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

Answer:

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

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.

$n=\frac{PV}{RT}=\frac{w}{M}$

rearrange the equation is as follows.

$w =\frac{MPV}{RT}.......................(1)$

Substitute the each given value in the equation(1)

$w_{O}_{2}= \frac{31.9988 \times 27.92 \times 10.00}{0.082 \times 298.15}= 365.17g$

Molar mass of oxygen = 31.9988 g/mol

Temperature = $=25^{o}C= 25+273.15=298.15K$

$w_{O}_{3}= \frac{47.9982 \times 2.72 \times 10.00}{0.082 \times 298.15}= 53.36g$

Mass % of Ozone:

$Mass\,perecentage\,of\,O_{3}=\frac{w_{O}_{3}}{w_{O}_{2}+w_{O}_{3}}$

$=\frac{53.36 \times 100}{418.53}= 12.75%$

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