Quick question about Le Chateliers

[member232]

---

Members do not see this ad.

The given reaction is:

2 NO2 (g) + H2O (l) <------> HNO2 (aq) + HNO3 (aq)

Question:

Which will shift the reaction to the left?

A. Addition of sodium hydroxide to the solution
B. Addition of manganese(II) chloride to the solution
C. Removal of nitrate from the solution
D. Removal of water from the solution

Answer: D.

Now, I understand why A, B, and C, don't work. But since in the reaction H2O is listed as a liquid, won't the addition or removal of a liquid (or solid) not affect the Qrxn?

i.e. Am I wrong to think that Keq = [HNO3] [HNO2] / [NO2]^2 for this reaction?

Any input would be very much appreciated 🙂

Thanks in advance!

 

[MedPR]

Remove water = reaction wants to have more water = reaction shifts to the side that has water on it (in order to produce water).

 

[member232]

But isn't water considered a solvent as it is a liquid?

According to a variety of sources, this is quoting Kaplan in particular, " The concentrations of pure solids and pure liquids (the solvent really) do not appear in the equilibrium constant expression, because their concentrations do not change in the course of the reaction"

I know removing a reactant would shift the reaction to the left, but water isn't really a reactant here...

 

[LoLCareerGoals]

I don't think MedPR's explanation is correct. I agree that liquids and solids are not supposed to be present in the equilib constant equation. At test speed we can eliminate A, B, C and just shrug and pick D.
For a real explanation, here is my thought process:
Like you said: Keq = [HNO3] [HNO2] / [NO2]^2
Actually for a gas we have to use partial pressure, but not important here.

Now if water is removed, concentration of aqueous compounds (the ones on the right) increases because same # of moles of are present in less water. If that is the case, reaction shifts left.

 

[member232]

I don't think MedPR's explanation is correct. I agree that liquids and solids are not supposed to be present in the equilib constant equation. At test speed we can eliminate A, B, C and just shrug and pick D.
For a real explanation, here is my thought process:
Like you said: Keq = [HNO3] [HNO2] / [NO2]^2
Actually for a gas we have to use partial pressure, but not important here.

Now if water is removed, concentration of aqueous compounds (the ones on the right) increases because same # of moles of are present in less water. If that is the case, reaction shifts left.

I didn't think that far... good catch, thanks for the explanation 🙂