I wanted to figure out a way to make calcium metal with potassium oxide. $\ce{K2O}$ undergoes disproportionation at $\pu{350-430 °C}:$

$$\ce{2 K2O -> K2O2 + 2 K}$$

Now, if I mixed in $\ce{CaO}$ (1:2 molar ratio with $\ce{K2O})$ it would react with the potassium formed in the first reaction, forming calcium metal and $\ce{K2O}$ that would feed back into the first reaction. Eventually I should be left with mostly $\ce{K2O2}$ and calcium metal, but since these two react together, this can't be as simple.

Reaction of these two would lead to the same mixture as I started with. Is there a way to figure out what the concentration of Ca will be when the reaction reaches equilibrium?

  • $\begingroup$ If i could reduce the pressure enough to lower the melting point of K2O2 below 430 degrees, the formation of calcium metal would be favoured since it would float to the top, correct? $\endgroup$ – Francis L. Jul 14 '19 at 16:31

T l, Dr: despite what you see from the aqueous electromotive series, potassium does not reduce calcium oxide in this setting.

You may be assuming that potassium, which lies above calcium in the electromotive series for aqueous solutions, therefore lies above calcium in all other settings regardless of chemical environment or temperature. It doesn't. The electromotive series we see can change in different environments and at different temperatures.

For instance, in this Ellingham-Richardson diagram you see calcium oxidation at a very low oxygen potential, corresponding to a great tendency towards oxidation and thus a very high position for calcium in the "electromotive series" for anhydrous oxide formation. Where is potassium? Much higher up in free energy (starting around -650 on the free energy scale), corresponding to a less stable oxide and thus a lower tendency for potassium oxidation. In the absence of water solvent and at elevated temperatures (or even without elevated temperature but still in the absence of water), potassium does not displace calcium, or even manganese, from oxide formation. The order is very different from the aqueous one.

Apart from lithium, all of the alkali metals are notoriously poor at oxide formation, compared with what we expect from aqueous solutions. Look again at the diagram; sodium isn't so hot compared with calcium, magnesium or aluminum either.

Calcium is made industrially by either electrolysis or reduction of calcium oxide with aluminum. In the latter reaction aluminum still forms a less stable oxide than calcium, but unlike potassium it can form ternary calcium-aluminum oxides that are still more stable than just calcium oxide. Thus part of the calcium from the lime is converted to these ternary compounds while the rest, not fitting in the ternary compounds, comes off as vapor.

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    $\begingroup$ Terrific answer (+1)! And a higher resolution version (easier on my eyes!) of the Ellingham-Richardson figure is available at web.mit.edu/2.813/www/readings/Ellingham_diagrams.pdf. $\endgroup$ – Ed V Jul 15 '19 at 13:24
  • $\begingroup$ Could potassium reduce a non-oxide, like CaCO3 or Ca(OH)2, despite sitting lower in the ellingham diagram? $\endgroup$ – Francis L. Jul 16 '19 at 12:41
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    $\begingroup$ @Francis changing the anion could change the series again, but hydroxides and carbonates probably get you nowhere since these compounds with calcium decompose to oxides upon heating. HSAB suggests that chlorides or heavier halides could work, but these are more amenable to an electrolytic route. $\endgroup$ – Oscar Lanzi Jul 17 '19 at 8:41

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