shifting equilibrium position - pure liquids TBR

[Meredith92]

I've tried to look this up on past threads but havent found a satisfactory answer. I know that pure liquids arent in the equilibrium reaction, but for some reason in TBR it says that by removing H2O(l) we shift the equilibrium.

The reaction is:
2NO2(g) +H2O (l)<-- ---> HNO2 (aq) +HNO3(aq)

The question asks:
Which will shift the reaction to the left?
A Addition of sodium hydoxide 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 from TBR " D is correct. Addition of sodium hydroxide to solution deprotonates H2O (HNO3 is a strong acid and has already fully dissociated), and thus shifts the reaction to the product side to re-establish equilibrium. Addition of manganese (II)chloride (MnCL2) to solution removes both NO2 and NO3 from solution through complexing of the ligands, so the reaction will shift to the right. Removal of nitrate (NO3) from the solution results in a shift to the product side. Removal of water (A REACTANT) results in a shift in the reverse direction (left) to re-establish the equilibrium"

The answer implies that water is an actual reactant involved with the equilibrium expression... but I thought pure liquids werent included.

Also should we always assume that transition metals such as manganese can always form ligands with ions that remove them from the solution?

-----
Edit: MY BAD! I found an amazing thread already on this...
BerkReviewTech came to the rescue as usual:

BerkReviewTeach
01-08-2013, 03:15 PM
This question can be answered with a slightly different perspective. The reaction is:

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

Removing a pure liquid in high enough concentration should not significantly impact the equilibrium. But, removing water does increase the concentrations of both products (HNO2(aq) and HNO3(aq)), which will knock the reaction out of equilibrium and result in excess products being present. The reaction must shift to the left to reduce the solute concentrations down and back to equilibrium.

It's very simlar to what happens if you were to decrease the volume of the container holding the following equilibrium system:

1 A(g) + 1 B(l) <=> 1 X(g) + 1 Y(g)

Reducing the volume will force the reaction to the left side of the reaction, where there are less gas molecules. If we simply make the system aqueous instead of a gas phase equilibrium, then a decrease the volume of water solvent in the solution holding the following equilibrium system:

1 A(aq) + 1 H2O(l) <=> 1 X(aq) + 1 Y(aq)

we'd get the same shift to the left.

The actual reaction is:

2 A(g) + 1 H2O(l) <=> 1 X(aq) + 1 Y(aq)

so, a change in the molarity of X and/or Y will cause a shift in the equilibrium and a change in the volume of the container will change the partial pressure of A and cause a shift in the equilibrium.

One really important point to make is that a reaction can shift while keeping the same Keq. The actual value of Keq won't change (except with temperature), but the concentration and partial pressures can change if a change to the system disrupts the equilibrium.

---

Members don't see this ad.

 

[Meredith92]

This is again a similar question.. and I want to make sure I understand the concept!

TBR 100. The amount of salt that dissociates into water is ALWAYS increased by which of the following changes?
A increasing temp
B decreasing temp
C increasing the amount of water
D decreasing the amount of water

I first chose A because I thought this reaction involves breaking ionic bonds of the salt and only forming weak interactions with the liquid, so more energy must be put in.

However the answer is C
TBR says
"Choice D is eliminated, because less solvent reduces the amount of a salt that can dissociate into a solution. Because a solvation reaction can be either exothermic or endothermic, the effect on the system by a change in temp varies, and is thus unpredictable. this eliminates choices A and B. in most cases an increase in temp results in an increase in the amount of salt that dissociates into solution, but there are some exceptions. Only choice C, increasing the solvent, always increases the amount of a salt that dissociates into solution. as more solvent is added, more compound can dissociate. be sure that you realize that this question is asking about the AMOUNT (in mass or moles) of salt dissociated, not the concentration ( molar solubility).

---> so just to make sure i understand the last sentence- the molar solubility remains the same, so therefore increasing volume (increasing the solvent) means (from moles/L solution) V increases so moles must increase to keep the molar solubility the same.

-- can we also use the same logic as Berkreviewtech? That by increasing volume we are only affecting the aqueous products, not the solid reactant.

So in summary of both of these posts... solvents CAN affect the equilibrium by changing the volume of only some of the reactants/ products (not all- or they would all be equally affected) BUT they are still not included in the actual K equilibrium equation. They can only shift it by affecting some concentrations

 

[brood910]

I've tried to look this up on past threads but havent found a satisfactory answer. I know that pure liquids arent in the equilibrium reaction, but for some reason in TBR it says that by removing H2O(l) we shift the equilibrium.

The reaction is:
2NO2(g) +H2O (l)<-- ---> HNO2 (aq) +HNO3(aq)

The question asks:
Which will shift the reaction to the left?
A Addition of sodium hydoxide 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 from TBR " D is correct. Addition of sodium hydroxide to solution deprotonates H2O (HNO3 is a strong acid and has already fully dissociated), and thus shifts the reaction to the product side to re-establish equilibrium. Addition of manganese (II)chloride (MnCL2) to solution removes both NO2 and NO3 from solution through complexing of the ligands, so the reaction will shift to the right. Removal of nitrate (NO3) from the solution results in a shift to the product side. Removal of water (A REACTANT) results in a shift in the reverse direction (left) to re-establish the equilibrium"

The answer implies that water is an actual reactant involved with the equilibrium expression... but I thought pure liquids werent included.

Also should we always assume that transition metals such as manganese can always form ligands with ions that remove them from the solution?

-----
Edit: MY BAD! I found an amazing thread already on this...
BerkReviewTech came to the rescue as usual:

BerkReviewTeach
01-08-2013, 03:15 PM
This question can be answered with a slightly different perspective. The reaction is:

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

Removing a pure liquid in high enough concentration should not significantly impact the equilibrium. But, removing water does increase the concentrations of both products (HNO2(aq) and HNO3(aq)), which will knock the reaction out of equilibrium and result in excess products being present. The reaction must shift to the left to reduce the solute concentrations down and back to equilibrium.

It's very simlar to what happens if you were to decrease the volume of the container holding the following equilibrium system:

1 A(g) + 1 B(l) <=> 1 X(g) + 1 Y(g)

Reducing the volume will force the reaction to the left side of the reaction, where there are less gas molecules. If we simply make the system aqueous instead of a gas phase equilibrium, then a decrease the volume of water solvent in the solution holding the following equilibrium system:

1 A(aq) + 1 H2O(l) <=> 1 X(aq) + 1 Y(aq)

we'd get the same shift to the left.

The actual reaction is:

2 A(g) + 1 H2O(l) <=> 1 X(aq) + 1 Y(aq)

so, a change in the molarity of X and/or Y will cause a shift in the equilibrium and a change in the volume of the container will change the partial pressure of A and cause a shift in the equilibrium.

One really important point to make is that a reaction can shift while keeping the same Keq. The actual value of Keq won't change (except with temperature), but the concentration and partial pressures can change if a change to the system disrupts the equilibrium.

V/P changes affect equilibrium only if there is difference in mole # on reactant + product sides of the reaction. In this case, we have 2 on the reactant side and 2 on the product side. So we cannot really apply the V change concept here.
Instead, like the original poster said, it is because of the molarity change.

 

[Meredith92]

okay gotcha! I think i used the term volume incorrectly.. i meant the amount of solvent (which affects molarity)

 

[Meredith92]

This is again a similar question.. and I want to make sure I understand the concept!

TBR 100. The amount of salt that dissociates into water is ALWAYS increased by which of the following changes?
A increasing temp
B decreasing temp
C increasing the amount of water
D decreasing the amount of water

I first chose A because I thought this reaction involves breaking ionic bonds of the salt and only forming weak interactions with the liquid, so more energy must be put in.

However the answer is C
TBR says
"Choice D is eliminated, because less solvent reduces the amount of a salt that can dissociate into a solution. Because a solvation reaction can be either exothermic or endothermic, the effect on the system by a change in temp varies, and is thus unpredictable. this eliminates choices A and B. in most cases an increase in temp results in an increase in the amount of salt that dissociates into solution, but there are some exceptions. Only choice C, increasing the solvent, always increases the amount of a salt that dissociates into solution. as more solvent is added, more compound can dissociate. be sure that you realize that this question is asking about the AMOUNT (in mass or moles) of salt dissociated, not the concentration ( molar solubility).

---> so just to make sure i understand the last sentence- the molar solubility remains the same, so therefore increasing volume (increasing the solvent) means (from moles/L solution) V increases so moles must increase to keep the molar solubility the same.

-- can we also use the same logic as Berkreviewtech? That by increasing volume we are only affecting the aqueous products, not the solid reactant.

So in summary of both of these posts... solvents CAN affect the equilibrium by changing the volume of only some of the reactants/ products (not all- or they would all be equally affected) BUT they are still not included in the actual K equilibrium equation. They can only shift it by affecting some concentrations

Can someone confirm my interpretation of the last sentence of tbr's explanation is correct? They say to understand the moles not concentration are changing