Does adding or removing pure liquids/solids affect equilibrium?

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In the discussion, it is clarified that adding pure solid or liquid, such as UO2(s) or H2O(l), does not affect the equilibrium of a chemical reaction, as long as the concentrations or partial pressures of other components remain unchanged. The example provided highlights that removing H2O(l) from the reaction involving NO2 and H2O can shift the equilibrium to the left, as it dilutes the products HNO2 and HNO3. The discrepancy arises from the different states of matter and their impact on equilibrium. Additionally, when writing the equilibrium expression, the concentration of water is typically omitted as it remains constant in dilute solutions. Overall, understanding these principles is crucial for accurately analyzing chemical equilibria.
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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.

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

The answer is the equilibrium is not affected.

HOWEVER,

I stumbled upon a thread on studentdoctor 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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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.
 
Brilliant, thanks!
 
@Yggg: I have another question.

How would you write the equilibrium equation? Would you take into account H2O in the denominator or no?
 
No, you would not need to include the concentration of water in the denominator. Assuming the solution is sufficiently dilute, the concentration of water will essentially be a constant ~55.6 M throughout the reaction.

The expression for the equilibrium constant would look something like:
K = [HNO2][HNO3]/(P_NO2)^2
 
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