Inert Gas and Le Chatliers

[AngieGarcia_4F]

Why does adding an inert gas not effect an equation?

[Vinita Saxena 2I]

An inert gas doesn't react with the reactants so it does not change the concentration or the reaction in any way since it also does not change the volume.

[Michael Nguyen 1E]

Although adding an inert gas will increase the pressure of a system, the partial pressures of the reactive gases are not changing because the volume has stayed the same. Therefore, K stays the same and there is no shift in the system.

[KBELTRAMI_1E]

Although adding an inert gas will increase the pressure of a system, the partial pressures of the reactive gases are not changing because the volume has stayed the same. Therefore, K stays the same and there is no shift in the system.

so do we not have to be concerned about the change in pressure either? will adding an inert gas break equilibrium?

[805307623]

Adding an inert gas to a system in equilibrium at constant volume doesn't effect the equilibrium. However, when adding an inert gas to a system in equilibrium at constant pressure the the total volume will increase.

[Luc Zelissen 1K]

If you had Cl- as an inert gas it would be Cl- on the reactant side and Cl- on the product side. If you add this into the chemical equilibrium constant equation, the concentration of Cl- will be on the top and bottom of the division sign, so they will simply cancel each other out.

[Ipsita Srinivas 1K]

Adding an inert gas to a system in equilibrium at constant volume doesn't effect the equilibrium. However, when adding an inert gas to a system in equilibrium at constant pressure the the total volume will increase.

I'm kinda confused here; won't the rate of the reaction change if there is an inert gas because the reactants now have to undergo collisions with the inert gas as well? Does that not affect the equilibrium?

[Vincent Leong 2B]

inert gases don't react with the reactants so they ultimately don't consume the partial pressure or the concentration of the reactants. The fact that partial pressure of the reactants (which directly impacts the K constant) is not effected, the K constant is therefore not effected.