Naked Science Forum

Non Life Sciences => Chemistry => Topic started by: amalia on 14/11/2019 12:04:26

Title: Can we displace a more reactive metal in a salt with a less reactive metal?
Post by: amalia on 14/11/2019 12:04:26
Olivier wrote to us to ask the following:
Is it possible to displace a more reactive metal in a salt with a less reactive metal? For example:

 KCl + Na → NaCl + K


Do you know the answer?
Title: Re: Can we displace a more reactive metal in a salt with a less reactive metal?
Post by: evan_au on 15/11/2019 01:35:09
In a solution, the salt ions are just floating around freely, so you can't really say whether it is NaCl or KCl.

However, if you insert an ionic species which is very insoluble, it will drop out of solution, leaving the more reactive ion still dissolved.

However (if I correctly recall my long-ago chemistry lessons), almost all chlorides and almost all sodium salts are extremely soluble. But I guess you could find one which is less soluble than others...

...or use a more exotic solvent than water!
Title: Re: Can we displace a more reactive metal in a salt with a less reactive metal?
Post by: chiralSPO on 15/11/2019 19:10:21
I believe that the OP's question is related to single displacement (redox) reactions. Given K and Na as reactant or product, I would recommend against the use of water as solvent (!).

As far as the question goes... yes, it is possible to displace more active metals with less, but only by "cheating."

For any chemical reaction to proceed, the change in free energy (Gibbs or Helmholtz, depending on whether pressure is constant or volume is constant... I will refer to Gibbs, ΔG, because constant pressure is more common) must be negative. If there isn't a negative ΔG for the desired chemical reaction, it won't go... unless it gets pushed--ie energy can be added to the system (in specific ways) to accomplish the reaction. Just as boulders don't typically roll up hill on their own, but with the aid of a bulldozer and some diesel, it is trivial to roll boulders up hill.

Under standard conditions (1 atm, and 273 K),
Na + KCl → K + NaCl has a ΔG of (430.8 kJ/mol) (459.2 kJ/mol) = +28.6 kJ/mol

This high ΔG indicates that the reaction is highly unfavorable. But if at least 28.6 kJ/mol of energy is added in the right way, the reaction can go forward.

At 800 C (1073 K), NaCl, KCl, and Na are all liquids, while potassium is a gas.
(487.9 kJ/mol) ( 514.7 kJ/mol) = +26.8 kJ/mol

This ΔG is almost as disfavorable, but there is still an equilibrium. Given the Botzmann distribution (https://en.wikipedia.org/wiki/Boltzmann_distribution) we can see that with a difference in energy of 26.8 kJ/mol at 1073 K, we could expect that on the order of 5% of K-Na pairs would have reduced K and oxidized Na. And because the vapor pressure of potassium is about 1.4 atm at this temperature while the vapor pressure of sodium is only about 0.5 atm at this temperature, we can expect that K would make up roughly 15% of the gas in the headspace (the other 85% being Na). Thus one can start with pure sodium metal, add liquid KCl, and expect that the distillate would include a significant fraction of potassium. The distillate can be collected and redistilled, to give effectively pure metallic potassium.

Also, a suitable electrolyzer could reduce K+ to K while oxidizing Na to Na+ by supplying the extra energy in the form of electric potential (voltage).

https://webbook.nist.gov/cgi/cbook.cgi?Formula=NaCl&Units=SI&cTC=on
https://webbook.nist.gov/cgi/cbook.cgi?Formula=KCl&Units=SI&cTC=on
https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19680015650.pdf
https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19650014783.pdf