[Kryptid (OP)]

Let's say that we have a box that is impermeable to all matter and energy. That is, nothing inside the box can interact with anything outside of the box and vice versa. In essence, the box represents a closed system. Therefore, the total amount of matter-energy inside of the box never changes. Inside of the box is a mixture of hydrogen and oxygen gas. The gas is ignited and all of the hydrogen and oxygen burns to form water vapor.

Since this is a closed system, all of the energy released by the reaction is still contained in the box and is not released. The energy required to break one water molecule down into hydrogen and oxygen is the same as the amount of energy released by the formation of one molecule of water from those two gases. Now for my question: is it possible, after some very long period of time, that the thermal energy in the box combined with the random molecular collisions between water molecules could eventually recreate the initial mixture of oxygen and hydrogen gas? After a trillion years? Or even after a googol years? What does this mean for entropy?

If it could eventually revert to its original state, what about more complex objects? If you burned a tree with oxygen in the box to form carbon dioxide, water and ashes, could the tree and oxygen eventually reform after a long period of time?

[DoctorBeaver]

Doesn't fusion require more energy than fission? If so, then the energy released by separating water into hydrogen & oxygen would be insufficient to re-combine them.

[Kryptid (OP)]

1) Burning hydrogen with oxygen is a chemical reaction, not a nuclear one.

2) The idea that fusion generates more energy than fission isn't exactly relavent here. The fission of uranium and fusing of hydrogen are both exothermic reactions. They are not related to one another in the sense that they are opposites (fusing hydrogen doesn't produce uranium nor does the fissing of uranium produce hydrogen). The opposite of hydrogen fusing into helium would be the fissing of helium into hydrogen. The fusion of hydrogen is exothermic and the fissing of helium is endothermic.

[DoctorBeaver]

Maybe fusion is the wrong word. I'm not a scientist and sometimes I say things wrong.

Maybe I should have asked if separating water into hydrogen and oxygen requires (not produces) less energy than re-combining them.

[Kryptid (OP)]

According to the conservation of energy, it must be the same. Otherwise, energy would be created or destroyed. Due to inefficiencies such as heat loss, it may require more energy to split the water using technology, but in a closed system, the matter-energy is still the same before and after a reaction.

[another_someone]

I agree that the energy must balance.  In the real world, the leakages of energy from the system would mean that in reality to need to put more energy in than you got out, because some of the energy you obtained was leaked out of the system.  In this gedankenexperiment we are assuming no leakage.

Ofcourse, the fact that you will obtain as much energy out of the combustion as you will put into the disassociation does not mean you will not need more on a transient basis to trigger the disassociation, even if that extra is energy is later recovered.

I think the probability that all of the water will disassociate into hydrogen and oxygen is highly improbably small, but arguable probably non-zero; but the reality will be that some proportion of the water will continually be disassociating.  The issues that are going to matter are the ambient temperature and pressure, and how evenly the energy is distributed in the box, because you will need enough energy in any one locality in order to feed the disassociation process in that locality, but any energy in one region would have to mean less energy in another region, and so less likelihood that there will be sufficient energy elsewhere to drive disassociation.

[Make it Lady]

I agree that the energy must balance.  In the real world, the leakages of energy from the system would mean that in reality to need to put more energy in than you got out, because some of the energy you obtained was leaked out of the system.  In this gedankenexperiment we are assuming no leakage.

Ofcourse, the fact that you will obtain as much energy out of the combustion as you will put into the disassociation does not mean you will not need more on a transient basis to trigger the disassociation, even if that extra is energy is later recovered.

I think the probability that all of the water will disassociate into hydrogen and oxygen is highly improbably small, but arguable probably non-zero; but the reality will be that some proportion of the water will continually be disassociating.  The issues that are going to matter are the ambient temperature and pressure, and how evenly the energy is distributed in the box, because you will need enough energy in any one locality in order to feed the disassociation process in that locality, but any energy in one region would have to mean less energy in another region, and so less likelihood that there will be sufficient energy elsewhere to drive disassociation.

This is the perfect answer. It is very unlikely that you will get all the water turning back into the original components due to the spread of energy throughout the box. You may get the odd water molecule breaking down but the whole lot is unlikely to get that intense energy needed to reverse the reaction completely.If there is a lot of energy present you will also get changes in both directions happening at once.

[lyner]

The energy released by the combination of Hydrogen and Oxygen molecules corresponds to a specific frequency - most of the photons released will interact with surrounding molecules by giving them Kinetic Energy and resulting in an increase in temperature. The spectrum of energy will be , essentially, black body radiation  corresponding to the final temperature after the combustion. The vast majority of the photons will have much lower energy than is required to cause dissociation of the H and O atoms (the specific frequency which was produced during combination) so the H2O molecules will stay as H2O. There is a small probability that some H2O molecules will dissociate, temporarily but they will re-combine soon.
So, effectively, one frequency has been spread out into a broad spectrum of frequencies; dis-order rules and the lower energy state will dominate.

[another_someone]

The vast majority of the photons will have much lower energy than is required to cause dissociation of the H and O atoms (the specific frequency which was produced during combination) so the H2O molecules will stay as H2O.

Probably so, but we have not been given enough information to say it is so.  We have been told that no energy can leek in or out of the box, but we have not been told how much energy was initially in the box (it must be a non-zero amount of energy, otherwise the hydrogen and oxygen would be in solid form and would be unable to mix).  Thus, without knowing the base energy level, we cannot say what the average energy in the final mix will be, only what the increment in energy would be (hence my caveat about the dependency on the ambient temperature and pressure).

[DoctorBeaver]

I'll just shut up then  []

[Make it Lady]

I'll just shut up then  []

Very wise! This was a physical chemistry question. I love a bit of subject interaction.

[lyner]

we have not been given enough information to say it is so.  We have been told that no energy can leek in or out of the box, but we have not been told how much energy was initially in the box

Yes, you are right; if the temperature were high enough it would all be ions and nothing would combine with anything. I sort of assumed STP to start with.

[Soul Surfer]

as you say the initial temperature of the hydrogen and oxygen gases in the box has not been specified.  let us assume that it is moderate say around room temperature and not very cold or very hot.  As the hydrogen and oxygen reacts to form water vapour the temperature will rise however if the temperature rises enough some of the water molecules will start to dissociate so eventually a stable state will be reached inside the chamber where the number of hydrogen and oxygen atoms forming water and liberating energy equals the number of hydrogen and oxygen atom dissociating and releasing energy.  That is the simplest model.

However the situation will be a bit more complex than this because hydrogen and oxygen generally form diatomic molecules the temperature will be so high that theres a fair chance that some of the atoms and molecules will be ionised and there will probably also be a few  HO molecules as well so the container will be filled with a mixture of  hydrogen atoms molecules and ions oxygen atoms molecules and ions HO molecules and ions H20 Molecules and ions and possibly even a few O3 (ozone) molecules ands ions all in a stable equilibrium

 

[Kryptid (OP)]

Assume the starting conditions are at STP.

[Soul Surfer]

As I said  I just forgot to specify the gas pressure as well.

[Kryptid (OP)]

What if quantum tunneling is taken into account as well? Remember, I'm talking about extremely long periods of time. Let's say we wait for a time period long enough for quantum tunneling of particles to become a significant factor.

*Oh, let's also say that particles cannot tunnel out of the box.

[Soul Surfer]

At STP hydrogen and oxygen to not spontaneously ignite but the reaction does not need much of a spark to start it so even if you did not start the reaction there would be the occsional cases where atoms reacted and the temperature would rise until the reaction wetnt rapidly to the stable configuration.  After that no change would take place  (subject to your total isolation conditions)

[Soul Surfer]

You call it a thought experiment but it is possible to carry it out.  If you made the box out of a highly refractory noble metal like platinum and measured the peak temperature of the gas in the box immediately after the explosion and then raised the box to that temperature the final conditions of your experiment would be realised and the composition of the equilibrium gas mixture measured.

[JP]

I think in any of the cases mentioned above, you'll always have a tiny probability of getting the box back to its initial state.  This is especially true since you're dealing with such tiny systems, small energies, and such long periods of time that the smallest probabilities become important.  This means you'll probably need to use quantum mechanics to get accurate predictions, and quantum mechanics allows weird things to happen: H20 molecules can spontaneously separate into hydrogen and oxygen even if they don't have enough energy to do so. 

What does this mean for entropy?

Entropy basically tells us what the most likely state for your box is: in this case, it is incredibly likely to stay as water, since it'll require a very unlikely series of events in order to revert back to hydrogen and oxygen.  If it takes 1 googol years in order to see the box change to hydrogen and oxygen, for the remaining 1 googol - 1 years you probably saw nearly all water in the box.  Entropy is still useful, since it tells you what you expect to see most of the time. 

[Bored chemist]

Eventually the mixture will reach an equlibrium state where the energy released from the reaction will raise the temperature to a point that the reverse reaction (reformation of H2 and O2) is as fast as the forward reaction.
That temperature is easy enough to find. It's the adiabatic flame temperature for H2 and O2; about 3080K
http://en.wikipedia.org/wiki/Adiabatic_flame_temperature


In principle the whole lot might revert to the starting material but the odds are against it.
Calculating thr real probablity isn't easy so in the great tradition of such things I will calculate them for something simpler.
Imagine a box with a sliding partition down the middle with nitrogen on one side (left) and argon on the other. They are at NPT and just to make thigs easier we can pretend the N2 molecules have the same weight as the Ar (theyr'e pretty close and itf you aren't happy with that we can put the box in orbit so they are weightless).
Then you take the partition away and wait.
After a while the gases mix. How likely are they to "unmix"?
Well, their positions are decided by chance and the chance of any one atom being in, say , the left hand side is 50%. Consider each atom or molecule in turn.
There's a 50% chance that the first nitrogen is in the left hand side. There's a 50% chance that the second nitrogen is too so ther's a 1 in 4 chance they both are. look at a 3rd nitrogen and you have 1 in 8. Keep playing this game for all the molecules in the box. For the sort of size of box you might carry groceries in there will be about 10^23 nitrogen molecules. That means the likelihood of them all being on one side of the box is about 1 in 2^(10^23)
We are talking about pretty thin odds here. A google years doesn't come close.
In case you wondered, I hadn't forgotten about the argon. The likelihood of them all being on the right side of the box is pretty much the same number so the probablity of both things happening is roughly the square of that probablity
I think that's about 1 in 2^(2*(10^23))

OK that's a very rough aproximation to the original question but the odds are so remote that if I'm out by a hundred orders of magnitude, it barely matters. I'd need to know what the box was made of but tunneling might be important. The suposed half life for the decay of protons ( something like 10^35 years) would certainly be a factor.