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Author: Subject: Make Potassium (from versuchschemie.de)
Pyro
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[*] posted on 5-11-2013 at 23:13


a gallon of model jet hydraulic fluid, I found it lying around. why exactly won't it work?

[Edited on 6-11-2013 by Pyro]




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[*] posted on 6-11-2013 at 08:55


If you look at its structure, you should be able to identify two functional groups that won't behave nicely in the presence of potassium metal.



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[*] posted on 10-11-2013 at 21:05


I didn't want to start a new thread and figured this would be a good place to ask and feel out any interests.

Is it possible to use potassium as a reducing agent to obtain other metals?
I've read about the Reike metal synthesis which yields highly reactive metals, but have not been able to find any detailed procedures or documentation. Could one use this method to make sodium for example? Why must the reaction be in tetrahydrofuran (THF) can other solvents be used? More information on this topic would be great.

Another method I've come across is the Hunter process that uses molten sodium to reduce titanium tetrachloride, could potassium be used in a similar fashion and with other metals besides titanium.

I have not been able to come across much information about using alkali metals as reducing agents, perhaps it's because it's impractical or impossible, but if it's the former than that shouldn't stop us from pursuing this idea, since we can make potassium quite readily these days. :D

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[*] posted on 23-11-2013 at 10:23


yes it is possible to reduce other less reactive metal ions to there cosponsoring metals using potassium, but for reducing things with alkali metals your better off using Li as it's a better reducing agent than K and it's far easier to get hold of :)
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[*] posted on 24-11-2013 at 09:56


Here we go again with the use of vague and pretty meaningless terms and ‘explanations’ that resemble old wives tales more than actual science.

Whether or not a particular chemical reaction (here a redox reaction) can proceed or not depends on a number of factors and cannot be summarised easily in a few sentences. It is not simply a matter of ‘reactivity’ or the half potentials series.

For reactions where no reaction products can leave the reaction mix, the change in Gibbs Free Energy ΔG (= G2 - G1) must be smaller than zero for the reaction to be able to proceed. Even then the reaction might need heating to allow it to start and sustain.

This can lead to seemingly ‘contradictory’ situations. For example, NaOH can be reduced by Al: NaOH + 2/3 Al == > Na + 1/3 Al2O3 + ½ H2 because ΔG < 0 (to be fair, hydrogen does leave the mix here)

But Na can also reduce AlCl3: AlCl3 + 3 Na == > Al + 3 NaCl, because ΔG < 0.

So which is the ‘best’ reducing agent here, Na or Al? It’s a meaningless question.

The Rieke metal preparations proceed simply because ΔG < 0. It so happens that THF is a suitable, aprotic solvent as a reaction medium.

In processes where one of the reaction products can be removed from the reaction mix, seemingly unlikely reactions become totally possible. Cs can be prepared from CsCl by reduction with Li, K or Ca:

CsCl + Li < === > Cs + LiCl at high temperature and under high vacuum, because the Cs metal then distils off, leaving the non-volatile CsCl, Li and LiCl behind.

Re. the availability of Li over K, that such a subjective matter it isn’t even worth squabbling about.

Quote: Originally posted by Squall181  

Another method I've come across is the Hunter process that uses molten sodium to reduce titanium tetrachloride, could potassium be used in a similar fashion and with other metals besides titanium.


Today the Kroll process uses Mg to reduce TiCl4. But 'Tetra' as it is fondly known in the industry is very hard to home produce and the reduction reaction requires pressurised vessels because of reaction heat generation and the volatility of TiCl4.


[Edited on 24-11-2013 by blogfast25]




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Galinstan
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[*] posted on 24-11-2013 at 11:36


When calculating gibbs free energy change temperature is already taken in to account so a reaction with G<0 will not require heating to happen but might require a change of temperature to achieve a measurable rate.
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[*] posted on 25-11-2013 at 05:58


Quote: Originally posted by Galinstan  
When calculating gibbs free energy change temperature is already taken in to account so a reaction with G<0 will not require heating to happen but might require a change of temperature to achieve a measurable rate.


You are completely and plainly wrong.

Reactions that have strong negative Delta G are plenty, yet most don't proceed without heating. Examples: carbon + oxygen, butane + oxygen, nearly all thermites.

By your account the world would be a constant fireball, with all these spontaneous combustions happening!

You want to read up on 'collision theory' and 'Arrhenius equation'. Most reactions with negative Delta G still have kinetic obstacles to overcome: they are the rule, not the exception.

The reaction that is the subject of this long thread is another point in case: at RT nothing happens.

[Edited on 25-11-2013 by blogfast25]




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[*] posted on 25-11-2013 at 06:43


You are correct just because a reaction has a negative Gibbs free energy change doesn't mean an OBSERVABLE reaction happens because the value of delta G is not proportionate to rate, so even though the reaction is thermodynamically favourable the rate might be so slow it appears as if no reaction at all is happening. and doesn't your example of thermites have more to do with surface contact than activation energy.
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[*] posted on 25-11-2013 at 12:31


Quote: Originally posted by Galinstan  
and doesn't your example of thermites have more to do with surface contact than activation energy.


In a sense, yes. But at RT the collisions between the reactant species are by far not energetic enough to be reactive (as opposed to merely elastic).

Thermodynamics (considerations of Free Energy) and Kinetics (considerations of mechanisms and rates) are two very different things and should not be conflated.




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[*] posted on 23-12-2013 at 17:34


I've noticed over the past few months that there is a curious brown 'fluff' floating alongside my potassium in it's kerosene-filled vial. The stuff seems to accumulate with time, and forms a colloid when shaken, discoloring the solvent to yellow-brown where it would normally be clear. This does not look like the brown potassium oxide I have dealt with before, and it may be some alkoxide, as I sometimes add a small amount of alcohol to the vial to keep the potassium shiny. Is this fluffy stuff a problem? The potassium metal itself is about the same size, and as shiny as ever.



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[*] posted on 23-12-2013 at 19:37


Impurities in the kerosene maybe.



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[*] posted on 24-12-2013 at 05:42


EC1:

I don't know but one thing is for sure: potassium is not meant to be kept under oil/paraffin/kerosene forever. Only properly glass ampouled material can be kept 'indefinitely'. Oiled alkali metals are meant to be used, not just put on display.




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[*] posted on 15-6-2014 at 11:28


I'll attempt this famous reaction soon, probably before the end of the week. Since then, I just wanted to share that the legendary Nurdrage's video isn't exactly gone :D http://v.youku.com/v_show/id_XNTAxODAzMTc2.html
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[*] posted on 15-6-2014 at 12:07


Quote: Originally posted by alexleyenda  
I'll attempt this famous reaction soon, probably before the end of the week. Since then, I just wanted to share that the legendary Nurdrage's video isn't exactly gone :D http://v.youku.com/v_show/id_XNTAxODAzMTc2.html


But unfortunately it is only a piece of the video. :( Anyone have the links to the rest of it? Thanks!
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[*] posted on 15-6-2014 at 14:44


Really, it was longer than 10 mins? Anyways, all the essential informations are there I think.
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[*] posted on 15-6-2014 at 18:02


Quote: Originally posted by alexleyenda  
I'll attempt this famous reaction soon, probably before the end of the week. Since then, I just wanted to share that the legendary Nurdrage's video isn't exactly gone :D http://v.youku.com/v_show/id_XNTAxODAzMTc2.html

Thanks! I've been looking for that for a while.




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[*] posted on 30-6-2014 at 21:10


Quote: Originally posted by alexleyenda  
I just wanted to share that the legendary Nurdrage's video isn't exactly gone :D


The legendary NurdRage isn't exactly gone, either:

<iframe sandbox width="560" height="315" src="//www.youtube.com/embed/gw9ZzvSiyAQ" frameborder="0" allowfullscreen></iframe>

And neither is his potassium video:

<iframe sandbox width="560" height="315" src="//www.youtube.com/embed/JAo2F9ymKXA" frameborder="0" allowfullscreen></iframe>




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[*] posted on 2-8-2014 at 22:16


i looked though the previous pages and found no evidence that anybody ever tried adding a tiny bit of I2 to activate the magnesium.
This would make sense as it's used to activate Mg for many other uses (grignard, drying EtOH,...)




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[*] posted on 3-8-2014 at 07:58


That is an interesting idea - if I still had any t-BuOH, I'd certainly try it.



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[*] posted on 3-8-2014 at 10:08


so would I but my lab is currently in boxes.

vigorous stirring with a magnetic stir bar will probably help the reaction along very well, coalescing the K as well as providing lots more surface area for the reaction.




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[*] posted on 3-8-2014 at 12:57


I2 isn't necessary if your magnesium is sufficiently blank. The way it initiates Grignards is quite different from what it can do here: a small amount of I2 will only 'clean' a small amount of Mg.

As regards the role of stirring, it's far from clear judging by the many attempts described in the thread. The boiling of the solvent seems to provide enough agitation for most people's purposes, it did for me. A very early attempt by 'len1' using high speed stirring in fact failed completely to produce any potassium.




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[*] posted on 3-8-2014 at 13:17


I still intend to try this experiment, the only thing I need to do is buy some quality Mg powder which I have still not got around to doing(I would buy tetralin also but that almost feels like cheating, the idea is to make it simple for everyone!) . If I can make it work without I2 then I'll try I2 with the other grades of Mg and see if something happens. Right now I can't be bothered since I already have commercial K anyways. Still a fascinating project to try though!

[Edited on 3-8-2014 by Mailinmypocket]
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[*] posted on 3-8-2014 at 18:43


@blogfast25: that is also all that is necessary to activate it.



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[*] posted on 4-8-2014 at 12:23


Quote: Originally posted by Pyro  
@blogfast25: that is also all that is necessary to activate it.


The difference is that in Grignards that practice is fairly proven where as here we really have no idea.

Here the first port of call is to try a 'good grade' of Mg and try it as such.




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[*] posted on 25-11-2014 at 11:01


How would propylene carbonate work as an alternative "solvent" for the reaction instead of tetralin? It isn't significantly cheaper to buy, but it is much easier and cheaper to make and it is also denser than potassium like tetralin.
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