Does adding or removing pure liquids/solids affect equilibrium?

In summary, according to Zumdahl's textbook, the addition of additional UO2 to the system does not shift the equilibrium of the reaction UO2(s) + 4HF (g) <-------> UF4(g) + 2H2O(g). However, in a separate case involving the reaction 2NO2(g)+H2O(l) <-> HNO2(aq)+HNO3(aq), adding more water will dilute the HNO2 and HNO3, causing the reaction to proceed to create more products and re-establish equilibrium. In general, adding a pure solid or liquid should not affect the equilibrium of a system, assuming it does not change the concentrations of other components. The
  • #1
MechRocket
14
0
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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  • #2
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.
 
  • #3
Brilliant, thanks!
 
  • #4
@Yggg: I have another question.

How would you write the equilibrium equation? Would you take into account H2O in the denominator or no?
 
  • #5
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
 

1. How does adding a pure liquid/solid affect equilibrium?

Adding a pure liquid or solid to a system at equilibrium does not affect the equilibrium position. This is because the concentration of the pure substance remains constant and does not change the overall concentration of the reactants or products.

2. Can adding a pure liquid/solid shift the equilibrium position?

No, adding a pure liquid or solid does not shift the equilibrium position. The position of equilibrium is determined by the concentrations of the reactants and products, and adding a pure substance does not change these concentrations.

3. Will removing a pure liquid/solid change the equilibrium constant?

No, removing a pure liquid or solid does not change the equilibrium constant. The equilibrium constant is a ratio of concentrations and does not depend on the physical state of the substances involved.

4. How does removing a pure liquid/solid impact the equilibrium reaction?

Removing a pure liquid or solid will not impact the equilibrium reaction. The reaction will continue to proceed in the same direction until the equilibrium constant is reached again.

5. Can removing a pure liquid/solid shift the equilibrium position?

No, removing a pure liquid or solid does not shift the equilibrium position. The position of equilibrium is determined by the concentrations of the reactants and products, and removing a pure substance does not change these concentrations.

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