Why does removing pure liquid from solution displace equilibrium?

many times we see water acting as the solvent in the solution. however, in this situation it is acting as a reactant. thus, if we remove a reactant then we're going to shift the equilibrium

That rule is for solvents. There are many liquids - bromine, short alkenes, etc. I have a feeling that if one of these was substituted for water, you'd have gotten this question right. Water is serving as a reactant, so it's concentration matters.

Monkeymaniac said:

Ah that makes sense. For the case of pure liquid solvent, I now understand why we can ignore it in calculating equilibrium shift. Would you guys please help me understand why we would also ignore a pure solid when determining equilibrium shift, like in salt dissolution reaction? That is, how can the concentration of the solid remain relatively constant? It seems that unlike pure liquid solvent, the solid's volume relative to acqueous components is a lot smaller.

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we don't include pure solids or liquids in an equilibrium expression. for your case above, water was acting as a reactant so therefore it must be included in our expression. there's an implicit assumption that pure liquids and solids aren't changing in concentration and therefore aren't included

loveoforganic said:

That rule is for solvents. There are many liquids - bromine, short alkenes, etc. I have a feeling that if one of these was substituted for water, you'd have gotten this question right. Water is serving as a reactant, so it's concentration matters.

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You mean the solvent in this case is not water but another liquid? That can't be the case since it says "(aq)" next to the reactants and products...

even if the solvent were not a reagent, removing it displaces equilibrium because the concentrations of all the species would change by that amount.

bleargh said:

even if the solvent were not a reagent, removing it displaces equilibrium because the concentrations of all the species would change by that amount.

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Right, but how do we know it's to the left?

we don't. in the op's scenario though, it's a reagent so we know direction.

if you write out the formula for Keq, though, you will see the concentration justification as well

bleargh said:

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we don't. in the op's scenario though, it's a reagent so we know direction.

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But according to what you wrote above, the reason for the equilibrium shift is because of the change in concentrations of the solutes, not the water.

if you write out the formula for Keq, though, you will see the concentration justification as well

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How? I don't see how this would affect it. In fact, changing the volume of solvent will change the concentrations of all solutes by the same factor, effectively canceling everything out. Basically, I don't understand why changing the volume of water will shift the equilibrium at all.

OP, can you reference a page number (and edition)?

anyone figure this out?

ishchayill said:

But according to what you wrote above, the reason for the equilibrium shift is because of the change in concentrations of the solutes, not the water.

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i was speaking in generalities at first. in certain situations, depending on the chemcial reaction, removing solvent would not displace equilibrium. the one cited by OP is one such example, so we need a further justification.

How? I don't see how this would affect it. In fact, changing the volume of solvent will change the concentrations of all solutes by the same factor, effectively canceling everything out. Basically, I don't understand why changing the volume of water will shift the equilibrium at all.

OP, can you reference a page number (and edition)?

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incorrect, to wit -

e.g.

NaCl --> Na + Cl (edit: ignore that NaCl here is a solid, i can't come up with a decomp. reaction off the top of my head)

where K=[Na][Cl]/[NaCl]. removing water from this and changing the concentration by equal factors would in fact not cancel out.

going back to the OP's problem:

2NO2(g)+H2O(l) (it's l not g) <-> HNO2(aq)+HNO3(aq),

K=[HNO2][HNO3]/[NO2]^2[H2O]

let's assume for the sake of easy calculation that initially at equilibrium, K=1, and all the concentrations are 1, including H2O. now we remove half the water, to raise concentration by 2.

Q=(2*2)/(2^2*0.5)=2

Since Q>K, we know we have too much product, so the equilibrium must shift to the left

How can you include [H2O] in the equilibrium expression? It's a liquid solvent and should be left out.

incorrect, to wit -

e.g.

NaCl --> Na + Cl

where K=[Na][Cl]/[NaCl]. removing water from this and changing the concentration by equal factors would in fact not cancel out.

going back to the OP's problem:

2NO2(g)+H2O(l) (it's l not g) <-> HNO2(aq)+HNO3(aq),

K=[HNO2][HNO3]/[NO2]^2[H2O]

let's assume for the sake of easy calculation that initially at equilibrium, K=1, and all the concentrations are 1, including H2O. now we remove half the water, to raise concentration by 2.

Q=(2*2)/(2^2*0.5)=2

Since Q>K, we know we have too much product, so the equilibrium must shift to the left

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dont we remove pure liquids and solids from the Keq however?

[Quote functions don't seem to be working today ...]

ishchayill said:

How can you include [H2O] in the equilibrium expression? It's a liquid solvent and should be left out.

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no only pure solids are left out. we take out the liquid solvent for most applications because for most purposes the change in [H2O] is negligible and thus it doesn't affect the math. in this case, though, we ARE changing its concentration in appreciable amounts and thus it needs to go back in

bleargh said:

no only pure solids are left out. we take out the liquid solvent for most applications because for most purposes the change in [H2O] is negligible and thus it doesn't affect the math. in this case, though, we ARE changing its concentration in appreciable amounts and thus it needs to go back in

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Ok pure solids are left out. So pure liquids are only omitted when? Now I became confused. I was under the assumption we always omit pure liquids. For example if this was Br(l) instead of H2O(l) would I include it or what

for all intents and purposes leave pure liquids out. note you don't even have to know anything about K for this problem, just LeChat's principle

bleargh said:

for all intents and purposes leave pure liquids out. note you don't even have to know anything about K for this problem, just LeChat's principle

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but I thought that with le chateliers principle if i add or remove a liquid/solid it wouldnt do anything. Is this wrong?

IntelInside said:

but I thought that with le chateliers principle if i add or remove a liquid/solid it wouldnt do anything. Is this wrong?

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yes.

Regarding leaving stuff out, whether changing the amount of fluid affects equilibrium, etc, etc: This is one of many situations where if you stop trying to come up with overgeneralized "rules" to compensate for your lack of understanding of the underlying concept, things will actually get much easier.

bleargh said:

yes.

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can you explain this then

IntelInside said:

can you explain this then

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explain to me why removing a liquid product wouldn't drive a reaction forward?

In ester formation by the condensation of an alcohol and a carboxylic acid, water is a product. This reaction is catalyzed by acid. In dilute acidic solution, carboxylic acid and alcohol (reactants) are favored. When you run it in a nonaqueous solvent in distill out the water as you run, the ester (product) is favored.

This is due to LeChatlier's (you're altering the concentration of water, a product in the reaction).