[Bublik]

Hi guys,
Let's say you have the reaction
A(aq) + B(aq)  C(l)

Since the product is a liquid, the equilibrium expression is just K_C=1/([A])

Now, let's say we had an equilibrium mixture, and then we added C(l). The equilibrium would shift left. However, when it shifts left, that means that [A] and increase and the equilibrium constant actually changes, which doesn't make sense.

So when should I actually ignore liquids and solids in equilibrium reactions? Is it only when the products (or reactants) consist of only liquid such as in this case?

[Borek]

let's say we had an equilibrium mixture, and then we added C(l). The equilibrium would shift left.

It would not.

[Bublik]

I see... OK I'm taking this out from the answer key of an assignment:

CO(g) + 2H₂ (g)  CH₃OH(l)

If you remove methanol from the reaction once it starts to form, will the equilibrium shift towards the reactants, product, or remain unchanged?

Answer: Products. Removal of a product once it begins to form leads to more product being generated.

But wouldn't this shift to the products side decreases the concentrations of the reactants and change the equilibrium constant as a result?

[Borek]

This is a poorly worded question. Reaction as written doesn't make sense - you can't make liquid methanol without producing saturated vapor first, so in fact the question asks about two reactions:

CO(g) + 2H₂(g) CH₃OH(g)

and

CH₃OH(g) CH₃OH(l)

[Bublik]

Thanks.

This was a question from this year's Canadian Chemistry Olympiad final selection examination. I was wondering about it for a while now. I'll conclude that an equilibrium can't shift by adding or removing a liquid from the reaction.

[curiouscat]

I'll conclude that an equilibrium can't shift by adding or removing a liquid from the reaction.

Be careful! Liquids can be misciblle.

A  + B  C

where A,B, C all liquids ans   C is miscible with A and B will shift on removing C

[Corribus]

Yes, because the activity of a liquid is 1 only when the liquid is pure.

[magician4]

I see... OK I'm taking this out from the answer key of an assignment:

CO(g) + 2H₂ (g)  CH₃OH(l)

If you remove methanol from the reaction once it starts to form, will the equilibrium shift towards the reactants, product, or remain unchanged?

Answer: Products. Removal of a product once it begins to form leads to more product being generated.

I.m.h.o. this answer is wrong

as long as there is any amount (no matter how large) of liquid methanol present, the concentration*⁾ of methanol-vapour in this situation will become a constant  (at a given temperature), as the liquid will "buffer" the gaseous conc. via vapour pressure

hence, I could take away any amount of liquid methanol from the situation (as long as I leave at least one drop behind) without changing the equilibrium with respect to the gas composition

  correct answer would have to be "unchanged"

look at it from another point of view: there is no such thing as an equilibrium
c (methanol)_(l)  c(methanol)_(g)

there only is the dependency of the vapour pressure of methanol from the temperature :
p_(partial)(methanol)_(g) = f(T)
(as long as there is any whatsoever large amount of liquid methanol around)


regards

Ingo

*⁾
more precisely: the partial pressure