Potassium transformation Process 2

[SFTB]

Hello,

I've been doing researches in patent database, Internet site from companies, chemistry books to retrieve information on the fabrication of KHCO3 and K2CO3.

KCl (Sylvite or potash) is found in nature.

TO produce potassium carbonate or bicarbonate, it is possible to use an amine and/or alcool solution +CO2 to directly transform KCL in to its bi/carbonates (Ref : US patent 4,010,243)

Otherwise, what is done in industry (Oxychem, Armand products - Muscle Shoals), is an electrolysis of KCl to obtain KOH ans then do a carbonation of it to obtain K2CO3 or KHCO3.

My question is : can someone explain me why it is impossible to tranforme directly CO2 and KCl to potassium bi/carbonate with only those reactives.

2 KCL + CO2 +H2O --> K2CO3 + 2 HCl
or
KCL + CO2 + H2O --> KHCO3 + HCl

is it a question of activation energy (which would explain why it is done by electrolysis) Any reference or comments would be useful.

 

[phex]

you sure that the direct proces is non catalytic? this appears to be something like "oldfashioned and known" against "new and unknown". Might be that the right catalyst has been found and the direct route became a viable possibility...

chris

 

[SFTB]

I not sure I understand waht you mean... Let me summarize.

The electrolysis of KCl to form KOH is done separatly because by-products of this reaction are pure Chlorine (Cl2) and pure Hydrogen(H2) So there is an economical aspect for doing this operation. Then, bi/carbonate products are made from carbonation. This has been known since the beginning of the 20th century.

Since 1980, several production process started using amine (like MEA) to allow direct reaction of KCL with CO2 to produce bi/carbonates. It seams to work well...

What I don't find in me research is : IS it possible do a direct réaction of KCl with CO2 without amine to creat bi/carbonates. I think it is impossible or it has a very poor yield of reaction. That would explain why some work with amine have been done.

I looking for a technical confirmation of this assumption I just made. (I have a small chemistry knowledge)

By non-catalytic, do you mean that you think the reaction KCl-CO2 could happen even if there was no amine acting as a catalitic ingredient?

 

[nbucska]

There is a wellknown reaction:
NaCl + CO2 + H4NOH ---> NaHC03 + H4NCL

Isn't this similar to the amine reaction ?

Here the solubility of the NaHC03 is much lower so it
precipitates...


<nbucska@pcperipherals.com>

 

[SFTB]

nbucska, I can see by your reply that you are familiar with the Solvay process. You are right with the formula. It is possible the produce massive quantities of NaHCO3 from CO2, NaCl and ammoniac (aqueus solution). In fact, all elements considered, NaHCO3 has the lowest solubility in water = result precipitation.

Event if K is similar to Na (halide), the same formula could be (replace Na by K in your equation):

KCl + NH4HCO3 ---> KHCO3 + NH4Cl

The problem with this reaction is that KHCO3 has higher solubility than NH4HCO3 (unlike NaHCO3 in the Solvay process reaction) . As a result, the solvant (water) gets saturated in NH4HCO3 before you've reached the point where KHCO3 would be suffisent to precipitate! That is why some scientists are working with an amine to increase the solubility so KHCO3 would be the least soluble molecule of the equation and would be easier to precipitate.

 

[nbucska]

Hi SFTB:
Returning to your question about:
2 KCL + CO2 +H2O --&gt; K2CO3 + 2 HCl
or
KCL + CO2 + H2O --&gt; KHCO3 + HCl
-------------------------------
What would prevent the reaction from
turning around, and forming KCl
with the loss of CO2 ?

I mean, you could maintain this reaction only if you
could remove the HCl to maintain neutrality.


<nbucska@pcperipherals.com>

 

[kenvlach]

Having neutralized mass quantities of acids, I would tend to agree with nbucska. But, I am not as familiar with potassium compounds as with sodium. So, rather than relying upon our opinions, consider the thermodynamics of the reactions.

2 KCl(s) + CO₂(g) +H₂O(l) --&gt; K₂CO₃ (s) + 2 HCl(g)
DG_{Rxn, 298K} = -1063.5 + 2 (-95.3) – 2 (-408.5) – (-394.4) – (-237.1)
= + 194.4 kJ/mol

2 KOH(s) + CO₂ (g) --&gt; K₂CO₃ (s) + H₂O(l)
DG_{Rxn, 298K} = -1063.5 –237.1 – 2 (-378.7) – (-394.4)
= -148.8 kJ/mol

The first reaction is thermodynamically unfavorable. The second is thermodynamically favorable. I have neglected solution effects in order to use standard DG_f values, but the RHS and LHS solution effects would somewhat cancel. Data from CRC Handbook of Chemistry and Physics, 79th Edn. (1998).

 

[SFTB]

Thanks to both of you nbucska & kenvlach, your answers have been really useful for me. The thermodynamic aspect is really what I was looking for!

Thanks again, I will also check the reference you gave me kenvlach.