Talk:Uranium trioxide/Archive 1

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Source scans

As mentioned on Wikipedia talk:Requests for mediation/Depleted uranium and related articles, there are now scans of new source articles in:

 http://www.bovik.org/du/scans/

--James S. 05:42, 23 February 2006 (UTC)

UO3 solid and gas

I moved the discussion of UO3 gas into comment temporarily until I can look into the literature. Anyway, gaseous UO3 seems to be a minor aspect of the basic chemistry of the material uranium trioxide, which is what readers will be seeking in this article. The next order of business might be to describe the solid state coordination environment of UO3 and its simple hydrates, as summarized in "Structural Inorganic Chem." --Smokefoot 18:10, 24 February 2006 (UTC)

UO3 molecular structure

Hi Nrcprm2026 it looks like you are editing UO3 article e.g. by inserting "UO3 bond angles are about 90 and 180 degrees, with bond lengths between 1.2 and 2.2 Angstroms; the narrower uranium ion is in the central position." The phrasing is unusually non-chemical, and more specifically the bond angles and distances also seem unexpected. I was wondering if this is a joke or something?I was going to edit this silly phrase but I get the feeling that there's a joke going on here. --Smokefoot 00:44, 25 February 2006 (UTC)

The joke is hereWikipedia:Requests for mediation/Depleted uranium and related articles/UO3 vapor and the punch line is hereWikipedia:Requests for mediation/Depleted uranium and related articles. --DV8 2XL 01:07, 25 February 2006 (UTC)

No joke; replacing. --James S. 09:31, 17 March 2006 (UTC)

DV8 2XL and Smokefoot are two different people

Why does Smokefoot claim DV8 2XL's edit? --James S. 05:16, 25 February 2006 (UTC)

Hey Nrcprm2026: Sorry you got me there, I am not a wiki-expert - and am making only slow progress with the syntax - so I dont know what a sockpuppet is, but I figure it is not a compliment. Now that I have your attention, what is a "narrow atom" and where did you get the structural data on UO3, like the extraordinarily short U-O distance? I am a boring inorganic chemist who just wants to contribute data and mainly reactions in context of other oxides, chlorides, and sulfides. I had to pull student from UO3 work because I thought that the history of that article became too strange. --Smokefoot 05:30, 25 February 2006 (UTC)

Are you the same person as DV8 2XL? And, what do you think the variance of the bond lengths are at STP? --James S. 05:49, 25 February 2006 (UTC)

No I am smokefoot and inorganicker, I have observed DV8 2XL making edits on UO3 that looked ok to me. At STP, UO3 is a rock, i.e. a polymeric coordination solid with a high coordination number. As others know better than me, UO3 is a particularly complex material since apparently several polymorphs exist. I was hoping to describe the structure in the UO3 article but this gas-phase stuff keeps appearing, and I am not familiar with the recent chemical physics of that molecule, if it even exits. Also it appears that UO3 is a politically lively topic, which really complicates any discussion of the geeky inorganic chem, so I am going to let you folks battle this one out. --Smokefoot 06:03, 25 February 2006 (UTC)

Given the good faith questions that this raises in a mediation of which I am a part, I think I'm going to ask for this question to be reviewed by an administrator with CheckUser privleges, although given the edit history it looks pretty obvious what happened here. What a tangled web we weave.... --James S. 08:10, 25 February 2006 (UTC)

In fact I insist that you do James; then I will expect your apology here and on smokefoots, mine and your talk pages --DV8 2XL 15:01, 25 February 2006 (UTC)

Why do you suppose Smokefoot described an edit that you made? --James S. 18:11, 25 February 2006 (UTC)

I suspect, James because it was right. Now, do the check user thing and apolgise to both of us. --DV8 2XL 19:42, 25 February 2006 (UTC)

A simple check on "User contributions" would have permitted an editor with good faith to assure themselves that these two editors are different people. I have had a CheckUser request run, and it shows that they edit from different countries. I simply cannot tolerate the harrassment of editors who are prima facie not involved in the dispute under mediation: given the choice between protecting the articles and taking the dispute back to the Arbitration Committee, I have chosen the latter. Physchim62 (talk) 14:44, 26 February 2006 (UTC)

If you think my suspicions were unfounded, then why did you run a checkuser to begin with? There is no question that Smokefoot described DV8 2XL's edit as his own, is there? --James S. 17:18, 26 February 2006 (UTC)

Hallelujah. And thanks for everything you've tried to do to bring reason and civility to this issue, Physchim62. Dr U 15:11, 26 February 2006 (UTC)

Combustion product of uranium

Some uranyl oxide gas is produced when uranium burns in air.

U₃O₈ -> UO₃ + U₂O₅ --James S. 09:33, March 11, 2006

• Rubbish. Stopped signing your posts James? --DV8 2XL 20:22, 11 March 2006 (UTC)

What equation do you think describes the reaction from Ackermann (1960)? When I edit from the library, I don't log in; so much for "supported and encouraged." --James S. 05:59, 12 March 2006 (UTC)

Well according to History for this page you were logged in. Never the less the saddest thing about this is that we probably see eye-to-eye on many aspects of the DU ammunition topic. In particular I absolutely agree that aerosoling uranium combustion products, and permitting spent ammo to corrode on the ground, is unlikely to have any beneficial effects on the environment or the health of those living there. But your increasingly shrill tones and insistence on the existence and the role of this gas serves only to cloud the issue and undermines your credibility. --DV8 2XL 11:34, 12 March 2006 (UTC)

Hm; I did forget to sign here on the talk page; I thought you were referring to article edits. Anyway, you avoided the question. What do you think is the reaction describing production of the gas? Why are you so upset about the existence? You might as well get over it; the Army has already decided that they had better get around to measuring the gas condensates pretty soon. --James S. 12:30, 12 March 2006 (UTC)

Ackermann (1960) contemplates an equilibrium:

n/3U₃O_8(s) + n/6O_2(g) <--> (UO₃)_n(g)

at the very high temperatures and very low pressures in the experimental apparatus they were using. You cannot extrapolate this to any potential reaction at the combustion temperature of uranium metal at 1atm. --DV8 2XL 13:37, 12 March 2006 (UTC)

So, based on Wilson (1961), what would you expect the distribution of the atomic weights of oxide particles including monomers to look like from a uranium flame at STP? It's not whether, it's how much, and you know it. --James S. 14:13, 12 March 2006 (UTC)

Undetectably small, undetectably short-lived, and undetectably close to the reaction site; if present at all. --DV8 2XL 14:29, 12 March 2006 (UTC)

I guess that depends on the resolution of your detector. You didn't answer the question. The question was about the probability distribution of the atomic weights of the oxide particles. Mean? Median? Normal? Lognormal? Uniform? Skew? Kurtosis? You wouldn't need to guess if you looked up the earlier peer reviewed citations. Or will you ask me to scan them in for you? --James S. 15:16, 12 March 2006 (UTC)

If you really understood Ackermann's work or the equilibrium equation (or in fact how probability theory is applied in chemistry) you would be asking this nonsensical question.

However if your ego will be sufficiently salved by having me admit that a few isolated molecules of UO₃ might exist (as a monomer) for a picosecond or two at the focus of a high kinetic energy, high temperature event with a uranium impactor, then fine. That they contribute to detectable UO₃ solid residues, or they are a factor in the exposure profiles of humans, or that they are in any other way an important phase of this molecule at STP, is not - just not - supported by anything or any reference you have put forward. --DV8 2XL 16:07, 12 March 2006 (UTC)

Is that your honest or facetious characterization of the combustion product distribution? I'm not saying UO₃ won't condense; even by glomming to other particles, but is "a few isolated molecules of UO₃ might exist for a picoseconds or two" where you really want to be standing on this question? Which of the most recent references I have put forward in this article have you read? --James S. 16:39, 12 March 2006 (UTC)

Yes that's the best your going to get out of me. This exchange has gotten tiresome and is now closed. --DV8 2XL 16:49, 12 March 2006 (UTC)

I removed the inappropriate references; the case is proven by Wilson (1961) quoting Ackermann (1960). --James S. 19:57, 12 March 2006 (UTC)

Gas

Nothing that you have posted in the way of reference speaks to this gas to anyone except you James. Nothing. Please leave this article as is untill you do find proper citations, and establish that the do in fact support your views. --DV8 2XL 14:29, 13 March 2006 (UTC)

On the contrary, the evidence is clear in Wilson (1961), at the bottom of page 213:

[Ackermann et al. (1960) show that] U₃O₈ crystals result from the two step process:

1/3 U₃O₈(s) + 1/6 O₂(g) -> UO₃(g) at T₁

UO₃(g) -> 1/3 U₃O₈ + 1/6 O₂(g) at T₂

Where T₂ < T₁

This isn't worth blanking talk pages during arbitration over, it's just a physical fact. --James S. 21:02, 13 March 2006 (UTC)

Bei 1200 bis 1800 K (not 300 K) verdampft festes U3O8 in Gegenwart von [molekularen] Sauerstoff zu gasfoermigem, vermutlich monomerem UO3. Gmelin Handbuch der anorganischen Chemiek_, 8th edition, volume U-C2 (1978), page 118,

B. Salbu, et al., in "Oxidation states of uranium in depleted uranium particles from Kuwait," Journal of Environmental Radioactivity, vol. 78, no. 2 (October 2004) pp. 125-135, found spectrographic evidence of hexavalent uranium ions and UO3 particles (no gas) in an enclosed pyrophoric uranium munitions burn.

R.L. Gilchrist, J.A. Glissmyer, and J. Mishima, "Characterization of Airborne Uranium from Test Firings of XM774 Ammunition," PNL-2944, Richland, WA: Battelle Pacific Northwest Laboratory, November 1979. The combustion products were described thusly, "About 75% of the airborne DU was U3O8, and 25% was UO2." NO UO3 http://www.gulflink.osd.mil/du_ii/du_ii_tabl1.htm http://www.gulflink.osd.mil/du_ii/du_ii_tabl2.htm No mention of "uranium trioxide" or "UO3" is made on either of those summary compilation pages.

The summary for M.A. Parkhurst, J.R. Johnson, J. Mishima, and J.L. Pierce, "Evaluation of DU Aerosol Data: Its Adequacy for Inhalation Modeling," PNL-10903, Richland, WA: Battelle Pacific Northwest Laboratory, December 1995, No mention is made of the uranium trioxide.

Mitsakou, et al., "Modeling the Dispersion of Depleted Uranium Aerosol," Health Physics, vol. 84, no. 4 (2003) pp. 538-544, and R.E.J. Mitchel and S. Sunder, "Depleted Uranium Dust from Fired Munitions: Physical, Chemical and Biological Properties," Health Physics, vol. 87, no. 1 (2004), pp. 57-67; neither find any form of hexavalent uranium at all.

I find no hint for gas at normal temperature in any of these articles! --Stone 10:49, 14 March 2006 (UTC)

The above-cited literature neglects UO3 at room temperature. So where is the proof for UO3 at room temperature and normal pressure? Ackermann and Nakajima are no proof one is high temperature the other low pressure. Mass spectroscopy is done at very low pressure! --Stone 16:50, 14 March 2006 (UTC)

I have added some additional references to the article to address these concerns. Indeed, before widespread availability of lasers, mass spectroscopy was done in near-vacuum. --James S. 11:42, 17 March 2006 (UTC)

What is lasers mass spectroscopy? Never heard of it? MALDI is ionisation! There are only three MS I know of section flight TOF and ion trap and all need Vakuum!--Stone 11:48, 19 March 2006 (UTC)

Raman spectroscopy [1] --James S. 19:24, 19 March 2006 (UTC)

Melting Point

With a melting point of 500°C there is no gas at 400°C during the synthesis!--Stone 16:50, 14 March 2006 (UTC)

The solid decomposes before it melts. If I knew the specific heat capacity, then we could learn the temperature at which the gas decomposes. It's above the burning temperature. --James S. 11:39, 17 March 2006 (UTC)

Mond process

Normal methode for purification is chemical transport reaction. Ackermann writes it in his paper that he uses it do purify the U3O8 and to get crystals. There is tonns of literature on chemical transport reactions. ( Holleman Wieberg, Lehrbuch der Anorganischen Chemie 1985, for the Mond process)--Stone 16:50, 14 March 2006 (UTC)

Please see below. --James S. 11:39, 17 March 2006 (UTC)

Decomposition of UO₃ to U₃O₈

I was looking at WebElements UO₃ page. Under melting point, it says decomposes to to U₃O₈ - does anyone know what temperature this happens at?

Following the link to the U₃O₈ page, I find the m.p. is given as 1150 °C; 1300 °C decomposes to UO₂. And UO₂'s m.p. is given as 2827 °C. It would be nice to have this information stated simply in a section or in the infobox. I'm scared to touch this article, though, for all the heated debate I've seen on the talk page and elsewhere.

If this info is accurate, there should be a sequence of events like this:

UO₃(s) → (at ?°C) → U₃O₈(s) → (at 1150°C) → U₃O₈(l) → (at 1300°C) → UO₂(s) → (at 2827 °C) → UO₂(l)

I'm finding this article a bit confusing to read. Can the basic chemistry (that all textbooks agree on!) be near the top of the article, leaving the row about UO₃ gas for the bottom of the page. I know I'm only a student and I don't have seven PhD's like most of you who contribute to this article, but I'm sure many, if not most, people who look for uranium trioxide on Wikipedia aren't experts and would prefer the simple information first.

Surprisingly enough, the temperature that you want is not known with certainty. The best I value I can see for the moment is somewhere between 150 °C and 600 °C, depending on your belief in different sources. U3O8 itself is reported to start losing oxygen at 800 °C (Gmelin): NIST data gives an equilibrium temperature of 182 °C, although other direct experimental studies suggest that solid UO3 (probably with oxygen defects) is at least metastable at higher temperatures. Physchim62 (talk) 17:09, 15 March 2006 (UTC)

Sections removed as unsupported

I found the following sections to be unsupported so I removed them:

Uranium trioxide molecules are a postulated intermediate in the chemical transport reaction forming U₃O₈ crystals. This reaction is closely related to other chemical transport reactions, for example the Mond process for purifying nickel

Solid UO₃ decomposes at temperatures greater than 150 °C releasing O₂ to afford high surface area UO₂

--James S. 23:28, 14 March 2006 (UTC)

What do you mean they are unsupported? James they are supported by your very own references. Maybe you should take the time to read them yourself. --DV8 2XL 01:05, 15 March 2006 (UTC)

"postulated" is too weak for the spectrographic detection cited. "Closely related" seems too strong -- the only relation seems to be metal reactions in a gaseous solution. (So, would all aqueous reactions be "closely" related because they involve liquid water -- water is more specific than metals in general?) The decomposition temperature of UO₃(s) has been claimed here, over the past month, at widely divergent temperatures, and I'd just like to get a source reference for that temperature, please. --James S. 16:29, 15 March 2006 (UTC)

You are not talking sense. Many editors from the Chem project have looked at your citations and found you in error. You are going to loose this fight for the simple reason that you are wrong. Why you continue to assert these untruths in the face of so much evidence that you are misinterpreting your own sources is beyond me. --DV8 2XL 17:00, 15 March 2006 (UTC)

Please try to remain civil and refrain from making personal attacks. I have added a link to the "keep a cool head" page to the top of this talk page which may help. --James S. 17:03, 15 March 2006 (UTC)

Uranium monoxide and other potential gas decompositions

Sometimes there's a section in this article saying UO₃ cannot decompose to UO + O₂ because UO is electrovalently impossible. However, in this reference from bovik.org, the very same page that gives the formula 1/3 U₃O₈(s) + 1/6 O₂(g) → UO₃(g) also says "small amount of UO(g), UO₂(g)" - therefore, whatever electrovalence has to say on the issue, UO seems to exist. —Preceding unsigned comment added by Benjah-bmm27 (talk • contribs)

Please have a look at the four thermodynamic parameters on page 98 (pdf page 116) of this reference. Those are up-to-date. Note the different sign for the free energy of formation of UO. I guess I meant impossible in the statistical sense, as in, very unlikely, as opposed to very likely for UO₃(g) and UO₂(g) formation. Do you agree? --James S. 16:29, 15 March 2006 (UTC)

Uranium monoxide is known in the gas phase (the sign of the free energy change is temperature-dependent!), but nobody is suggesting that it is a major product. There are other decomposition pathways for gaseous UO3 which do not involve UO; 2UO3 → 2UO2 + O2, for exmaple, without even considering solid state interactions. Physchim62 (talk) 17:17, 15 March 2006 (UTC)

This is fine, I totally agree. I just felt the phrase 'electrovalently impossible' could be replaced with something else. Thanks for explaining, guys. Ben 17:27, 15 March 2006 (UTC).

At what temperature are gas collisions likely to provide 620 kilojoules per mole? --James S. 21:59, 15 March 2006 (UTC)

That depends on the relative contributions of enthalpy and entropy. Still, as I said above, nobody is suggesting that uranium monoxide is a major product. NIST data suggest that it is very difficult to produce. Physchim62 (talk) 13:14, 16 March 2006 (UTC)

See also Specific heat capacity. --James S. 12:05, 17 March 2006 (UTC)

Guido M, Balducci G (1991). "Identification and stability of U₂ O₂ U₂ O₃ U₂ O₄ gaseous oxides molekules". J Chem Phys. 95: 5373–5376.

Most stable in gasphase is UO₂--Stone 14:16, 16 March 2006 (UTC)

Ackermann (1960): pressures of sublimation detection contrast with derived parameters

Please try to keep from confusing the pressures of UO₃(g) sublimation in near-vacuum for late-1950s spectrography used to derive thermodynamic formation data with UO₃(g) formation from ionic plasma after combustion. The two are entirely different. For one thing, sublimation is a surface effect (and surface is proportional to the inverse of the cube of particle size -- and Ackerman et al. (1960) used micron-scale particles) while gas formation from ionic plasma is a dissolved solution reaction, with the maximum possible surface area in effect. The thermodynamic energies of formation for UO₃(g), UO₂(g) -- both negative net -- and UO(g) -- which is positive -- are shown in the OECD reference which Cadmium found. Thank you. --James S. 17:11, 15 March 2006 (UTC)

Sorry. I'm only doing my best. Ben 17:16, 15 March 2006 (UTC).

Sublimation is only kinetically related to surface area, not thermodynamically: the equilibrium vapour pressure is independant of the surface area of the particles. You can't beat thermodynamics by increasing the surface area!

Why don't you calculate a figure for the highest possible concentration of uranium trioxide gas—you now have all the data you need to do so. Physchim62 (talk) 13:21, 16 March 2006 (UTC)

The surface is irrelevant for the equilibrium only the speed for finding the equilibrium is faster with larger surface.--Stone 14:08, 16 March 2006 (UTC)

Lets try it:

The Born-Haber cycle is important for the rest of the discussion.

If there are to ways to get from reactants to the product, the sum of energy of all reaction steps is the same.

Sodium reacts with water giving sodium hydroxide solution and hydrogen the hydrogen is reacted with chlorine and then dissolved in the sodium hydroxide solution resulting salt water. (the energy is XXX kJ/mol

Sodium reacts with chlorine to form Sodiumchloride and than dissolved in water resulting salt water. (the energy is YYY kJ/mol)

With the Born-Haber cycle XXX = YYY kJ/mol.

This is fact and there is no way around this fact! --Stone 14:08, 16 March 2006 (UTC)

Sorry, I'm confused. Could you please start article(s) on the cycle? Links fixed; will read. --James S. 20:12, 17 March 2006 (UTC)

Oxidation

Uranium is oxidized to Uranium trioxide and than sublimated to form Uranium trioxide gaseous. AAA kJ/mol

Uranium is evaporated and than oxidised in gas phase to Uranium trioxide gaseous. BBB kJ/mol

Uranium and oxygen form a plasma and while cooling oxidised in gas phase to Uranium trioxide gaseous. CCC kJ/mol

AAA=BBB=CCC

It does not matter which way to get to the Uranium trioxide gaseous all energies are the same. --Stone 14:08, 16 March 2006 (UTC)

AAA is not a number because solid decomposition occurs before sublimation. --James S. 12:00, 17 March 2006 (UTC)

Ok! But the other numers are all the same and with this there is no UO3 in gas phase because the energy gained by condensation is high! --Stone 20:45, 17 March 2006 (UTC)

Uranium trioxide gaseous

The reaction from Uranium trioxide gaseous to Uranium trioxide solid is a simple sublimation reaction. DDD kJ/mol

The way this sublimation is going makes no difference for the energy. Energy for sublimation is DDD and for condensation is –DDD.

If the energy for sublimation is large the energy for condensation is also large too. The low vapour pressure at 2300K of 0.34 Pa indicates clearly that the energy for sublimation is high. With this high energy the energy for condensation is also high.

Statement: The energy for sublimation is high. (or not?)

This lead to fast condensation to the solid Uranium trioxide.

If there is a compound with a low vapour pressure or high energy for sublimation and forms a gas as a predominant species over a broad range of temperature and pressure this would harm hurt or kill the Born Haber Cycle and a perpetuum mobile would be in accessible in an easy way.

There is no chance that with 0.34 Pa at 2300K (or is this number also under discussion?) that there is more Uranium trioxide is in gas phase at lower temperature. Condensation reaction takes place at any particle in the atmosphere till equilibrium is reached. This equilibrium has to be on the side of Uranium trioxide not the gas phase. --Stone 14:08, 16 March 2006 (UTC)

What are the mean distances a uranium trioxide molecule in air would traverse, linear and total? --James S. 12:03, 17 March 2006 (UTC)

Within the dense cloude of an Uranium fire mm not more than cm.--217.185.17.145 11:04, 19 March 2006 (UTC)

Source? Show your work? --James S. 19:27, 19 March 2006 (UTC)

You have not showed any proof for your gasous UO3 will be in atmosphere for weeks either! So what! There are only three MS articles showing evidence for UO3 and all other show nothing, even Gmelin does not give anything about it.--Stone 07:17, 20 March 2006 (UTC)

Please review the diffusion and settling rates of particles on the scale of five angstroms with the CRC Handbook's particle dispersoid chart. Water condenses, too, but still in many places the humidity remains above zero. Gmelin specifically states that the "taking up of oxygen by U₃O₈" is "not infrequently ignored," citing Ackermann (1960.) Your edits show that you have either not read or that you have failed to understand the texts to which you refer. --James S. 21:15, 20 March 2006 (UTC)

Please try to keep from confusing the pressures of UO3(g) sublimation in near-vacuum for late-1950s spectrography used to derive thermodynamic formation data with UO3(g) formation from ionic plasma after combustion.

The reaction is somewhat different but the reaction which is important is the reaction back to the solid phase which is the same and hase the same energy. It is not important where you come from! --Stone 11:53, 19 March 2006 (UTC)

Mond process

Chemicla transport reactions of Chlorides Fluorides and oxides are often used for purification so: The mond process was used by Ackermann to purify the U3O8! So why is it not a chemical transport reaction which is closely related to the mond process? greizbirnbamholastauden --Stone 10:26, 16 March 2006 (UTC)

Could you please explain why you think the mond process is more relevant than any other gasous solution reaction?

What does "greizbirnbamholastauden" mean? --James S. 11:54, 17 March 2006 (UTC)

Mond process is the only chemical transport reaction I found in the Wikipedia. Sublimation is a physical gas phase transport reaction. Where the solide has the same composition than the gas phase and the then forming gasphase.

Chemical transport reaction: the substance undergoes a chemical reaction to a other species and at a point with a different temperature the reversereaction takes place and the same substance as in the beginning is there.

For me reading the chemical transport reaction is what takes place in the reaction chamber discribed by Ackermann. And for not having a chemical transport reaction the link to the[[Mond process] is the best we have. --Stone 18:45, 17 March 2006 (UTC)

Are we missing the point here?

This is an encyclopedic article on uranium trioxide. That, under laboratory conditions, or in the center of a flame some molecules with enough kinetic energy to be theoretically defined as a gas may exist for a moment is really not germane to a general treatment of this compound. Only one editor wishes to keep this discussion in this article to support a contention elsewhere on Wikipedia that this gas is a factor in human exposure to uranium. If it is to remain here then this sub topic must be given a full treatment detailing exactly and in no uncertain terms the physical conditions that this gas forms, it's relative volumes, and it's short lifespan; or the entire passage removed as non notable. --DV8 2XL 15:15, 16 March 2006 (UTC)

I guess we are. We can say that the compound has such-and-such a vapor pressure (the 1960 Ackermann paper gives an equation) and that it is monomeric in the gas phase. Anything beyond that is more trivia and less useful fact. UO₃ will form when uranium metal combusts in air and it will rapidly attain equilibrium with (i.e. convert to) U₃O₈ as the combustion products cool down. 129.215.195.81 20:09, 24 March 2006 (UTC)

Which has been the whole argument from the beginning of this farce. There is one editor that wants to make a case that clouds of this gas are floating about the battlefield after the use of depleted uranium munitions and that it is responsible for undetectable exposure to uranium in the human body. in his own words:

" UO3 gas remains dissolved in the atmosphere for weeks, but as a monomolecular gas is absorbed immediately upon inhalation, leading to accumulation in tissues including gonocytes (testes) and white blood cells, but virtually no residual presence in urine other than what might be present from coincident particulate exposure."

Make of that what you will. --DV8 2XL 20:28, 24 March 2006 (UTC)

Discussion from mediation

UO3 vapor; Total inhalation exposure; Teratogenicity; Neurotoxicity; Carcinogenicity; and other questions from "Can the value of a human poison be known without knowledge of its long-term effects?" to "Does the oxygen gradient in a fire modify the effective surface area of burning particles by a scalar value?" (history.) --James S. 08:54, 17 March 2006 (UTC)

References needed

Reference needed for:

The uranium trioxide is shipped between processing facilities in the form of a UO₃ gel. In the jargon of the uranium refining industry, the chemical solution containing the concentrated uranium trioxide is called "OK liquor". Upon heating, this material liberates ammonia, giving UO₃

Take your pick - Google: "OK liquor" +uranium --DV8 2XL 09:47, 20 March 2006 (UTC)

Looks Good! --Stone 20:00, 20 March 2006 (UTC)

Reference needed for:

UO₃ bond angles are about 90 and 180 degrees, with bond lengths between 1.5 and 2.2 Angstroms; the narrower uranium ion is in the central position.

That is from Simon Cotton's book, if I remember correctly. Do your sources show different bond angles and/or lengths, or do you have any other reason to doubt them? Verification of sources is a shared responsibility, and I have done my part. --James S. 23:13, 20 March 2006 (UTC)

Gaseous monomeric UO₃ is produced by combustion of uranium metal in air from 2200-2800 Kelvin Ackermann RJ, Thorn RJ, Alexander C, Tetenbaum M (1960). "Free Energies of Formation of Gaseous Uranium, Molybdenum, and Tungsten Trioxides". J Phys Chem. 64: 350–355.{{cite journal}}: CS1 maint: multiple names: authors list (link) THE REFERENCED ACKERMANN STARTS FROM U3O8 NOT URANIUM

Indeed, but as the French thermodynamic data shows, UO₃ formation liberates more energy than any other oxide. The source-supported facts that U₃O₈ is the dominant (75%) particulate combustion product, combined with the fact that combustion takes place well above the 2000 Kelvin which Ackermann claims is the top of the range of the UO₃(g) production fully supports the existence of the gas as a combustion product. --James S. 23:13, 20 March 2006 (UTC)

As combustion product UO3 may be pressent, nobody will disagree with it, but condensation to U3O8 and UO3 is the only way the stuff can go from there! To fly around for weeks is not possible!--Stone 08:14, 21 March 2006 (UTC)

Mouradian EM, Baker L jr (1963). "Burning Temperatures of Uranium and Zirconium in Air". Nucl Sci Engen. 15: 388–394. THERE IS NO STATEMENT OF UO3 IN THE ARTICLE --COMBINATION OF THE ARTICLES GIVES A GOOD SUGGESTION; BUT AN EXTRAPOLATION FROM 1000-1300°K FROM ACKERMANN TO 2200°-2800°C IN MOURADIN IS NOT ALLOWED IF OXYGEN LOSS IS POSSIBLE-SECOND POINT: THE PRESSURE OF 10 EXP -7 ATM AT 1300°C EXTRAPOLATION TO 300°K GIVES 10 EXP -53 ATM WHICH IS REALLY NOT MUCH

First of all, it's 1000-2000 Kelvin for the Ackermann reaction. The Mouradian and Baker (1963) paper is cited strictly for the combustion temperature curve in Figure 6. --James S. 23:13, 20 March 2006 (UTC)

So there is no paper indicating that at 2000°C the UO3 is the major product in combution or? Akermann states that at his condition the pressure of UO3 is somewhat 10 -7 atm.--Stone 08:14, 21 March 2006 (UTC)

The production of UO₃ gas vapor is "not infrequently ignored" (Gmelin vol. U-C1, p. 98). UO₃(g) molecules condense. IF THE "not infrequently ignored" UO3 WOULD BE A MAYOR POINT AS MAYOR PRODUCT IF URANIUM IS BURNING IN OXYGEN. IT WOULD BE MENTIONED IN THE PAPER: Salbu B, Janssens K, Lind OC, Proost K, Gijsels L, Danesi PR (2005). "Oxidation states of uranium in depleted uranium particles from Kuwait". Journal of Environmental Radioactivity. 78: 125–135.{{cite journal}}: CS1 maint: multiple names: authors list (link)

"Thus, environmental or health impact assessments for areas affected by DU munitions should take into account the presence of respiratory UO2, U3O8 and even UO3 particles, theircorresponding weathering rates and the subsequent mobilisation of U from oxidized DU particles." THE OTHER OXIDES ARE ALSO NOT VERY VOLATILE AND THEY ARE FOUND IN THE RESIDUE, So WHERE HAS AL THE UO3 GONE.

It condenses into U₃O₈ crystals, film, and very small, nearly molecular particles. Still it travels through the air further and faster than the large, partially-burned uranium dioxide dust, which, remember, is only 25% of the particulate combustion product. Did you even read what you copied? "assessments for areas affected by DU munitions should take into account the presence of respiratory ... UO₃ particles...." --James S. 23:13, 20 March 2006 (UTC)

Why should a UO3 gas travel faster than a small nearly molecular particle? Brown molecular movment is fully controled by statistic collision of particles and gives a summ of movment vectors of 0. The gas diffusion is slow either. Convection from the fire and wind are the only transport reactions at all. So why insist on the fast traveling molecules which are in fact not mor volatile than the molecules.--Stone 08:14, 21 March 2006 (UTC)

Kinetik or thermodynamic? Thermodynamical U3O8 would be the major product, but with fast cooling UO3 should also form in significant amounts as kinetic product of condensation reaction. Studies show that nearly no UO3 is found when DU burns (Kuwait studies).

If you would stick to the respiratory UO₃ particles everything would be OK. They disolve more readyly in the lung and have a bigger effect than the unsoluable U3O8 and UO2, but the gas phase is simply wrong.

Wilson WB (1961). "High-Pressure High-Temperature Investigation of the Uranium-Oxygen System". J Inorg Nuclear Chem. 19: 212–222. WILSON STATES: "U3O8 to UO3, are not included, since they are beyond the compositional range of the present investigation." WHY CITE HIM THEN.

Yes, and then it goes on to show their production at the bottom of page 213. I put the Wilson (1961) paper online in full, and everyone can see it. --James S. 23:13, 20 March 2006 (UTC)

But only as citing the ackeman paper which is already cited. Cite cited stuff is nort OK with science! primary literatur OK secondary literature OK but to cite primary and than cite secondary which only cites back to the primary only giving an impression that there is not enough hard date from research to support the fact!--Stone 08:14, 21 March 2006 (UTC)

Guido M, Balducci G (1991). "Identification and stability of U₃O₃U₃O₃ and U₃O₃ (g) gasous oxides molecules". J Phys Chem. 95: 5373–5376. GUIDO STATES THAT UO2 IS THE PREDOMINANT SPECIES IN THE MASS SPEC FOLLOWED BY UO THAN THE UO3: "corresponding to corrected ion intensities, in pA, of 1102, 20702, 285, 0.6,0.45, and 0.6 for the UO₊, UO₂₊, UO₃₊, U₂O₂₊, U₂O₃₊, and U₂O₄₊ ions, respectively" WHY IS THAN UO2 THE MAIN COMPOUND IN THE GASPHASE?

Beats me, that isn't consistent with the thermodynamic data, the particulate measurement data, or anything else. I don't recall reading that paper. Where did you find it? I would like to read that in context. --James S. 23:13, 20 March 2006 (UTC)

The paper makes assumptions on the different MS papers and gives a good clue what MS of UO3 is about. If you like to read it no problem!--Stone 08:14, 21 March 2006 (UTC)

Individual UO₃ gas vapor molecules will not decompose below the burning temperature of uranium in air, because uranium monoxide requires additional energy to form, as does the release of O₂ by a single UO₃ molecule. (Hoekstra and Siegel 1958; Wanner and Forest (2004) p. 98.) Other literature: Nakajima K Arai Y (2001). J Nuclear Mater. 294: 250–255. {{cite journal}}: Missing or empty |title= (help); Text "title Mass-spectrometric investigation of) UO₃ (g)" ignored (help) Chatillion C, Defoort F, Froment K (2005). "Mass spectrometric critical assessment of thermodynamic data for UO₃ (g)". J Phys Chem Solid. 66: 379–382.{{cite journal}}: CS1 maint: multiple names: authors list (link)

Intrinsic stability of molecules is only important if no other molecules are pressent which is not true for air. The oxygen and the carbondioxide and the other molecules are available and wil react with the U03 which has not enough coordinated oxygen atomes arround to be satisfied.

THE NITROGEN COMPOUNDS SHOULD GO TO THEIR OWN PAGE, AS THE URANIUM TRIOXIDE IS NOT THE MAIN PRODUCT IN AIRIAL COMBUSTION OF URANIUM. IT SHOUL GO TO THE U3O8 WHICH IS THE MAIN PRODUCT OF COMPUSTION. Uranium-nitrogen salts UN_x, where x usually is one, form above 800 degrees Celsius. S. Cotton (1991) Lanthanides and Actinides (New York: Oxford University Press) also writes, on page 127: "Aerial oxidation of any uranium compound eventually results in the formation of a uranyl compound."--Stone 20:00, 20 March 2006 (UTC)

Nitrogen compounds forming at the combustion temperatures are pertinent to discussion products produced of burning in air, which is 80% nitrogen. I don't think there is a Uranium-nitrogen combustion products article yet. --James S. 23:20, 20 March 2006 (UTC)

Than but it on the page for Du or somewhat else but this page is about UO3 not the combustion products of Uranium.--Stone 08:14, 21 March 2006 (UTC)

Copied from ArbCom page by DV8 2XL - of more value here

Born Harber cycle

Please leave these calculations alone, if you want to comment please do so in the sub section marked comments. I am making a point of doing the calculations in plain sight.

For the conversion of uranium metal to the monoatomic gas at STP, DeltaH = 533.0 +- 8.0 KJ mol-1

If we assume the reaction 2U + 3O2 to make 2UO3 then we can assume that the oxygen atoms (arrnaged in molecules are already in the gas phase. So no heat is required to form the oxygen gas.

Using Hess's Law, as the Delta H of formation of UO3 gas (from metallic U) is listed at -799.2 KJ mol-1, we can assume that the energy for the conversion of the monoatomic gas to the UO3 gas is 1332.2 KJ mol-1. This has allowed us to get to the point where we can draw a simple Born Haber cycle. This is rather mundane.

The next step is more interesting, assume that a uranium fire occurs in air (20% O2 and 80% N2). For the purposes of the calculation assume that the uranium oxide product is in the form of a gas.

2U + 3 O2 + 12 N2 --> 2 UO3 + 12 N2

We can expect that the reaction will generate 799.2 KJ mol-1

If we assume that no heat is transfered (adiabatic) system the it is possible to eastimate the flame temperture. I hope to goodness that this does not start a flame war, dire pun

The molar heat capacity of UO3 gas listed in my fav. source is 64.5 (plus minus 2) J K-1 mol-1

[2] states that the molar heat capacity of nitrogen is 20.8 J K-1 mol-1

So the heat capacity of 2 UO3 + 12 N2 will be (2 x 64.5) + (12 x 20.8) J K-1 mol-1

Which is 189.3 J K-1 mol-1 for each mole of uranium which burns. Call this Cx

Each mole of uranium which burns to make the gas, will give out 799.2 KJ (Delta H)

T = 298 K + (Delta H/Cx) = 298 + 4222 K = 4520 K

This is going to be very hot, and could cause uranium oxide gas to form for a moment. As the mixture is cooled by mixing with more air, the uranium oxide will then condense. The calculation which I have done is for a rare case of uranium being vapourised and then oxidised, this would happen in reactive spluttering where in an argon/oxygen gas at low preasure uranium atoms are spalled off of a surface by plasma discharge. So I think that it is reasonable to assume that uranium trioxide gas can be formed (for a moment) by burning uranium under a set of idealised conditions, but I expect that it will rapidly condense onto a surface.

Anyone who wants to is welcome to check my maths:) Cadmium

Comments on the Born Harber cycle

Flame temperature data. --James S. 20:14, 17 March 2006 (UTC)

2570K is the highest combustion temp that I can see in this reference. Of what pertenance to Cd's calculations can this be? At best it would seem to imply that any UO₃(g) that's formed from compustion in open air wouldn't make it past the flame boundary before condensing, if that. --DV8 2XL 21:55, 20 March 2006 (UTC)

The point is that it's above the 2000 K point at which U₃O₈ continues to burn. --James S. 23:23, 20 March 2006 (UTC)

Not at 1 atm. - it decomposes at 1573.15 kelvin - to UO₂ which is a solid to 3100.15 kelvin. Numbers aren't there James. --DV8 2XL 00:28, 21 March 2006 (UTC)

If that were the case, then UO₂(s) would comprise more than 25% of the particulate combustion product, and U₃O₈ less than 75%. What source are you using? --James S. 00:37, 21 March 2006 (UTC)

A lot of the final product will come from the cooler parts of the combustion enveolpe - this is more due to the physics of the flames geometry than the chemistry. I got the numbers from my Rubber Bible. --DV8 2XL 02:13, 21 March 2006 (UTC)

So, what does your rubber bible say that U₃O₈ decomposes into, and what source does it give for that temperature? --James S. 16:41, 21 March 2006 (UTC)

The decomposition product is UO₂: the temperature is consitent with the NIST thermodynamic data for the two species (even if there are far too many sig. figs.!). Physchim62 (talk) 16:31, 22 March 2006 (UTC)

Kinetic and thermodynamic products at different reaction conditions can differ a lot! Thermodynamic reactions can be favoured but because of the reaction partners or a sharp drop in temperature where the activation energy for the exothermic reaction is no longer present they can be kinetically hindered. So U3O8 or UO3 what are they kinetical or thermodynamic products and at what temperatures under what conditions. Without all the data, asuming which of them is in the burningproducts in gas phase of DU, is speculative or even missleading! The only thing which is shure is that nobody ever detected UO3 over burning DU (Yes Ackermann showed it but only as best presumption he could make, because he had no MS of his gas phase). Only as endproduct of the burning as solide. So state this and leave the rest as speculation, what it realy is!--Stone 09:59, 21 March 2006 (UTC)

Which is why this whole line of inquiry is perilously close to violating WP:NOR, Like Stone is saying; withhout hard clear references we are moving into specualtion. --DV8 2XL 15:15, 21 March 2006 (UTC)

In fact, Salbu et al. (2005) did detect UO₃ in combustion residues. This is not speculation or original research, the fact of UO₃(g) is supported by an abundance of sources from the scholarly and peer-reviewed scientific literature. The irony of being asked for "hard clear references" is palpable. It is clear from the edits to this page that those wishing to downplay or suppress the reality of uranyl compounds as aerial combustion products of uranium have no better familiarity with the scientific literature than some of them do with the English language. It is not my responsibility to prove verification of sources I have already verified, and I doubt it is my responsibility to try to understand the tortured ambigious spellings and grammar as have recently become so plentiful here. However, that doesn't mean that I won't continue to try, because I think this is important, and the practice is good preparation for real-time debate, and it's like shooting fish in a barrel. --James S. 16:41, 21 March 2006 (UTC)

In fact, Salbu et al. (2005) did detect UO₃. This is fully right and this fact was never disputed! But this is not a gas but a solid particles as you can read in the abstract! So this is no proof for the gas, give proof for gas at normal conditions!--Stone 16:58, 21 March 2006 (UTC)

Actually, this is disputed: Salbu's results are equally, if not more, consitent with U₃O₈ as the solid product formed from the munitions dump fire. Physchim62 (talk) 16:31, 22 March 2006 (UTC)

Again, Wilson (1961) p. 213. The statistical dynamics involve the number of gas molecules unable to glom onto others before reaching open air; this number is some fraction, not zero. --James S. 17:58, 21 March 2006 (UTC)

The pressure of Iron over a pice of iron is not zero eiter but nobody would say it is a amount which has any significance. Even above ice there are a lot of molecules of water present. If going 200°C below freezing point like on titan you find no water in the atmosphere at all. mare than 500°C below the point you can detect it you will find molecules in gas phase but if you inhale all the air from Uranium trioxide for years you would not get more than drinking a glas of mineral water. --Stone 08:15, 22 March 2006 (UTC)

You said something about raman spectroscopy. This is not a MS methode at all it si a spectroscopy methode determine the fibrations and flexing in a molecule, but it is not suitable for determining the molecular mass of molecule. But if you have a good raman spectra of a UO3 at standart conditions this would also qualify as source for UO3 gas. --Stone 08:15, 22 March 2006 (UTC)

Please stop quoting willson, because if you dont stop I have to give back the compliment you made: Do you ever read what you are quating?

Willson has done no research on UO3 he only quotes the ackeman paper nothing more. --Stone 08:53, 23 March 2006 (UTC)

You mean, other than actually burning uranium and measuring the gas condensates by observation of crystals? --James S. 10:17, 23 March 2006 (UTC)

The page 213 you always revere to is basically a quote. And the rest of the paper deals with the real burning products of uranium not mentioning UO3 as significally imortant any more.--Stone 10:42, 23 March 2006 (UTC)

And Salbu's similar description of UO₃-rich condensates? --James S. 11:25, 23 March 2006 (UTC)

As stated in some older publications U3O8 was contributed as UO2+2(UO3) so U03 rich says nothing. U03 solide is not ever questioned the gas is the problem! --Stone 11:35, 23 March 2006 (UTC)

Salbu's description were of particles in which the oxidation state of uranium was +5.5±0.5: this is more consistent with U₃O₈ (oxidation state +5.33) than UO₃ (oxidation state +6). This also fits with the normal description of U₃O₈ as the product of burning uranium in the presence of excess oxygen. Salbu also looked for uranyl carbonate, which would be formed by weathering of U₃O₈: they were unable to verify or to rule out its presence. Physchim62 (talk) 12:55, 23 March 2006 (UTC)

Ackerman equation

Lets start wit the equation it self: can be found in [3]

log p= (((1.821 ± 0.075)· 10⁴)/T) + 6.84 ± 0.58 log p= (((1.856 ± 0.016)· 10⁴)/T) + 6.928 ± 0.103

I choose the first equation because it gives the higher pressure for UO3 gas.

log p= (((1.821 ± 0.075)· 10⁴)/T) + 6.84 ± 0.58

Lets state 300K as the temperature (26°C or so)

putting in the numbers gives:

log p = -53.86

put both sides in the power over ten (native speaker needed)

p = 10^-53.86 atm

p = 1.38 · 10^-54 atm

convert from atm to Pa (1 atm = 101325 Pa)

p = 1.3985 · 10^-49 Pa (N/m²)

Assuming that the laws of ideal gas are good for this problem

(which they are not, because the the condensation and the interaction between the molecules give even lower pressure for real gas at lower temperature than expected by the ideal gas euqaution)

p V= n R T

n = (p · V) / (R · T)

• P is the pressure (pascal)
• V is the volume (cubic meter) lets assume we look at one cubic meter m³
• n is the amount of gas (mole)
• R is the ideal gas constant (joule per kelvin per mole) 8.314472 J · K^-1 · mol^-1
• T is the absolute temperature (kelvin)

n =(1.3985 · 10^-49 N/m² · 1 m³)/ (8.314472 J · K^-1 · mol^-1 · 300 K)

n = 5.57 · 10^-53 mol

so this is the number of mol per cubic meter of uranium trioxide.

1 mol = 6.023 · 10²³ particles

So to get the number of particles in one cubic meter simpl multipie the numbers.

6.023 · 10²³ 5.57 · 10^-53 = 3.35 · 10^-29 particle per m³

Other way to point this numer is you need 2.98 · 10²⁸ m³ to find on single atom of UO3.

Volume of earth 1.0832×10¹² km³ = 1.0832×10²¹ m³ so even if the whole earth would be atmosphere there would be no molecule founnd in it. 600 times earth volume and you find your UO3.

Where is your calculation on this topic?

--Stone 08:53, 23 March 2006 (UTC)

Multiply by the ratio of surface area in plasma to that of Ackermann's micron-scale particles. --James S. 11:25, 23 March 2006 (UTC)

...and the result is the same. Physchim62 (talk) 11:39, 23 March 2006 (UTC)

The pressure is absulutly independent on the surface! It is only the equilibrium reached faster if the surface is higher! A bound of sugar dissolves in a litre of water faster if you grind it, but the amount of sugar in the water will be in the end the same. Mass action there is no surface in this law for solution and condensation. --Stone 11:46, 23 March 2006 (UTC)

Yup, the equilbrium equation tells you the concentration at which just as much of the gas is condensing (per unit surface area) as evaporating. That concentration is independent of the actual surface area available -- once that level is reached, no more gas evaporates on average, because for every molecule that escapes, another one sticks.

It shouldn't be surprising to anyone that a uranium oxide would have low vapor pressure at room temperature: it is both massive and partially ionic (so that electrostatic forces should help it stick to things). zowie 15:43, 23 March 2006 (UTC)

Under the above theory of sublimation-as-combustion, the nearly exclusive combustion product would be uranium(0) metal. Sorry, that is incorrect. Please try again. --171.64.133.83 19:08, 23 March 2006 (UTC)

You're babbling gibberish James. It's over. --DV8 2XL 19:15, 23 March 2006 (UTC)

Please try to refrain from making personal attacks. Let me put this in simple terms: The above discussion, using sublimation equilibrium production, which was used to derive thermodynamic parameters of formation, as evidence of what occurs during cooling after combustion is completely misleading, for the following reasons: First, the weak vapor pressures indicate that less than 1% of the ionized uranium metal would form into UO₂, also essentially no U₃O₈, and essentially no UO₃ in any phase. But we know for a fact that when uranium metal burns, it does form oxides, and does not remain in the metalic state. Therefore, trying to use sublimation effects to predict combustion products is misleading and wrong. It is so sad that so many people persist, but it is so easy to show how wrong it is that I really don't care whether you keep doing it or not. I do agree that it is over: The intellectual and moral bankruptcy shown so plainly in these attempts to pretend sublimation is combustion represent a serious behavior problem on the part of those, including nuclear power and military DU proponents, who are embarrased by the fact that uranium burns to produce poison gas. --James S. 17:52, 24 March 2006 (UTC)

This is not an equation for the formation of UO3, which has nothing to do et all with how much UO3 is in the air after cooling. This is an equation for the condensation and sublimation of UO3. If there is a small amount of UO3 formed (It does not matter how much) after equilibrium you have this pressure of UO3 in the air after all. The formation is done a chemical reaction with a chemical equation this is simple physics!--Stone 06:59, 24 March 2006 (UTC)

The thermodynamic parameters show that UO₃ is a much more likely combustion product than UO₂, which represents 25% of the particulate product. When uranyl oxide gas escapes into open air, the dispersion and settling rates for a 5-6 Angstrom molecule indicate that it might not reach equilibrium (and condense somewhere) for weeks -- or more. And after it eventually does, it is most likely to condense on to dusts, where the uranyl ion is still very soluble upon inhalation, and remains suspended in the atmosphere for additional time. So it is still toxicologically equivalent to soluble uranyl gas even after much of it settles. Do you have any reasons to the contrary other than your own opinions followed by exclamation marks, such as actual reputable sources from the medical or scientific literature? --James S. 17:52, 24 March 2006 (UTC)

O.K.

So is there any shred of a doubt in anyone's mind that UO₃(g) is NOT a factor in uranium combustion in open air? Can we put this to rest, and remove its mention from the article now? --DV8 2XL 19:11, 23 March 2006 (UTC)

Absolutely not. Wilson (1960), Cotton (1991), and the spectrographers following Wilson, all make it abundantly clear that uranium trioxide gas is an aerial combustion product of uranium. You have shown no evidence to the contrary, and your attempts to use sublimation proportions to represent combustion product formation are easily shown as abject fraud. --James S. 17:52, 24 March 2006 (UTC)

You have had every one of your arguments shredded. People have gone to great lengths to explain the science and do the calculations to demonstrate that your contention that this gas is a factor is without foundation. You have not defended yourself with any arguments other than references that further prove you wrong. There is no scientific or rational reason that this topic should be covered in this article. You continued forcing of this issue only serves to underline your own ignorance of science and your contempt for the process here at Wikipedia. --DV8 2XL 18:18, 24 March 2006 (UTC)

Who has countered the equations of Wilson (1961)? Who has countered the plain statement of Cotton (1991)? Have you personally even gone to the library once to research these questions? Simply saying sublimation formation is the same as combustion formation is absurd, no matter how many times it is repeated, or by how many people. --James S. 19:26, 24 March 2006 (UTC)

Following WP:BOLD I have reworded the section to include a discussion of UO₃ but to downplay the gaseous phase (which has been shown to not be persistent). zowie 18:30, 24 March 2006 (UTC)

With all due respect Zowie the article is about UO₃. The Issue is if it exists in a gaseous form and if that gas is formed in any meaningful amounts when uranium metal burns in air. I believe that the discussions above indicate that it doesn't. Any mention of this gas in this article cannot meet Wikipedia's criteria for notability or verifiability and in fact violates the WP:NOR rule because of that. --DV8 2XL 19:03, 24 March 2006 (UTC)

Your beliefs are unsupported by sources. Look at the thermodynamic enthalpies of formation: UO₃ is a much more likley product than UO₂, which is 25% of the solid product. You haven't shared a speck of evidence that the crystalization observed by Wilson (1961) and Salbu et al. (2005) isn't according to the equation given by Wilson. You would rather point to a discussion of vapor pressure? Water vapor condenses at STP, too. Will you claim, therefore, that steam "cannot meet Wikipedia's criteria for notability or verifiability" as a product of boiling water? --James S. 19:26, 24 March 2006 (UTC)

Hmmm... I guess I should chime in then to your original comment -- I have not yet seen (on this page) a convincing argument that UO₃ is not important in uranium combustion -- only that it is not an important final product of that combustion. I thought that part was resolved. I guess I'll step back from this one -- feel free to revert... :-P zowie 19:20, 24 March 2006 (UTC)

I see the problem now - James seems to think that the laws of thermodynamics do not apply to combustion or that UO₃, once formed, will not condense (or both). 129.215.37.180 02:40, 25 March 2006 (UTC)

How long do molecules stay in air?

The amount of dust makes a huge difference, but doesn't affect the solubility very much. There are probably some in your lungs right now. I am having trouble figuring out exactly how many because some of the statistics involve catastrophic distributions (lognornals on integers with deviations or variances greater than their means or modes) and I do not know how to transit confidence interval bounds through them yet. --James S. 20:39, 24 March 2006 (UTC)

So I see you are using the term 'dust' now. Are you ready to put the gas issue to rest and admit that you are and have been in error about it since the beginning> --DV8 2XL 20:48, 24 March 2006 (UTC)

No, the amount of dust already in the air makes a huge difference in how long the individual gas molecules stay in the air, but it doesn't change the fact that they do. When they condense on dust, they can have proportionally more soluble uranyl ion than the same amount of uranium in U₃O₈ aerosol crystals. The temperature at which U₃O₈ stops burning is high, but not as high as the original flame. I see no reason to believe that at least 7% of metalic uranium involved in ordnance combustion does not remain in the form of UO₃ gas molecules for more than a week. I have noticed the conspicuous absence of any reports of the mass of the solid particulates in relation to the mass of the burned metal, from Mishima's work through to the present. --James S. 21:26, 24 March 2006 (UTC)

Ridiculous nonsense from beginning to end. You have no idea what you are talking about, none whatsoever. You haven't understood any of the calculations that have been done up page at your insistence have you?--DV8 2XL 21:38, 24 March 2006 (UTC)

On the contrary, I have. I have been asking for accurate computations, not insufficiently-sourced edits confusing combustion with sublimation. Please refrain from making personal attacks. --James S. 22:36, 24 March 2006 (UTC)

It's not a personal attack to point out that someone is wrong, or confused. You are in the wrong here James, plain and simple, you have been told over and over again by everyone that has looked at this subject. We have all taken great pains to look over your source material and point out where you have gone wrong in your reasoning and understanding of the fundamentals, yet you still claim that your thesis is the right one and that it is properly sourced when it has been clearly demonstrated erroneous. Your are not a scientist, not by station, not by knowledge, not by conduct. You are now simply just another pathetic crank. --DV8 2XL 22:51, 24 March 2006 (UTC)

DV82XL, I once accidentally drove you to a wiki vacation, and for this I am sorry; it was unintentional. However, as a neutral observer in this debate, I find that James S. seems to be able to advance a calm and level-headed argument, whereas the other side seems to descend into name-calling and insults to drive a point home. Please try to maintain a collegial attitude.

As to the science: I know for a fact that lamp black has an absurdly tiny vapor pressure, yet it appears as a combustion product in any candle flame. It will darken cathedral ceilings tens of feet above the candles. You can smell an alter from the door. To change the analogy: have you ever been in a Hindi store that smelled of incense? The smoke from incense is mostly particulate, and that part which is not particulate has a tiny vapor pressure; yet the room can be very pungent. As to the arguments about equilibrium above; I find them to be bogus: the atmosphere is not in eqilibrium. I can smell my neighbors barbeque from a block away. I can smell a forest fire from a hundred miles away. Jungle fires in Malaysia can turn the daytime sky hazy on the other side of the planet: just a few summers ago, I think it was fires in Borneo or Sumatra that drifted smoke over Mexico and Southern US. James S. is making claims about combustion products. They may well be wrong, but a simple application of common sense indicates that they are not obviously wrong; to me "common sense" is on the side of James S. If you wish to debunk his arguments, you have to get a lot more sophisticated than the level of argumentation I see here. linas 03:58, 25 March 2006 (UTC)

First I don't believe your are sorry at all for driving me off - in fact I believe that you have entered this argument at this late date to try and repeat it.

I have called James out for what he is; he has shown absolutely no understanding of what many people have gone to the trouble to show him by calculation, and more telling he has not provided counter arguments using calculation himself, nor are you. He is obstreperously holding on to a position in the face of a monumental amount of proof that he is wrong. Because like you he cares nothing of science, truth or reason beyond the needs of a political agenda.

You also know absolutely nothing about the science or you would be able to follow the arguments that have been posted above. What you are talking about has absolutely nothing to do with the issues here at all since we are discussing the formation of one species, not the distribution of combustion products. And your arguments and examples are even more ridiculous than his --DV8 2XL 05:10, 25 March 2006 (UTC)

Whoa. Time to slow down a bit. DV82XL, I like you and we've had several good interactions before now. But it's definitely time for Wikipedia:A nice cup of tea and a sit down. I'm a pretty competent guy (at least I like to fool myself into thinking so) and I've done my share of equilibrium calculations, but I'm still having trouble following the arguments on this page. James may or may not be out to lunch here, but Linas was pointing out (and now I am too) that you are not making the point you think you are making - at least not as well as you seem to think you are.

I chimed in a bit on the equilibrium calculation above, which seemed to be about whether UO₃ would remain in the gaseous state once formed -- but I'm not clear how we got from that back to UO₃ not forming at all from combustion. It would be helpful (to me at least) if you would please explain what is going on and what points you are trying to assert and defend. I would summarize what I think is going on, but that might not be helpful. Could you keep it to just a few brief paragraphs listing what you see as the main problems with the article as it stands, or what you would like to be in the article in DV8XL's Republic?

It would also be helpful to me if James would summarize what he is trying to assert and defend, in the same way -- I'm trying to understand what is happening, but to be frank I'm sort of lost with the way the arguments have morphed.

Also, for the moment, would you guys please try to keep from wrangling or directly attacking one another? We'll get this all sorted out, I'm sure -- after all it took months but we finally got Electric Universe concept to a somewhat calm state. zowie 05:58, 25 March 2006 (UTC)

History of the page

DV82XL, thanks for the references (on my talk page) to prior history. Again, I'm a little puzzled what is going on here. In the past there seems to have been a strong argument over whether depleted uranium is a hazard, and there was much spirited debunking! On this page, that debate seems to have taken the form of an argument over whether uranium trioxide gas is produced by combustion of depleted uranium in air. That notion (that uranium trioxide gas is a final combustion product) seems to have been debunked rather thoroughly.

But I do not see anything that indicates UO₃ is not produced at all by combustion of metallic uranium - clearly that will take a trip to the library to check some of the references in the debate.

Meanwhile, granting (for the sake of argument only) that UO₃ might actually be produced by combustion (albeit perhaps in solid form as a kind of soot, or perhaps as a low-quantity side product compared to the full oxide), I don't see anything else in the article that seems out in left field. Is that the only thing you're currently concerned about in this particular article (never mind what went on over depleted uranium)?

Cheers, zowie 14:55, 25 March 2006 (UTC)

The issue is and remains the production of gas, not soot, not particulates. Gas. Gas that remains uncondensed for an extended period of time. James says it exists, thermodynamics and the Gas laws say it doesn't, at least not in open air combustion.

However this is all behind us now, as I have joined up with James to make sure that his version of uranium combustion gets the wider distribution in Wikipedia that it deserves. You see as much as I think he is a deluded crank, he has my respect because he has the strength of his convictions and is willing to go to the wall for them. This is a refreshing difference to the cowards that have written long involved arguments here in talk debunking his ideas, but can't bring themselves to support me with a revert or two. Or I might add those that arrive late tisk-tisking me about my demeanor here without even becoming familiar with the issue at hand. --!DV8 2XL 15:12, 25 March 2006 (UTC)

Why do you say that the thermodynamics don't support the gas formation? The enthalpy of formation is almost twice that of the dioxide. Where do you think the U₃O₈ comes from, in light of the fact that its decomposition temperature is below that of Baker and Mouradian's flame temperature? --James S. 15:17, 25 March 2006 (UTC)

Why do you care what I think any more James? I'm on your side now. I'm not going to fight with you anymore about this because it is like talking to a wall. You've won. Congratulations! Now let's get to work and make sure we don't miss a single topic where this gas and uranium toxicology can be mentioned. --DV8 2XL 15:24, 25 March 2006 (UTC)

Guilty as charged, DV82XL, on the tsk-tsking... Sorry if I seemed overbearing. But the current article no longer seems to be about the same controversy that I hear on the talk page. zowie 15:25, 25 March 2006 (UTC)

UO3 Gas. My personal conclusion.

• The formation of UO3 is simply impossible from sublimation. This is shown by the calculations above. (Even James when I read the coments right)
• James still claims that if it is produced by combustion the whole calculation is worth nothing.
• James insists on "In the air tonight, and for weeks"
• Can ANYBODY make clear that the formation of UO3 does not change the following equilibrium calculations.

Simple way to pu it: If water vapor is produced by a fire or other combustion the condensation rate and temperature is the same as if you produce the vapor by subliamtion or boiling. The Born haber cycle states this and makes nothing other possible.--Stone 06:35, 27 March 2006 (UTC)

To the lamp black argumentation! You do exact the same like James! The lamp black is carbon produced by sublimation and part oxidation of organic materials. This is not C (gas). C gas is not stable in any conditions in air. So claiming that C gas is in the air for weeks is simply false! But you can do it as long its not in the carbon page. The lamp black you get from Borneo or from your oil lamp is graphit particles nothing more and nothing less. Particles can travel the world for years nobody will deny the fact and your obvious was never denied, but: THERE ARE NO GAS MOLECULES AROUND! The condensation takes place at temperatures where the equilibrium is reached fast and all the UO3 condenses and it forms small particles which can fly around similar to the gas molecules, but they are no gas and will never be.--Stone 06:46, 27 March 2006 (UTC)

TO James. Where is the literature profe for the 5-6 Angström particles and their non condensation you claim earlier!--Stone 06:46, 27 March 2006 (UTC)

Introduction, toxicology and structure

Introduction

At the moment, the introduction stands as:

Uranium trioxide (UO₃), also called uranyl oxide and uranic oxide, is the hexavalent oxide of uranium. The toxic, teratogenic, and radioactive solid may be obtained by heating uranyl nitrate to 400 °C, or burning uranium in air, producing a gas which condenses at standard temperature and pressure.

In one of its common solid phase forms, uranyl oxide is a yellow-orange powder.

I was reading Greenwood & Earnshaw (p. 1268) - here is their introduction to uranium trioxide:

The only anhydrous trioxide is UO₃, a common form of which (γ-UO₃) is obtained by heating UO₂(NO₃)ˑ6H₂O in air at 400°C; six other forms are also known.

Can we put in the fact that it is γ-UO₃ that is obtained by heating uranyl nitrate at 400°C? --Ben 13:11, 25 March 2006 (UTC).

Sure! --James S. 13:47, 25 March 2006 (UTC)

I think we should move the words toxic and teratogenic to a subsection, because it's confusing to have physical and chemical properties listed in between the effects of UO₃ on humans. To those of you who are thinking 'but UO₃ is so toxic and teratogenic, we must make this clear right at the start of the article', I would ask you to take a look at isotretinoin. I found this from the article on teratogenesis - where it lists isotretinoin as a very strong teratogen. Even though its teratogenicity is very important, it is not discussed in the introduction in its article, but in a subsection further down.

I think this layout would help make the UO₃ article more NPOV, as it would have the same emphasis on tetatogenicity as other articles. I am refraining from making this change immediately because of the great deal of dispute that exists surrounding this article. --Ben 13:11, 25 March 2006 (UTC).

I think the teratogenicity is the most notable aspect of this compound. This is one of only three major accidental teratogens, the others being thalidomide and Agent Orange; please see those to see how they handle the teratogenicity. --James S. 13:47, 25 March 2006 (UTC)

What does "accidental teratogenicity" mean? Uranium trioxide certainly isn't the only or strongest teratogen. As Ben said, the teratogenicity of retinoic acid is well known, diethylstilboestrol has made headlines, and in the laboratory one sometimes encounters polycyclic linear aromatics and lead salts, substances that women who do not take contraceptives are banned from working with. 129.215.37.180 14:53, 25 March 2006 (UTC)

Structure

I've added a PNG of a UO₃ molecule. I'm not sure if it is right. I know that UO₃ is mostly encountered as a solid and as such has a crystal structure not represented by my diagram. I'll see if I can make a model of that too. I'm having trouble finding details of the structure of solid UO₃ - there are seven forms apparently. If anyone can send me a pdb, mol, xyz or other file that iMol can read of the crystal structure of UO₃, I'll happily make a PNG of this and upload it here.

Sorry for the long comment. I'm hoping for a happy ending to the wars on this article! --Ben 13:11, 25 March 2006 (UTC).

I left a comment on your talk page with a link to a pencil diagram I copied on to a photocopy I had handy.

As in the NIST diagram, the bond angles are 90 and 180 degrees. The uranium ion has a narrow radius near the central position, and the lengths are flexible according to the infrared spectra. However, there is an X-ray spectrography method as well.

I, too, hope that everyone settles down. --James S. 13:47, 25 March 2006 (UTC)

The problem with that NIST diagram is that you can't trust it! NIST draws sulfur trioxide as a T-shape, whereas it is really trigonal planar.

Ben 14:04, 25 March 2006 (UTC)

Not sure if a structural diagram makes sense. The most commonly encountered form is the solid, which is polymeric. 129.215.37.180 14:55, 25 March 2006 (UTC)

That's fine. Just let me know, perhaps on my talk page, if diagrams are wanted.

--Ben 15:31, 25 March 2006 (UTC)

A stucture of the solid makes perfect sense because it is the only predominate phase in the normal conditions. I have the structural data of the predominant forms of beta and gamma Uraniumoxide found in literature. To put up a gasphase structure would be not very good, because this could mislead readers to the assumption that mono molecular structure contributes to any major form at normal condition!

Acta Cryst. (1963). 16, 993 The Crystal Structure of γ-UOs BY R. ENGIVKANN AND P. I~. DE WOLFF

Acta Cryst. (1966). 20, 292 The Crystal Structure of High-Pressure UO3* BY S. SIEGEL, H. HOEKSTRA AND E. SHERRY Acta Cryst. (1966). 21, 589 The Structure of β-UO3* BY P. C. DEBETS --Stone 06:55, 27 March 2006 (UTC) UO₃ vs. O₃U

I put back in the NIST reference that was kindly added earlier, as NIST is widely recognized as an authoritative source. If they refer to the molecule as O₃U then it is fair to say that at least a significant part of the scientific establishment calls it that. zowie 16:28, 26 March 2006 (UTC)

I'm not going to change it (sorry I removed O₃U the first time) because a couple of people want it on this article. However, just because NIST's Chemistry WebBook says something, that doesn't mean at least a significant part of the scientific establishment agrees. NIST is widely recognised, but it doesn't follow that everything they publish is authoritative. Look at their page on ammonia for example. No-one ever refers to ammonia as H₃N. The use of the Hill System formulae isn't widespread throughout Wikipedia, and it doesn't really add much to articles, but if you guys are really keen on it, maybe you could start a dicussion at the WikiProject on Chemistry.

All the best

--Ben 16:49, 26 March 2006 (UTC)

Who the hack is NIST? No book I ever had in my hands was having this kind of nomenclatura, why should we put it in wen there is only one source for it. --Stone 07:58, 27 March 2006 (UTC)

NIST are the National Institute of Standards and Technology in Boulder, CO. Chemical Abstracts use the name notation in their printed indices. OK, to write down the formula like this doesn't add new information but won't hurt either. No particular preference. 129.215.37.180 12:13, 27 March 2006 (UTC)

Where is the reference for the structure? This looks not very good for me! Hybridisation and the lone-pairs should also be shown, to get a picture if this structure could be right. Gaspahse spectroscopy for structural purpose was in no paper I found on my search for uraniumoxide gas. Even the the MS-spectroscopy papers do not state any structural data. --Stone 08:03, 27 March 2006 (UTC)

I have removed the structure. When a casual reader sees the diagram he might think that the compound might be molecular when in fact it's a solid and the monomer exists only at very high temperatures. 129.215.37.180 12:18, 27 March 2006 (UTC)

OK, Thanks! The structures in solide indicate UO₂^s+ UO₄^2- as a good guess for the gamma form. --Stone 16:09, 27 March 2006 (UTC)

All of the well-developed chemical compound articles show a single molecule diagram in the chembox. The source is the Cotton (1991) book, and other actinide chemistry books, including the Gmelin Handbook, although I didn't scan those pages in. I replaced the diagram, the word "hexavalent," and the source-supported discussion of why UO₃(g) doesn't decompose by itself: the entropy of formation of UO as well as the Hoekstra and Siegel's U.N. conference paper saying UO, when it is prodded into existence, will absorb O₂ both support the restored paragraph. --James S. 20:30, 27 March 2006 (UTC)

The sentence should either read "hexavalent uranium compounds" or "uranium(VI) compounds". I'd vote for the oxidation number in numerals myself, no one spells oxidation numbers out in words any more. The structural formula as a single molecule is misleading, the casual reader might think that the compound is a monomer when in fact it is a coordination polymer. A description of the coordination polyhedra or a diagram of the solid structure would make sense. James, if you want the structural diagram back in, please provide better arguments. "The others do, too" isn't an argument, besides the others don't, at least not for salts. 129.215.37.8 01:46, 28 March 2006 (UTC)

The structure of crystaline solides like Magnesium oxide and other oxides are the chemicals compared with uranium trioxide. This is no salt like structure.--217.185.48.220 09:28, 28 March 2006 (UTC)--Stone 09:29, 28 March 2006 (UTC)

Sorry man, this is a misunderstanding. What I meant to say with "salt" is that solid UO₃ is a coordination polymer, not a molecular solid. I never insisted that its structure is in any way like rock salt. 129.215.37.8 13:38, 28 March 2006 (UTC)

Sorry, I understand what you mean, but before somebody writes something after misunderstanding it would be a good start for another discussion.--Stone 15:42, 28 March 2006 (UTC)

The paragraph on the stability of UO3 is at least misleading and could be written better. It's true that UO3 is (thermodynamically) stable towards decomposition towards UO2 and O2, and the solid is (kinetically) stable towards loss of oxygen at lower temperatures. However, the 1960 Ackermann paper has solid facts that should go in: gaseous UO3 is in equilibrium with U3O8 and molecular oxygen, and at higher temperatures the equilibrium is towards UO3.

structure

molecule structure

The structure from Gmelin of the molecular UO3 is from deep temperature matrix experiments. UO3 is incorporated of a solid argon matrix which is inreactive against nearly everything at low temperatures. The Raman and IR spectroscopy date suggest that the linear UO₂²⁺ with a O^2- on the side which gives T-shaped molecule. The trigonal planar (D_3d) structure was also proposed but only favoured by calculations. The T-shape is only estimated by the symmetry of the vibrations shown in Raman and IR. (Comment for James: Raman is no MS spectroscopy, it is comparable to IR spectroscopy) The deep temperature argon matrix experiments are the state of the art experiment for instable molecules to be detected. In the matrix the molecules can not react with each other because they are separated by inreactive noble gas molecules und the necessary activation energy is not available at deep temperatures. (kinetically and thermodynamically hindered reaction conditions) The concentration of molecules can be high, because of that Raman and IR are able to detect them.

The calculations are from the Pykko paper, right? No idea how good his basis set was (and with uranium one must take relativistic effects into account!), besides C_2v is D_3d minus the threefold, and I have no clue how easily the molecule is deformed when in a solid matrix. Anyway, I have added a paragraph on the gas-phase structure now.

Structure in solid state

X-ray crystal structure is only possible for the solid and shows for gamma UO3 two types of uranium atoms the hexa (octaedric) coordinated in. The structure can described as UO₂²⁺ UO₄^2-. In beta UO3 three types of uranium atoms are present. Two forming Uranyl with four oxygens in bigger distance. One with a octaedric surrounding and two with a octaedric surrounding connected over one oxygen. Which gives a 2(UO₂²⁺) UO₄^2- U₃O₇^2- as a good suggestion.

I have now put the sum formula for the gamma phase in the infobox. Not great, but better than having a structural formula for the single molevule as before. A diagram of the solid state structure would be best. 129.215.195.81 18:25, 28 March 2006 (UTC)

Ackermann: These observations suggest the possibility that UO3 is the principal species in equilibrium with U3O8 under the experimental conditions. This is weak not a clear statement, at best.

Well, at least they have a vapor pressure. It's a start. 129.215.195.81 18:25, 28 March 2006 (UTC)