5G 9

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KNguyen_1I
Posts: 101
Joined: Sat Aug 17, 2019 12:16 am

5G 9

Postby KNguyen_1I » Tue Jan 07, 2020 8:09 pm

"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, predict 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."

I got b (the partial pressure stays the same), c (their ratios stay the same), and d (because d is the equilibrium constant equation) with the reasoning that as the conditions do not change, the ratios of partial pressure and mols would stay the same. Can anyone corroborate or qualify this?

Hannah Lee 2F
Posts: 117
Joined: Thu Jul 11, 2019 12:15 am

Re: 5G 9

Postby Hannah Lee 2F » Tue Jan 07, 2020 8:35 pm

(a) The amount of O2 @ eq will be different in the containers. Because container 2 has a higher amount of initial reactant (0.50 mol O3), it will subsequently have a higher amount of product (mol O2) at equilibrium. This is because K must remain constant across both containers, so if the rxn starts off with more reactant (higher [R]), it must result in a higher amount of product ([P]) to maintain the same K.

(b) The partial pressure of O2 will be higher in container 2. K = partial pressure of reactant / partial pressure of product, or [R] / [P].
K must remain constant, and the amount of reactant (and thus the partial pressure of reactant, which makes up the denominator) is larger in container 2, so the numerator must also be larger to make up for the larger denominator.

(c) The ratio is different because it doesn't take into account the stoichiometric coefficients for the ratio of products / reactants, so it doesn't align with the equilibrium constant K.

(d) The ratio is the same because it describes K, which will be constant across both containers. Each rxn has its own characteristic K, and this value can only be changed by changing the temperature (which is not done in this question).

(e) This ratio is merely the inverse of K (a constant) so it will be the same.


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