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I have a compound potassium oxide($\ce{K2O}$) and I am trying to convert it into potassium nitride ($\ce{K3N}$). Here are a few possible reaction methods :

  • Split potassium oxide into the constituent elements potassium and oxygen (Not very good as a lot of energy is required)
  • $\ce{K2O + N2 -> K3N + O2}$ (however this won't happen(?) because nitrogen has a lower electronegativity value than oxygen does)

and so on.

Does anyone know how can I convert the oxide to the nitride, releasing oxygen in the process, by not having to use a lot of energy-requiring processes?

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    $\begingroup$ The second reaction can happen- Thermodynamics are altered by concentration differential. Lots of Nitrogen and little K2O. It will need catalysing to break the nitrogen to atoms which is a kinetics problem. Potassium Nitride is not all that stable. Just look at Sodium Nitride in wikipedia. $\endgroup$ Commented Nov 10, 2013 at 10:12
  • $\begingroup$ @user2617804 How would I be best to begin on doing more research for the process that you mentioned above? $\endgroup$
    – user2117
    Commented Nov 10, 2013 at 10:15
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    $\begingroup$ @user2617804 - Potassium nitride $\ce{K3N} \ne$ potassium azide $\ce{KN3}$ $\endgroup$
    – Ben Norris
    Commented Nov 10, 2013 at 20:03

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Do you really want to go there?

Based on this entry from Wikipedia the compound is synthesized and stable only below room temperature, and on top of this requires elemental potassium:

This compound was produced by the reaction of potassium metal and liquid nitrogen at 77 K (−196.2 °C; −321.1 °F) under vacuum:[1]

$\ce{6K + N2 → 2K3N}$

This compound decomposes back into potassium and nitrogen at room temperature.

You may want to check that the target is actually $\ce{K3N}$ versus the more plausible $\ce{KN3}$.

Cited Reference

  1. Fischer, D.; Cancarevic, Z.; Schön, J. C.; Jansen, M. (January 2004). "Zur Synthese und Struktur von K3N". Zeitschrift für anorganische und allgemeine Chemie (in German). 630 (1): 156–160. https://doi.org/10.1002/zaac.200300280
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