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math
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Quote: Originally posted by math  | 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 |
Maybe a better questions would be:
how can I calculate the quantity of water at X temperature to be used to keep the whole reaction at, say, 50°C maximum?
Thank you 
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watson.fawkes
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| Quote: Originally posted by unionised | 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. [...] Secondly, the singlet
state also doesn't react with hydrogen. It would , in due course lose that energy- probably by collision.
[...]
OK I forgot that the singlet state is also a radical |
Ignorance confirmed.
There are lots of states of oxygen, not just "triplet" and "singlet". Since you have persisted in your lack of self-knowledge of your ignorance, I
went and found data. For a rather complete list, see Krupenie, P.H., The spectrum of molecular oxygen, J. Phys. Chem. Ref. Data, 1972, 1, 423. There's
a copy on the NIST site and a reference threadlet in References here. Table 2 of that paper lists nineteen (19) of them. That table also lists
O2+ states, which are relevant to O2 since they correspond to "Rydberg" states, where one electron is highly excited
but not quite ionized.
Furthermore, even restricting attention to the lowest states of oxygen, there are two singlet states, not just one. Quite apparently you have not
bothered to read the Wikipedia page on singlet oxygen, which I referenced before, so there's the link—read it this time. You wouldn't persist with saying "the" singlet state
consistently if you had done so earlier. That page shows both of these singlet states quite well, along with the (reduced) spectroscopic symbols.
O2 has a pair of degenerate 1πg orbitals. There are three occupancy modes of two electrons in these states; one is the
ground state triplet, the other two are the singlets. And these three all have the same occupancy signature in the MO when you ignore degeneracy. That
signature is (3σg)2(1πu)4(1πg)2.
The excited states I was referring to upthread have occupancy signature
(3σg)2(1πu)4(1πg)1(3σu)1;
there are two such states. It turns out, though, that there are two other signatures:
(3σg)2(1πu)3(1πg)3, with six states, and
(3σg)1(1πu)4(1πg)2(3σu)1,
with eight states. I'm still plowing through that reference paper. It's 120 pages long, filled with results, using notation I'm not facile with.
Nevertheless, there are lots and lots of transitions in this molecule in the visible and near UV. Many are forbidden transitions with small cross
sections, albeit, but apparently not all are. There are fourteen subsections on the various spectroscopic systems (roughly, related transition in some
band), and hundreds to thousands of individual lines. I would like to think that I have enough humility that I won't call a potential effect
"inconceivable" before I have even a basic understanding of the system I'm thinking about. You are, yet again, confusing atomic
absorption with bulk gas absorption, and also confusing necessary conditions with sufficient ones. Atomic absorption is a necessary condition for bulk
gas absorption. It is not always sufficient. If, for example, the natural decay lifetime of a state is significantly shorter than the transit time
across the mean free path of the molecule, then you can have an atomic absorption line without a significant bulk gas absorption line (they'll pretty
much always be detectable with sensitive instruments). I do not know whether this pertains to molecular O2 or not, but I am quite sure that
conflating these two concepts is bogus. Yes ... yes it is.
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unionised
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A few weeks ago I was at the Kennedy space centre.
They have lots of stuff there for tourists to gawp at and helpful tour guides to answer questions.
He looked bored so I pointed at the rocket engine that they had on display and asked if he could point out the spark plug.
He couldn't- but he agreed that there must be one somewhere because mixtures of hydrogen and oxygen don't spontaneously ignite.
He did point out that there are flares lit in the vicinity of the rockets to light any leaks before they build up and become a bigger problem (most
things in the vicinity of a space rocket are built to stand up to a flash fire)
It's often sunny in Florida.
NASA don't think that sunlight ignites air/ hydrogen mixtures.
For the record, I'm fully aware that I don't know the details of the transitions.
I don't think they matter.
Do you, for example, accept that oxygen, as we breathe it, does not have any significant absorptions in the visible region of the spectrum?
I know that there is the pair of transitions to triplet states You may remember that I gave the wavelengths corresponding to both of them which makes
this assertion "Furthermore, even restricting attention to the lowest states of oxygen, there are two singlet states, not just one. " a bit redundant.
I know there are. I gave their excitation energies earlier when I was pointing out that they are not in the infra red. They are, in fact, in the
visible range . That was my point, if they actually happened to any significant extent, then oxygen would be green. It isn't.
When I referred to the singlet state I was referring to the whole bunch of excited states- I didn't bother to specify rotational or vibrational levels
either.
I grant you it's sloppy language but when it comes down to it, they don't matter.
Do you think that there are any excited states of oxygen which can be reached from the ground state by a single photon transition involving EM
radiation that is present in sunlight at ground level and which react with hydrogen?
BTW, re "I'm still plowing through that reference paper."
Why?
"You are, yet again, confusing atomic absorption"
Bollocks. I never stated what was absorbing: the law works perfectly well with molecules or atoms since it's a restatement of the conservation of
energy. I just pointed out that if a photon goes in and out then it can't do anything. Photons don't have a big repertoire: they get absorbed or they
don't. If they don't get absorbed they don't have any effect.
"Atomic absorption is a necessary condition for bulk gas absorption."
That must be a real pig for the IR and µwave spectroscopists since they never deal with atoms- only molecules.
"Nevertheless, there are lots and lots of transitions in this molecule in the visible and near UV."
There are indeed, but not many of them start from the ground state which is (fairly nearly) the only occupied state at room temperature.
Specifically there are the two I mentioned (split into many sub levels by vibrational and rotational excitations).
And the interesting things about them are that they are still weak as hell (oxygen still isn't green)
and the radiation needed to induce them has already been seriously attenuated by the passage through a big thick cloud of exactly the right stuff to
absorb them- to whit , oxygen.
Those excited states don't react with hydrogen.
BTW, feel free to actually answer the questions I asked earlier.
Here they are again
Why do they call hard UV "vacuum ultraviolet"?
Why is that radiation not present in sunlight?
(I guess I should have specified sunlight near ground level.)
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watson.fawkes
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Won't bother. You don't want to engage what
I've said and repeat yourself. No sense wasting time with you.
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Poppy
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What if the baloon is hit by the stream jet of a very high energty cosmic ray particle/ radiation, that would focus and cause the baloon to explode,
maybe?
This is the way I use to get cancer...
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arsphenamine
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ditto, speaking only for self. | Quote: | | There are lots of states of oxygen, not just "triplet" and "singlet". |
So, there I was, thinking I'd
casually model singlet vs. triplet total energies at a moderate basis set and theory level.
Typically, without exactly specifying the valence spin states, you get a Δ E that's 10-15 kcal too high.
The less costly correlation methods want to select the highest energy spin orbits or a weighted average of all spin combinations.
After you've specified your spin state, every ab initio QC package out there has its own obtuse nomenclatural conventions.
Yeah, you can do closed-shell computations on an open shell triplet without a peep of complaint from the application.
| Quote: | | For a rather complete list, see Krupenie, P.H., The spectrum of molecular oxygen, J. Phys. Chem. Ref. Data, 1972, 1, 423. There's a copy on the NIST
site and a reference threadlet in References here. |
Thanks.
Several NB's ...
The Jablonski Diagram, an agreeable shorthand for spectroscopically-observable transitions between different spin states,
by its very existence confirms that absorption and emission spectra need not be the same or even complementary, viz.,
an absorption spectrum is not an emission spectrum.
It has been silently amended in practice to include extreme rotation, translation, bend+stretch+torsion changes from cavitation states,
given the industrial advent of sonochemical engineering and hydrodynamic cavitation reactions.
Last N.B. -- Some lab reactions produce 1O2 with peroxide and hypochlorite, O2-2 and OCl-.
Both species are produced in the upper atmosphere. The chloride comes from freons.
I will accept that, in a pure sense, ambient UV radiation will not initiate O2+H2 combustion.
To my mind, the likelihood of 'adventitious' mediating substances (metal ions, chromophoric triplet sensitizers,etc) re-opens this line of inquiry.
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unionised
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Simply not true. I quoted you about half a dozen times in my last post alone.
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