Question 11.11 (Sixth Edition)

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Steve Magana 2I
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Joined: Fri Sep 28, 2018 12:24 am

Question 11.11 (Sixth Edition)

Postby Steve Magana 2I » Sat Jan 12, 2019 11:14 pm

Question: A 0.10-mol sample of pure ozone, O3, is placed in a sealed 1.0-L container and the reaction 2 O3(g) <---> 3 O2(g) is allowed to reach equilibrium. A 0.50-mol sample of pure ozone is placed in a second 1.0-L container at the same temperature and allowed to reach equilibrium. Without doing any calculations, predict which of the following will be different in the two containers at equilibrium. Which will be the same? (a) Amount of O2; (b) concentration of O2; (c) the ratio [O2]/[O3];
(d) the ratio [O2]^3/[O3]^2; (e) the ratio [O3]^2/[O2]^3. Explain each of your answers.

I know this is a pretty lengthy question, but I'm having trouble with predicting the answers without doing any calculations. If anyone could help please, that would be great. Thank you!

Matthew Choi 2H
Posts: 59
Joined: Fri Sep 28, 2018 12:18 am

Re: Question 11.11 (Sixth Edition)

Postby Matthew Choi 2H » Sat Jan 12, 2019 11:38 pm

a/b) If you start with more moles of O3 then the partial pressure of O3 will increase. You know that the Kp must remain constant so, looking at the equation for Kp = (PO2)^3/(PO3)^2 you can reason that the partial pressure of O2 will also increase to keep the ratio the same. The only way that the partial pressure for O2 can increase (since the temp. and volume remained the same) is for the number of moles of O2 to increase.

c) The ratio of partial pressures given in part c does not match the ratio for the Kp, therefore, the ratio is subject to change as the number of moles of O3 initially change.

d) This ratio is the ratio for the Kp expression so it will be the same.

e) This is the inverse of the Kp expression so it will also be the same.


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