[mafagafo]

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.
We start with an equilibrium of:
    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:
[tex]4Ag(s)+O_2(g)\rightarrow2Ag_2O(s)[/tex]
That gives me*
[tex]K_c=\frac{1}{[O_2]}[/tex]
[tex]K_p=\frac{1}{P_{O_2}}[/tex]
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.

[mjc123]

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]

Thanks. I misunderstood Kc then.