5G.9

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RRahimtoola1I
Posts: 102
Joined: Fri Aug 09, 2019 12:15 am

5G.9

Postby RRahimtoola1I » Thu Jan 09, 2020 11:13 am

Can someone explain how to understand this question?

"A sample of ozone, O3, amounting to 0.10 mol, is placed in a sealed container of volume 1.0 L and the reaction 2 O3(g) S 3 O2(g) is allowed to reach equilibrium. Then 0.50 mol O3 is placed in a second container of volume 1.0 L at the same temperature and allowed to reach equilibrium. Without doing any calculations, pre- dict which of the following will be different in the two containers at equilibrium. Which will be the same? (a) Amount of O2; (b) partial pressure of O2; (c) the ratio PO2/PO3; (d) the ratio (PO2)3/(PO3)2; (e) the ratio (PO3)2/(PO2)3. Explain each of your answers."

Thank you.

Haley Dveirin 1E
Posts: 101
Joined: Sat Jul 20, 2019 12:17 am

Re: 5G.9

Postby Haley Dveirin 1E » Thu Jan 09, 2020 11:45 am

a. the second situation will have more O2 because the only thing changing is the moles of O3 (not the volume of the flask) so the concentration of O3 is greater so the concentration of O2 must also be greater. For b. the partial pressure of O2 will be higher in the second situation because if there is a higher concentration of O2 in the second situation and the volume of both situations are the same there will be higher pressure. c. they are both the same because the ration between O3 and O2 within each situation is the same. d. they are the same because d is just the reciprocal of c.

Ashley Fang 2G
Posts: 102
Joined: Fri Aug 30, 2019 12:17 am

Re: 5G.9

Postby Ashley Fang 2G » Thu Jan 09, 2020 12:04 pm

PO2/PO3 will be different because the ratio of the reactant to the product is different at equilibrium point for different amounts of starting reactant.
Only (PO2)^3/(PO3)^2 or (PO3)^2/(PO2)^3 are guaranteed to be the same because they are the equilibrium constant and the inverse of it.

DesireBrown1J
Posts: 98
Joined: Wed Sep 18, 2019 12:18 am

Re: 5G.9

Postby DesireBrown1J » Thu Jan 09, 2020 4:32 pm

Ashley Fang 2G wrote:PO2/PO3 will be different because the ratio of the reactant to the product is different at equilibrium point for different amounts of starting reactant.
Only (PO2)^3/(PO3)^2 or (PO3)^2/(PO2)^3 are guaranteed to be the same because they are the equilibrium constant and the inverse of it.


For c, would the 2nd container with 0.50 mol O3 have a larger ratio of PO2/PO3 because it has a greater overall partial pressure?

Ashley Fang 2G
Posts: 102
Joined: Fri Aug 30, 2019 12:17 am

Re: 5G.9

Postby Ashley Fang 2G » Mon Jan 13, 2020 9:30 am

DesireBrown1J wrote:For c, would the 2nd container with 0.50 mol O3 have a larger ratio of PO2/PO3 because it has a greater overall partial pressure?


I'm not completely sure as I think we would need more information in order to calculate the ratio PO2/PO3


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