Question: Equilibrium and Partial Pressures

[Pantaloons]

Hey Mcathelon athletes,

 

Say you had a chemical reaction at equilibrium, taking place in a sealed container:

X(g) + Y(g) <--> 3Z(g)

And you pumped Helium into the container.

What would happen to the reaction?

 

+++

I realize that Helium, being an inert Noble gas, would not react with any of the species and thus not affect the equilibrium; however, shouldn't Helium also affect the partial pressures of the gases in the container?

 

And if it affects the partial pressures, would Helium therefore shift the reaction equilibrium towards the left?

 

Thanks!

[dan0105]

I realize that Helium, being an inert Noble gas, would not react with any of the species and thus not affect the equilibrium; however, shouldn't Helium also affect the partial pressures of the gases in the container?

 

And if it affects the partial pressures, would Helium therefore shift the reaction equilibrium towards the left?

 

Thanks!

 

Ya you would think that, but think about what you have. You are in a sealed container so what you are really doing is forcing some more gas in which doesn't play a role in the eqb. Now, what is the partial pressure of any given substance in the mix? Its still the same, because the partial pressure of each gas is just what that gas (ALONE) contributes to the pressure of the container. (So sure there is more total pressure due to the helium, but the partials don't change). So if helium isn't in the eqb equation, and partial pressures don't change, nothing actually changed!!!!

 

The story ISN'T the same however if volume is allowed to change when the helium is added, read more about it:

http://mymcat.com/w/index.php?title=Equilibriums#Effect_of_adding_an_inert_gas

[Pantaloons]

Ahh, interesting!

I think I'm understanding now.

 

So, when you're pumping He into the rigid container, the total pressure increases; however, the mol fraction for each gas decreases (since there are more mols in total). Therefore, overall, the partial pressure of each gas stays the same?

 

I think i get this now. Thanks so much!

[dan0105]

Hmm, that still doesn't sound quite right.

 

Consider something like N2 and H2 and O2 in a container.

The Ptotal = ppN2 + ppH2 + ppO2, which if we put fake numbers in we could pretend it were: 10 = 5 + 2.5 + 2.5

 

Now when He is added to the system but the volume is constant, then total pressure increases, to something like say 12, but the original partial pressures don't change we just added an extra one, so we get 12 = 5 + 2.5 + 2.5 + 2 (for the He)

 

Using those partial pressures then in the eqb equation will result in the exact same thing.

[The Law]

Think of it this way, when you add the helium it is not going to react with anything. We know that the total pressure will go up, but total pressure itself does not affect equilibrium. Volume is what affects equilibrium, or the partial pressures of the gasses. The volume of the container remains constant, so the partial pressure of each individual gas remains constant.

 

Yes, you are right - the mole fraction will change (it will go down b/c total moles went up). This causes p.p. to remain the same:

 

Partial pressure = mole fraction x total pressure

 

If the total pressure goes up and the mole fraction goes down proportionally, then the partial pressure will remain the same. This is what happens when an inert gas is added. This is why inert gases will not affect equilibrium - the partial pressure stays the same.

 

Now, you must be careful. If the container is not rigid, initially, the total pressure of the container increases and will cause an increase in volume. This increase in volume, will then cause a corresponding decrease in the total pressure of the system... so in this case:

 

Partial Pressure = mole fraction (an amount that decreased because the total moles increased) x total pressure (an amount that also decreased because volume went up)

 

So in this case, the partial pressure goes down. The volume goes up.

 

As in anytime volume increases, this will shift equilibrium to produce more of the side with more moles.

 

Hope this helps...

[Pantaloons]

Thanks everyone for the responses!

 

Best of luck on this home stretch before test day!

[Steve-0]

As a side note... if you take the original equation:

 

X(g) + Y(g) <--> 3Z(g)

 

When an inert gas is added, the total pressure of the system goes up (as has been said above). Since the pressure has increased, the system will react in a way to decrease the effect of this change according to Le Chateliers priciple. In this case, the reaction in the left direction will become more favorable because less molecules of gas are present on the left than the right. Am I right?

[Pantaloons]

Nope, afraid not.

Volume, not total pressure, affects equilibrium.

If you work out the partial pressures, you'll notice that they don't change when an inert gas is added (given constant volume).

 

See Law's explanation.

[dan0105]

Exactly, you've got it.

Forget Steve-o. (Steve-o, read the whole thread again)

[The Law]

As a side note... if you take the original equation:

 

X(g) + Y(g) <--> 3Z(g)

 

When an inert gas is added, the total pressure of the system goes up (as has been said above). Since the pressure has increased, the system will react in a way to decrease the effect of this change according to Le Chateliers priciple. In this case, the reaction in the left direction will become more favorable because less molecules of gas are present on the left than the right. Am I right?

 

Total pressure isn't affecting the equilibrium. It is the partial pressures that affect it. Check out the answer I provided above and try to work out what I said on paper - it should clear things up.

[gamma]

will the question always specify a change in volume?

 

also, if in a question we were asked to determine which partial pressure changes the most after an inert gas is added + volume changes, and we were given quantitative values, we'd just decrease the total pressure by the same factor that volume increases and multiply by the newly-decreased mole fraction, n'est pas?