Le Chatlier's: Inert Gases

[justadream]

TBR says that in reactions that occur in rigid containers, inert gases do not causes shifts in a reaction either way (neither to the right nor to the left).

But doesn't adding in an inert gas increase the pressure?

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[Czarcasm]

TBR says that in reactions that occur in rigid containers, inert gases do not causes shifts in a reaction either way (neither to the right nor to the left).

But doesn't adding in an inert gas increase the pressure?

I believe the way it was explained to me, was that for a given reaction (where each reactant and product has a given partial pressure), adding an inert gas (which is not part of that reaction) would not effect it's equilibrium. Even though the total pressure of the system would increase, each relative partial pressure would be proportional the same I guess, and so there would be no propensity to shift in either direction (considering it didn't experience a stress). Probably just a fact worth remembering I guess.

 

[justadream]

@Czarcasm

I take your point about the relative ratios of the reactants being the same but I'm still confused because TBR also says

"if inert gas is added to an expandable container, it can disrupt the equilibrium, because the partial pressures of the component gases are changed"

If you have an expandable container, adding inert gas STILL should not affect the relative ratio of the reactants. But here, TBR is saying equilibrium does change?

 

[techfan]

In an expandable container the volume changes when you add the inert gas. PV=nRT so the partial pressure=nRT/V and you are messing with concentration.

Though it's hard to imagine in a theoretical sense think about this reaction:

2CO(g) + O2(g) <---> 2CO2(g) if the reaction takes place in a rigid container with a fixed lid, adding He (an inert gas) wouldn't shift the reaction either way (He wouldn't react with either O2 CO or CO2) and the reaction would stay the same. The pressure the wall sees (since pressure=F/A) would increase, but that wouldn't shift the reaction. Now put that reaction in a balloon (where the walls can expand and volume can increase), when you put the He in there it bounces against the walls and increases the volume. An increase in volume tends to shift the reaction to the side with more molecules (entropy favors disorder when it can have it and more volume means there is more space for the molecules to jump around).

 

[Chrisz]

@Czarcasm

I take your point about the relative ratios of the reactants being the same but I'm still confused because TBR also says

"if inert gas is added to an expandable container, it can disrupt the equilibrium, because the partial pressures of the component gases are changed"

If you have an expandable container, adding inert gas STILL should not affect the relative ratio of the reactants. But here, TBR is saying equilibrium does change?

If you add an inert gas into a expandable container, it will increase the volume of the container, the Qp will change in response to the expanded volume. The reaction will adjust such that the number of gas molecules increase and the volume occupied per unit of reaction molecules will decrease. This is Le chatel principle.

Another way of visualizing this question is to use equilibrium constant approach
Assuming the reaction is initially at equilibrium,
********** [Pco2]^2
Qp=Kp =-------------------
********** [(Pco)^2*(PO2)]

********** (Nco2RT/v)^2
********= ---------------------------------
********* (NcoRT/v)^2*(No2RT/v)

*******=cv^3/v^2=cv

we have derived Qp=Kp=cv for the initial equilibrium.
Now what if we increase v? If we increase v, we will increase Qp, and Qp >Kp.
Then the reaction will adjust to the left to increase number of gas molecules.

I do not recommend the second approach for MCAT purpose, Le chatel is much quicker. The second approach can be used to verify the answer you get when you practice on le chatel principle. Once you build up your confidence in applying le chatel principle, you will not bother to verify your answer with equilibrium contsant approach.

 

[Lighthill-FFT]

I never really liked L'Chatlier's Principle. It always seemed really heuristic to me until I learned statistical mechanics. But for your personal edification:

Consider the definition of pressure. The macroscopic definition is force per area. Under the macroscopic definition, your concerns are completely valid: it appears that increased pressure would lead to increased intermolecular events, leading to an increased rate of reaction. However, we have to analyze the assumptions made when using the macroscopic definition. Yes, force per area increases but it increases differentially: by introducing a new element to the enviorment, we change the microscopic enviorment making the macroscopic definition useless at giving us info about whats going on on a microscopic scale (IE chemical reactions).

Now we consider the microscopic definition: pressure is generated by the total change of momenta in the particles against a standard. Its really hard to measure this for a bunch of reasons, so we choose something that scales proportionally with it: the total number of molecular collisions that occur averaged over a long enough time to get convergence. If we add an inert gas under this definition, the results are more clear: the total number of collisions increases because the total number of gas molecules increase. However, the rate of reaction doesn't increase at all because most of the new collisions are with inert particles.

We can now return to L'Chatlier's Principle with a deeper understanding. I like to think of L'Chatlier's principle as Newton's First Law (every action has a reaction) except on a macroscopic chemical scale. We cannot just look at the pressure for the reasons I described above. We look at partial pressures because, roughly, partial pressures define the number of collision events contributed by each chemical species.

TLDR; Yes the pressure changes, but the new pressure and the old pressure represent different things physically so we cannot look at total pressure. We instead look at partial pressure, which of course does not change in an ideal world.