[karmax]

Question:

The equilibrium constant for the reaction:

2SO₂ (g) + O₂ (g) <---> 2SO₃ (g)

is 0.14 atm^-1 at 900K. If a reaction vessel is filled with SO₃ at a partial pressure of 0.10 atm and with O₂ and SO₂ each at a partial pressure of 0.20 atm, is the reaction at equilibrium? If not, in what direction does it proceed?

So, I know how to check if a reaction is at equilibrium/what direction it must proceed to get to equilibrium. What confuses me is the changing pressure... I have no idea how that affects equilibrium. Any help on how to tackle this problem is much appreciated!

[Fluorine]

What does increased pressure do to the gas molecules?
Would it bring them closer together or farther apart?
How would this directly relate to the rate of reaction?

If you can answer these, which way is the equilibrium going to be forced - why?

[karmax]

What does increased pressure do to the gas molecules?
Would it bring them closer together or farther apart?
How would this directly relate to the rate of reaction?

If you can answer these, which way is the equilibrium going to be forced - why?

Increased pressure makes the gas molecules closer together.. and thus increases the rate of reaction. So in order to reduce pressure, the equilibrium would have to shift to the right (since its the side that has fewer molecules). Is that right?

[Fluorine]

There you go, it'll shift to the right.

However I would say it goes to the right because it has more pressure not because of the quantity. The molecules will collide more often and have better chance of forming the product SO₃ due to being forced closer. This is how I recall it.

[Borek]

What confuses me is the changing pressure... I have no idea how that affects equilibrium.

Le Chatelier's principle. But there is no changing pressure in this question, or at least it is not something that should be taken into account when looking for an answer.