Gchem equil

[chiddler]

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Why don't solids affect equilibrium with gases?

i forget

 

[syoung]

me too. i just remember that they just DONT contribute to the Eq equation.
something to do w/ it's density?

 

[folktale]

me too. i just remember that they just DONT contribute to the Eq equation.
something to do w/ it's density?

But then why would aqueous solutions still be part of the eq equation? Only pure solids and liquids are excluded.

 

[typicalindian]

Isn't it because the equilibrium expression only includes free concentrations of each reactant, and solids are not freely floating around in solution so they can't be counted?

 

[MedPR]

I think it's because solids are not available to react. If you want a precipitate to participate in the reaction, you have to make it soluble first.

 

[gettheleadout]

So in a gaseous equilibrium the values factored into the equilibrium constant expression are partial pressure values right? A solid or liquid has no partial pressure because it has no pressure because it's not a gas, so it cannot be factored in.

With aqueous equilibrium it's a bit less obvious: the values in the k expression here are molar concentrations right? Molarity is moles solute over liters solution, so even though there is a certain mass (number of moles) of solid sitting in the solution, it doesn't exist as solute, so it cannot have a concentration and cannot be factored in. Same for pure liquids in solution.

 

[chiddler]

thanks very much

 

[SaintJude]

But then why would aqueous solutions still be part of the eq equation? Only pure solids and liquids are excluded.

Ah, wished I had seen thread earlier. So only aqueous solutions and gases are part of the equation.

But still, once adding a solid can we make a production about a shift caused due to Le Chatelier's principle? I suspect, no...but I'm not sure.

 

[milski]

So in a gaseous equilibrium the values factored into the equilibrium constant expression are partial pressure values right? A solid or liquid has no partial pressure because it has no pressure because it's not a gas, so it cannot be factored in.

With aqueous equilibrium it's a bit less obvious: the values in the k expression here are molar concentrations right? Molarity is moles solute over liters solution, so even though there is a certain mass (number of moles) of solid sitting in the solution, it doesn't exist as solute, so it cannot have a concentration and cannot be factored in. Same for pure liquids in solution.

Partial pressures are proportional to molar concentrations, so it really does not matter if you're using one or the other. You are correct for the rest of it - they take a constant activity number of 1 (with the proper units), so you can ignore them in the equilibrium equation.

 

[milski]

Ah, wished I had seen thread earlier. So only aqueous solutions and gases are part of the equation.

But still, once adding a solid can we make a production about a shift caused due to Le Chatelier's principle? I suspect, no...but I'm not sure.

Adding a solid can make a reaction possible (as opposed to not being there). It will not shift the equilibrium if some of the solid was already present in the solution. General stoichiometry concerns still apply, if the reaction goes to completion.