HW 11.11

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hannaluong4E
Posts: 22
Joined: Fri Sep 25, 2015 3:00 am

HW 11.11

Postby hannaluong4E » Fri Nov 06, 2015 8:18 pm

A .1 mol sample of pure ozone, O3, is placed in a sealed 1 L container and the reaction 2O3 -> 3O2 is allowed to reach equilibrium. A .5 mol sample of pure ozone is placed in a second 1 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?

I don't understand why (a) the amount of O2 is greater in the second container and (c) why the ratio [O2]/[O3] will be different.

kshalbi
Posts: 68
Joined: Fri Sep 25, 2015 3:00 am

Re: HW 11.11

Postby kshalbi » Fri Nov 06, 2015 11:43 pm

The amount of O2 is greater in the second container because you are starting with .5 moles of pure ozone which is more than .1 moles of it. Equilibrium reactions try to get rid of the change and in this case, to get rid of all the moles of O3, equilibrium will favor the forward reaction and produce more moles of O2 than the first container just because you started with more moles of O3. The ratio [O2]/[O3] will be different because it is not the ratio for equilibrium since the superscripts are missing.


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