[jackfiore]

I'm confused.

In the contact process: 2SO₂ +O₂   2SO₃

If the pressure increases, then the equilibrium will shift to oppose the change.

So it will shift such that the pressure is reduced.

Pressure is inversely proportional to Volume: so as V increases, P decreases.

Therefore the reaction goes to the side where there is a bigger volume.

There are 3 moles of gas on the LHS but only 2 on the RHS. The more moles of gas, the bigger the volume.

So the equilibrium shifts to the side with the bigger volume : LHS.

But that's not right is it? What went wrong? I think something went wrong with my reasoning.

New: Perhaps it has something to do with the fact that P is inversely proportional to V only holds true if the same amount of particles is present. Here, the more moles, the more particles, and the more particles, the more collisions = higher pressure. But the volume is still increasing as the number of moles increases, so presumably this means that the volume increase by increase of moles is not as significant as number of particles increasing as the number of moles increases?

[Borek]

Pressure is inversely proportional to Volume: so as V increases, P decreases.

Assume constant volume. If volume changes it usually means process is isobaric (constant pressure) and constant pressure doesn't influence the equilibrium.

[bakerbg]

Hey Jackfiore! one way i know how to increase the pressure of a system in equilibrium is to decrease it's volume - so I believe you are on the right track there.

Now if the molecules find themselves in a more confined space, under increased pressure they have to do something to counter it - Le Chateliers principle. There are just too many molecules in the reduced space. Since we are talking about a reversible reaction, the system will just simply combine more SO[2] with O2 to produce less SO3. There will now be less molecules to collide with each other and the walls of the container resulting in less pressure in the system.

So, increasing the pressure of the system causes it the favor the reaction that causes a reduction in pressure. hope I made things a little clearer...

[Denu]

When you are observing the effect of a factor on the rate of the reaction, you have to keep the other factors constant. Denial of doing so will lead to confusion; the kind you're experiencing right now. In other words, to observe effect of pressure on the rate of the reaction, you have to keep the volume of the container vessel constant. The gas molecules will occupy any space available to them, so the volume stays constant. You can then change pressure as you wish, and observe the effects on the rate of the reaction, and, if any, on the value of Kp.

Furthermore, if you want to observe the effect on the rate of the reaction by altering the volume, you have to keep the other factors constant, which in this case is pressure.

In Biology, we use control experiments to be certain that the effect of a factor on a specific observation is caused by the factor under study, and not another factor.

After reading this, you should be able to answer your question just fine.