Like N2.
Br2 is brown. Br2 is not significantly dissociated near room temperature and pressure. Therefore, if you fill a small part of a long, transparent tube with saturated Br2 vapour, and then let that vapour mix with air in the rest of the tube:
  • the pressure/volume in the tube should not change
  • if the temperatures of vapour and air are initially equal, they should not change on mixing
  • the intensity of colour seen across the tube should dilute
  • but the intensity of the colour seen along the tube should remain unchanged

If, however, you fill the small part of the tube with N2O4 vapour, and do the same:
there would be equilibrium
N2O4<->2NO2
the vapour of NO2 would be brown, like Br2
but:
mixing the vapour with air would move the equilibrium further towards NO2;
if the tube is held at constant temperature, the gas inside should expand;
if entry of heat is hampered, the vapours should cool;
the intensity of colour, as viewed along the tube, should increase.


[Edited on 11-7-2024 by chornedsnorkack]

[Edited on 11-7-2024 by chornedsnorkack]

[Edited on 11-7-2024 by chornedsnorkack]

[Edited on 11-7-2024 by chornedsnorkack]

[Edited on 12-7-2024 by chornedsnorkack]
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