5G.9

[Shanzey]

For part c, why does the ratio of P(O2)/P(O3) change between the two experiments? I understand that the equilibrium constant K= (P(O2)^3/P(O3)^2) does not change, but why would the ratio P(O2)/P(O3) based off of the amount of moles present?

[Chem_Mod]

The equilibrium constant is what remains the same in both reactions. It is a number describes the point at which the forward and reverse reactions are equal, using a different ratio that is not the equilibrium constant or a manipulation of the equilibrium constant like (K^-1), would not result in a value that is equal in both reactions.

[Daria Azizad 1K]

There would be different amounts of O2 and O3 because of the different concentrations. This ratio does not refer to the eq. constant, so they would not be the same.

[Shutong Hou_1F]

P(O2)/P(O3) isn't the equilibrium constant of this reaction, but P(O2)^3/P(O3)^2 is. Remember that a^2/b^3 = 5 doesn't mean that a/b = 5. Because the equilibrium concentration is different in these two cases, only equilibrium constant would be the same when comparing these two conditions.