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!

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]

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.

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

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.)

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.

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]

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.)