Question

An evacuated steel vessel with a volume of 10.00 liters is filled with 32.00 atm ofpure O2 at 25°C. An electric discharge is passed through the vessel, causing someof the oxygen to be converted into ozone. As a result, the pressure inside thevessel drops to 30.64 atm at 25°C. Calculate the final percent by mass of ozonein the vessel

Answer 1

The final mass percent of ozone in the vessel is 12.75%.

Step-by-step explanation:

The chemical reaction of formation of ozone from oxygen is as follows

`3O_(2) \rightarrow 2O_(3)`

`P_(initial): 32\,\,\, 0` (Here, molar mass of oxygen molecule = 32 g/mol)

`\Delta P: -3x \,\,\,+2x`

`P_(final):32-3x\,\,\, 2x`

From the given, at the end of the total pressure

`P_(final)` = 30.64

The gas filled with 32.00 atm.

`P_(total)= P_(O)_(2) + P_(O)_(3) = 30.64 = 32 -x`

`x=32-30.64=1.36\,atm`

Substitute the "x' value then we get the pressure of each gas.

`P_(O)_(2)=32-3(1.36)= 27.92\,atm`

`P_(O)_(3)=2(1.36)= 2.72\,atm`

Let's calculate the weight of the each gas.

`n=(PV)/(RT)=(w)/(M)`

rearrange the equation is as follows.

`w =(MPV)/(RT).......................(1)`

Substitute the each given value in the equation(1)

`w_(O)_(2)= (31.9988 * 27.92 * 10.00)/(0.082 * 298.15)= 365.17g`

Molar mass of oxygen = 31.9988 g/mol

Temperature =
`=25^(o)C= 25+273.15=298.15K`

`w_(O)_(3)= (47.9982 * 2.72 * 10.00)/(0.082 * 298.15)= 53.36g`

Mass % of Ozone:

`Mass\,perecentage\,of\,O_(3)=(w_(O)_(3))/(w_(O)_(2)+w_(O)_(3))`

`=(53.36 * 100)/(418.53)= 12.75%`

Therefore,Final percent by mass of ozone in the vessel is 12.75%.