H2O: H3O+ and OH- formation

[Genecks]

So, a question popped into my head.

 

Maybe I'm not understanding the dynamics and mechanics of the H2O molecule, but I thought I would pose the question.

 

Why does water trade-off the H+ ion to another water molecule while in equilibrium? Why? Does that not take energy? And as such, shouldn't that eventually decrease the energy in a system, thus reducing water's ability to maintain an equilibrium?

 

I believe the modern belief is that water at equilibrium is always trading off the H+ ion, but that takes an amount of energy, right? Bonds and broken and re-created, right?

Edited December 20, 2009 by Genecks

[farmboy]

In such a system only a tiny number of the water molecues will have dissosciated, so the reaction does lie heavily in favour of H2O. However, even though most of the water remains water due to the high energy barrier, every now and again a molecule will get the energy it needs to dissosciate. In this instance you have to remember that the equilibrium reached will involve the atmosphere etc. In a completely closed system, any energy given out as bonds are broken etc. will remain in the system, just in another form. The total energy of the closed system would be a constant.

[Genecks]

So, could we say that in a closed system that H20 does not dissociate?

[Psycho]

So, could we say that in a closed system that H20 does not dissociate?

No, as in a theoretical closed system there would be no energy loss or gain either, the dissociation would remain constant with different water molecules dissociating in the same manner as normal as the energy level would remain constant.