Explaining the Effects of Removing UO2 on a Chemical Equilibrium

AI Thread Summary
Removing UO2 from the reaction system UO2(s) + 4HF(g) <-> UF4(g) + 2H2O(l) has distinct effects on chemical equilibrium. When some UO2 is removed, it does not affect the equilibrium because solids do not influence the position of equilibrium as long as they are present. However, if all UO2 is removed, the system shifts to increase the concentration of HF and decrease UF4 to re-establish equilibrium, as UO2 is necessary for the reaction to proceed. This shift occurs because the absence of UO2 means the reaction cannot produce UF4, prompting changes in the concentrations of the gaseous reactants. Understanding these dynamics is crucial for analyzing chemical equilibria involving solids.
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Homework Statement


Suppose the reaction system UO2(s) +4HF(g) <-> UF4(g) +2H2O(l)
is sealed in a container and allowed to equilibrate at a particular temperature.

Explain what happens to the nature of each substance if you:
a. remove some UO2 from the system
b. remove all UO2 from the system


Homework Equations





The Attempt at a Solution


I know that UO2 is solid and that it does not affect the equilibrium but in my answers it says that a) will have no effect while b) will increase [HF] and decrease [UF4]...why is that? I thought that all solids will have no effect no matter how much you take out..Thanks.
 
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Soilds have no effect as long as they exist. If the soid doesn't exist, system will shift the equilibrium till the solid appears.
 
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