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math
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hydrogen susceptibility to sunlight question
Hello,
wikipedia states that
Hydrogen gas forms explosive mixtures with air if it is 4–74% concentrated. The mixtures spontaneously explode by spark, heat or sunlight.
So if I was about to fill a balloon with hydrogen gas made by the reaction of NaOH with aluminium in a glass bottle, by directly attaching the balloon
at the neck of the bottle, does it mean that inside there will be stored an air-hydrogen mixture and that exposing said balloon to sunlight it'll
explode instead of just flying up?
What'd be a simple fix to this issue, if present? Should I let some of the reaction going, and then attach the balloon to the bottle, so that the air
has been purged out?
Thank you 
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woelen
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I have had hydrogen/oxygen mixes in sunlight many many times and never had an explosion. I think that this is one of Wikipedia's bad errors. I know of
H2/Cl2 mixes which can explode in sunlight, but I've never heard of this property for H2/O2 mixes.
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vmelkon
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I have made hydrogen and air mixtures. I had to apply a spark to cause it to explode. Besides, I think if you do it right, you can have your balloon
with more than 95% hydrogen.
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ChemistryGhost
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Maybe Wikipedia confused hydrogen molecules with hydrogen ions.
"Imagination is more important than knowledge" ~Einstein
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AJKOER
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My research suggests that Hydrogen is most easily ignited with 29% H2 in air (H2/O2 ratio 2 to 1), but explosion occurs at 41% H2 in air (H2/O2 at
3.5). The lower explosion limit in pure O2 is 10%, but ignition is, in general, more difficult at LEL (and, in my opinion, most likely lower than 9/1
in sunlight). Source: http://arhab.org/pdfs/h2_safety_fsheet.pdf ). In fact, another web thread, one contributor suggests in sunlight (no documentation) an 8 to 1 ratio
(Hydrogen to pure oxygen).
Perhaps, before calling Wiki in error on this point, someone should place a calculated amount of NaOCl, NaOH, Aluminum and H2O2 (added last) in
several partially crushed one liter perfectly clear plastic bottles (to allow various amounts of H2/O2 mixtures) in direct (and/or diffused) sunlight.
Allow several days for a reaction/explosion to take place.
Intended preparatory reactions:
NaOCl + H2O2 --> NaCl + H2O + O2 (g)
2 Al + 2 NaOH + 2 H2O → 2 NaAlO2 + 3 H2 (g)
[Edited on 27-8-2012 by AJKOER]
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unionised
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Neither hydrogen not oxygen has any absorptions that are accessible to visible light or even to the UV components of sunlight.
Light that's not absorbed doesn't have an effect.
http://en.wikipedia.org/wiki/Photoelectrochemical_processes#...
So, wiki says that wiki is wrong.
Incidentally, if I were planning to demonstrate this I'd avoid using hypochlorite because adding chlorine (whic is known to be photochemically active)
would mess up the results.
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AJKOER
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To avoid the possible argument relating to small amounts of photochemically active Cl2O or Cl2 from the Bleach initiating any explosion, I agree it
would be better to use MnO2 and H2O2 as the oxygen generator, else one would have to repeat the experiment in the event of a positive result without
the use of NaOCl.
In my opinion, the main question is whether a single hydrogen bond can be loosened by sunlight to start a chain reaction with O2 resulting in an
explosion.
__________________________________
{EDIT] I just find a reference, exploring the explosive chain reaction of H2 and O2. The reference states: "Book considers the example in which the
initiation reaction is photo initiated: H2 --> 2 H rate=v=intensity of radiation=I...."
See page 11 at link:
https://docs.google.com/viewer?a=v&q=cache:13Gw2yMjahIJ:www.southalabama.edu/chemistry/barletta/notes_ch26.ppt+Hydrogen+Oxygen+explosion+ratio+'ch
ain+reaction'&hl=en&gl=us&pid=bl&srcid=ADGEESjn-2m4R3wlmjkJnXNLr0J5FtTE0AnodggTbJCKd7uuIfAS0_tNHZ4wsSOiqXb1qKFUdlBhOkgUAc42rT5D11D4AWd
6JpoGJy6USWujq3JRfxfL9prx4qRmwKnzOkVY-g0oNRom&sig=AHIEtbRtaIfoCdd7YgPYPgyYbZ28zmBHaQ
However, in my opinion, this summary homework reference to "Chapter 26: Chain Reactions" of the unspecified book does not provide a definite yes
answer as the Termination steps (in the H2/O2 chain reaction) may be significant.
[Edited on 27-8-2012 by AJKOER]
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unionised
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Well, I agree with this bit
" the main question is whether a single hydrogen bond can be loosened by sunlight to start a chain reaction with O2 resulting in an explosion."
and I don't see how the answer can be yes.
People use deuterium lamps as a source of UV these days because they are brighter, but they used to use hydrogen lamps.
The UV (at least down as far as 185nm) escaped from the lamp.
So the hydrogen couldn't be absorbing it.
So we know that hydrogen doesn't absorb UV with a wavelength more than 185 nm (it could be smaller).
Oxygen is a slightly better candidate.It does have some absorptions at longer wavelength but they lead to a triplet state rather than to dissociation
and they are very weak (for most cases you can ignore them- oxygen isn't green.)
The UV cut off for dissociation is about 195nm.
Now lets think about sunlight.
It comes from the sun with plenty of hard UV.
But - here's the point- any wavelengths short enough to be absorbed by O2 will be absorbed by O2 in the atmosphere before they get to the ground.
So, no light that's absorbed by oxygen reaches us (because the atmosphere would have absorbed it already)
And no light that could be absorbed by hydrogen could get through, because it's even more readily absorbed than the light that would split oxygen.
So, the light that we see isn't absorbed by either gas.
How can it have an effect?
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Morgan
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Static occurrence
http://www.h2incidents.org/incident.asp?inc=281
[Edited on 27-8-2012 by Morgan]
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AJKOER
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Unionised:
I had to amend my comment based on a recent find.
I am now more in the Yes camp, but not completely.
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unionised
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Morgan,
Good point.
Whatever the pros and cons of the sunlight might be, static is a serious threat.
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AJKOER
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OK, I found the reference I alluded to previously. If correct, it does shed light on the H2/O2 photo initiated chain reaction. To quote Jon Richfield:
"If you put nice, cold, dry, clean O2+H2 in a clean glass vessel in the dark under low pressure, with no catalysts and radiation anywhere, you might
have to wait a few lifetimes for much H2O or H2O2 to show up, let alone a major reaction, flame or explosion.
However, the more you increase the temperature and pressure and radiation, the more spontaneous reaction proceeds, and the greater the prospect of
ignition. Sprinkle a bit of Pd powder, or even put in a strip of Pd metal, and you can be pretty sure of reaction, in fact you might be asking for
trouble.
Just plain dry O2 + H2 in bright sunlight will react slowly (H2+Cl2 would explode) and O2 + H2 at 100C to 200C also react slowly, but significantly.
Get the picture? A spark or an accident could set it off, but there is slow reaction under a wide range of circumstances. If you are talking about
storing the mix together for geological timespans, you might as well label your stores as H2O.
You will be right for almost the whole time."
Link: http://www.last-word.com/content_handling/show_tree/tree_id/...
There is also a reference (see http://jcp.aip.org/resource/1/jcpsa6/v7/i8/p710_s1?isAuthori... ) that supports the non-explosive composition reaction. To quote: "The kinetics of
the upper explosion limit and the nonexplosive thermal reaction of H2 and O2 uniquely demonstrates the formation of HO2 in collisions H+O2+M, its
destruction on surfaces and its ability to propagate chains by gas phase reactions with H2. Below 500° HO2 gradually ceases to be a chain carrier in
the gas phase; assuming its further reactions to be heterogeneous, involving self‐neutralization and reaction with H2, the kinetics of the
mercury‐photosensitized H2☒O2 reaction are well interpreted." See also: http://pubs.rsc.org/en/content/articlelanding/1958/tf/tf9585...
From Jon Richfield, my take is that a photo-initiated chain reaction may proceed slowly forming H2O and usually not explosively. However, the reaction
mix may be more sensitive especially with increased pressure and heat to, for example, static electric denotation and the presence of metals.
Personally, I am comfortable with this scenario and how it integrates known accidents with comments made. The answer, in my opinion, is a slightly
cautious no with respect to the possibility of explosion.
[Edited on 27-8-2012 by AJKOER]
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unionised
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My take on it is that Jon Richfield is clearly talking through his hat if he thinks ordinary sunlight, here at ground level will do that.
In the presence of mercury (or chlorine) the reaction is plausible in sunlight; but oxygen+ hydrogen don't absorb daylight so it can't affect the rate
of reaction. It's slow at room temperature- possibly so slow that initiation by cosmic rays is faster than the thermal pathway but the idea that light
which isn't absorbed, somehow supplies initiation energy is, lets be clear about this,
A breach of the conservation of energy.
Of course, if you are talking about some putative reaction during the creation of the earth - and, in particular, in the higher reaches of the
atmosphere where there's lots of hard UV it's a different story. But that's got little or nothing to do with the OP
[Edited on 27-8-12 by unionised]
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watson.fawkes
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There's a lot of "inconceivable" in this thread. Quote: Originally posted by unionised  | | Neither hydrogen not oxygen has any absorptions that are accessible to visible light or even to the UV components of sunlight. |
That's not true. The transition between triplet (ground state) and singlet oxygen is in the infrared. While this isn't the best
example (forbidden transition in isolation, i.e. in gas phase), it's illustrative of the transition energies between molecular orbitals. These lower
energies, however, do not to bond dissociation nor even ionization, but rather to excited molecular states. Almost all of these excited states are
radicals, with lone electrons in some previously-unoccupied molecular orbitals. The transition of H2 to its first excited state, with one
electron in the σ* state, is at 109 nm, in the UVC range. I couldn't quickly locate the data for the analogous energy for the
excitation in O2 for its transition π2p* → σ2p*, but it's a
longer wavelength than that.
Furthermore, radiation absorption isn't the only way to get excited states. There's a certainly steady-state population of excited states from the
distribution of velocities and occasional molecular collisions at high energies. There's more of these at higher temperatures, so it's unwise to
assume that the kinetic chain that could initiate an explosion is the same one that happens at a shock front. | Quote: Originally posted by unionised | So, no light that's absorbed by oxygen reaches us (because the atmosphere would have absorbed it already)
And no light that could be absorbed by hydrogen could get through, because it's even more readily absorbed than the light that would split oxygen.
So, the light that we see isn't absorbed by either gas. |
This argument is utterly bogus, because every
absorption line is also an emission line. In the atmosphere, molecules do absorb some of the radiation, whereupon they go into an excited state. They
then quickly emit radiation, returning to the ground state.
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arsphenamine
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Direct transition from triplet to singlet is up against a large energy barrier:
singlet 1O2 is ~95 kJ/mol higher in energy than 3O2.
The explanation is quantum chemical, involving selection rules for spin, parity, symmetry, etc.
1O2 generation is usually catalyzed by a metal ion or a dye.
The catalyst may be biogenic/endogenous.
Yeah, you over there with the blue eyes; I'm tokkin ta you.
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watson.fawkes
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Yeah, I hinted at that with
the phrase "forbidden transition". It's the reason that singlet oxygen has such a long lifetime in the gas phase. For such a transition to occur, it (less or more) has to do an internal-double
transition, which is a very much less likely occurrence. The lifetime is around six or more orders of magnitude larger when it's an isolated molecule
in the gas phase than when it's interacting with other molecules of other species.
My point, however, was to estimate the transition energies of O2 excited states, and the singlet state transition gives a lower bound.
Since it's in the infrared, it fails to provide a lower bound that's above the visible band. Indeed, since the first H2 excited state
operates at lower principal quantum numbers, and thus at higher transition energies, it's a pretty good guess that all or most of the O2
electronic transition energies are somewhere in the bands between infrared and UVC, that is, right where solar output is good.
If I had proper MO data, I would have preferred to have just cited it.
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AndersHoveland
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I have never read anything about sunlight being able to ignite mixtures of hydrogen and oxygen.
But shorter frequency UV light does cause a photochemical reaction that can lead to ignition if there is enough intensity of UV light.
| Quote: |
A sharp wavelength dependence on the amount of incident laser energy necessary to ignite a premixed flow of H2/O2 at atmospheric pressure is reported.
This wavelength dependence exhibits a spectral profile similar to the two-photon fluorescence excitation curve for flame oxygen atoms, and the
respective peaks correspond to exactly the same wavelength near 225.6 nm. This similarity clearly indicates that oxygen-atom production and subsequent
excitation is an important step in the efficient (∼0.5-mJ) laser ignition of H2/O2 flows in this wavelength region. In addition, the dependence
of the incident laser energy on the equivalence ratio reveals that the most efficient ignition occurs far into the fuel-lean region
|
Platinum gauze can also catalyze a reaction between hydrogen and oxygen at room temperature. During the reaction, the platinum can reach an
incandescent heat, at which point it can ignite the gas mixture. It takes a long time for the reaction to proceed to the point of ignition, so for
demonstration purposes the platinum is always heated before being placed into the stream of hydrogen. As the catalytic action becomes much faster when
the platinum is heated, this greatly reduces the necessary time it takes for the platinum to reach a glowing heat.
[Edited on 29-8-2012 by AndersHoveland]
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math
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Thank you all for the explanations 
Thinking about the reaction between aq. NaOH and Al, I was thinking if adding a large excess of water (possibly as heat moderator in the exothermic
reaction) would impair the amount of H2 produced.
I think it won't, given H2 extremely low solubility in H2O, but my main concern is that the reaction would proceed so slow that the amount produced is
lost through the balloon rubber at the same rate.
Maybe it's overthinking, but I'd like to use another refrigeration method other than ice.
Thank you
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arsphenamine
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| Quote: Originally posted by watson.fawkes | My point, however, was to estimate the transition energies of O2 excited states, and the singlet state transition gives a lower bound.
...
If I had proper MO data, I would have preferred to have just cited it. |
What is proper MO data for you?
This is interesting stuff.
(edit)
good gloss on 3O2 at Chemogenesis web book
[Edited on 30-8-2012 by arsphenamine]
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watson.fawkes
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Well, to start with, one with all the numbers in place, principally energies
for each molecular state. When I was looking this up, I saw plenty of MO energy diagrams that were qualitative in nature, intended to teach concepts,
particularly the difference between bonding and anti-bonding orbitals. (This includes the link you cited, which I saw in my earlier search.) A proper
diagram would have an upper bar at the top that represents the ionization potential, conventionally set at zero, yielding the convention that the
energy of any particular orbital state is a negative number. In addition, there would be a number of orbital states that are unoccupied in the ground
state but are present for excited states, usually at least two or three such. In addition, atomic spectroscopy diagrams tend to group the angular
momentum quantum number in columns, so that it's easier to see some the selection rules.
Now I did find this page of physical data from NIST, but I couldn't make much sense about it in a few minutes. There's lots of literature citations, but I don't
have immediate access to the literature, and I haven't spent the time working through it. This one shows some promise about being a decent summary of
the observed data: Krupenie, P.H., The spectrum of molecular oxygen, J. Phys. Chem. Ref. Data, 1972, 1, 423.
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unionised
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It seems a little knowledge is a dangerous thing.
"That's not true. The transition between triplet (ground state) and singlet oxygen is in the infrared. "
Kind of. There are two such transitions. About 759 to 765 nm and 687 to 692nm.
Something with a significant absorption at those wavelengths would be green.
You would think someone might have mentioned that.
Oh!, hang on. I did.
Then there's the other thing about them. As Watson helpfully points out.
"These lower energies, however, do not to bond dissociation nor even ionization"
So, once you fill in the missing word, that wraps it up for them being a contributor to the initiation of a free radical reaction. (Unless you are
talking about two photon events, but those are so rare that the generation of radicals from cosmic rays would outpace them.)
So far so average.
"These lower energies, however, do not to bond dissociation nor even ionization, but rather to excited molecular states. Almost all of these excited
states are radicals,"
Not exactly.
The ground state for oxygen is a di radical.
The excited states (the singlet ones) are not radicals.
To be fair, you would have been right for most diatomic molecules.
Then there's "the transition of H2 to its first excited state, with one electron in the σ* state, is at 109 nm, in the UVC range. "
Well, OK I agree with that- hydrogen doesn't do a lot photochemically until you get to 109nm.
I had pointed out that the first absorption must be below 185nm but I hadn't cared how much further down the wavelength scale it was because there's
no UV that short in sunlight.
"Furthermore, radiation absorption isn't the only way to get excited states. "
Quite right!
"There's a certainly steady-state population of excited states from the distribution of velocities and occasional molecular collisions at high
energies. "
Indeed!
Yes, you can get significant populations of higher excited states by heating the gas.
But that's relevant to the question of "can I set off Hydrogen and oxygen with a lit match?"
If there were enough thermally generated free radicals then the reaction wouldn't wait for the sunlight- the materials would ignite on mixing.
That's what happens with fluorine and hydrogen. The F F bond is weaker so there's a much better chance of thermal energy breaking the bond and
starting the reaction (there's also a bigger energy release to help propagate the reaction)
There's another point about thermal energy. It makes molecules bump into one another.
So that brings us to "...every absorption line is also an emission line. In the atmosphere, molecules do absorb some of the radiation, whereupon they
go into an excited state. They then quickly emit radiation, returning to the ground state."
Close, but no cigar. There's another fate for excited states.
Collisional deactivation is the process where two molecules (one in an excited state) collide and bounce of each other but they leave faster than they
came in, and the energy of the excited state is transfered to kinetic energy (and vibrational+ rotational).
That's quite common.
The laws of thermodynamics mean that the sun (which is quite hot) wants to warm up the (rather cooler) atmosphere, and so it does.
It means that if there's a transition that can absorb light it does so overall.
That's why we are screened from UV by the ozone layer. It's why hard UV is sometimes called vacuum UV and it's pretty much the same reason that people
talk about the greenhouse effect. It's even why those wavelengths in the visible/ near IR are demonstrably missing from sunlight (ask Mr Fraunhofer).
They are not strong enough transitions to make a difference in a balloon full of gas, but if you have an entire atmosphere of them you can see them.
What it isn't, is "Bogus".
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AJKOER
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| Quote: Originally posted by unionised | My take on it is that Jon Richfield is clearly talking through his hat if he thinks ordinary sunlight, here at ground level will do that.
|
For the record, Jon Richfield is a teacher with a specialty in Entomology.
I did like his ability to translate some of the literature, obviously outside his realm of specialization, an attribute of a good teacher at that.
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watson.fawkes
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Apparently you don't know what a radical is when you get away from the standard examples. All it means is unpaired electrons. The excited state of O2 where the two
highest-energy electrons occupy (1) one of the two π* MO's and (2) the σ* MO is, indeed, a radical state. Both
electrons are unpaired in their respective MO states. It's also a diradical, since there are two unpaired electrons. Unlike the diradical of the
ground state of triplet oxygen, this diradical state does not have the same angular momentum degeneracy. It's a reactive state, much more so that
ground-state O2.
What you were claiming before is that light not absorbed by the atmosphere can't be chemically reactive. That claim is bogus. | Quote: Originally posted by unionised | So, no light that's absorbed by oxygen reaches us (because the atmosphere would have absorbed it already)
And no light that could be absorbed by hydrogen could get through, because it's even more readily absorbed than the light that would split oxygen.
So, the light that we see isn't absorbed by either gas.
How can it have an effect? |
The state of oxygen I mentioned has transition line to the ground state somewhere
in the visible or near UV. (I wish we knew the figure; it would clarify a few issues.) That transition line does not necessarily lead to a strong
atmospheric absorption band, however. While there are other ways of discharging an excited state to ground state, such as collisions, it does not mean
that every such excited state has a fast enough alternative to emission that would lead to an atmospheric absorption band. A bulk absorption band
requires that there by some other pathway, and that pathway must be fast enough to outcompete the natural emission rate (see Fermi's golden rule) enough to see some bulk absorption instead of mere re-emission.
One thing that's should be clear at this point is that if sunlight can ignite H2-O2 mixtures, it's not going to be mediated by
an excited-H2 state. The transition line is in the upper end of the UVC band, and there's no significant UVC flux (at sea level) that would
lead to direct excitation. The excited states of O2 are low enough energy that they aren't going to excite H2 by a collision
process. (To be sure, the rate isn't exactly zero, but if you want to compute the full partition function for this system and provide an upper bound
on the rate, be my guest.) So I would have to say that excitation of O2 to the state I named above seems like the best candidate at this
point for the initiation point of a plausible mechanism.
Just to be complete, I would hardly guess that sunlight is capable of igniting mixtures reliably, nor at every concentration of the explosive range.
In addition, I would have to guess that details of weather, latitude and altitude, all of which bear on atmospheric attenuation, are also at play. Nor
do I omit the possibility that the reports have omitted the use of solar concentrators, in which case it could simply be a temperature effect as
reported.
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arsphenamine
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General notes on O2 --
The 1O2 emission spectrum is typically reported as 1270 nm or 7875 cm-1, putting it in the near infra red region,
typically where we see only overtones of organic compounds (if at all).
The 3O2 emission spectrum is reported as 680 nm or deep red and visible.
Oxygen pressure measurement using singlet oxygen emission
The best single reference I can find is
Physical Mechanisms of Generation and Deactivation of Singlet Oxygen
Schweitzer, Claude; Schmidt, Reinhard
Chemical Reviews, 2003, 103 (5), pp 1685–1758
DOI: 10.1021/cr010371d
The DOI is incorrectly reported by Wiley&Sons, and you should look on the ACS site first.
In more practical terms, you need a photosensitizer to generate singlet O2. Porphyrins generally work, especially those with a bound platinum in the
center, but iron, magnesium, or copper will do. Fe,Mg,Cu - porphyrins correspond to heme, chlorophyl, and octopus bloods.
[Edited on 31-8-2012 by arsphenamine]
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unionised
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The ground state of oxygen is a radical, and, at ordinary temperatures it doesn't react with hydrogen.
Absorption of visible light can produce the singlet state but that doesn't practically cause a reaction for two reasons.
Firstly the transition is forbidden. If I shone a beam of light at the right frequency through a balloon full of oxygen practically all of it would go
straight through.
A very small fraction of it would be absorbed.
Oxygen still isn't green.
Secondly, the singlet state also doesn't react with hydrogen. It would , in due course lose that energy- probably by collision.
Now the transition is very weak but it does happen.
If you have a broad band source- like the sun and you look at the spectrum through a whole atmosphere's worth of oxygen, the absorption does matter.
There's not much left of the relevant wavelengths.
So the wavelengths of light that might get absorbed (to some small extent) but which don't cause the reaction (because the singlet state is also inert
to hydrogen) have already been largely absorbed and are not there (as witnessed by Mr Fraunfofer).
OK I forgot that the singlet state is also a radical- but the point remains it's a poor absorber on a balloon sized scale and a much better one on an
ocean of air sized one so there's not much chance of it triggering a balloon full of hydrogen and oxygen because it's not there.
Even if it was there it doesn't produce a reaction.
I know that the singlet state has some more accessible absorption bands, but then you are talking about a two photon process and, as I pointed out,
those are rare. So rare as to be unimportant compared to, for example cosmic ray interactions.
As for "What you were claiming before is that light not absorbed by the atmosphere can't be chemically reactive. That claim is bogus"
Light that isn't absorbed can't cause a reaction.
http://en.wikipedia.org/wiki/Stark%E2%80%93Einstein_law#Star...
Why do they call hard UV "vacuum ultraviolet"?
Why is that radiation not present in sunlight?
[Edited on 31-8-12 by unionised]
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