The following reaction is used in some self-contained breathing devices as a source of O2.
4KO2(s) + 2C02(g)---> 2K2CO3(s) + 3O2(g)
Suppose that a sample of CO2(g) is added to an evacuated flask containing KO2(s) and equilibrium is established. The equilibrium partial pressure of CO2(g)is found to be 7.20×10−2atm.
What is the equilibrium partial pressure of O2?
What is the total gas pressure at the equilibrium?
To find the equilibrium partial pressure of O2, we need to use the stoichiometry of the reaction. According to the balanced equation:
4KO2(s) + 2CO2(g) → 2K2CO3(s) + 3O2(g)
We can see that for every 2 moles of CO2, we get 3 moles of O2. Therefore, the ratio of the equilibrium partial pressures of CO2 to O2 is:
2CO2 : 3O2
Given that the equilibrium partial pressure of CO2 is 7.20×10^(-2) atm, we can use this ratio to find the equilibrium partial pressure of O2.
Let's assume the equilibrium partial pressure of O2 is x atm. Now we can set up the proportion:
(7.20×10^(-2) atm) / (2CO2) = (x atm) / (3O2)
Cross-multiplying, we get:
(7.20×10^(-2) atm) * (3O2) = (2CO2) * (x atm)
Simplifying, we get:
(21.6×10^(-2)) atm * O2 = (2CO2) * x atm
Dividing both sides by (21.6×10^(-2)) atm, we get:
O2 = (2CO2) * x / (21.6×10^(-2))
Now, substitute the value of CO2, which is 7.20×10^(-2) atm:
O2 = (2 * 7.20×10^(-2) atm) * x / (21.6×10^(-2))
Simplifying further, we get:
O2 = (14.40×10^(-2) atm) * x / (21.6×10^(-2))
O2 = 0.6667x atm
Therefore, the equilibrium partial pressure of O2 is 0.6667 times the value of x.
To find the total gas pressure at equilibrium, we need to consider all the gases present in the system. In this case, the total gas pressure will be the sum of the partial pressures of CO2, O2, and any other gases involved.
Given that the equilibrium partial pressure of CO2 is 7.20×10^(-2) atm, and we have already determined the equilibrium partial pressure of O2 to be 0.6667x atm, we can calculate the total gas pressure.
Total gas pressure = Partial pressure of CO2 + Partial pressure of O2 + Other partial pressures
Assuming there are no other gases involved in this reaction, we can simplify it to:
Total gas pressure = Partial pressure of CO2 + Partial pressure of O2
Substituting the given values:
Total gas pressure = 7.20×10^(-2) atm + 0.6667x atm
We don't have a specific value for x, but we know that the equilibrium partial pressure of CO2 is 7.20×10^(-2) atm. We can substitute this value to find:
Total gas pressure = 7.20×10^(-2) atm + 0.6667 * (7.20×10^(-2)) atm
Simplifying, we get:
Total gas pressure = 7.20×10^(-2) atm + 4.79944×10^(-2) atm
Total gas pressure = 11.99944×10^(-2) atm
Therefore, the total gas pressure at equilibrium is 11.99944×10^(-2) atm.