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Author: Subject: KCl+Li does react and isolates K metal but why not KCl+Mg ?
metalresearcher
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[*] posted on 4-8-2023 at 09:12
KCl+Li does react and isolates K metal but why not KCl+Mg ?


Recently I heated up KCl + Li metal in a retort to 800 C and captured K vapor in mineral oil condensing to small beads of K metal.
The chemistry behind is that K has a much lower boiling point (768 C) than Lithium so K vapor is driven off from the reaction vessel and can be captured in a receiver under mineral oil which I did successfully.

Following this rule: one of the reaction results being driven off, because magnesium has about the same boiling point as Li and K has a much lower bp, it should isolate K metal in the same way, leaving liquid MgCl2 in the vessel.

2KCl + Mg => (?) 2K + MgCl2
150 24

I wanted to try this on a smaller scale: I bent a 10mm steel tube about 90 degrees, squeezed one end and welded it watertight on that end. Then I filled in a ratio 25 : 4 (== 150 : 24) 0.4g Mg lathings and 2.5g KCl and forced it to the closed end. I put a blob of Kaowool on the open end to prevent too easy accessibility to ambient air and oxygen.
I heated it to 1000 C for a few minutes (the closed end was yellow hot), so I expected K2O fumes or even small flames on the open end.
But, nothing happened.
To confirm this, I let it cool and removed the blob and added water into the open end and no reaction of possible K metal inside the tube. Conclusion: NO reaction happened.

Why ?
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clearly_not_atara
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[*] posted on 4-8-2023 at 09:35


Enthalpy of 2 KCl is probably much lower than MgCl2. For KOH vs MgO, compatibility of Mg2+ in crystal structure with O2- overrules naïve electron-orbital considerations in determining the low energy equilibrium, but MgCl2 would feature tiny Mg2+ holding onto two Cl- balloons — hardly favorable!



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RustyShackleford
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[*] posted on 4-8-2023 at 14:28


KCl: ΔHf(l) = -421 kJ/mol
Mg: ΔHf(l) = 4.79 kJ/mol | Li : ΔHf(l) = 2.38 kJ/mol
MgCl2 : ΔHf(l) = -601 kJ/mol | LiCl : ΔHf(l) = -390.76 kJ/mol
K: ΔHf(g) = 89 kJ/mol

ΔHr (Mg+2KCl=> 2K + MgCl2) = 414.21 kJ
ΔHr (2Li +2KCl=> 2K + 2LiCl) = 233.76 kJ

Formation by Mg requires almost twice as much energy, waay less favorable. a few minutes of heating would surely produce only a trace. Does the Li route even make any that fast? i thought the distillation took a few hours in that case?

[Edited on 4-8-2023 by RustyShackleford]
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[*] posted on 4-8-2023 at 15:03


RS: note that the enthalpy difference for the Li rxn in your calculations is 80% just the enthalpy of vaporization of K! So if we are boiling K then we should be most of the way there. But boiling K is obviously a little dangerous.



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[*] posted on 4-8-2023 at 15:43


further inquiry to the NIST database reveals the real issue, entropy .
ΔS for the Li reaction is 117.84 J/mol-K, ΔH being 117.65 kJ/mol yields spontaneous reaction at ~ 1000K
ΔS for the Mg reaction is 241.77 J/mol-K, ΔH being 414.21 kJ/mol yields spontaneous reaction at 1717K, much above the 1000C OP mentioned.

Funnily enough this temperature criteria can be seen in codys video when he did the Na reaction (for which the reaction is spontaneous above 1108K): https://youtu.be/PkTaz3UEThY?t=176
He didnt get any product at 1400F (1033K), but as soon as he changed to 1550F (1116K) he got the potassium. He also failed the Mg reaction with 1600F (1144K)


[Edited on 5-8-2023 by RustyShackleford]
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[*] posted on 4-8-2023 at 16:34


This may be useful for the math
https://www.sciencemadness.org/whisper/viewthread.php?tid=15...

For the reaction at 1000c
Mg + 2KCl = MgCl2 + 2K

ΔH  234.7308  kJ/mol
ΔS  34.5013  J mol/K
ΔG  190.8055  kJ/mol 




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4-8-2023 at 20:09
metalresearcher
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[*] posted on 4-8-2023 at 22:37


Quote: Originally posted by RustyShackleford  
further inquiry to the NIST database reveals the real issue, entropy .
ΔS for the Li reaction is 117.84 J/mol-K, ΔH being 117.65 kJ/mol yields spontaneous reaction at ~ 1000K
ΔS for the Mg reaction is 241.77 J/mol-K, ΔH being 414.21 kJ/mol yields spontaneous reaction at 1717K, much above the 1000C OP mentioned.

Funnily enough this temperature criteria can be seen in codys video when he did the Na reaction (for which the reaction is spontaneous above 1108K): https://youtu.be/PkTaz3UEThY?t=176
He didnt get any product at 1400F (1033K), but as soon as he changed to 1550F (1116K) he got the potassium. He also failed the Mg reaction with 1600F (1144K)


[Edited on 5-8-2023 by RustyShackleford]


Where did you find this ? Would be an easy way to calculate whether a reaction is viable. I checked nist.gov, but could not find enthalpy or entropy tables.

[Edited on 2023-8-5 by metalresearcher]
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[*] posted on 5-8-2023 at 02:48


Quote: Originally posted by metalresearcher  


Where did you find this ? Would be an easy way to calculate whether a reaction is viable. I checked nist.gov, but could not find enthalpy or entropy tables.

[Edited on 2023-8-5 by metalresearcher]


https://webbook.nist.gov/chemistry/

Click one of the search options, e.g. by formula, find your compound, then under "Gas phase thermochemistry data" or
"Condensed phase thermochemistry data" you can find standard formation enthalpies and entropy. There is also tables/graphs if you click the options further down the page.

[Edited on 5-8-2023 by RustyShackleford]
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