Sciencemadness Discussion Board - Make Potassium (from versuchschemie.de)

My DOT 3 was opened 3 or 4 years ago and left [capped] on a shelf since. No doubt it's wet.

Has naphthalene (moth balls) received any attention? SG ~ 1.14; BP ~ 218*C. No -OH groups to tackle.
Tom

Yes, it is hygroscopic.

I wanted to try naphthalene after Nurdrage’s success with Tetralin ® but it turned out my [url=http://www.sciencemadness.org/talk/viewthread.php?tid=15348#pid199586]’naphthalene’ wasn’t naphthalene[/img]. The idea kind of got forgotten after that. But it’s worth pursuing, IMHO. Just make sure you have the realm McCoy (do an MP test…)

Quote: Originally posted by Twospoons

I would expect the metallic K balls to be attracted into the E-field, creating a bit of a squeeze that would help them coalesce. They wouldn't have to be floating (ending the need for a high density HC) as the E-field would provide all the necessary driving force.
Sorry I can't help with a practical test.

Just ran a simulation with FEMM4.2 - there will be a small force attracting the K balls into the E-field.

[Edited on 1-5-2011 by Twospoons]

That would work if the K was indeed electrically charged. That seems to be the case with many emulsions. But with K?

What’s FEMM4.2?

Worth considering…

m1tanker78 - 2-5-2011 at 07:05

Blog, I tried sodium in naphthalene; briefly outlined here. It's inconclusive for K but thought you might be interested in the results with Na. Based on that, I believe K will float but won't easily coalesce.

Tom

Twospoons - 2-5-2011 at 14:31

That would work if the K was indeed electrically charged. That seems to be the case with many emulsions. But with K?

What’s FEMM4.2?

Worth considering…

It works even if the K is not charged, as it is a conductive metal floating in a dielectric sea. As the K moves into the high field region the field energy drops, as the K 'shorts out' the dielectric. The reduction in field energy is what gives rise to the force.
FEMM4.2 is a Finite Element Analysis program for 2D magnetic, electrostatic, and thermal simulations.

watson.fawkes - 2-5-2011 at 18:27

Actually, after refreshing the concept of surface tension or better, surface energy (dW/dS, the work dW needed to increase the surface area by dS), it’s clear that adding a surfactant to the solvent phase will actually stabilise the ‘K/solvent emulsion’, thereby reducing the K coalescence rate. Much like what happens when you add some detergent to an oil/water emulsion.

In reality we’d have to increase the surface tension, thus favouring minimisation of surface area, ergo larger globules… That’s where other solvents, perhaps dioxane or THF might come into it…

I've found a small section in Physical Chemistry of Surfaces titled "Stabilization of Emulsions by Solid Particles" (p 510 in 6th ed). Given the reports herein of the MgO dust on the surface of the K globules, I'd guess that it's these MgO particles that are stabilizing the interface. The use of a surfactant would be to affect the contact angle between the MgO particle and the K globule. If this contact angle goes to zero, the MgO particles will not adhere to the K.

Sedit - 2-5-2011 at 18:59

Wouldn't just the small, slow addition of EtOH to the melted globs lower the surface tension by cleaning it off while generating Alkoxides? I doubt much would be lost and I would think it would allow for nice blobs in the end.

watson.fawkes - 2-5-2011 at 19:06

Quote: Originally posted by Twospoons

I had a rather left field idea today. It may be possible to get the K to coalesce by applying a very strong electric field - something on the order of 10kV/cm.
Could be done with a neon sign transformer, rectified, with electrodes consisting of sealed glass tubes with wire inside, spaced a few cm apart

I had the exact same idea; I should have mentioned it before.

The physics is that conductors self-polarize in an electric field. This is similar to Gauss's law, except here the charge is on the surface and also creating a counteracting electric field until the internal field inside the conductor is zero, at which point no more charge moves. In the present case, a pair of adjacent globules will develop polarizations along the line of (+ -)(+ -), so that the globules are attracted to each other at their poles. This creates a macroscopic pressure of the droplets against each other, a force that might overcome whatever is holding the droplets apart.

The practical difficulties of this instrument are not trivial, although they should be manageable.

The glassblowing required isn't too hard for a glassblower, if you are or have access to one. The electrode tips should be arranged so that the greatest field is between the ends; this will concentrate the available polarization in one place. The ends will look sort of like pincers.
If you want a wand-shaped device, packing the lead wires with a relatively high dielectric-constant material will also help, essentially by sucking up the field so that the highest remains at the ends. Suitable packings can be liquid, like ethylene glycol, or solid, like TiO2; you could even use water. Remember that the packing goes in the space between the wires, so a liquid packing requires wire support.
The power supply for this doesn't need to be particularly high current, so I'd avoid a neon-sign transformer, simply because they create a fairly big hazard when your insulation fails. Much better is a Cockcroft-Walton generator, because they can be designed with high voltages and low internal charge capacity, so their shock hazard is limited. Some experimentation will be necessary to see just what voltages are required, but I don't imagine it will be more than a few kV over, say, a 5 mm gap.

Twospoons - 2-5-2011 at 20:26

Quote: Originally posted by watson.fawkes

I had the exact same idea;

Must be a good one then!
The simulation I ran gave forces around 100th of a newton with fields of ~10kV per cm, for a metal ball ~5mm diameter.
I suggested a NST as its a very easy way to get to 20kV. And no more dangerous than playing with blobs of molten K

Multipliers are good too- just watch out for the stored charge.

Hell, a Van deGraaf might work, should you have access to one.

The glass work is simple - a 5mm tube will seal on its own if you melt the end in a propane flame.

watson.fawkes - 3-5-2011 at 04:02

Quote: Originally posted by Twospoons

Must be a good one then! [...] I suggested a NST as its a very easy way to get to 20kV. And no more dangerous than playing with blobs of molten K

Multipliers are good too- just watch out for the stored charge

It's not apparent to me that it's such a good idea. Putting a 7 kV potential inside your hand is not something that the HV crowd does in any kind of regular basis. You have to plan for failure, particularly in this case. It's not like glass never breaks in the lab. I'd say that this tool is significantly more dangerous, at least at first blush, than molten K.

As far as supplies, the difference between a multiplier and a neon sign transformer is that a multiplier has a bit of stored charge, which will dissipate, and the NST has no stored charge, and will keep pumping current indefinitely. You absolutely need an inherently current-limiting supply for this tool to be anywhere near safe. Using an HV transformer pretty much negates that from the start.

Upon further consideration, I'd heartily suggest putting the entire multiplier in the wand on the far side of the handle, so that only the supply voltage of 600 V - 1 kV or so is in the handle. This complicates construction, of course, but seem like a much better idea than running flexible HV cables and having the wand just be passive conductors. At 20 kV potential, it's not an insignificant problem just to source adequate insulation for your wire. In order to avoid corona and arcing inside such a relatively small space, it's necessary to use some kind of potting. Oil or paraffin wax would both work, it seems. You're still going to end up with a thick handle.

Perhaps a much better option is to make this not-a-hand-tool at all. The advantage of a hand tool is that you can wave it around and provide mechanical motion easily. With a fixed tip, you have to move the K, rather than vice-versa, which leads to a whole host of other problems.

The upshot is that this should under no circumstances be anyone's first HV project. There's nothing here which is simultaneously safe and easy.

blogfast25 - 3-5-2011 at 07:04

Sedit:

That’s already been tried (but with IPA), with very mixed results. And it consumes potassium.

Watson and Twospoons:

If Twospoons’ simulation is correct (0.01 N, for 10 kV/cm for a 0.5 cm (5 mm) diameter K ball) then this method should work extremely well: the actual weight of such a ball (unsuspended) is only about 0.0005 N (mass = 0.05 g)! Unhindered by any other forces acting on it (that’s not the case, I know) it would undergo an acceleration of no less than about 200 m.s-2 ! That really sounds too high to me…

Way to go if it's true! :cool:

Later on today I'll be testing the more mundane option of using xylene as a coalescing liquid...

[Edited on 3-5-2011 by blogfast25]

blogfast25 - 3-5-2011 at 10:51

Xylene (racemic mixture) didn’t really do the trick either. Four pieces of potassium (ca. 5 mm) were cut from a larger piece and added to about 20 ml cold xylene in a 100 ml conical flask. The ensemble was then heated on an electrical hot plate to about 100 C.

At first the pieces were reluctant to assume their normal spherical shape, presumably because of surface impurities forming some sort of ‘skin’.

1 drop of methylated spirits was the added and the pieces immediately took on their ball shape with a silver mirror finish. But no amount of teasing them together could persuade them to merge. I gave up after about 15”.

Next up should be either dioxane or ‘ethyl ester of DOT 4’…

Twospoons - 3-5-2011 at 16:46

Electrostatic coalescence taken to this thread

I think the forces were more like 0.001N. And they vary according to position in the field.

m1tanker78 - 3-5-2011 at 18:21

BlogFast, what did the K do in the xylene? Based on the reported densities, they should have hovered around. What exactly did the drop of methylated spirits do to make 'em ball up? Someone else a few pages back reported success with dioxane so I don't see why it won't work.

Tanker

blogfast25 - 4-5-2011 at 04:13

Tanker:

Well, the balls weren’t really balls to begin with, as I’d cut off random shapes from a large blob. I fully expected them to assume their spherical shape on melting but they didn’t really: they just kind got more rounded. They were typically blue skinned and I thought the skin was kind of retaining their cut shape. When I added just the one drop of methylated spirits, a bit of reaction showed up and then in seconds they went nice and silvery and popped into their spherical shape. But no floating and no merging either.

Later on I’m gonna try to and press them into each other with a test tube and some suitable ramming tool: just trying to press the oil in between them out and the potassium metal into itself…

I’ve no dioxane at hand so’ll probably try and make some from glycol.

*************

The test by squeezing the solid balls together wasn’t 100 % successful either but it worked more or less.

4 balls of about 5 mm diameter were loaded to a standard size test tube and just about covered with kerosene.

A ‘ramming rod’ was fashioned which fitted snugly into the tube while allowing enough daylight for the kero to evade the rod. Pressure was thus applied on the balls, deforming them like hard putty and pushing them into roughly a hemisphere (the shape of the bottom of the test tube). But shaking the tube clearly showed the potassium hadn’t merged yet. So I rammed again and left the ramming rod in place while putting the tube into a water bath of about 70 C.

There partial merge occurred: two pieced welded together quickly while two others were creeping up in the daylight between the ramrod and the tube wall. I hasten to add that this effect was non-capillary: the daylight was far too large for that.

Adding the two renegade pieces to the molten blob and a bit of skilful ramming later all potassium was merged into one single ball.

This method certainly has potential (with some improvements) to fairly quickly merge multiple smallish balls together, say 1 - 5 mm. That size often takes an age to merge into larger units too…

[Edited on 4-5-2011 by blogfast25]

m1tanker78 - 6-5-2011 at 12:58

BlogFast, I thought you'd been using kerosene or oil to merge the potassium nuggets all along (separate from the slag). Does the t-alcohol's -OH react with K to form alkoxide? If so, this could be what's making the nuggets stubborn to merge. I had to 'wash' my post-brake fluid Na several times in hot oil before they'd all coalesce. The alk skin seems to linger, causing problems with merging.

Could be the reason for stubborn K in xylene as well?

EDIT: Never mind, I had another look at your proposed reaction scheme and saw that it does produce alkoxide. I'm going to try the strainer method (in the refining thread) and see if that makes things any easier.

Tank

[Edited on 5-6-2011 by m1tanker78]

blogfast25 - 6-5-2011 at 13:39

BlogFast, I thought you'd been using kerosene or oil to merge the potassium nuggets all along (separate from the slag). Does the t-alcohol's -OH react with K to form alkoxide? If so, this could be what's making the nuggets stubborn to merge. EDIT: Never mind, I had another look at your proposed reaction scheme and saw that it does produce alkoxide. I'm going to try the strainer method (in the refining thread) and see if that makes things any easier.

Tank

[Edited on 5-6-2011 by m1tanker78]

Yes, the catalyst reacts with the K but remember that there isn't much catalyst (0.1 mol per mol of K), so that at the end of the reaction all t-alcohol is present as K t-alkoxide. This does not impede coalescence in any way. And presence of the alkoxide cannot be avoided anyway.

I've already removed the solvent (containing the alkoxide) and replaced it with fresh solvent: it makes no difference: K is difficult to get to coalesce, period.

[Edited on 6-5-2011 by blogfast25]

m1tanker78 - 6-5-2011 at 14:10

I also removed the Na from -OH and alkoxide source. My point was that the alkoxide (presumably) lingers on the surface of the Na. It doesn't matter if I take the oil to 100 degrees or 300 degrees. As the nuggets become larger, the 'skin' becomes more visible when molten. I never had so much trouble coalescing even the dirtiest Na before.

Quote:

K is difficult to get to coalesce, period.

I disagree (but can't put it to test). I would argue that K that's never been in contact with -OH will merge relatively easily. Like you said, it's not an option here but the distinction should be made, IMO.

Did you ever get around to trying dioxane (from glycol)??

Tank

blogfast25 - 7-5-2011 at 06:15

It’s simply not true, Tank: in the several tests I’ve carried out of late all my coalescing experiments were with pieces of K with perfectly clean skin (bar that thin blueish film you always get with freshly cut K), cut from a much larger blob (about 2 cm wide) and they show clearly show that that K too is hard to get to clump together, no difference.

It’s mentioned also higher up by people like len1 as a property intrinsic to K, not to how it was produced. It has no ‘memory’ of the latter.

Dioxane? Give us a chance, man! I’ve yet to synthesise some because it’s quite expensive. That’ll be sometime next week. Synth will be time consuming to get rid of the last bits of water/acid in it. And I don't expect miracles from dioxane eithr, just an improvement, see also the patent itself.

The other solvent that interests me is the ‘ethyl ester of brake fluid’ but I’ve no glacial acid at hand. Because of the multiple ether functions in the brake fluid an esterified version would be a bit ‘dioxane like’…

[Edited on 7-5-2011 by blogfast25]

m1tanker78 - 7-5-2011 at 09:24

BlogFast: I don't know how the following translates into chemical/scientific terms so bear with me a little bit.

I have no reason to doubt that the film that forms on post-BF treated Na is sodium alkoxide so I'm running on that basis. This alkoxide film is 'repelled' by the molten sodium and by the oil. If either of those were to fail then the film wouldn't be a .... well, film. With that in mind, if you take some sodium (assume K too) nuggets, cut them in pieces and bring them to MP in a HC, the alkoxide film is going to 'extend' to cover the entire metal/HC interface once again. You'll now have fragments of the stubborn metal (counter intuitive to the goal which is merging).

Pressing the solidified K obviously disturbs the [immobile] film so that's why some of your pressed K successfully merged.

According to the net, alkoxide begins to decompose from 100 to 300 degrees C. But would it truly decompose in contact with a reactive metal?? I believe not, especially not at higher temps.

This is my basis for wanting to differentiate coalescing alkali metals produced by catalytic t-alcohol and alkali metals in the general sense. This may also help me further understand what the hell's taking place in this alcohol-catalyzed reaction. As I said before, I never recall having much trouble merging Na that has never seen brake fluid which reinforces the above stated. Sodium oxide forms a not-so-subborn crust or easily diffuses into hot oil then settles out after cooling.

++++++++++++++++++++

Two relatively easy experiments pop into mind with K. One is to take two large-ish spheres of K and cut out a cube with each (or somehow 'peel' the outer layer). Immediately place the cubes in HC and attempt to coalesce.

The other is very dangerous but may be worth a try sometime. Take 2 or 3 K globs and give each one a very quick dip in ice water. Quickly plunge them in HC, preferably with an intermediate dip in a non-flammable HC.

Both of these procedures should produce K that's mostly free of alkoxide, only ordinary oxides, and will probably merge much more easily.

Just thinking out loud here....

Tank

blogfast25 - 8-5-2011 at 08:30

Tank:

You claim that you have ”no reason to doubt that the film that forms on post-BF treated Na is sodium alkoxide” yet I see no reasons to underpin that claim.

Let’s start with the potassium alkoxides of t-butanol and 2M2M (the t-alcohols we know work). These do not ‘form a film’. We know these to be effective because they HAVE to form alkane-soluble potassium alkoxides. Without alkoxide solubility, no sustained reaction could take place. So forget about a film of K alkoxide on the potassium. It ain’t there.

How about the sodium alkoxide formed when sodium reacts with the brake fluid polyether ester hydroxyl functionality? Well, I can’t vouch for it but in all likelihood the resp. alkoxide will be highly soluble in the brake fluid. Why not test this with a nice, clean cut piece of sodium and some fresh (and dry) DOT 3? The potassium dissolves without leaving a trace in DOT 4.

The experiments you prescribe are interesting but IMHO simply not necessary. ‘Cleaning’ the potassium with a quick dunk in ethanol has already been tried: it makes no difference. Freshly prepared, cooled potassium (using our method) looks precisely like you would expect it to: typically blueish.

You’re looking too far a field. I’d really try and esterify the brakefluid if I was you, then remove all traces of excess acid and water and see where that leads.

The real causes of the difficult coalescence of K and Na are:

1. Surface tension not high enough: much higher surface tension would favour small globules merging into larger ones because for the same volume, one larger globule has less surface area than several small ones. Hence looking at more polar solvents like dioxane, which have higher surface energy.

2. The almost weightlessness of the metals when suspended insolvents of similar density. Imagine how quick mercury would coalesce: the globules congregating at the lowest point of the container would exercise considerable force on each other. Not so with K and Na...

[Edited on 8-5-2011 by blogfast25]

m1tanker78 - 8-5-2011 at 12:17

BlogFast, I'd already dissolved Na in 'dry' DOT 3 (thanks to my neighbor - had an unopened bottle and donated to me). Indeed, it fizzled away in the warm BF without a visible trace. I see your point about alkoxide being soluble in BF. With that out the way, I still maintain that there's a stubborn 'film' on my post-BF Na which apparently is NOT alkoxide and makes coalescing said Na much more difficult.

A few points to keep in mind before dismissing the film claim:

I. Na treated with BF was finicky to merge even after 5 rinses in hot, clean mineral oil. I discarded the oil after each cycle - replenishing with fresh oil.

II. Na that has never been in contact with BF merges extremely easily under oil in spite of it pulling into a tight ball and having surface oxides/crust.

III. Correct me if I'm wrong. Rinsing/dipping K in ethanol isn't much different from dipping in BF in terms of avoiding -OH. Both provide a primary alcohol function, no? If so, the ethanol dip you mentioned was irrelevant, possibly counter-productive if my theory is correct.

I haven't had a chance to look over the complete ingredient list for the particular BF's I've used. There may be something else (unrelated to alcohol function) that causes the annoying film. I recall a similar effect when I experimented with a synthetic heavy oil.

If I can rig something up to safely make an attempt at 'brake fluid ester', I'll give it a try. I need to read up on the process and try to adapt.

Tom

[Edited on 5-8-2011 by m1tanker78]

blogfast25 - 9-5-2011 at 03:13

Again though, the ethoxides and methoxides of K and Na are highly soluble in the respective alcohols. You can even test that with their hydroxides:

(K, Na)OH(s) + (Et, Me)OH(l) === > (K, Na)(Et, Me)O(sol) + H2O(l)

The pure alcohols act as weak acids also towards these strong hydroxides and they dissolve completely, as alkoxides.

For the esterification of DOT 3/DOT 4 you need some glacial acetic acid and some conc. H2SO4.

Another potential coalescing fluid I’ll be testing shortly is isopropyl myristate:

http://en.wikipedia.org/wiki/Isopropyl_myristate

Very OTC (but not cheap), from Dentyl pH. With that end ester group this may have enough polarity to work. I’m drying some now.

[Edited on 9-5-2011 by blogfast25]

m1tanker78 - 9-5-2011 at 13:54

DentylPH, huh? May be worth the investment if a) it works and b) it's at least somewhat resistant to decomposition at elevated temperatures. That being the case, it could be reused at least several times.

Brake fluid ester: I have conc. sulfuric (~98%) but no glacial acetic. Still, I did some tinkering with first blending H2SO4 with BF, then blending that with mineral oil that contained a few small sodium pieces and some block residue. Adding a little bit of H2SO4 to BF at RT reduced reactivity with sodium by half compared to sodium in straight BF in a side-by-side (at RT). I was anticipating an angry reaction with the acid blend but the contrary resulted.

No gas evolved from BF + H2SO4 however, a little warming did occur. Not sure how big a part moisture in the BF played. Hot (Mineral Oil + (BF + H2SO4)) spared and cleaned the small sodium pieces but created some black sludge which hampered my attempts to coalesce. I suppose some of the glycol component was carbonized (dehydrated).

Could IPM be a candidate for conversion to a t-alcohol?

Tank

blogfast25 - 10-5-2011 at 05:08

Could IPM be a candidate for conversion to a t-alcohol?

Tank

Yes, like all fatty acid esters (IPM = the isopropyl ester of myristic acid - tetradecanoic acid), the process is described on the thread:

http://www.sciencemadness.org/talk/viewthread.php?tid=15171&...

Quote:

And isopropyl myristate as the organic phase of ‘Dentyl’ mouth wash can be double Grignarded (that becomes expensive) with MeI to 2-methyl-2-alkanol or with EtBr to 3-ethyl-3-alkanol. Very long chain alcohols…

But I’ve no real experience with Grignards. Also, my gut feeling is that the resulting alcohol may already be too long to be very effective. C6, C7, maybe even C8 would be ideal, I think.

So, it's a bit low on the list of priorities...

[Edited on 10-5-2011 by blogfast25]

Random - 26-5-2011 at 08:44

Beeswax should contain ester between CH3(CH2)28CH2-OH and CH3(CH2)14COOH. This is very long chain alcohol, so could that be substitute for those tertiary alcohols. Maybe there could be some other use too as it's easy to prepare with hydrolysis.

blogfast25 - 26-5-2011 at 11:23

Quote: Originally posted by Random

Beeswax should contain ester between CH3(CH2)28CH2-OH and CH3(CH2)14COOH. This is very long chain alcohol, so could that be substitute for those tertiary alcohols. Maybe there could be some other use too as it's easy to prepare with hydrolysis.

These super long alcohols are almost certainly too long for my purpose (solubility becomes an issue). Also, primary alkoxides don’t appear to be stable enough at around 200 C.

To separate them from the beeswax mix is also very difficult.

LanthanumK - 1-6-2011 at 12:17

Why do these threads inevitably drift?

Megamarko94 - 4-7-2011 at 07:03

Did anyone have any progress on this preparation?

maxidastier - 12-7-2011 at 05:38

Did anyone try it with Dioxane meanwhile?

Bitburger - 5-8-2011 at 10:22

Thermodynamically it makes sense though: NaOH HoF 298 K = - 426 kJ/mol, KOH HoF 298 K = - 425 kJ/mol, MgO HoF 298 K = - 606 kJ/mol:

NaOH (KOH) + Mg - Na (K) + MgO + ½ H2 HoR 298 = -602 + 426 = -176 kJ/mol. Quite a bit of heat... This would be typical of quite a slow reaction which is what we're seeing...

It can be shown that the Rielke reactions take place too with the same logic.

Other example: reduction of AlCl3 with K (or Na) but also reduction of NaOH with Al powder or Mg...

If the t-butyl alcohol somehow solubilises the KOH then that's half the work done...

[Edited on 2-12-2010 by blogfast25]

Where did you find these values, references?
How did you done this calculation?
This reaction is not done under standard conditions, so delta S and delta G are totally different from delta S° and delta H° and so deltaG°.
Have you calculated the phase transitions?

blogfast25 - 5-8-2011 at 10:37

Values of HoF in standard conditions can be found in NIST Webbook.

The calculation is elemental, basic undergraduate stuff but you need to understand Hess' Law.

The correction for higher temperature can be shown to quite neglible in this case (and many like it). Total Delta H at 200 C is almost identical to Delta H at RT, perhaps 10 - 20 % difference...

Omniata - 21-8-2011 at 07:52

Just out of curiosity more than anything, don't flame me for asking but I've noted a few things reading this thread in its entirety and would like a few answers to things I've been thinking about, or a discussion at the very least...

I used to do quite a lot of Grignard chemistry some time ago and a few ideas came to light...

1) Has anyone tried using Iodine as an activator for the Magnesium? Potassium reacts with Magnesium Iodide to form Reike Magnesium (far more reactive)

3) The product of reacting t-BuOH with Potassium is, obviously, Potassium tertiary butoxide. As this can be readily purchased and is a salt/solid, would it not be more efficient considering the reaction conditions to add the Potassium salt as opposed to the alcohol thus reducing the amount of water or hydrogen produced?

Therefore, the reaction could be begun by heating the KOH and Mg to about 100°C to begin the dehydration then as the temperature starts to rise add a crystal/bead of Iodine to activate the Mg further, then add the t-BuOK salt to the reaction...

Just an idea, I know the process works as it stands but one of the limiting factors seems to be creating a "dry" reaction... So far this appears to have only been achieved by adding extra Mg or using a high boiling point solvent...

[Edited on 21-8-2011 by Omniata]

[Edited on 21-8-2011 by Omniata]

blogfast25 - 21-8-2011 at 12:17

Omniata:

The idea of adding the catalyst as a K-salt probably works but potassium t-butoxide is far more expensive (and harder to get) than t-butanol...

Omniata - 21-8-2011 at 13:19

Omniata:

The idea of adding the catalyst as a K-salt probably works but potassium t-butoxide is far more expensive (and harder to get) than t-butanol...

I'm inclined to disagree...

It would cost me £25 for 500g of t-BuOK and £80 for 500ml of t-BuOH...

It's also far easier for me to get the K salt, hence my query...

blogfast25 - 22-8-2011 at 03:49

Well, try it by all means, it should work. It would further confirm the reaction path theory we've developed if it did work...

Have you compared price on mol/mol basis? And where would you buy the K t-butoxide?

[Edited on 22-8-2011 by blogfast25]

Omniata - 22-8-2011 at 09:56

Apollo scientific stock it but can be "haggled" with...

Another supply was through eurolabsupplies ...

I'd like to know about using Iodine as a catalyst...

I'm setting up to do a run in a few weeks, just wanted to see if others had tried these options first before I go ahead blind...

blogfast25 - 22-8-2011 at 10:12

Thanks!

Iodine? I understand the idea but from my experience with this method most Mg powders work well, no need for iodine activation. Considering there is a lot of alkali present it may not work as it (allegedly) does with Grignards. I use a decent reagent grade Mg, not even very fine.

The patent also mentions reducing butoxides, rather than straight alkali metal hydroxides (look it up near the top of this thread, if you haven't already)...

There is also this very related thread, in case you've missed it:

http://www.sciencemadness.org/talk/viewthread.php?tid=15171

[Edited on 22-8-2011 by blogfast25]

Omniata - 22-8-2011 at 11:24

The other thing that caught my attention with the halogens was Reike Magnesium...

It's produced by reacting an Mg halogen, eg: MgI2, with K under specific conditions, wiki link

This could promote the reaction considerably...

blogfast25 - 22-8-2011 at 11:50

The other thing that caught my attention with the halogens was Reike Magnesium...

It's produced by reacting an Mg halogen, eg: MgI2, with K under specific conditions, wiki link

This could promote the reaction considerably...

Rieke has been touched upon briefly higher up. That reacion is almost the opposite of what we're doing here. Not sure how that 'could promote the reaction'...

Omniata - 22-8-2011 at 13:22

I was thinking in terms of the reactivity... MgI would be produced then consumed by the pure K thus producing highly active Mg... For a small amount of I it should drive the reaction as the active Mg would drive the parent reaction...

Although not a healthy thing to do my other consideration was to using a VERY small amount of Mercury to activate the Mg but then it would form an amalgam with the K

blogfast25 - 23-8-2011 at 04:12

Hmmm... my personal feeling is that more is to be gained by looking at the catalyst, in particular for producing sodium. As the patent indicates, that's a very long process with t-butanol. Several of us here believe it may be due to poor solubility of the sodium alkoxide and that longer chain 2-methyl-alkan-2-ols might resolve that problem. t-alcohols are of course made (among other ways) by means of Grignard alkylation (see relevant thread).

Try perhaps simply replication one of our runs? We could do with some fresh activity here!

Omniata - 23-8-2011 at 09:38

I'm not too sure that it will just be down to the solubility of the Sodium alkoxide...

Sodium has a significantly higher melting point to the tune of almost 40°C greater than potassium and a higher density by 0.1gm/cm³, that's approximately 10% greater than Potassium...

Also the properties of Sodium, shear stresses, etc, are significantly greater compared to potassium...

This would to me indicate that to produce Sodium from the alkoxide you'll need a higher boiling point solvent and more aggressive conditions in the favour of the pure metal to drive the reaction, ie: as I'm sure you already know, conditions that the alkoxide doesn't like and the pure metal does...

In layman's terms from my point of view the production of pure Sodium is far more energetic than Potassium, I'm not talking reactivity here, just that it would seem to take a lot more energy to "construct" solid sodium than Potassium...

Additionally Sodium can form a negative ion due to the reduced nuclear shielding compared to Potassium and this may interfere with the reaction...

Edit: Also as Mg is higher in the periodic table than K and so selectively more reactive due to nuclear shielding, it may require Beryllium by theory to produce the Sodium metal by this method? Just speculation of course...

[Edited on 23-8-2011 by Omniata]

blogfast25 - 23-8-2011 at 09:57

In layman's terms from my point of view the production of pure Sodium is far more energetic than Potassium, I'm not talking reactivity here, just that it would seem to take a lot more energy to "construct" solid sodium than Potassium...

Additionally Sodium can form a negative ion due to the reduced nuclear shielding compared to Potassium and this may interfere with the reaction...

Edit: Also as Mg is higher in the periodic table than K and so selectively more reactive due to nuclear shielding, it may require Beryllium by theory to produce the Sodium metal by this method? Just speculation of course...

[Edited on 23-8-2011 by Omniata]

At 200 C reaction temperature, the MP of solid Na is almost certainly not at play here: the metal is well above its MP...

As regards the energies and reactivities:

for 2 MOH(s) + 2 Mg(s) == > 2 M(l) + 2 MgO(s) + H2(g)

... the reaction enthalpy for M=K and M=Na is almost identical, because the Heats of Formation of NaOH and KOH are almost identical. And not much different for Rb and Cs. These small differences are the unlikely to be the cause of sodium being slower to produce. Lower concentration of the Na alkoxide would lower the reaction speed (see proposed reaction mechanism higher up in the thread), at least that would fit what we know.

Of K and Na, K is the more reactive one here, with reactivity increasing further to Rb and Cs (see for instance their reactivities towards water). But that difference doesn't seem to play here because the reaction is driven by lattice energy of the precipitating MgO and by the escaping hydrogen.

Be is unaccessible to almost anyone, at least in significant quantities. Also it may not form stable enough alkoxides

[Edited on 23-8-2011 by blogfast25]

Omniata - 23-8-2011 at 10:28

I'm aware of the reactivity of the alkali metals, I'm fully aware of that from studying high school chemistry, albeit many moons ago, and they aren't at dispute here...

My query was with relevance to the positioning of Mg relative to K in the periodic table and the structural properties of Na vs. K.

blogfast25 - 23-8-2011 at 11:20

My query was with relevance to the positioning of Mg relative to K in the periodic table and the structural properties of Na vs. K.

Neither of these things are really relevant here. What matters for any chemical reaction is the change in Gibbs Free Energy ΔG = ΔH - TΔS when going from left to right (state 2 - state 1). ΔG < 0 means the reaction is thermodynamically speaking possible, ΔG > 0 not.

Secondly, even if ΔG < 0, that still doesn’t mean the reaction will proceed spontaneously in the given conditions because there may be thermal hindrances to be overcome. That is actually the case here: simply heating a mixture of KOH and Mg in a non-polar, aprotic solvent to 200 C does not cause reaction. That’s where the catalyst comes into it: it lowers the thermal hindrance so that that stuff can go on at about 200 C.

That’s why looking at potentially better catalysts is of real interest here.

blogfast25 - 24-8-2011 at 04:36

Be careful, if a nuclear property is measurable it is applicable...
That is default by basics laws of physics...

[...]

Sometimes it's good practice to take a step back and look at the whole picture, not just what you want to see

, or what you think you are seeing...

[...]

No offence intended...

Hmmm… the only nuclear property that is at play here is the atomic number because it determines the quantum mechanics of the electron cloud of the elements involved and thus their chemistry. This kind of chemistry is determined by the properties of the electron cloud, not by nuclear properties.

The whole picture? I don’t see you looking at the whole picture at all. Your dismissal of the thermodynamics of the reaction is telling.

As regards:

”I started the "fresh look" based on looking through the smokescreen and what was actually taking place, NOT what was being SEEN by blinkered eyes!!!”

Wow! And what have you actually contributed with this ‘open mind’ of yours? ‘Blinkered eyes’? Sure, it’s those blinkered eyes of many here that have carried this project forward!

No offence was taken or intended here but you need to get down and dirty and run some tests before you can make much of any kind of claim. You’re looking through ‘the smoke screen’? O-kay: what have you seen so far?

watson.fawkes - 24-8-2011 at 06:12

Be careful, if a nuclear property is measurable it is applicable...
That is default by basics laws of physics...

No. Such a difference might be applicable, and it might not be applicable. Not every difference leads to a distinction. The essence of good scientific theory is the elimination of extraneous causes, or at the very least ranking causes in order of significance of their effects.

Quote:

We're not discussing Gibbs free energy or thermodynamics, we're looking at basic principles and analysing the results...

You'll be doing bad science if you simply toss away what others have already been thinking about this system. What you are saying, with this comment, is that you are ranking whatever is in your head above everything discussed before. This is pure arrogance, and I don't see yet even a shred of justification for it.

Quote:

Sometimes it's good practice to take a step back and look at the whole picture, not just what you want to see

, or what you think you are seeing...

In general, that's true. Yet all I've seen is that you've observed differences only, without positing any mechanism about how these differences would be significant.

All the preceding is the polite and precise way of stating what I first thought upon reading your post: Put up or shut up.

blogfast25 - 24-8-2011 at 06:38

Definitely better put than I did, dear Watson.

ScienceHideout - 24-8-2011 at 09:15

Sorry: too lazy to read all of the posts over again- but did we try this with toluene? It could work- toluene is very versatile.

redox - 24-8-2011 at 09:16

Toluene as a solvent? It boils a 110 or so. That is much too low a temperature for this reaction.

[Edited on 24-8-2011 by redox]

ScienceHideout - 24-8-2011 at 09:22

What's the opposite of boiling point depression? I can't think of the word... we could mix the toluene with other stuff and make it boil at a higher temp.

Omniata - 24-8-2011 at 09:50

Quote: Originally posted by ScienceHideout

What's the opposite of boiling point depression? I can't think of the word... we could mix the toluene with other stuff and make it boil at a higher temp.

The only way you could make toluene boil at a higher temperature is to increase the pressure...

Aryl compounds aren't a good idea though due to the reactivity towards K

blogfast25 - 24-8-2011 at 12:04

Toluene isn't a good idea full stop. Ordinary kerosene does just fine, as do several other high boiling, inert, aprotic solvents.

Ordering D70

Takron - 22-9-2011 at 21:02

There are 35 pages to go through to see if this has been said already, but I found the site they bought the D70 at and it's only 7EU for a liter, which is like 10 or 12 bucks US I think. I don't know what the shipping is yet but they ship worldwide and to individuals if I read it right. I'm not sure on the hazard charges if there are any. I looks like though that getting D70 is totally doable.

Link Here

blogfast25 - 23-9-2011 at 04:28

Quote: Originally posted by Takron

There are 35 pages to go through to see if this has been said already, but I found the site they bought the D70 at and it's only 7EU for a liter, which is like 10 or 12 bucks US I think. I don't know what the shipping is yet but they ship worldwide and to individuals if I read it right. I'm not sure on the hazard charges if there are any. I looks like though that getting D70 is totally doable.

Link Here

Still, why invest so much money in a 'fancy' solvent when a good quality lamp oil (kerosene) or high paraffin will do?

Kremer has a US shop advertised, so you may get the Shellsol from there, if you insist on having it...

[Edited on 23-9-2011 by blogfast25]

AndersHoveland - 23-9-2011 at 09:23

D70 petroleum solvent has longer molecular chains, which means it can be heated to a higher temperature with less danger of catching fire. The hydrocarbon chains in keresene are typically typically 6-14 carbon atoms long. Those lighter chainer are going to vaporize out if it is strongly heated, which is generally a very bad thing.

Shellsol D70 boils at around 200degC, if I remember correctly. It contains about 60% hydrocarbons and 40% cycloalkanes (saturated hydrocarbons in a ring). It often also contains traces of benzene (only around 1-2%).
The molecular chains in D70 typically contain baround 14 to 15 carbon atoms.

[Edited on 23-9-2011 by AndersHoveland]

blogfast25 - 23-9-2011 at 11:56

Not really, Anders. My lamp oil (deodorised kerosene) boils also at around 200 C and makes perfect potassium. Read the thread and you'll see there's really nothing much special about Sh. D70. I have it and used it at first, now I just don't spend my money on it anymore. Others have used candle wax (high paraffin) and other hydrocarbon based solvents. As long as it's inert and has a sufficiently high boiling point it works. That's what this thread clearly shows, among other things...

Remember, this needs to be done under complete reflux, so some volatility of the solvent is simply not an issue.

Neil - 26-10-2011 at 11:15

"A method of preparing magnesium alcoholates Mg(OR)2 (where R=alkyl residue having 2 to 10 carbon atoms) is described. Addition products of 1,3-dienes to magnesium metal in a polar, aprotic solvent are reached with an alcohol R—OH. "

Looked relevant.

Attachment: Magnesium alcholates.pdf (50kB)
This file has been downloaded 843 times

blogfast25 - 26-10-2011 at 11:48

Definitely. Thanks.

Neil - 26-10-2011 at 12:27

It strikes me that no one mentions using petroleum jelly (Vaseline)

http://www.bmed.mcgill.ca/REKLAB/manual/MSDS/Materials%20Lis...

The above MSDS gives a boiling point of 343oC

blogfast25 - 26-10-2011 at 12:31

Actually, it has been mentioned by Nurdrage and me. The problem is that it resolidifies to a waxy solid, making it difficult to recover the metal, in particular the fines, which get stuck in there.

Why is everybody always pickin' on my kerosene, that's what I want to know!

Neil - 31-10-2011 at 16:18

Go Kero go, no hate here. There was a lot of not being able to get a suitable solvent in the thread, Vaseline is about as OTC as you can get.

Preparation and thermal decomposition of Magnesium Alcoholates.

Attachment: magnesium alcholates decomp points.htm (81kB)
This file has been downloaded 1002 times

let me know if it does not open, I ripped it from a host site.

MyNameIsUnnecessarilyLong - 31-10-2011 at 16:54

Has anyone made cesium with this method yet? Do you guys think paraffin oil/mineral oil would work ok?

[Edited on 11-1-2011 by MyNameIsUnnecessarilyLong]

Neil - 31-10-2011 at 16:59

Kerosene is just the trade mark name for paraffin oil.

AndersHoveland - 1-11-2011 at 14:20

The problem is that [the petroleum jelly] resolidifies to a waxy solid, making it difficult to recover the metal, in particular the fines, which get stuck in there.

Could not the petroleum jelly ("Vasoline") be dissolved away with kerosene (or even more volatile petrol) after the reaction?

Petroleum jelly would seem advantageous over directly using kerosene because there would not be danger that it would catch fire when being heated.

blogfast25 - 3-11-2011 at 07:27

Probably. What's to be gained by this complication though?

Sedit - 3-11-2011 at 08:06

Quote: Originally posted by AndersHoveland

The problem is that [the petroleum jelly] resolidifies to a waxy solid, making it difficult to recover the metal, in particular the fines, which get stuck in there.

Why wouldn't there be? Its still an oil albeit a very viscous one.

http://www.youtube.com/watch?v=cwW1VG3w3yc

Neil - 3-11-2011 at 08:09

Using a reaction flask run the reaction with Vaseline thinned with a little kero, let cool and gel.

Scrap the gel out into a strainer and proceed with M1tankers method in the K/Na refining thread?

Seems like it could simplify things, from my peanut gallery seat.

Neil - 3-11-2011 at 08:19

Quote: Originally posted by Sedit

Quote: Originally posted by AndersHoveland

The problem is that [the petroleum jelly] resolidifies to a waxy solid, making it difficult to recover the metal, in particular the fines, which get stuck in there.

Why wouldn't there be? Its still an oil albeit a very viscous one.

http://www.youtube.com/watch?v=cwW1VG3w3yc

It burns well once you get it going, I've used it cut with kerosene to make fuels with great success.

Getting it to burn is just more difficult, it needs to really boil to burn or have a wick, when it does go it burns like boiling paraffin wax. Like paraffin it is just harder to get it to burn alone then it is with kerosene.

petroleum jelly may be more dangerous then kerosene for no other reason then if it does ignites it will react much more violently with water as it will be much hotter then liquid kerosene can get.

AndersHoveland - 3-11-2011 at 16:13

Lamp oil which is commonly available for outside torches could also be used. It can be purchased at a camping store.

Neil - 3-11-2011 at 16:28

Lamp oil = Kerosene...

blogfast25 - 4-11-2011 at 06:56

Ergo: Kerosene = Lamp oli...

Can we move on now?

oldtimer - 10-11-2011 at 18:40

This thread has kept me from sleeping early the last 3 nights, it was some of the most fascinating reading I found for a long time - both because of the subject, but even more so seeing knowledge evolve, people contributing, errors made and discovered. Thanks to everyone.

A small contribution concerning the solvents:

This Shell site shows the available grades of Shell's saturated, low aromatic solvents, and the geographic regions in which they are sold. The D70 is only available in Europe and Africa. Most likely, it is produced in a European refinery. There is a D80 grade, with a slightly higher boiling range, that might be a replacement and is available in America.

A little googleing revealed that other oil companies offer very similar products.

Knowing that oil companies are not only competitors, but actually cooperate in a lot of fields, I would not be surpised if Shell's and Exxon's D80 run out of exactly the same pipe in a refinery in the U.S., and that one could find identical products also from other suppliers. It might be worth the effort having a talk with the local sales office or distributor of one of the oil co's. One might end up buying a 5 gal can - at the same price as 1/2 l of a reagent grade hydrocarbon. But at least you buy something that according to it's spec sheet should work.

Lamp oils etc. are certainly the same stuff - one just does not know which stuff exactly. And the "official" solvent is definitely free of citronella.

condennnsa - 11-12-2011 at 00:22

I ran out of KOH , but have a lot of KNO3
what is the best way to make KOH from KNO3?

I figured, I'd add a stoichiometric amount of 35% H2SO4 to KNO3, then boiling the solution to dryness outside should boil off all the nitric acid and leave me with K2SO4 / KHSO4
Then dissolving this back in water and mixing with a slurry of excess Ca(OH)2, stirring it for a couple of hours , will the calcium precipitate as CaSO4? leaving the KOH in solution

I know that from potassium carbonate , CaCO3 precipitates with no problem, but I'm not sure about the sulfate since wikipedia lists CaSO4's solubility at some 0.2g/100ml , much more than CaCO3 ?... thanks

blogfast25 - 12-12-2011 at 10:22

It's one thing to make fairly weak solutions of KOH, quite another to try and obtain solid KOH which is very hard to do.

Go and buy some more KOH and don't waste your time trying to convert KNO3...

Neil - 23-12-2011 at 05:18

soulless spam ghoul reported.

garry21 - 23-12-2011 at 09:45

I guess i could use Magnesium Ribbon
to remplace Magnesium turnings
wonder if the wield is better when using Sodium Hydroxide

Arthur Dent - 24-12-2011 at 05:48

@ garry21

Magnesium ribbon has a large surface area, and the fine oxide formed on the surface might impede the reaction, I believe it was determined in previous posts that a source of Magnesium devoid of surface oxidation would react more readily with the potassium hydroxide. Some have even suggested grinding a Magnesium block on a file under mineral oil to ensure less oxidation on the surface of the metallic particles.

I have some magnesium ribbon also and never thought of using it for that reaction... perhaps it could work. Still haven't tried it even though I have gathered all of the necessary reagents and glassware.

Also, it was determined that this particular experiment doesn't appear to work with Sodium, only Potassium.

Robert

[Edited on 24-12-2011 by Arthur Dent]

[Edited on 24-12-2011 by Arthur Dent]

blogfast25 - 24-12-2011 at 08:36

Quote: Originally posted by Arthur Dent

@ garry21

Magnesium ribbon has a large surface area, and the fine oxide formed on the surface might impede the reaction, I believe it was determined in previous posts that a source of Magnesium devoid of surface oxidation would react more readily with the potassium hydroxide. Some have even suggested grinding a Magnesium block on a file under mineral oil to ensure less oxidation on the surface of the metallic particles.

I have some magnesium ribbon also and never thought of using it for that reaction... perhaps it could work. Still haven't tried it even though I have gathered all of the necessary reagents and glassware.

Also, it was determined that this particular experiment doesn't appear to work with Sodium, only Potassium.

Robert

[Edited on 24-12-2011 by Arthur Dent][Edited on 24-12-2011 by Arthur Dent]

A couple of corrections.

Just about any old magnesium should work but the greatly reduced surface area for ribbon should make this reaction a slow boat to china. The state of 'passivation' of the Mg may play some part but it's far from clear. All Mg is passivated to some degree anyway, the element being very reactive (electropositive) towards oxygen.

Accoding to the patent, sodium synthesis using this method also works but it takes much more time. That is why some of us are looking into tertiary alcohol catalysts with much longer chain lengths: the sodium alkoxides of these are believed to be more soluble in paraffinic solvents. No definite conclusions have yet been formulated though.

[Edited on 24-12-2011 by blogfast25]

Arthur Dent - 24-12-2011 at 08:49

I stand corrected. I'll put away my ribbon.

I'll look into a Mg block then.

Have a great Xmas break, blogfast25!

Robert

blogfast25 - 24-12-2011 at 11:02

U 2!

Some pencil sharpeners are made of Mg (sounds crazy but it's true). Grinding even under water should be good enough for this reaction.

condennnsa - 28-12-2011 at 08:25

nurdrage put out his video on potassium : http://www.youtube.com/watch?v=gHgyn-wsxFw&feature=g-u&a...

blogfast25 - 28-12-2011 at 12:48

Excellent video! As per usual from NR…

A few comments.

Tetrahydronaphthalene (trademark: Tetralin) is a very expensive solvent (Sigma Aldrich $417/20 L) and few, if any, will sell it to hobbyists. I wonder also whether the cost is justified here. Earlier up this thread NR claimed he’d done a reaction with Tetralin as solvent that took only an hour to complete but that doesn’t seem to be the case here. Also the claimed faster coalescence of the formed potassium isn’t really that apparent. I remain happy to use the much more OTC and much cheaper kerosene (lamp oil).

NR remains dedicated to the rather ritualistic addition of the catalyst in stages, as prescribed by the source patent but my own experiences with adding all the catalyst at once and at the very beginning (so-called ‘one pot’) are very positive and indistinguishable from the ‘phased addition’ results.

I’m surprised to see some of his potassium globules go black while the reaction is still ongoing. Could this be the result of prolonged exposure to the solvent, which contains double bonds? I've never experienced that with kerosene or Shellsol D.

[Edited on 28-12-2011 by blogfast25]

mr.crow - 28-12-2011 at 15:37

Quote: Originally posted by condennnsa

nurdrage put out his video on potassium : http://www.youtube.com/watch?v=gHgyn-wsxFw&feature=g-u&a...

Incredible! I'm glad he got the video going

White Yeti - 29-12-2011 at 12:26

Has anyone calculated the enthalpy change of:
2Mg + 2KOH ---> 2K + 2MgO + H2 ?

I calculated it and came up with +790KJ/mol. That's quite an energy barrier! And yet the synthesis seems to work...

Can anyone double check the enthalpy change of this reaction? I may have made a mistake (the first time I calculated this enthalpy change I made a mistake).

blogfast25 - 29-12-2011 at 14:03

WY:

That's wrong. It's been done higher up: about - 200 kJ/mol of K, off the top of my head. Per mol of K it's basically:

HoF(298, MgO) - HoF(298, KOH) = - 602 - (-425) = - 177 kJ/mol of K (@ 398 K). The escaping hydrogen helps drive the equilibrium further to the right.

There is a kinetic barrier, even at 200 C but it is overcome by the catalyst: that's what catalysts do.

[Edited on 29-12-2011 by blogfast25]

neptunium - 30-12-2011 at 07:41

in the old days (1810's) they used to prepare K by passing molten KOH on iron turnings heated over 1000C in the canon of a riffle. one end sealed by a mercury bath. on the other side, pure potassium metal would come running down .
i might try to prepare large quantity of K that way one day...sounds like a fun challenge

White Yeti - 30-12-2011 at 10:13