SplendidAcylation
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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:
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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bnull
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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.
Quod scripsi, scripsi.
B. N. Ull
P.S.: Did you know that we have a Library?
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DraconicAcid
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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]
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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Precipitates
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| 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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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 
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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