Complex Equilibria Question

[booji]

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I came across a complex equilibrium question while studying and after reading the explanation, I am a little puzzled. I am hoping that you guys would be able to help. The question reads as follows:

Which will shift reaction 4 to the let?
A. Addition of NaOH to the solution
B. Addition of Mn(II) Chloride ot the solution
C. Removal of NO3- from the solution
D. Removal of H2O from the solution.

Reaction 4 is a part of a complex equilibrim given below:

2 NO2 + H2O <-- --> HNO2 + HNO3 Rxn 4
MnCl2 + 2 NO2 - <-- --> Mn(NO2)2 + 2Cl- Rxn 5
MnCl2 + 2 NO3- <-- --> Mn(NO3)2 + 2Cl- Rxn 6

Personally, I had narrowed down my choices to C and D, and eventually chose C due to the following reasoning:

Removing nitrate from Rxn 6 would mean that the equilibrium of rxn 6 would shift to favor the reactants (meaning more NO3- would be produced). Since NO3- is coming from the dissociation of HNO3, increase in NO3- concentration would shift that dissociation equilibrium to favor the production of HNO3 would in turn shift Rxn 4's equilibrium to the left.
I was also thinking that D could be a more direct answer since removal of H2O will infact shift rxn 4's equilibrium to the left; but eventually I decided against it because H2O was in the standard state and hence would not affect the equilibrium.

The correct answer is apparently D. Although I can follow why it is D, I still thought that H2O would not affect the equilibrium because its a liquid (in fact there was a previous question in which there was a similar scenario and when I choose the liquid substance the answer explanation claimed that solids and pure liquids will not affect the equilibrium). Thus the root of my question is, when should we consider pure solids and liquids, and when should be ignore them when considering the Le Chatelier's principle and equilibrium shifts.

Thanks,

 

[singkri]

For eg, in ester formation removal of H2O from the pdt side does speed up the forward reaction, doesnt it?

I think unless it is explicitly mentioned that h2Ois a liquid in a particular reaction, you can assume it is in vapour form??

Not sure, but we do sometimes dehydrate the product to speed up the reaction.

 

[booji]

Singkri,

Thanks for your help. Unfortunatley, yes H2O is specifically mentioned to be in liquid form.

 

[scottj72]

The way I see it you really need to focus on RXN 4 reactants and products.

2 NO2 + H2O <-- --> HNO2 + HNO3 could also be written as
2 NO2 + H2O <-- --> [H+] [NO2-] + [H+] [NO3-]

Which will shift reaction 4 to the let?
A. Addition of NaOH to the solution

[OH-] from the NaOH can react with H+ to make more water and push the reaction to the right
B. Addition of Mn(II) Chloride ot the solution

Obviously will cause RXN 5&6 to move to the right which also consumes more of the products of RXN 4, so pushes RXN 4 to the right
C. Removal of NO3- from the solution

This may push RXN 6 to the left as you mentioned but for RXN 4 you are still removing the product, so it also pushes RXN 4 to the right.

D. Removal of H2O from the solution.

Here H2O is one of the reactants. If you remove a reactant it will have to push to the left. I believe thi would be a reaction between a gas and a liquid.

 

[scottj72]

The way I see it you really need to focus on RXN 4 reactants and products.

2 NO2 + H2O <-- --> HNO2 + HNO3 could also be written as
2 NO2 + H2O <-- --> [H+] [NO2-] + [H+] [NO3-]

Which will shift reaction 4 to the let?
A. Addition of NaOH to the solution

[OH-] from the NaOH can react with H+ to make more water and push the reaction to the right
B. Addition of Mn(II) Chloride ot the solution

Obviously will cause RXN 5&6 to move to the right which also consumes more of the products of RXN 4, so pushes RXN 4 to the right
C. Removal of NO3- from the solution

This may push RXN 6 to the right as you mentioned but for RXN 4 you are still removing the product, so it also pushes RXN 4 to the right.

D. Removal of H2O from the solution.

Here H2O is one of the reactants. If you remove a reactant it will have to push to the left. I believe thi would be a reaction between a gas and a liquid.

 

[booji]

The way I see it you really need to focus on RXN 4 reactants and products.

2 NO2 + H2O <-- --> HNO2 + HNO3 could also be written as
2 NO2 + H2O <-- --> [H+] [NO2-] + [H+] [NO3-]

Which will shift reaction 4 to the let?
A. Addition of NaOH to the solution

[OH-] from the NaOH can react with H+ to make more water and push the reaction to the right
B. Addition of Mn(II) Chloride ot the solution

Obviously will cause RXN 5&6 to move to the right which also consumes more of the products of RXN 4, so pushes RXN 4 to the right
C. Removal of NO3- from the solution

This may push RXN 6 to the right as you mentioned but for RXN 4 you are still removing the product, so it also pushes RXN 4 to the right.

D. Removal of H2O from the solution.

Here H2O is one of the reactants. If you remove a reactant it will have to push to the left. I believe thi would be a reaction between a gas and a liquid.

Thanks for your response. According to Zumdahl (a famous gen chem textbook), "Removal of a substance will result in the reaction moving in the direction that forms more of the substance. Remember that this is a change in concentration so changing the amount of a pure solid or liquid which is already in equilibrium should not make the reaction consume the additional solid or liquid. Example, if solid NaCl is already in equilibrium with a saturated solution of NaCl, adding more pure NaCl solid will not cause additional dissolving. Conversely, removal of some of the solid NaCl which is in equilibrium with a saturated solution of NaCl will not cause further crystallization of Nacl from the saturated solution."

So therefore, according to Zumdahl, since the reaction 4 is already in equilibrium, removal of H2O would not cause a shift to the left. Is this the definitive answer though?

 

[RPedigo]

Thanks for your response. According to Zumdahl (a famous gen chem textbook), "Removal of a substance will result in the reaction moving in the direction that forms more of the substance. Remember that this is a change in concentration so changing the amount of a pure solid or liquid which is already in equilibrium should not make the reaction consume the additional solid or liquid. Example, if solid NaCl is already in equilibrium with a saturated solution of NaCl, adding more pure NaCl solid will not cause additional dissolving. Conversely, removal of some of the solid NaCl which is in equilibrium with a saturated solution of NaCl will not cause further crystallization of Nacl from the saturated solution."

So therefore, according to Zumdahl, since the reaction 4 is already in equilibrium, removal of H2O would not cause a shift to the left. Is this the definitive answer though?

Zumdahl is saying adding more NaCl to that saturated solution wouldn't change the equilibrium, because it wouldn't change the concentration of sodium chloride in the solution-- its saturated, so adding more won't do jack to the equilibrium. If something is saturated at, say, 1.00M, if you add more, the solution will still be 1.00M, so Le Chatlier's principle doesn't apply.

However, usually water is a pure solvent, and we don't include it in the equilibrium expression. Here, it appears that it is a direct reactant, and is indeed consumed in the course of the reaction; I believe that's their reasoning here. Typically water just kicks it and watches a reaction happen; here, it's an active player.

Alternatively, A B and C are really, really, really wrong, so choice D is the only one that's not absolutely completely incorrect. It sucks, but it is the best answer, because literally A B and C will definitely 100% shift it to the right, which is not what we want.

 

[scottj72]

Well.. Not seeing the whole context of the exert, but here is my take on what you posted.

"Removal of a substance will result in the reaction moving in the direction that forms more of the substance."

So this supports the removal of H20 shifting the equation to the left.

I believe there rest of what you wrote is referring to aqueous solutions because there you would be dealing with concentrations. I am pretty sure RXN 4 involves NO2-gas and H2O. There are no concentrations, soluability,saturation, dissociation or anything like that to deal with. Just water reacting with a gas.

 

[booji]

Thank you guys for the clarification. I appreciate your help as always.

Booji