# How is the rate of appearance of O2 related to the rate of disappearance of KO2?

Potassium superoxide (KO2) is a yellow solid that decomposes in moist air. It is a rare example of a stable salt of the superoxide anion. It is used as a CO2 scrubber, H2O dehumidifier, and O2 generator in rebreathers, spacecraft, submarines, and spacesuits.

## Reaction of KO2 with water

When KO2 comes in contact with water, it undergoes disproportionation to potassium hydroxide (KOH), oxygen (O2), and hydrogen peroxide (H2O2):

\begin{aligned} & 2 KO_2 + H_2O \rightarrow 2 KOH + \frac{3}{2} O_2 \\ & KO_2 + H_2O \rightarrow KOH + \frac{1}{2} H_2O_2 + \frac{1}{2} O_2 \end{aligned}

The rate of appearance of O2 is proportional to the rate of disappearance of KO2, since both reactions involve a 1:1 stoichiometric ratio between these two species. The rate law for the first reaction can be written as:

$$\text{rate} = k [KO_2] [H_2O]$$

where k is the rate constant, and [ ] denotes the concentration. Similarly, the rate law for the second reaction can be written as:

$$\text{rate} = k’ [KO_2] [H_2O]$$

where k’ is another rate constant. The overall rate of O2 production can be obtained by adding the rates of both reactions:

$$\text{rate}_{O_2} = \frac{3}{2} k [KO_2] [H_2O] + \frac{1}{2} k’ [KO_2] [H_2O]$$

Reaction of KO2 with CO2

Another way that KO2 can produce O2 is by reacting with carbon dioxide (CO2), which is a common pollutant in closed environments. The reaction can be written as:

\begin{aligned} & 2 KO_2 + CO_2 \rightarrow K_2CO_3 + \frac{3}{2} O_2 \\ & 4 KO_2 + 4 CO_2 + 3 H_2O \rightarrow 4 KHCO_3 + 6 O_2 \end{aligned}

The first reaction is favored at low temperatures, while the second reaction is favored at high temperatures.  The rate of appearance of O2 is also proportional to the rate of disappearance of KO