# Where does Oxygen come from in KO2 reaction with CO2?

$\ce{KO2 + CO2}$ gives $\ce{K2CO3 + O2}$

I think it should be from $\ce{KO2}$ as $\ce{2KO2}$ can split to $\ce{K2O}$ and $\ce{3/2O2}$. This $\ce{K2O}$ being unstable reacts with $\ce{CO2}$ to form $\ce{K2CO3}$.

Only problem is, is the first step possible? Is $\ce{O2}$ released from $\ce{KO2}$ or $\ce{CO2}$?

• Homework is a deprecated tag! Dec 20 '17 at 6:55
• KO2 can decompose, but as Nilay Ghosh said, the product is not K2O; also, even that requires heating, which is simply not there. Also, K2O is perfectly stable by itself (though it would surely react with CO2, given a chance). Other than that, you are right. Dec 20 '17 at 7:07

Yes, the oxygen indeed comes from the decomposition of potassium superoxide ($\ce{K2O}$) but the other decomposition product is potassium peroxide ($\ce{K2O2}$) and not potassium monoxide($\ce{K2O}$).(Here)

$$\ce{2KO2 ->[290°C, vacuum] K2O2 + O2~~~~~~~~~~(1)}$$ Now, this potassium peroxide reacts with carbon dioxide to form potassium carbonate and oxygen.(Here)

$$\ce{2K2O2 + 2CO2 → 2K2CO3 + O2~~~~~~~(2)}$$

Adding $\ce{(1)\times2}$ and $\ce{(2)}$, we get full reaction. (Here)

$$\ce{4KO2 + 2CO2 → 2K2CO3 + 3O2}$$

If reaction proceeds in aqueous medium, it gives potassium hydrocarbonate and oxygen.(Here)

$$\ce{4KO2 + 4CO2 + 2H2O → 4KHCO3 + 3O2}$$

• chemiday.com is by no means a reliable source of information. Dec 20 '17 at 10:00
• "If the reaction proceeds in an aqueous medium"? Doesn't potassium oxide react violently with water? Dec 20 '17 at 12:54