How to think about increased temperature effect on water autoprotolysis equilibrium constant?

[zenterix]

Homework Statement

The density of liquid water decreases as the temperature increases from 25C to 50C. Will this effect cause $K_2$ to increase or decrease? Why?

Relevant Equations

I thought I knew the answer to this question until today.

Apparently not.

From the van't Hoff equation we have

$$\ln{\left ( \frac{K_p(T_1)}{K_p(T_2)} \right )}=\frac{\Delta H^\circ_{rxn}}{R}\left ( \frac{1}{T_2}-\frac{1}{T_1} \right )\tag{1}$$

So for $T_2>T_1$ we have $K_p(T_1)<K_p(T_2)$.

which is to say that the equilibrium constant increases for the higher temperature and the reaction shifts towards products (ie ions, hydronium and hydroxide).

Does the analysis above even take into account something like the change in density of water?

Today I learned that the water autoprotolysis constant is

$$\mathrm{K_{auto}=\frac{[H_3O^+][OH^-]}{[H_2O]^2}}\tag{2}$$

and

$$\mathrm{K_w=K_{auto}[H_2O]^2=[H_3O^+][OH^-]}\tag{3}$$

Which of these constants is the one that is considered in (1)?

 

[Philip Koeck]

Today I learned that the water autoprotolysis constant is

$$\mathrm{K_{auto}=\frac{[H_3O^+][OH^-]}{[H_2O]^2}}\tag{2}$$

In the equilibrium expression for a solution you only need to include the concentration of solutes.
The solvent doesn't matter. It has "activity" equal to 1.
So the density change of the solvent due to temperature change has very little effect, I would say.

 

[Borek]

In the equilibrium expression for a solution you only need to include the concentration of solutes.
The solvent doesn't matter. It has "activity" equal to 1.

That's only an approximation.

Imagine equimolar mixture of water and ethanol. Which one is solvent, and which one has the activity of 1?

 

[Philip Koeck]

That's only an approximation.

Imagine equimolar mixture of water and ethanol. Which one is solvent, and which one has the activity of 1?

Good point!

I'm not sure about your example though. There's no reaction going on there, is there?
So I can mix water and ethanol at any ratio and it'll stay that way if nothing evaporates.
It won't move towards a chemical equilibrium.
Or am I missing something?

Not really my field, but I like to learn!

 

[Borek]

There's no reaction going on there, is there?

Doesn't matter - I could go for other examples of things possibly reacting (like equimolar mixture of acetic acid and ethanol), but activity in general is much more universal concept, not limited to chemical processes (and don't get me started on what is chemical reaction vs what is physical process as this classification is artificial - makes some sense when you start learning, but gets more and more blurry once you get into fine prints).