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Author: Subject: KCl + Ca does release K metal, but KCl + Mg not ?
metalresearcher
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[*] posted on 5-8-2024 at 09:18
KCl + Ca does release K metal, but KCl + Mg not ?


In the recent past I successfully isolated potassium metal by heating Lithium + KCl in a retort to 800 C and captured balls of K under mineral oil.
Now I saw this video: https://youtu.be/Ckp22QCsFyM where the guy uses Calcium instead of Li and it went successfully. I don't have Ca metal but I do have Mg Metal, so I gave it a try.
Unfortunately nothing happened.

Ca (solid / liquid) + 2KCl (liquid) <=> CaCl2 (liquid) + 2K (vapor)

Weird, because Li, Ca and Mg have all three a much higher boiling point (1100-1400 C) than K (768 C) and that is why the first two can isolate K despite these metals being more 'noble' than K.
The reason is that the K is formed as a vapor which is driven off and that shifts the equilibrium to the right, favoring K metal.

As told earlier I got it from Li metal and the guy in the video from Ca metal.
I used 1.8g Mg metal and put it under 11g of KCl and heated it to 800..1000 C. At that temperature both Mg and KCl are liquid (but don't mix like oil and water) and have intimate contact with each other. I used the same setup: a stainless steel retort with attached a steel tube ending in a mineral oil bath.

Why does the Mg not react, unlike Ca or Li ?
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[*] posted on 5-8-2024 at 10:12


maybe because the reactivity series:

"In the reactivity series, the metals with the highest propensity to donate their electrons to react are listed first, followed by less reactive ones. Therefore, a metal higher on the list can displace anything below it.

Metals: Cs, Rb,K, Na, Li, Ba, Sr, Ca, Mg, Al, Cr, Mn, Zn, Fe, Cd, Tl, Co, Ni, Sn, Pb, H, Sb, Bi, As, Cu, Hg, Ag, Pd, Pt, Au, Ir, Rh, Os.

This order represents also, as regards the metals, the relative power of displacing other metals from salt solutions; so that any chosen metal can precipitate from solutions of equivalent strength any metals that follow it in the list.

From a compound of an element with an electro-negative atom or group, a more electro-positive element will displace a less electro-positive. Thus, as before shown, zinc, which is more electro-positive than copper or silver will displace these metals from solutions of their salts, whilst copper will displace
silver from solutions of silver salts.

I know this is for solution, but in your case one the element is removed (for example K by evaporation). So maybe the above does not apply.




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metalresearcher
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[*] posted on 5-8-2024 at 10:26


RU_KLO: That is indeed the case. Normally, even Li or Ca releasing electrons from K+ in a salt solution (mix of molten KCl +LiCl) would violate this rule.
But, thanks to evaporation the K can be freed.

BTW, in the meantime I used the same retort with 1g Li and 10g KCl which did capture balls of K metal under the oil. So the retort is not the issue.
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Rainwater
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[*] posted on 5-8-2024 at 17:54


Running some numbers through a calculator gives the following results I find interesting
$$ Li + KCl \rightarrow K + LiCl $$
T =1173.15 Kelvin
ΔH 114.3529 kJ/mol
ΔS 32.7189 J mol/K
ΔG 75.9687 kJ/mol
The calculator says this reaction will not work at this temperature.
But you have some spicy silverish metal that says it does.

Now this calculator has an unknown error with a complex exponential relationship to temperature and entropy of the compounds.
The further away you get from stp the more the error.
Lets pretend its linear so i dont have to do a lot of math.
Lets also assume that at 900c(1173.15K) the error is +76kJ/mol.
So the ΔT = 1173.15-273.15 = 900K
76/32.719/900 = -2.58*10-3 per degree K
And ill just refer to that as the "CF" correction factor.

$$ 2KCl + Ca \rightarrow 2K + CaCl_2 $$ gives me
ΔH 265.0564 kJ/mol
ΔS 87.9058 J mol/K
ΔG 161.9297 kJ/mol
And applying the correction factor it looks like 162+(87.9*CF*900) = = -42
So the reaction should work. And at a lower temperature that lithium.
Likly around 766c if this inaccurate math is accurate enough.
Not really hot enough to distill the product.

$$ 2KCl + Mg \rightarrow 2K + MgCl_2 $$ gives you
ΔH 399.3628 kJ/mol
ΔS 61.8395 J mol/K
ΔG 326.8158 kJ/mol
And after the correction factor
182
So even after fudging the numbers around, its not looking good for Magnesium

$$ 3KCl + Al \rightarrow 3K + AlCl_3 $$
ΔH 981.8593 kJ/mol
ΔS 301.1643 J mol/K
ΔG 628.5484 kJ/mol
My favorite reagent. Al, not looking so good.
After the CF
O wait.628−(301×0.00258×900) = -70.9
Damit man, throw a bit in there and make me look a fool. That would be cool.if it did work.
Bit i dont have high hopes.
This correction factor is not accurate once I started changing cation groups.
I have only confirmed it to work within the same column of the periodic table, and at best it is an educated guess.

[Edited on 6-8-2024 by Rainwater]




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chornedsnorkack
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[*] posted on 6-8-2024 at 10:58


What is your source for delta-S?
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[*] posted on 6-8-2024 at 11:23


Quote: Originally posted by Rainwater  
Running some numbers through a calculator gives the following results I find interesting


Which site did you use ? I cannot find any site.
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[*] posted on 6-8-2024 at 11:37


The calculator is from
https://www.sciencemadness.org/whisper/viewthread.php?tid=15...

Thermodynamic values are copied from
data taken from https://www.drjez.com/uco/ChemTools/Standard%20Thermodynamic... March 6 2022




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[*] posted on 6-8-2024 at 12:33


I calculated

2KCl+Ca→2K+CaCl2 ΔG=0 at 1085K

2KCl+Mg→2K+MgCl2 ΔG=0 at 1523K

3KCl+Al→3K+AlCl3 ΔG=0 at 1895K

This is calculated from NIST values. No correction factor was used.

I believe the reason it works so much lower for Ca than Mg is mostly about the stability of CaCl2 (ΔH=-774kj/mol) compared to MgCl2 (ΔH=-601kj/mol). I speculate that the chlorines bind to the metal center more strong in Ca because it has a larger ionic radius so the chlorines do not have as much repulsive force against each other




[Edited on 6-8-2024 by walruslover69]
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[*] posted on 6-8-2024 at 18:13


At first i was trying to figure why my excell sheet table was so far off. But then i realized some of the data on the NIST website is in kcal, and not kJ. Theres a button to click to swap back to Kj
Im not sure if/how that conversion would effect the math.
I like their tempature variation plots. Showing how the entropy and empathy change with temp.




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[*] posted on 7-8-2024 at 05:33


Quote:
empathy change with temp.

LOL!
Some key data:
KCl:
https://en.wikipedia.org/wiki/Potassium_chloride
standard enthalpy -436 kJ/mol
melts at 770 C
LiCl:
https://en.wikipedia.org/wiki/Lithium_chloride
standard enthalpy -408 kJ/mol
melts at 610 C
CaCl2:
https://en.wikipedia.org/wiki/Calcium_chloride
standard enthalpy -796 kJ/mol
melts at 775 C
MgCl2:
standard enthalpy -642 kJ/mol
melts at 714 C
This makes the standard reaction enthalpies at +25 C, standardized for 1 mol K:
KCl+Li=LiCl+K-28 kJ/mol
KCl+1/2Ca=1/2CaCl2+K-38 kJ/mol
KCl+1/2Mg=1/2MgCl2+K-115 kJ/mol
As you heat the system, the enthalpy changes with temp. There are continuous enthalpy changes with differences of heat capacity between sides of reaction. There are also step changes of reaction enthalpy - modest ones when a reagent melts (or has a solid phase change) and a big one when a reagent boils.
But the reaction entropy will stay relatively small as long as both sides are solid or liquid - and change a lot if a gas appears on one side.
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[*] posted on 7-8-2024 at 10:31


Quote: Originally posted by chornedsnorkack  
Quote:
empathy change with temp.

LOL!

This makes the standard reaction enthalpies at +25 C, standardized for 1 mol K:
KCl+Li=LiCl+K-28 kJ/mol
KCl+1/2Ca=1/2CaCl2+K-38 kJ/mol
KCl+1/2Mg=1/2MgCl2+K-115 kJ/mol
As you heat the system, the enthalpy changes with temp. There are continuous enthalpy changes with differences of heat capacity between sides of reaction. There are also step changes of reaction enthalpy - modest ones when a reagent melts (or has a solid phase change) and a big one when a reagent boils.
But the reaction entropy will stay relatively small as long as both sides are solid or liquid - and change a lot if a gas appears on one side.


These three reactions you specify all release energy at 25 C as the enthalpies are negative. At room temperature ??
So heating up the system might even release more enthalpy ....
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[*] posted on 7-8-2024 at 12:59


Quote: Originally posted by metalresearcher  
Quote: Originally posted by chornedsnorkack  
Quote:
empathy change with temp.

LOL!

This makes the standard reaction enthalpies at +25 C, standardized for 1 mol K:
KCl+Li=LiCl+K-28 kJ/mol
KCl+1/2Ca=1/2CaCl2+K-38 kJ/mol
KCl+1/2Mg=1/2MgCl2+K-115 kJ/mol
As you heat the system, the enthalpy changes with temp. There are continuous enthalpy changes with differences of heat capacity between sides of reaction. There are also step changes of reaction enthalpy - modest ones when a reagent melts (or has a solid phase change) and a big one when a reagent boils.
But the reaction entropy will stay relatively small as long as both sides are solid or liquid - and change a lot if a gas appears on one side.


These three reactions you specify all release energy at 25 C as the enthalpies are negative. At room temperature ??
So heating up the system might even release more enthalpy ....

No, they spend energy:
the enthalpy of formation of KCl is -436, that of LiCl is -408, that is where -28 comes from
etc.
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[*] posted on 7-8-2024 at 18:37


Aside from my spell checking abilities, and a thought experment.
Quote:
Why does the Mg not react, unlike Ca or Li ?

There is some or many principles preventing this reaction.
But thermodynamics can be used to make a prediction about the conditions required to force a reaction to proceed.
The problem with using thermodynamics is the conditions at which the initial values are obtained.
Some compounds have almost no information available.
Some reactions are to complex and not fully written in a way that this calculation would be practical(a.k.a black powder)
And sometimes an over worked and half zombied lab tech mistakenly introduces some kind of error into the data.
I like accurate temperature and got a few pricy things that give me high percision and accuracy around 1100-1400c. This is why I believe that closly matched reactions can be accurately predicted with an error correction technique like I mentioned above.
When doing a reaction, i use these calculations as a method of determining consistency with previous runs, tracking reaction progress, and on more than one occasion, identify reagent contamination.
Speaking of high temps, KCl + Al, 3 hours at 1400c.
I got some very pure kcl that distilled up my calorimeter and prevented the lid from not breaking when opened with great effort.




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[*] posted on 8-8-2024 at 11:09



Now I did another test in the hunt for alkali metals (K).

I used K2CO3 (liquid) + 2C (solid) => 2K (vapor) + 3CO (gas)

I had to crank up the temperature to 1200 C because this shifts to the right only above 1100-1200 C.
I used powdered charcoal and dried potash in mass ratio 1:6 (24:140).
It bubbled vigorously in the receiver of the distiller filled with mineral oil, probably because of first expanding and then released CO.
But the oil got black as some finely divided C entered it, so I have to let it settle overnight and hope a bit of K metal reached the oil.
After finishing I heated the tube to allow possible condensed and frozen K metal let melt and drip to the oil.
After cooling I opened the retort and poured some water in it, no K remains were in it (it was way above the boiling point of K: 768 C).
But pouring water in the tube resulted in small explosions and purplish flames, so there was still some K metal in it.
And the retort was not leaking: pouring water in the retort kept inside, so the weld on the bottom survived the extreme heat. It is all made of stainless steel 316L welded with 316 sticks.

Picture 1: furnace running, picture 2 closeup of the bum of the retort at max temperature.


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[*] posted on 8-8-2024 at 11:58


Ironcarbide and Potassiumhydroxid works at 500C. Ironcarbid was made with iron fillings or powder and heated with tar.
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