When to include water in equilibrium?

[rocketbooster]

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I know this is fundamental knowledge and I definitely knew the answer to this a month ago.

anyways, is water only excluded from equilibrium in the equilibrium constant? water is INCLUDED when considering changes in equilibrium in accordance to Lechatlier's principle, right?

 

[VanBrown]

umm.. I wouldn't put it into a rate equillibrium ever but I would keep it in mind for solubilities.

I think the major thing to remember is that water is almost always a solvent and will have a molarity that is like 55.5M on both sides of the equillibrium equation so it will essentially cancel.

However, with solubility... adding or taking water away will effect where a solution will be saturated. Like if you add water to a saturated solution of Ca(OH)2 more of the Ca(OH)2 will dissolve.... the equilibrium doesn't change but the amount dissolved does...

 

[rocketbooster]

umm.. I wouldn't put it into a rate equillibrium ever but I would keep it in mind for solubilities.

I think the major thing to remember is that water is almost always a solvent and will have a molarity that is like 55.5M on both sides of the equillibrium equation so it will essentially cancel.

However, with solubility... adding or taking water away will effect where a solution will be saturated. Like if you add water to a saturated solution of Ca(OH)2 more of the Ca(OH)2 will dissolve.... the equilibrium doesn't change but the amount dissolved does...

yep, that's what i meant. doesn't affect the equilibrium, thus doesn't affect the rate. i know it how it affects solubility. again, doesn't change the Ksp, only changes the amounts dissolved/concentration of solute.

what i was referring to was in Le Chat. if water is a reactant or product. if it's a reactant in the chemical equation and water is removed, then the equilibrium will be shifted to the left to produce more water. to, water does affect equilibrium shifts by Le Chat. the actual equilibrium constant will not be changed. also, in rate laws, water/liquids and solids aren't included, which is why the Keq, Ksp, etc. do not change.

just doublechecking.

 

[tncekm]

Yeah, I believe that in an "aqueous solution" there is such a high concentration of water relative to the reactants and products of interest that the concentration never changes (for all practical purposes) and thus doesn't change anything.

 

[Venicat]

We don't consider a pure substance like H2O in an equilibrium expression because it has an activity of 1, and concentrations are used an approximation of activity in equilibrium expressions. It's a time waster to multiply by 1.

 

[ParthVader]

We don't consider a pure substance like H2O in an equilibrium expression because it has an activity of 1, and concentrations are used an approximation of activity in equilibrium expressions. It's a time waster to multiply by 1.

Well its because when writing the equilibrium expression you divide everything by its standard state to get rid of the unit problem. For aqueous solutions standard state is 1.0 M so you divide those by 1.0M, but for water standard state is 55.5 M, and pure water is always 55.5M so dividing 55.5 M by 55.5 M you just get 1, and you can leave pure water out of the equilibrium constant.