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blogfast25
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Thanks sciecesquirrel for the link. Live and learn...
Looks like we've really covered quite a lot of theoretical ground here. Someone here (moi!) will be mightily disappointed if all this turns out to be
a dud... ![:o](./images/smilies/shocked.gif)
A vessel filled with an ‘incompressible’ fluid is sealed off with a movable, massless piston of surface area S. We apply a force F to the piston:
pressure inside the cylinder now increases by P = F/S. Lets assume the piston also moves by a small amount Δs (on application of the force F),
the work done is W = Δs x F. Is this the Free Energy increase of the system, Watson?
[Edited on 7-12-2010 by blogfast25]
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condennnsa
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Well is there anyone here that has all the required reagents to give it a shot?
I ordered the shellsol, i'm working on the t-butyl, but it'll take some time .
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woelen
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Quote: Originally posted by Pok ![](images/xpblue/lastpost.gif) |
I know science. I know that reproduction is neccessary. But if I just do it again and again, this wouldn't help you. Some of YOU has to do it
(scientifically - which means: in exactly the same way I did it.) | With this I
absolutely agree. Someone else than you should be able to reproduce the experimental outcome.
Quote: Originally posted by Pok ![](images/xpblue/lastpost.gif) | Tell you how to do it? I thought you already have every detail. Look at versuchschemie.de and here. I made it like the patent with modifications that
I already told you. You should easily be able to repeat this synthesis.
I hope you know what "exactly" means: 50ml, 25cm-glaspipe with wet toilet paper (moisten it sometimes to prevent dehydration), balloon at the top with
tiny hole, zero oxygen input, Mg filings made from 99-100% Mg made by filing with a medium file. and so on It might be that I really
forgot one or two details. If this is the case, I would apologize - just ask! (I'll be back in 24 hours) | With this I cannot agree. If the experimental outcome really depends on how you filed your pieces of magnesium, if it only works
with magnesium from exactly that piece of scrap which you obtained from eBay, if it only works with a 25 cm glass pipe and with a volume of 50 ml,
then it is not good at all.
You also made modifications (such as using ten times smaller amounts) and you thought out your own setup, etc. Scientific reproduction and exact
writeup is important, but it must be within reason.
Does the color of the balloon also affect the outcome. Please specify the color in the form of an RGB color, like #rrggbb ![:D](./images/smilies/biggrin.gif)
Next weekend I'll try the series of experiments I wrote in my previous post. If all that fails, then I'll even consider buying ShellSol D70 (it is not
expensive, EUR 7 for 1 liter at Kremer pigments and there is no shipping issue with this). But if even with these things it still doesn't work, then
I'm inclined to think it is either fake, or you were extremely lucky and indeed won the jackpok.
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woelen
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Quote: Originally posted by metalresearcher ![](images/xpblue/lastpost.gif) | Quote: Originally posted by woelen ![](images/xpblue/lastpost.gif) | but I do have high grade paraffin oil (colorless viscous liquid, perfectly odorless and completely free of acidic and unsaturated ingredients).
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Where can I get this oil ? Or can I melt candles which is also paraffin ?
[Edited on 2010-12-7 by metalresearcher] |
This oil can be purchased at drugstores. It also is available online in the Netherlands.
http://www.drogisterij.net/diversen-schoonmaak-Chempropack-p...
I am very happy with the quality of this oil. It is clear like water, has no color and no smell and it is really inert. I store Na-metal, K-metal and
Li-metal under it and I also use it sometimes in distillations for making seals really airtight. Bromine only discolors very slowly when added to this
oil, indicating that it does not contain unsaturated hydrocarbons.
[Edited on 7-12-10 by woelen]
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ScienceSquirrel
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It sounds like it is this stuff, it is used in creams and all sorts of things, so any pharmacy (chemist) should stock it or you could get it from a
place that sells massage oils or such like.
http://www.henriettesherbal.com/eclectic/bpc1911/paraffin2.h...
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Pok
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Quote: Originally posted by woelen ![](images/xpblue/lastpost.gif) | If the experimental outcome really depends on how you filed your pieces of magnesium, if it only works with magnesium from exactly that piece of scrap
which you obtained from eBay, if it only works with a 25 cm glass pipe and with a volume of 50 ml, then it is not good at all.
You also made modifications (such as using ten times smaller amounts) and you thought out your own setup, etc. Scientific reproduction and exact
writeup is important, but it must be within reason.
Does the color of the balloon also affect the outcome. Please specify the color in the form of an RGB color, like #rrggbb ![:D](./images/smilies/biggrin.gif)
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The experiment won't depend on so many things. But if you want to exclude sources of errors you should follow my description. Why my filing
description? Because I saw the difference in Mg particle size between my and lens Mg filings. Why 25cm pipe? Too short: t-butanol can't condense
totally and will be lost. Too long wouldn't be a problem. Why 50 ml? The relation of volume of air above your reaction vessel and reaction volume in
very small dimension might be so large, that enough air is available to oxidise the K. I also don't think that you can come to such a point. Smaller
reaction volume should be ok (5 ml, too). Why balloon? Because someone did it without ANY inert atmosphere (open pipe) - this is just one possible way
to exclude air. Maybe just one thing: the purity of your t-butanol (I used > 99,3%) and Shellsol D70 can be of crucial importance. BTW: my ballons
were black. This may influence the absorbtion of sunlight and so the reflux of t-butanol if it gets to warm by the sun .
But if you want to call me a liar you have to do it in my way. Why trying so many possibilities? Just try 1: my. If it fails: i lied.
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condennnsa
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i see on wikipedia, and also other sources (Ullmann's encyclopedia of industial chem for example) that
"tert-Butanol is used as an octane booster for gasoline, as an oxygenate gasoline additive,"
You won't get a lot of butanol in gasoline, but then a little is all is needed, right?
If this really proves itself to be a reliable procedure, it might be a source for folk who can't get any.
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watson.fawkes
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Quote: Originally posted by blogfast25 ![](images/xpblue/lastpost.gif) | A vessel filled with an ‘incompressible’ fluid is sealed off with a movable, massless piston of surface area S. We apply a force F to the piston:
pressure inside the cylinder now increases by P = F/S. Lets assume the piston also moves by a small amount Δs (on application of the force F),
the work done is W = Δs x F. Is this the Free Energy increase of the system, Watson? | Not exactly. You
can have a change in energy with no displacement change in an incompressible fluid. Incompressible is an idealization, but it doesn't affect the math.
One way of approaching this with a "constitutive relation", a proposition that pressure in material X is a function (only) of its density and internal
energy: p = p<sub>X</sub>( ρ, e). For most materials, such functions are monotonic in internal energy; when e increases, so does
p.
You can see this use of pressure as an energy density in Euler's equations, in the divergence normal form equation for conservation of energy. More elementary, but also perhaps useful, is Hyperphysics on pressure. I will freely admit that I didn't find any specifically targeted references for this topic.
In more general equations of motion, which might include reactions, the internal energy can be interconverted between chemical species.
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Pok
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some more pictures of the 300ml experiment.
![](http://www.versuchschemie.de/upload/files3/67979825_4948.jpg)
beginning at about 100-180°C - massive evolution of H2. Al-foil prevents heat loss. Toilet paper at stopper prevents cooling water from above toilet
paper to come in contact with hot parts. Toilet paper around pipe (here longer than 25 cm) wetted for cooling/reflux. balloon at the end (fixed with
cellotape) with an invisible small hole.
![](http://www.versuchschemie.de/upload/files3/20682940_4948.jpg)
T-Butanol not added yet. Picture of massive H2 evolution. Shiny: Mg filings. lumps: KOH.
![](http://www.versuchschemie.de/upload/files3/85505940_4948.jpg)
At t-butanol addition.
![](http://www.versuchschemie.de/upload/files3/94447720_4948.jpg)
After t-butanol addition. Thick fogs of Shellsol and/or t-butanol above the mix. First K bullets visible.
You can remember these pictures if you want to reproduce the experiment and compare them with the things you will see in your experiment.
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blogfast25
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@ Watson:
Yep, pressure as energy density. Fluid mechanics (after 30 year of hibernation) is slowly waking up in that bunch of neurons up here.
At the molecular/atomic level, repulsion between orbitals...
[Edited on 7-12-2010 by blogfast25]
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blogfast25
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@ pok:
I small a rat: you never told us the black balloon was shiny or matt! And the toilet paper: recycled or embossed? ![:P](./images/smilies/tongue.gif)
And still radio silence from our other German friends, so much for 'Deutsche grundlichkeit'... ;-)
[Edited on 7-12-2010 by blogfast25]
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Pok
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Quote: Originally posted by blogfast25 ![](images/xpblue/lastpost.gif) | @ pok:
I small a rat: you never told us the black balloon was shiny or matt! And the toilet paper: recycled or embossed? |
Oh sorry. The toilet paper was recycled as I don't want to kill trees for my ass .
If you use normal toilet paper the synthesis will not work: Recycled toilet paper can take up more water. With normal toilet paper you will not gain
potassium but gold. And that's really not the goal here. ![:D](./images/smilies/biggrin.gif)
Germans aren't grundlich. I had to make the chaotic background black in the picture here http://www.versuchschemie.de/upload/files3/67979825_4948.jpg , because I'm so very un-grundlich.![:D](./images/smilies/biggrin.gif)
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MagicJigPipe
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I will try this around the 23rd of December if no one has had success by then. I have all the materials required but I lack one thing right now: the
time. Due to some plans that I have that is the soonest I can try (not to mention finals).
Even if someone has tried it by then I will probably still attempt it. I have decided to use VM&P Naptha the first time. If that fails, I will
attempt to use use dodecane. KOH and t-butanol will both be reagent grade.
I'm looking forward to it!
"There must be no barriers to freedom of inquiry ... There is no place for dogma in science. The scientist is free, and must be free to ask any
question, to doubt any assertion, to seek for any evidence, to correct any errors. ... We know that the only way to avoid error is to detect it and
that the only way to detect it is to be free to inquire. And we know that as long as men are free to ask what they must, free to say what they think,
free to think what they will, freedom can never be lost, and science can never regress." -J. Robert Oppenheimer
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mr.crow
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Thank you Pok, you are awesome
Double, double toil and trouble; Fire burn, and caldron bubble
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Sedit
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Sorry pok im not trying to offend in anyway but I do see a serious flaw that I can not explain and its something thats bugging me. I would love to
reproduce the experiment but with the holidays ahead I just can not.
However to explain why I see a problem look at the volume of Mg in the pictures you just posted at the start of the experiment. Now at the end there
is a huge reduction in the amount of metal in the experiment presumably converted over to K. However I do not see any precipitate which there should
be a LARGE amount of if logic serves me correctly. Not only just large but given the fineness and fluffyness of inorganics precipitated in a non polar
solvent my instincts would tell me that that entire flask should be filled with a white precipitate yet as the reaction proceeds all im seeing is a
conglomeration of metal into shinny spheres and a total reduction in reactants volume with no explination as to where they are going. It also appears
as though the KOH flakes are untouched as though the metal has melted around them.
Like I said im not trying to make an enemy here Im just curious as to why there does not appear to be any MgOH precipitate of any kind. Perhaps the
tert-butanol is making them soluble for all I know but as the reaction is outlined something appears amiss.
Knowledge is useless to useless people...
"I see a lot of patterns in our behavior as a nation that parallel a lot of other historical processes. The fall of Rome, the fall of Germany — the
fall of the ruling country, the people who think they can do whatever they want without anybody else's consent. I've seen this story
before."~Maynard James Keenan
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ScienceSquirrel
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We are mad scientists and it would be nice if we could express our opinions in a way suited to the Common Room of the University Of Mad Science (
Chemistry School ).
I have my doubts about this reaction; I cannot explaln the results published by Pok and no one has managed to reproduce his results to date.
But that does not mean to say that he is wrong or a liar and I think he deserves a respectful hearing.
Science should be a creative activity carried out in a spirit of constructive criticism, I hope that we can try to be true to that ideal.
PS: I reserve my right to spear the odd troll! ![:D](./images/smilies/biggrin.gif)
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Sedit
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Quote: Originally posted by ScienceSquirrel ![](images/xpblue/lastpost.gif) | We are mad scientists and it would be nice if we could express our opinions in a way suited to the Common Room of the University Of Mad Science
But that does not mean to say that he is wrong or a liar and I think he deserves a respectful hearing.
Science should be a creative activity carried out in a spirit of constructive criticism, I hope that we can try to be true to that ideal.
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This is my intent completely. Parts of the reactions progress confuse me and do not appear to proceed the way they should so I ask Pok and all for
that matter, Where is the Hydroxide or oxide precipitate that should be created in abundance. I am at a lose as to what motivation Pok would have to
lie so im left to believe this really happened. The question is why can no one reproduce it and where is the missing mass.
Knowledge is useless to useless people...
"I see a lot of patterns in our behavior as a nation that parallel a lot of other historical processes. The fall of Rome, the fall of Germany — the
fall of the ruling country, the people who think they can do whatever they want without anybody else's consent. I've seen this story
before."~Maynard James Keenan
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Pok
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Quote: Originally posted by Sedit ![](images/xpblue/lastpost.gif) |
However to explain why I see a problem look at the volume of Mg in the pictures you just posted at the start of the experiment. Now at the end there
is a huge reduction in the amount of metal in the experiment presumably converted over to K. However I do not see any precipitate which there should
be a LARGE amount of if logic serves me correctly. Not only just large but given the fineness and fluffyness of inorganics precipitated in a non polar
solvent my instincts would tell me that that entire flask should be filled with a white precipitate yet as the reaction proceeds all im seeing is a
conglomeration of metal into shinny spheres and a total reduction in reactants volume with no explination as to where they are going. It also appears
as though the KOH flakes are untouched as though the metal has melted around them.
Like I said im not trying to make an enemy here Im just curious as to why there does not appear to be any MgOH precipitate of any kind. Perhaps the
tert-butanol is making them soluble for all I know but as the reaction is outlined something appears amiss. |
Ok. It's easy to explain. The Mg(OH)2 or MgO which you are missing is in the form of a quite hard (but porous) light gray crusts (see here: http://www.versuchschemie.de/upload/files3/2272042_4948.jpg ) - the shiny balls are K, all the rest (lumps of gray stuff) is the Mg(OH)2 or MgO!!
You think: so much Mg at the beginning can't be decreased to such small amounts of hydroxide or oxide? The reason is: The Mg filings are highly
irregular (due to filing), highly voluminous. What you can see are really only some grams of Mg. The product (hydroxide/oxide) is highly compact. So
its volume seems to have decreased unimaginable. - For imagination: If you burn voluminous Mg filings and crush the product into powder, the powder
will have a much lesser volume. You will (probably) only get a dust-like precipitate of Mg products if you stirr continiously. I simply didn't. I hope
this is OK for understanding.
PS: Its 4 am here and I have to write an exam today so see you in a few hours (I'm not going away )...
[Edited on 8-12-2010 by Pok]
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Pok
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No reproduction of my results? Because nobody has tried it yet (as I described it)!
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ScienceSquirrel
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Some people are intending to try it exactly as you have described it.
The best way to prove your point is to hep them to succeed.
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bbartlog
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Quote: | no one has managed to reproduce his results to date. |
You mean, besides the guys that filed the patent? And the other guy on versuchschemie who saw some half-millimeter globules of potassium when he tried
to reproduce the experiment? I think at this point the skepticism is overblown. Really, you're basically saying that someone filed a patent that was
not just in error or overbroad, but a complete fabrication, *and* then someone else created an elaborate hoax based on it, just for kicks. We already
know that the patent has some issues (presenting calculated amounts as measured ones, suggesting that isopropanol could be used as well as
tert-butanol), but that is hardly the same thing as assuming that the authors just made the whole thing up.
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aonomus
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Here is a suggestion, if the sludge is indistinguishable between potassium metal and magnesium metal (and its oxides/hydroxides) check the mass
balance. If you can account for the loss of hydrogen that should indicate reaction progression.
If anyone has a scale with good resolution, they should be able to see if there is any change in mass based simply on the release of hydrogen (which
is an indicator of reaction completeness). In an ideal world there would be no solvent loss, but its worth a shot. Maybe better accuracy would be had
by filtering the whole mass and washing with hexanes to clean off solvent residues.
The other thought would be to measure the release of hydrogen from the reaction based on volume and calculate backwards to see how many mmol of KOH
reacted.
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Arthur Dent
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Quote: Originally posted by blogfast25 ![](images/xpblue/lastpost.gif) |
Cool for the heptane, unfortunately it is totally UNSUITABLE for this purpose: way too low boiling point…
t-butanol, $36 for 500 ml? Very, very good, go for it and sell some back to those here who can’t get any!!!
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Yeah I know. I just mentioned it because I discovered this inexpensive heptane source as I was reading the MSDS of various "over the counter" solvents
to find an equivalent to Shellsol...
As for the t-butanol, if a member of this forum succesfully reproduces the experiment, i'll buy a bottle and split in 2 or 3 among members residing in
Canada if I find people that will be willing to share the costs (I think shipping/hazmat fees to the US would be outrageously expensive, so I'd rather
keep this north of the 42nd.).
Anyway, I believe that this experiment requires only minute amounts of t-butanol, a syringe's worth?
Robert
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woelen
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Quote: Originally posted by Pok ![](images/xpblue/lastpost.gif) |
The experiment won't depend on so many things. But if you want to exclude sources of errors you should follow my description. Why my filing
description? Because I saw the difference in Mg particle size between my and lens Mg filings. Why 25cm pipe? Too short: t-butanol can't condense
totally and will be lost. Too long wouldn't be a problem. Why 50 ml? The relation of volume of air above your reaction vessel and reaction volume in
very small dimension might be so large, that enough air is available to oxidise the K. I also don't think that you can come to such a point. Smaller
reaction volume should be ok (5 ml, too). Why balloon? Because someone did it without ANY inert atmosphere (open pipe) - this is just one possible way
to exclude air. Maybe just one thing: the purity of your t-butanol (I used > 99,3%) and Shellsol D70 can be of crucial importance.
| Yes, these sound reasonable. No air may come into the reaction vessel, the initial amount of oxygen must be
minimal (this can also be achieved by flushing the air away with nitrogen, or any other inert gas, or even hydrogen). I have reagent grade
tert-butanol, so that should be no issue.
The key may even be that the tube is short, in the start there is some tert-butanol, lateron it is gone. This is a good point and I'll take this into
account in my experiments next weekend (one variation with long tube and one variation with short tube).
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Eclectic
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Did we decide that k-1 kerosene with the volatiles distilled off and butoxyethanol as the alcohol were not worth trying?
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