Has there been an improved understanding of water ?

[exchemist]

No problem with this attitude per se, but there certainly IS a problem with an attitude that fails to learn what is already known - from basic physics and chemistry - before attempting to do so.

The physical state of a substance (whether it is solid, liquid or gas) is a BULK property, resulting from the attractive forces BETWEEN MOLECULES in bulk. To give you an idea of the scales involved, 18g of water has ~ 6 x 10^23 molecules in it, so 1 cc has ~ 3 x 10^22 molecules. Since a nanometre is 10^-7 cm, 1 cubic nanometre of water contains 10^-21 of this amount, in other words about 30 water molecules. Bulk properties only have meaning for amounts of substance where the number of molecules on or close to the surface is negligible compared to the number within the body of the substance and large enough for statistical mechanics to govern its behaviour - say for assemblies of a billion or so molecules or more. This would correspond to a water droplet about 0.03 microns across.

So an experiment designed to react individual atoms of H and O (to produce the H-O radical for example) is not going to provide any information at all about the physical state of any hypothetical fluid made of H-O radicals - or indeed anything else.

On the other hand, if you make a lot of H-O radicals and put them together, they will react (with considerable generation of heat) to produce water (H-O-H) and O=O (oxygen) and/or possibly some H-O-O-H (hydrogen peroxide). This is because the entity H-O is unstable, due to oxygen having an unfilled valence shell of electrons - it's what we call a "free radical". We know this: it's just chemistry.

I can only repeat what I and others have said: water is liquid at NTP due the unusual strength of the intermolecular attractions, which are called "hydrogen bonds".

There do seem to be some aspects of H-bonding that are worthy of further study and it is true that H-bonding arises due to the particular properties of H and O atoms, namely the small size of the H atom and the ability of the non-bonding electrons in the valence shell of bound oxygen to line up directionally (in what we call "lone pairs"). If it is that you wish to study, I suggest reading about the theory of hydrogen bonding - it's a very interesting topic, in my opinion.

But do not fantasise about supposed "mysteries" regarding why water is a liquid, or supposedly mysterious properties of hydrogen and oxygen. That's just Bermuda Triangle stuff.

Supplementary: just found this browsing the internet, which actually has estimations of rate constant of the OH + OH -> H₂O + O reaction, which appears to be the dominant decomposition route:http://www.nist.gov/data/PDFfiles/jpcrd9.pdf

If you know some reaction kinetics, you can work out the rate of decomposition as a function of temperature and pressure, from this. Suffice to say it is pretty rapid. So you can't make a bottle of liquid hydroxyl radicals. As we have all been trying to explain.

 

[river]

No problem with this attitude per se, but there certainly IS a problem with an attitude that fails to learn what is already known - from basic physics and chemistry - before attempting to do so.

Sure

The physical state of a substance (whether it is solid, liquid or gas) is a BULK property, resulting from the attractive forces BETWEEN MOLECULES in bulk. To give you an idea of the scales involved, 18g of water has ~ 6 x 10^23 molecules in it, so 1 cc has ~ 3 x 10^22 molecules. Since a nanometre is 10^-7 cm, 1 cubic nanometre of water contains 10^-21 of this amount, in other words about 30 water molecules. Bulk properties only have meaning for amounts of substance where the number of molecules on or close to the surface is negligible compared to the number within the body of the substance and large enough for statistical mechanics to govern its behaviour - say for assemblies of a billion or so molecules or more. This would correspond to a water droplet about 0.03 microns across.

My idea though has nothing to do with , statistical mechanics

So an experiment designed to react individual atoms of H and O (to produce the H-O radical for example) is not going to provide any information at all about the physical state of any hypothetical fluid made of H-O radicals - or indeed anything else.

You misunderstand my experiment

My experiment is not about individual reactions

It is about how each atom of H and O react to extreme low temperatures separately

In otherwords in completely separate experiments , at extreme low temperatures

Down to where both become liquid , naturally , -256C and -236C

 

[exchemist]

Sure


My idea though has nothing to do with , statistical mechanics




You misunderstand my experiment

My experiment is not about individual reactions

It is about how each atom of H and O react to extreme low temperatures separately

In otherwords in completely separate experiments , at extreme low temperatures

Down to where both become liquid , naturally , -256C and -236C

OK. I think, in view of the misunderstandings, it would be good if you could describe the experiment you have in mind, in a few lines.

The questions I would have are firstly, how will you get atomic hydrogen, (that is H as opposed to H₂), and atomic oxygen, (O, as opposed to O₂) down to low temperatures without the atoms combining to form the normal diatomic molecules?

And secondly, what properties would you then hope to measure and what do you think that would that tell you?

 

[river]

OK. I think, in view of the misunderstandings, it would be good if you could describe the experiment you have in mind, in a few lines.

To have a closed chamber of a single of H

The questions I would have are firstly, how will you get atomic hydrogen, (that is H as opposed to H₂), and atomic oxygen, (O, as opposed to O₂) down to low temperatures without the atoms combining to form the normal diatomic molecules?

Your having a rough time understanding my experiment

They are not together ever

Bringing down both H and O too their cold liquid states are entirely two separate experiments

 

[river]

So why do both experiments ?

Because if either show that in fact H and O show liquid characteristics then we know that both atomically have within them the ability to manifest a liquid state on their own

The investigation then become How and Why , a single atom of either can do so

Which implies that the macro state of each is just coming to a critical mass too the point of observation , only

 

[exchemist]

So why do both experiments ?

Because if either show that in fact H and O show liquid characteristics then we know that both atomically have within them the ability to manifest a liquid state on their own

The investigation then become How and Why , a single atom of either can do so

Which implies that the macro state of each is just coming to a critical mass too the point of observation , only

I think I understand what you propose: 2 separate low temperature experiments, in one of which you prepare a closed chamber containing H (by which I assume you mean separate atoms of hydrogen), whose properties you propose to explore and a second one in which you attempt the same thing for O (meaning separate atoms of oxygen). Correct?

But how will you set up a chamber with separate H atoms in it? Or one with separate O atoms in it?

Surely this is impossible, because H atoms will react (rapidly and with release of a lot of energy) with each other, to produce diatomic molecules of H₂. The same will happen with the O atoms.

How will you prevent this from happening?

 

[Russ_Watters]

Sounds to me like river is saying a single atom, alone in a chamber....not being aware that a single atom cannot ever be described as a "liquid".

 

[exchemist]

Sounds to me like river is saying a single atom, alone in a chamber....not being aware that a single atom cannot ever be described as a "liquid".

That's what I tried to explain, a few posts back, but I seemed not to have hit the mark.

Let's see: there's clearly a big misunderstanding somewhere and I think we are close now to nailing what it is.

 

[wellwisher]

Oxygen and hydrogen in the liquid states exists as two atom molecules, O2 and H2. These do not exist as single H and O atoms. Single H and O atoms is a higher energy state such on the surface of the sun.

One important reason these molecules need to get so cold to liquify, is connected to the additional degrees of freedom associated with being composed of two atoms. A single atom can rotate and translate (move in a line). The binary O2 and H2 can also do this, while also being able to bend and vibrate at the bond. This is easier for H2, because O2 has a double bond.

What this brings to the table is the ability to store structural entropy, making it harder to form the order; liquid. For entropy to increase, energy needs to be absorb. In the case of these molecules, energy is absorbed into these extra expressions of freedom. We really need to lower the temperature to remove the energy within the structural entropy, to form the limited freedom of the liquid. Picture the molecule wiggling in many ways making it hard to pack them until the worms stop moving so much.

Entropy wants to increase but to make these liquids we need to lower the entropy but extracting the energy that is defining the entropy. At the same time, they have all this freedom going on always trying to increase.

 

[river]

Sounds to me like river is saying a single atom, alone in a chamber....not being aware that a single atom cannot ever be described as a "liquid".

Understood

Nevertheless , the experiment could show that a single atom could produce a liquid like state

 

[river]

I think I understand what you propose: 2 separate low temperature experiments, in one of which you prepare a closed chamber containing H (by which I assume you mean separate atoms of hydrogen), whose properties you propose to explore and a second one in which you attempt the same thing for O (meaning separate atoms of oxygen). Correct?

But how will you set up a chamber with separate H atoms in it? Or one with separate O atoms in it?

Surely this is impossible, because H atoms will react (rapidly and with release of a lot of energy) with each other, to produce diatomic molecules of H₂. The same will happen with the O atoms.

How will you prevent this from happening?

From my research homolytic fission should separate H2 into neutral atoms

http://en.wikipedia.org/wiki/Homolysis_(chemistry)

 

[river]

An example of the homolytic process

http://www.avogadro.co.uk/light/fission/bondfission.htm

 

[Russ_Watters]

Understood

Nevertheless , the experiment could show that a single atom could produce a liquid like state

No.

 

[exchemist]

From my research homolytic fission should separate H2 into neutral atoms

http://en.wikipedia.org/wiki/Homolysis_(chemistry)

Sure, but how to split diatomic hydrogen (or oxygen) molecules into separate atoms is not the issue. That's dead easy. The problem is, having split them, how to keep them separate, when the instant one atom encounters another they will bond together once more. That is what I asked you: how will you stop this happening, long enough to measure any properties on "bulk" atomic hydrogen or oxygen.

If, as you now suggest to Russ Watters, measurements would be made on an isolated atom, then the point in my earlier post, also reiterated by Russ Watters, applies. The physical state of a substance is a bulk property. If you do not have a billion or so enough molecules or atoms present and colliding with each other in normal thermal motion, you cannot define a physical state.

It seems as if you do not know what it means for a substance to be liquid. Let me summarise. It means the molecules or atoms comprising it are in thermal motion, colliding and rebounding from each other, but their thermal energy (that is, their chaotic, statistical motion) is not enough to achieve the "escape velocity" needed to overcome the intermolecular attraction between them. Evaporation from a liquid (i.e. the change from liquid to gas) is all about achieving this escape velocity of some molecules from the bulk of the others. With only one atom or molecule you have no intermolecular or interatomic attraction and so there is nothing to "escape" from.

So how on earth are you going to determine whether a single atom, not colliding with anything, is "solid-like", "liquid-like" or "gaseous-like"? It's an utterly meaningless idea.

 

[wellwisher]

To create hydrogen atoms you need to add energy to break the bond within H2. If you maintain this level of energy, as the background energy, as the hydrogen atoms try to recombine, they will quickly reverse allowing hydrogen atoms to remain. But since the difference between hydrogen atoms and hydrogen molecules is the sharing of electrons, this high energy state will have hydrogen atoms ionized into higher energy levels like on the surface of a star. We call this the fourth state of matter; plasma. But plasma is not a liquid.

One way to make liquid hydrogen atoms is with pressure. If we pressurize hydrogen enough, such as within the cores of huge planets, the hydrogen will form a metal.

Metals have all the atom arranged in a crystal lattices, sharing their outer electrons. In the case of hydrogen, we have single hydrogen atoms in a lattice of the metal, sharing its one electron. These also form liquid metal states.

Presently known "super" states of matter are superconductors, superfluid liquids and gases, and supersolids. Egor Babaev predicted that if hydrogen and deuterium have liquid metallic states, they may have quantum ordered states which cannot be classified as superconducting or superfluid in the usual sense. Instead, they may represent two possible novel types of quantum fluids: "superconducting superfluids" and "metallic superfluids". Such fluids were predicted to have highly unusual reactions to external magnetic fields and rotations, which might provide a means for experimental verification of Babaev's predictions. It has also been suggested that, under the influence of magnetic field, hydrogen may exhibit phase transitions from superconductivity to superfluidity and vice-versa

 

[river]

Sure, but how to split diatomic hydrogen (or oxygen) molecules into separate atoms is not the issue. That's dead easy. The problem is, having split them, how to keep them separate, when the instant one atom encounters another they will bond together once more. That is what I asked you: how will you stop this happening, long enough to measure any properties on "bulk" atomic hydrogen or oxygen.

Introduce a barrier

If, as you now suggest to Russ Watters, measurements would be made on an isolated atom, then the point in my earlier post, also reiterated by Russ Watters, applies. The physical state of a substance is a bulk property. If you do not have a billion or so enough molecules or atoms present and colliding with each other in normal thermal motion, you cannot define a physical state.

True but the idea is to find how the atom of both behaves and the properties of both at extreme temperatures

It seems as if you do not know what it means for a substance to be liquid. Let me summarise. It means the molecules or atoms comprising it are in thermal motion, colliding and rebounding from each other, but their thermal energy (that is, their chaotic, statistical motion) is not enough to achieve the "escape velocity" needed to overcome the intermolecular attraction between them. Evaporation from a liquid (i.e. the change from liquid to gas) is all about achieving this escape velocity of some molecules from the bulk of the others. With only one atom or molecule you have no intermolecular or interatomic attraction and so there is nothing to "escape" from.

Sure

I'll refrase , change in atomic properties

So how on earth are you going to determine whether a single atom, not colliding with anything, is "solid-like", "liquid-like" or "gaseous-like"? It's an utterly meaningless idea.

By the study of both atoms , individually

Its not a meaningless idea , I'm simply curious as to how the atom reacts on its own and the interesting thing is that nobody knows

 

[origin]

Sure, but how to split diatomic hydrogen (or oxygen) molecules into separate atoms is not the issue. That's dead easy. The problem is, having split them, how to keep them separate, when the instant one atom encounters another they will bond together once more. That is what I asked you: how will you stop this happening, long enough to measure any properties on "bulk" atomic hydrogen or oxygen.

Introduce a barrier

Well what a genius you are! Any idea how to do this??

If, as you now suggest to Russ Watters, measurements would be made on an isolated atom, then the point in my earlier post, also reiterated by Russ Watters, applies. The physical state of a substance is a bulk property. If you do not have a billion or so enough molecules or atoms present and colliding with each other in normal thermal motion, you cannot define a physical state.
True but the idea is to find how the atom of both behaves and the properties of both at extreme temperatures

It seems as if you do not know what it means for a substance to be liquid. Let me summarise. It means the molecules or atoms comprising it are in thermal motion, colliding and rebounding from each other, but their thermal energy (that is, their chaotic, statistical motion) is not enough to achieve the "escape velocity" needed to overcome the intermolecular attraction between them. Evaporation from a liquid (i.e. the change from liquid to gas) is all about achieving this escape velocity of some molecules from the bulk of the others. With only one atom or molecule you have no intermolecular or interatomic attraction and so there is nothing to "escape" from.
Sure

I'll refrase , change in atomic properties

So you you don't know what it means for a substance to be liquid. The liquid state is dictated by the chemical properties not the atomic properties.

So how on earth are you going to determine whether a single atom, not colliding with anything, is "solid-like", "liquid-like" or "gaseous-like"? It's an utterly meaningless idea.

By the study of both atoms , individually Its not a meaningless idea , I'm simply curious as to how the atom reacts on its own and the interesting thing is that nobody knows

Oh yes it is meaningless. But I have an answer for you. By looking at a single H atom and a single O atom we can deduce that the at all temperatures and all pressures the individual atoms will be in a gaseous state. Since they by defintion will not be in a liquid or solid state with other atoms. Tells you alot, huh? If you weren't so dense you would have realized that the links I gave you had your answer for Hydrogen, you see metallic hydrogen means that it is bonded as H and not H2. At higher pressures you will have liquid metallic H which is essentially a liquid of single H atoms, but again you were not even able to understand that.

 

[river]

Originally Posted by exchemist
Sure, but how to split diatomic hydrogen (or oxygen) molecules into separate atoms is not the issue. That's dead easy. The problem is, having split them, how to keep them separate, when the instant one atom encounters another they will bond together once more. That is what I asked you: how will you stop this happening, long enough to measure any properties on "bulk" atomic hydrogen or oxygen.

Originally Posted by river
Introduce a barrier

Well what a genius you are! Any idea how to do this??

Not yet but I'm sure its possible , don't you ?

 

[river]

Originally Posted by exchemist
If, as you now suggest to Russ Watters, measurements would be made on an isolated atom, then the point in my earlier post, also reiterated by Russ Watters, applies. The physical state of a substance is a bulk property. If you do not have a billion or so enough molecules or atoms present and colliding with each other in normal thermal motion, you cannot define a physical state.
True but the idea is to find how the atom of both behaves and the properties of both at extreme temperatures

It seems as if you do not know what it means for a substance to be liquid. Let me summarise. It means the molecules or atoms comprising it are in thermal motion, colliding and rebounding from each other, but their thermal energy (that is, their chaotic, statistical motion) is not enough to achieve the "escape velocity" needed to overcome the intermolecular attraction between them. Evaporation from a liquid (i.e. the change from liquid to gas) is all about achieving this escape velocity of some molecules from the bulk of the others. With only one atom or molecule you have no intermolecular or interatomic attraction and so there is nothing to "escape" from.

Sure

Originally Posted by river
I'll refrase , change in atomic properties

So you you don't know what it means for a substance to be liquid. The liquid state is dictated by the chemical properties not the atomic properties.

I do I just imaged the atom producing a drop like condensation only atomically

Anyway I did refrase

 

[river]

Originally Posted by exchemist
So how on earth are you going to determine whether a single atom, not colliding with anything, is "solid-like", "liquid-like" or "gaseous-like"? It's an utterly meaningless idea.

Originally Posted by river
By the study of both atoms , individually Its not a meaningless idea , I'm simply curious as to how the atom reacts on its own and the interesting thing is that nobody knows

Oh yes it is meaningless. But I have an answer for you. By looking at a single H atom and a single O atom we can deduce that the at all temperatures and all pressures the individual atoms will be in a gaseous state. Since they by defintion will not be in a liquid or solid state with other atoms. Tells you alot, huh? If you weren't so dense you would have realized that the links I gave you had your answer for Hydrogen, you see metallic hydrogen means that it is bonded as H and not H2. At higher pressures you will have liquid metallic H which is essentially a liquid of single H atoms, but again you were not even able to understand that.

I'm NOT interested in atoms of H but of ATOM of H , do you understand ?