Problem 9.124 Ozone (O3) can be prepared in the laboratory by passing an electri

Question

Problem 9.124

Ozone (O3) can be prepared in the laboratory by passing an electrical discharge through oxygen gas:
3O2(g)2O3(g).
Assume that an evacuated steel vessel with a volume of 10.0 L is filled with 32.55 atm of O2 at 25 C and an electric discharge is passed through the vessel, causing some of the oxygen to be converted into ozone. As a result, the pressure inside the vessel drops to 28.65 atm at 25.0 C.

Part A

What is the final mass percent of ozone in the vessel?

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Problem 9.124

Ozone (O3) can be prepared in the laboratory by passing an electrical discharge through oxygen gas:
3O2(g)2O3(g).
Assume that an evacuated steel vessel with a volume of 10.0 L is filled with 32.55 atm of O2 at 25 C and an electric discharge is passed through the vessel, causing some of the oxygen to be converted into ozone. As a result, the pressure inside the vessel drops to 28.65 atm at 25.0 C.

Part A

What is the final mass percent of ozone in the vessel?

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Incorrect; Try Again; 4 attempts

Explanation / Answer

Ptotal = O3 + O2

initially

P-O2 = 32.55

P-O3 = 0

final

P-O2 = 32.55 - 3x

P-O3 = 0 + 2x

and

Ptotal = 32.55 - 3x + 0 + 2x = 28.65

x = 32.55 -28.65 = 3.9

so

P-O2 = 32.55 - 3*3.9 = 20.85 atm

P-O3 = 0 + 2*3.9= 7.8 atm

Total P = 28.65

PV = nRT

n = PV/(RT) = 28.65*10/(0.082*298)) =11.724

of which

mol O3 = 7.8 /(28.65)*11.724 = 3.19 mol of O3

mass = mol*MW = 3.19*48 = 153.12 g of O3

mol of O2 = 20.85 /28.65*11.724 = 8.53212 mol of O2

mass = mol*MW = 8.53212*32 = 273.02784g of O2

O3 = 273.02784 / (273.02784+153.12 ) * 100 = 64.06% of O3

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