I really need help Ozone gas (O 3 ) is commonly injected into drinking water (al
ID: 479184 • Letter: I
Question
I really need help
Ozone gas (O3) is commonly injected into drinking water (also swimming pools and hot tubs) as a disinfectant. Pure ozone decays rapidly, so it cannot be stored for long periods; as a result, it is typically generated onsite by exposing dried air to an electrical discharge that causes gaseous oxygen to be converted to ozone by the reaction: 3 O2(g) à 2 O3(g). Typically, about one-seventh of the oxygen in the gas is converted to ozone in this step.
A water treatment plant for a city with a population of 1 million people must produce on the order of 400,000 m3 per day of clean water. If ozone is injected into the water at a dose of 1 mg/L, how much air (1.0 atm total pressure, 15oC, 21% O2) must be processed per day to provide the ozone?
Explanation / Answer
Ans. Given,
Stoichiometry of reaction: 3 O2(g) --------> 2 O3(g)
Conversion rate = 1/7th of total volume of O2 is converted into O3
Water required to be treated = 4.0 x 105 m3 /day
= 4.0 x 108 L /day ; [1 m3= 103 L]
Concentration of O3 = 1 mg/L of water
Now,
Total amount of O3 required per day = [O3] x total volume of water treated in 1 day
= (1 mg/L) x 4.0 x 108 L
= 4.0 x 108 mg
= 4.0 x 105 g ;[1 mg = 10-3 g]
Number of moles of O3=Mass of O3 (in g) / Molar mass of O3
= 4.0 x 105 g / (47.9982 g mol-1)
= 8.334 x 103 moles
Now,
Using ideal gas equation Ideal gas Law for O3: pV = nRT - equation 1
Where, p = pressure in atm = 1.0 atm
V = volume in L = ?
n = number of moles = 8.334 x 103 moles
R = universal gas constant= 0.0821 atm L mol-1K-1
T = absolute temperature = 288.15 K ; [150C = 288.15 K]
Or,
1.0 atm x V = (8.334 x 103 moles) x (0.0821 atm L mol-1K-1) x (288.15 K)
Or, 1.0 atm x V = 197.15839641 x 103 atm L
Or, V = (197.15839641 x 103 atm L) / 1 atm
Hence, V = 1.9715839641 x 105 L - statement 1
That, 1.9715839641 x 105 L O3is required to be produced per day.
Part 2: Calculating total amount of Air needed to be processes
Let the total amount of air processed per day = A liters
Amount of O2 in total air = 21% of A liters
= 21/100 A L = 0.21 A L ; [21 % = 21/100]
Conversion rate = 1/7th of total volume of O2
So, amount of O3 produced from 0.21 A L O2= (1/7) x 0.21 A L
= 0.03 A L - statement 2
Comparing statement 1 and 2-
1.9715839641 x 105 L = 0.03 A L
Or, A L = (1.9715839641 x 105 L) / 0.03
= 6.571946547 x 106 L
Hence, A (L) = 6.571946547 x 106 L
Therefore, total volume of air processed per day = 6.571946547 x 106 L