Ozone is depleted in the stratosphere by chlorine from CF₃Cl according to this set of equations:
CF₃Cl + UV light → CF₃ + Cl
Cl + O₃ → ClO + O₂
O₃ + UV light → O₂ + O
ClO + O → Cl + O₂
What total volume of ozone at a pressure of 25.0 mmHg and a temperature of 225 K is destroyed when all of the chlorine from 15.0 g of CF₃Cl goes through 10 cycles of the given reactions?

Question

Ozone is depleted in the stratosphere by chlorine from CF₃Cl according to this set of equations:

CF₃Cl + UV light → CF₃ + Cl

Cl + O₃ → ClO + O₂

O₃ + UV light → O₂ + O

ClO + O → Cl + O₂

What total volume of ozone at a pressure of 25.0 mmHg and a temperature of 225 K is destroyed when all of the chlorine from 15.0 g of CF₃Cl goes through 10 cycles of the given reactions?

Accepted Answer

Hey everyone. So you have chlorine from floor to floor methane ozone, we have a series of reactions below. and at 32 mm of Mercury, 288 Kelvin. We assume that all the chlorine from 16.5 g of methane undergoes 15 cycles. Or as calculate the total volume of ozone that is consumed in ozone is 03. There for one cycle we need to convert from grams of chlorophyll and methane thomas of ozone Start giving 16.5 g of course truffle or methane. And in one mole of course our floor methane of the Mueller mass Which is 12.011 plus three, 18. plus 35 .453 Mr. give us 104 0.46 gramps. And for one mode of course to florida, methane, We have two moles of ozone in one cycle And give us 0.3159. Malls of ozone dynamic is the ideal gas law equation. To find the volume, it's gonna be P. V. Equals Nrt our pressure. It's 32 millimeters of mercury. We're looking for the volume number of moles, 0.3159. And our constant Can be 0.08 - one. Later. Sounds atmosphere about about moles. I'm kelvin for the temperature Gonna have 288 Melvin. And now we need to convert the pressure from millimeters of mercury to atmosphere 32 kilometers of mercury. We have 760 millimeters of mercury in one atmosphere, You need to give us 0.04 21 atmosphere. You know, if you just plug in the values to the equation to get 0.04 atmosphere times of volume, It's going to be equal to 0.31 59 balls Time 0.08 21. Their sounds atmosphere about about malls. I'm Calvin. I'm Sergeant 88 Calvin committed by both sides by 0.04 atmosphere. We're gonna get the volume. It's gonna be 177 0.4 leaders. But that is in one cycle, We need to determine the volume in 15 cycles. They're going to take 177.4 leaders in one cycle Most about 15 cycles. And this gave us 2661 leaders, Which is 2.7 Times 10 to the 3rd Leaders. Thanks for watching my video and I hope it was helpful.

Verified Solution

Key Concepts

Ozone Depletion Mechanism

Ideal Gas Law

Stoichiometry of Chemical Reactions