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Does adding or removing pure liquids/solids affect equilibrium?

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FROGGBUSTER

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edit: Answer in post #5 & #9.

According to Zumdahl's textbook, it doesn't.

In the book, there is a question that asks which way the equilbrium will shift if additional UO2(s) is added to the system.

UO2(s) + 4HF (g) <-------> UF4(g) + 2H2O(g)

The answer is the equilibrium is not affected.

HOWEVER,

I stumbled upon a thread from here in which the original poster claims he/she has seen a problem in which removal of H2O(l) would shift the reaction to the left.

For the following reaction at equilibrium

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

which will shift reaction to the left?

Why the discrepancy?

Link here: http://forums.studentdoctor.net/showthread.php?p=9828806#post9828806
 
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SteveJMarist

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According to Zumdahl's textbook, it doesn't.

In the book, there is a question that asks which way the equilbrium will shift if additional UO2(s) is added to the system.



The answer is the equilibrium is not affected.

HOWEVER,

I stumbled upon a thread from here in which the original poster claims he/she has seen a problem in which removal of H2O(l) would shift the reaction to the left.



Why the discrepancy?

Link here: http://forums.studentdoctor.net/showthread.php?p=9828806#post9828806


Seems like your link answers your question. H2O is a reactant in this case, not a solvent. Thus, it is included in the equilibrium constant and also if it was removed it would shift the reaction to the left.
 

FROGGBUSTER

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Thanks for replying.

Why would adding additional UO2(s) to the first system not affect the equilibrium though?

edit: H2O is the solvent though isn't it? HNO2 & HNO3 are aqueous.
 

FROGGBUSTER

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I think I figured it out, or more accurately, someone figured it out for me, lol.

I posted this question on another forum and got this response:

In the second case, adding more water to the system will dilute the HNO2 and HNO3 (since these are aqueous). Therefore, to re-establish equilibrium, the reaction will proceed to create more products.

In general, however, adding more pure solid or liquid to a system should not affect the equilibrium of a system, assuming the addition of the solid or liquid does not change the concentrations/partial pressures of the other components of the system.
[/quote]

Because H2O is a solvent for the 2 products, increasing H2O would decrease the concentrations of both. Thus, the equilibrium would shift to the right to restore balance.

It's always so simple after you finally understand it, lol.
 
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FROGGBUSTER

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Because H2O is a solvent for the 2 products, increasing H2O would decrease the concentrations of both. Thus, the equilibrium would shift to the right to restore balance.

It's always so simple after you finally understand it, lol.

BUt your question asked us to rationalize why it would shift to the left![/QUOTE]

Well it's the same thing. Adding more water dilutes the solution and decreases the concentration of the products, so the equilibrium shifts right. Taking away water would concentrate the solution and increase the concentration of the products, so it would shift left.

I think the guy just read it incorrectly but the logic jives.

Thanks for the responses though.
 

FROGGBUSTER

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I've decided to follow up with another clarifying post.

Something to keep in mind is that the only thing that can change K is temperature. At a certain temperature, K will always be the same value; it doesn't depend on concentrations of products and/or reactants.

UO2(s) + 4HF (g) <-------> UF4(g) + 2H2O(g)

In the first example in the original post, adding solid will not affect the equilibrium because we're assuming the amount of solid you add doesn't significantly affect the volume/pressure/concentrations of the gases in the container.

K = [UF4]*[H2O]^2 / [HF]^4. This does not change upon adding a negligible amount of solid, and the reaction does not shift.

For the following reaction at equilibrium

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

which will shift reaction to the left?

In the second example, K = [HNO2]*[HNO3] / [NO2]^2.

H2O is NOT in the equilibrium equation. [H2O] is constant no matter if you add or remove H2O. Concentration is an intensive property.

When you remove H2O from the solution, again you assume that it is a negligible amount of H2O you are removing compared to the volume the gas takes up. Therefore, the pressure/volume/concentration of NO2 does not change.

However, removing water does increase the concentration of HNO2 & HNO3. Q, the reaction quotient, is now greater than K.

This means we have to shift to the left to once again reach the value of K, our equilibrium.

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I hope that was helpful! I'm going to verify this with some smarter people I know lol and I'll post back/edit if they disagree.
 
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