Chapter 7, Problem 1DE

(a)

Interpretation Introduction

To determine: The test of the given hypothesis on the alkali metals; also the factors favoring the superoxide formation.

Answer to Problem 1DE

Solution: The oxidation of rubidium and cesium in presence of air forms superoxide. The bigger size of cation is a favorable factor for superoxide formation.

Explanation of Solution

The alkali metals such as rubidium and cesium which has lower first ionization energies can be used to test the given hypothesis. The larger size of cation results in the easy removal of electron due to which ionization energy of the atom decreases. Therefore, the oxidation of rubidium and cesium metal in presence of oxygen results in the formation of superoxide which confirms the statement that the lower ionization energy favors the superoxide formation.

The larger size cation can accommodate the larger size anion around it. Therefore, the bigger size of cation is a property which favors the superoxide formation.

Conclusion

The oxidation of rubidium and cesium in presence of air forms superoxide. The bigger size of cation is a favorable factor for superoxide formation.

(b)

Interpretation Introduction

To determine: The experiment to determine the potassium peroxide reacts with ${\text{H}}_2\text{O}\left(g\right)$ and ${\text{CO}}_2\left(g\right)$ or not; also the products of the reaction.

Answer to Problem 1DE

Solution: The color of potassium superoxide salt changes from light yellow to white on reaction with water and carbon dioxide. The products of the reaction with water is $\text{KOH,}{\text{KHO}}_2$ and oxygen gas and the products of reaction with ${\text{CO}}_2\left(g\right)$ is ${\text{K}}_2{\text{CO}}_3$ and ${\text{O}}_2$ .

Explanation of Solution

The potassium superoxide is a yellow color salt. The reaction of potassium superoxide with water results in the change of its color from yellow to white. In the same way the potassium superoxide changes its color instaneously on coming in contact with water. This experiment confirms that the potassium superoxide reacts with carbon dioxide and water.

The reaction of potassium superoxide with water is given as,

${\text{2KO}}_2\left(s\right)+{\text{H}}_2\text{O}\left(g\right)\to \text{KOH}\left(s\right)+{\text{KHO}}_2\left(s\right)+{\text{O}}_2\left(g\right)$

The reaction of potassium superoxide with carbon dioxide is given as,

${\text{4KO}}_2\left(s\right)+2{\text{CO}}_2\left(g\right)\to 2{\text{K}}_2{\text{CO}}_3\left(s\right)+3{\text{O}}_2\left(g\right)$

Conclusion

The color of potassium superoxide salt changes from light yellow to white on reaction with water and carbon dioxide. The products of the reaction with water is $\text{KOH,}{\text{KHO}}_2$ and oxygen gas and the products of reaction with ${\text{CO}}_2\left(g\right)$ is ${\text{K}}_2{\text{CO}}_3$ and ${\text{O}}_2$ .

(c)

Interpretation Introduction

To determine: The experiment to determine the reaction given in part (b) important in the firefighters breathing masks or not.

Answer to Problem 1DE

Solution: The green canister present in the breathing apparatus contains potassium superoxide as oxygen generator.

Explanation of Solution

The firefighter breathing apparatus contains a green canister which contains the chemicals which are used in the breathing apparatus. This is present at the base of the breathing device, The oxygen generated by the green canister is done by using potassium superoxide present in the green canister. This confirms that the reaction given in part (b) is important in the firefighters breathing apparatus.

Conclusion

The green canister present in the breathing apparatus contains potassium superoxide as oxygen generator.

(d)

Interpretation Introduction

To determine: The experiment to determine the percentages of ${\text{KO}}_2\left(s\right)$ and ${\text{K}}_2\text{O}\left(s\right)$ in the given mixture.

Answer to Problem 1DE

Solution: The reaction of ${\text{KO}}_2\left(s\right)$ in vacuum at $290°\text{C}$ results in the formation of ${\text{K}}_2{\text{O}}_2$ which given percentage of ${\text{KO}}_2\left(s\right)$ which is again heated to $530°\text{C}$ forms ${\text{K}}_2\text{O}\left(s\right)$ and the difference of both will give the percentage of ${\text{K}}_2\text{O}\left(s\right)$ .

Explanation of Solution

The potassium superoxide on removal of oxygen at $290°\text{C}$ in vacuum produces ${\text{K}}_2{\text{O}}_2$ which is done as a experiment to determine the percentage of ${\text{KO}}_2\left(s\right)$ in the given mixture. The mass of product formed will give the mass of ${\text{KO}}_2\left(s\right)$ in the mixture.
The above reaction is still goes on heating will gives the formation of ${\text{K}}_2\text{O}\left(s\right)$ from the ${\text{K}}_2{\text{O}}_2$ which gives mass of ${\text{KO}}_2\left(s\right)$ and ${\text{K}}_2\text{O}\left(s\right)$ both in the given mixture which can be used to determine the mass of ${\text{K}}_2\text{O}\left(s\right)$ in the mixture.
After determining the mass of ${\text{KO}}_2\left(s\right)$ and ${\text{K}}_2\text{O}\left(s\right)$ , their percentages in the given mixture can be calculated.

Conclusion

The reaction of ${\text{KO}}_2\left(s\right)$ in vacuum at $290°\text{C}$ results in the formation of ${\text{K}}_2{\text{O}}_2$ which given percentage of ${\text{KO}}_2\left(s\right)$ which is again heated to $530°\text{C}$ forms ${\text{K}}_2\text{O}\left(s\right)$ and the difference of both will give the percentage of ${\text{K}}_2\text{O}\left(s\right)$ .