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Le Chatliers principle and liquids/solids

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GomerPyle

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I didn't think solids or liquids played a role in le chatliers principle, but according to aamc chem self assessment number 34, increasing the amount of reactant water pushes the equilibrium to the side with no water. I am a little confused because I thought solids/liquids didn't play a role in le chatliers.

Also - do solids and liquids play a role when determining enthalpy or entropy of a reaction? What about with free energy? I know they don't appear in equilibrium equations, nor the rate laws. I just want to fully understand when we don't include them!

Thanks
 

sazerac

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I don't have the question in front of me, but I suspect there are some solutes in a water solvent in your reaction. Don't focus on the water as H2O. Focus on how adding more solvent affects the concentration of the reactants that matter, and how that affects the equilibrium of the reaction.
 

The_Sunny_Doc

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If you have a reaction in a dilute solution (tons of h2O floating around) then adding a little water won't shift the rxn much. However, if water is a reactant (say you have a super concentrated acid solution), adding or removing water will cause a shift.

It's like those reversible orgo reactions. You add a ton of H2O (technically, aqueous acid) to remove a ketal protecting group because h2O is in the products, so to get the ketone on the left side of the arrow to reform, you dump a bunch of product (H2O) in there. Same with other reversible acid catalyzed rxns like Fischer esterification iirc. Just think of the analogous examples when you approach the questions. :)

PS there's a thread concerning this topic here:
http://forums.studentdoctor.net/archive/index.php/t-975455.html
 

GomerPyle

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The one thing I have a problem with is - how do you know whether a liquid (water) is a reactant or not? If it is in the equation in the first place, it has to be a reactant, doesn't it?

If water is a reactant and OH is the product, it is safe to say water is the reactant and increasing water will increase OH. But when do you know it is NOT a reactant? Also, how would you know whether the reactant mixture is dilute or saturated, and where do you draw the line to determine whether water is included in the equilibrium expression or not?
 

The_Sunny_Doc

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The one thing I have a problem with is - how do you know whether a liquid (water) is a reactant or not? If it is in the equation in the first place, it has to be a reactant, doesn't it?

If water is a reactant and OH is the product, it is safe to say water is the reactant and increasing water will increase OH. But when do you know it is NOT a reactant? Also, how would you know whether the reactant mixture is dilute or saturated, and where do you draw the line to determine whether water is included in the equilibrium expression or not?

You would know whether to account for H2O by seeing whether the water molecules interact with all of the reactants/products the same way (i.e. the H2O decreases the conc'n of the others equally) or whether the addition of H2O decreases just one of the reactants or products, disrupting the RATIO of concentrations.

So, when water is a reactant and a solvent, the amount of water present is going to change so little that H2O would not change the balance of products and reactants. You can tell when water is the solvent by looking for the words "aqueous solution"or "dilute aqueous acid." Only a few h2O molecules out of a billion are gonna dissociate, so your balance of products and reactants isn't gonna change much.

However, in an esterification, the reactants are a carboxylic acid and an alcohol, and there is no solvent. Since there aren't H2O molecules present in the first place, you have to account for the impact of water.
 
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