[tanne]

Can someone please help me with this one question? Which values from the reaction equations do I use?


Ozone is depleted in the stratosphere by chlorine from CF3Cl according to these equations:

CF3Cl + UV light   CF3 + Cl

Cl + O3   ClO + O2

O3 + UV light   O2 + O

ClO + O   Cl + O2

What volume of ozone at pressure 29.0mmHg and temp 216K can be destroyed when all of the chlorine from 10.0g of CF3Cl goes through 10 cycles of the reactions above?


Here are the converstions I got so far:

P= 29.0mmHg = 0.0382atm
T= 216K
molar mass of CF3Cl = 104.46mol
10gCF3Cl = 0.0957molCF3Cl

I know I'll use the ideal gas law to solve for the volume of O3. I'm just confused on how to fit the data from the equations in.

[sjb]

Can someone please help me with this one question? Which values from the reaction equations do I use?


Ozone is depleted in the stratosphere by chlorine from CF3Cl according to these equations:

CF3Cl + UV light   CF3 + Cl

Cl + O3   ClO + O2

O3 + UV light   O2 + O

ClO + O   Cl + O2

What volume of ozone at pressure 29.0mmHg and temp 216K can be destroyed when all of the chlorine from 10.0g of CF3Cl goes through 10 cycles of the reactions above?


Here are the converstions I got so far:

P= 29.0mmHg = 0.0382atm
T= 216K
molar mass of CF3Cl = 104.46mol
10gCF3Cl = 0.0957molCF3Cl

I know I'll use the ideal gas law to solve for the volume of O3. I'm just confused on how to fit the data from the equations in.

Well, from your equations, can you write a balanced equation for the decomposition of ozone by Cl^. ?

[tanne]

I figured it out. Thank you for your help. It wasn't necessary to write a balanced equation for the decomposition.

I was able to figure out by looking at the equations that 2 atoms of ozone are destroyed in 1 cycle. So I converted 10g of CF3Cl to moles of CF3Cl to moles of Cl. Then you can multiply this by the ratio from the equations (2 moles of O3:1 mole of Cl) where you get moles of O3. You can use this value in the ideal gas law and then multiply the answer by 10 cycles.

Thanks.