# Le Chatelier's principle and Pressure!

#### Jumb0

5+ Year Member
Why does changing the total pressure of a system at equilibrium via reduction of the vessel volume by way of piston compression cause a shift in the equilibrium, but addition of an inert gas does not?

CO2(g) + C(s) -------> 2 CO(g)

If you injected an amount of helium gas into a reaction vessel of the above reaction at equilibrium, there would be no shift in equilibrium, even though the mole ratio of gases in uneven between products and reactants. You would think that the resulting total pressure increase would shift the reaction towards the reactant side a.k.a. the side with less moles of gas...but this does not happen.

However, if you compress the reaction in a piston, the shift does occur.

What gives?

#### wuhsabee

##### BEARS. BEETS. BATTLESTAR GALACTICA.
5+ Year Member
Scenario #1: Increasing Pressure
- When you increase pressure, it typically means decrease volume (Boyle's Law: P1V1 = P2V2).
- If you decrease the volume, that means there is less space available. Since you have different moles of gas on both sides, Le Chat would shift the direction to counteract the stress that is going on with less space.
- With less room available, the equilibrium shifts to the side that takes up less space / less mols of gas. In which case, it will shift towards the reactants, since there is only 1 mol of gas (CO2) there, vs. 2 mols of gas (2CO) in product side.

Scenario #2: Adding an inert gas
- Inert gases don't participate in a lot of chemical reactions.
- When you add inert gases, and the vessel remains at a fixed volume, keep in mind that the Keq = [CO]^2 / [CO2]. If the vessel remains at a fixed volume, that means the molarity doesn't change. If molarity of reactants & products don't change, and we add something that is not part of the reaction at all (in this case, the inert gas), it won't affect the Keq formula, as it only takes into account the [products]/[reactants] originally present in the reaction. Therefore, we're still at equilibrium, and there is no shift.

OP

#### Jumb0

5+ Year Member
Scenario #1: Increasing Pressure
- When you increase pressure, it typically means decrease volume (Boyle's Law: P1V1 = P2V2).
- If you decrease the volume, that means there is less space available. Since you have different moles of gas on both sides, Le Chat would shift the direction to counteract the stress that is going on with less space.
- With less room available, the equilibrium shifts to the side that takes up less space / less mols of gas. In which case, it will shift towards the reactants, since there is only 1 mol of gas (CO2) there, vs. 2 mols of gas (2CO) in product side.

Scenario #2: Adding an inert gas
- Inert gases don't participate in a lot of chemical reactions.
- When you add inert gases, and the vessel remains at a fixed volume, keep in mind that the Keq = [CO]^2 / [CO2]. If the vessel remains at a fixed volume, that means the molarity doesn't change. If molarity of reactants & products don't change, and we add something that is not part of the reaction at all (in this case, the inert gas), it won't affect the Keq formula, as it only takes into account the [products]/[reactants] originally present in the reaction. Therefore, we're still at equilibrium, and there is no shift.
Thank you! That makes a lot of sense.
I guess I made the very silly error of thinking that the addition of the inert gas would somehow reduce the volume in the vessel. I pictured the addition of the the inert gas into the vessel like the addition of baseballs into a cardboard box and made the conceptual error of assuming that there would then be less volume for the species of the equilibrium reaction to react in. While there is some truth to this in the sense that the mean free path between ALL THE PARTICLES in the vessel has been reduced (since there are now more particles per unit area), the fact of the matter is that the inert gas will intersperse itself evenly among the reaction particles, thus not changing the distance between the SPECIES OF THE EQUILIBRIUM RXN. Indeed, the only way to reduce the distance between the species of the reaction would be to decrease the volume of the container.