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### Topic: Chemical equilibrium problem  (Read 4197 times)

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#### mafagafo

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• Posts: 20
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##### Chemical equilibrium problem
« on: May 06, 2014, 12:50:38 PM »
Original:
(ITA-SP) Dentro de um forno, mantido numa temperatura constante, temos um recipiente contendo 0,50 mol de Ag(s), 0,20 mol de Ag2O(s) e oxigênio gasoso exercendo uma pressão de 0,20 atm. As três substâncias estão em equilíbrio químico. Caso a quantidade de Ag2O(s) dentro do recipiente, na mesma temperatura, fosse 0,40 mol, a pressão, em atm, do oxigênio no equilíbrio seria:

Useful info:
T is constant.
V is constant.
0.50 mol of Ag(s)
0.20 mol of Ag2O(s)
0.20 atm of O2(g)

The question is: "if the amount of Ag2O(s) were 0.4 mole (at this same temperature), the pressure of O2 would be:"
a)  0.10
b)  0.20
c)  0.40
d)  sqrt(0.20)
e)  0.80

I can get the equation:
$$4Ag(s)+O_2(g)\rightarrow2Ag_2O(s)$$
That gives me*
$$K_c=\frac{1}{[O_2]}$$
$$K_p=\frac{1}{P_{O_2}}$$
As shown, the equilibrium constant depends only on the concentration (or pressure) of oxygen. Therefore, varying the amount of Ag2O, does not change the oxygen pressure.
* corrected.
« Last Edit: May 06, 2014, 02:54:47 PM by mafagafo »

#### mjc123

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##### Re: Chemical equilibrium problem
« Reply #1 on: May 06, 2014, 01:11:52 PM »
All solid phases are assumed to have an activity of 1. Therefore Kc = 1/[O2] and doesn't depend on the amount of Ag2O.
Generally, changing the amount (as distinct from the concentration) of one phase does not affect the position of equilibrium. (For homogeneous equilibria the two go hand in hand, but not heterogeneous.)

#### mafagafo

• Regular Member
• Posts: 20
• Mole Snacks: +1/-0
##### Re: Chemical equilibrium problem
« Reply #2 on: May 06, 2014, 01:19:55 PM »
Thanks. I misunderstood Kc then.