partial pressures

[Savannah Mance 4G]

In 5G.9 it says in two containers, one with 0.1 mol of O3 and the other with 0.5 mol of O3 with the rxn 2O3(g)<--->3O2(g) taking place will ratio of PO2/PO3 be the same (part b). I understand that the ratio won't be the same for the two containers because the partial pressures of the O3 as well as the O2 won't be the same in the two containers. But the next part asks if the ratio (PO2)^3/(PO2)^2 will be the same and the answer is yes but I don't understand why. Can someone please explain why?

[KSong_1J]

The ratio will be the same because (PO2)^3/(PO3)^2 is the formula for its equilibrium constant, K, which always stays the same whether more moles are added or not

[205007651]

Just to add on to the previous reply, the only reason k will change will be because of a change in temperature.

[Alondra]

K is a constant. I believe a change in pressure does not affect K but a change in temperature can.

[RasikaObla_4I]

The ratio (PO2)^3/(PO3)^2 represents the equilibrium constant K so it stays the same. As a result changing the concentrations of the products and reactants would not change the constant.

[kevinolvera1j]

A change to concentrations does not change the equilibrium constant. That is why K remains the same even when concentrations increase or decrease. What should be noted is that K is only for reactions at equilibrium, what that means is that if you add or take away products or reactants and therefore change the concentrations, you will have to wait for the system to reach equilibrium again.

[Shrayes Raman]

The second part ratio represents K and K will be the same regardless of quantity

[Eunice Nguyen 4I]

The ratio represented the equilibrium constant K, so it'll remain the same.