Equilibrium Shifts with Water

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brlin

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In the following reaction:
2NO2(g) + H2O (l) <-> HNO2(aq) + HNO3(aq)

Why does removing water shift the equilibrium to the left? I understand how it should work by Le Chatelier's principle, but I was also under the impression that pure solids / liquids don't affect equilibrium? Or does water not could as a pure liquid because it technically dissociates into H+ and -OH ions?
 
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Correct me if I'm wrong but I thinks its because water itself is a reactant in this reaction. Usually its just a medium for reactions to occur
 
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Removing water would not shift the equilibrium.

Keq = [HNO2][HNO3] / [NO2]^2

The fact that water is actually reacting in this example is irrelevant; water is a pure liquid and is not factored into the equilibrium expression because it cannot be given a concentration.

For the equilibrium to shift in any given direction (thereby necessarily altering the values of the concentrations present in the Keq expression) one of those concentrations present must change (forcing the others to change to accommodate).

Removing water in bulk could decrease the total volume of the solution, but because this would equally decrease the concentrations of everything present in the Keq expression, this would not cause an equilibrium shift.
 
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I think adding water DOES shift the equilibrium because like monkeyjack said, it is a reactant. Now, if you were to recalculate your equilibrium expression, you would not include water since it is a liquid, but the concentrations of reactants/products would be different. Think about temperature. You don't actually include temperature in your Keq expression, but if you change it, concentration of your reactants and products change, so even though it isn't present in the equation itself it still can shift equilibrium.
 
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cheesier said:
I think adding water DOES shift the equilibrium because like monkeyjack said, it is a reactant. Now, if you were to recalculate your equilibrium expression, you would not include water since it is a liquid, but the concentrations of reactants/products would be different. Think about temperature. You don't actually include temperature in your Keq expression, but if you change it, concentration of your reactants and products change, so even though it isn't present in the equation itself it still can shift equilibrium.
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Temperature is unique, its the only thing that changes Keq without factoring into the equation, because Keq is defined at a certain temperature.

Consider a dissolution reaction: MX2 (s) <--> M2+ (aq) + 2X- (aq)

The Keq for this reaction (Ksp in this case) is Ksp = [M2+][X-]^2

The solid MX2 is left out because it is a pure solid even though it is very much a reactant. As proof, if you remove it from the equation you have no reactants and thus no reaction.

Nevertheless, adding solid MX2 doesn't change the reaction because it's not in the solubility product. You can add as much solid MX2 as you want to a saturated solution of it and the equilibrium isn't going to change.
 
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getthelead, I had the exact same thoughts as you on this question. But according to BR, "Removal of water (a reactant) results in a shift in the reverse direction (left) to re-establish equilibrium)"

From what I know, liquids and solids only effect equilibrium if equilibrium is not yet reached and the liquid/solid is a limiting reagent. But I don't think that's the case for this situation. So I am still very confused about BR's answer.

It's Section III, Passage XII, question #81 for reference
 
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cheesier said:
I think adding water DOES shift the equilibrium because like monkeyjack said, it is a reactant. Now, if you were to recalculate your equilibrium expression, you would not include water since it is a liquid, but the concentrations of reactants/products would be different. Think about temperature. You don't actually include temperature in your Keq expression, but if you change it, concentration of your reactants and products change, so even though it isn't present in the equation itself it still can shift equilibrium.
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That doesn't make sense, if its a reactant you have to include it in the equilibrium expression after all it is prod over reactants. The question lies if water is a reactant or not.
 
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justhanging said:
That doesn't make sense, if its a reactant you have to include it in the equilibrium expression after all it is prod over reactants. The question lies if water is a reactant or not.
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No, water is clearly a reactant in this case. This can be proven by observing that NO2 alone cannot undergo the reaction without a source of hydrogen. This is as simple as determining what goes into the equilibrium expression.

Keq for a reaction in aqueous medium is Kc, where the values in the expression are concentrations. Solids and pure liquids are left out due to more complicate activity properties, but basically because they can't be expressed in terms of concentration.
 
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cheesier said:
I found a source, and it seems BR is correct. Very tricky.

http://www.newton.dep.anl.gov/askasci/chem07/chem07454.htm
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Equilibrium constant expressions do not usually include solvents because the standard states are chosen to have unit activity for pure solids and liquids. So changing the amount of a pure solid or pure liquid does not enter into the equilibrium constant expression... *unless* the liquid is a *solvent* in the reaction. In that situation, it matters whether the solution is dilute or concentrated. If the solution is very dilute we can treat the solvent as a "pure liquid" and neglect it from the equilibrium constant expression. However, if the solution is very concentrated then the activity of the solvent is no longer equal to one, and then it needs to be explicitly included in the equilibrium constant expression.

Here is an example. The reaction of acetic acid with water is HAc + H2O <-> Ac- + H3O+. When the acid is dilute we can get away with writing the equilibrium constant as Ka = [Ac-][H3O+]/[Hac]. However, when the acid is very concentrated, reaction of the acid with the water it is dissolved in measurably changes the concentration of water in the solution, and so the full equilibrium constant formula K = [Ac-][H3O+]/[Hac][H2O] must be used.
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This is really interesting. How do we know when a solution is concentrated enough to consider this? Also why the heck isn't this explicitly stated in chemistry/MCAT books?

Remember that the equilibrium constant is equal to the concentrations of the substances raised to their stoichiometric coefficients. Since pure liquids and solids do not have concentrations, they do not enter into the equation. So, as reactants and products, pure liquids and solids do not affect the chemical equilibrium.

However, water can act as a solvent. So if there are more moles of reactant in aqueous solution form tan there are of the products, than adding water will preferentially dilute more moles of reactants, this would preferentially slow down the forward reaction, and more reactants are produced than products.
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I sort of alluded to this earlier, because in this case dilution (or removal of water) will not preferentially shift the equilibrium because there are an equal number of moles of reactants and products (unless we're talking about a beaker of water and products reacting with gaseous NO2 just in a cloud above it...maybe we are, I don't know).

(1) Add acetone, or most any other pure water soluble organic compound to an aqueous solution of a water soluble ionic salt greatly reduces the water solubility of the salt.

(2) Conversely, adding a pure water soluble ionic salt to an aqueous solution of an organic compound reduces its water solubility. In the "chemical" jargon, this is called the>"salting out effect". Starting with a false premise will lead to incorrect consequences. You need to correct your original premise.
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These are both irrelevant because neither case involves a pure solid or pure liquid, both are soluble and would become aqueous, thus possessing concentrations.

You can definitely change the equilibrium of a reaction by adding water or a solid. The equilibrium is based on the ratio of concentrations for each component in the reaction vessel. This is the Keq value for the system. So if the addition of water or solid changes the concentration ratio the equilibrium will adjust. This is shown by using Le Chatelier's principle in chemistry classes. When the concentration is adjusted the equilibrium will shift in predictable a manner to react to the change. The times when this addition would not change the equilibrium would be when the solid/liquid does not interact with the solution itself or does not change the ratios between concentrations. If the addition of water simply dilutes all of the concentrations the same then the overall equilibrium will not shift as the overall ration does not change.
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Same comment as the second quote.
 
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Another way to think about it is this: both the water and the aqueous components contribute to the liquid phase. If water is consumed, there is more aqueous solution and if it was 50/50 water/aqueous solution and it shifts to 25/75 water/aqueous solution the effective concentration of water in the liquid phase has changed, therefore you could include it. For instance, if in a liter of solution you have appreciably more/less than 55.5 moles of pure H20 (water's normal concentration), it seems you would include it. That seems to be kind of what the site states.

I don't know if this is a correct interpretation, but it makes sense to me.

Edit: Same doesn't apply for solids because their concentration is a function of their density, which does not change during the reaction.
 
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I'd like to add at this point that I haven't read the passage, but TBR explicitly says in the Equilibrium chapter, "Do not include solids or pure liquids in the Keq expression, only solutes (for Kc) and gases (for Kp).

This seems like a weird technicality for TBR to test then, since we all agree that for water to affect the equilibrium position in this situation it MUST be present in the Keq expression.
 
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For MCAT purposes, if a reaction is in equilibrium, changing the concentrations of liquids or solids won't cause a shift. If the reaction is not in equilibrium, changing the concentrations of liquids or solids may cause a shift.
 
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H2O is clearly a reactant here. Removing it will shift the EQ to the left.

Let's say no water is present. Can you even write down the reaction? No.
Thus, water must be present in Keq.

A different example: HNO3 = H+ + NO3- (this process normally takes place in water but we do not really need water to write down the reaction, thus Keq willl not have water in this case)
 
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ChemTutor said:
H2O is clearly a reactant here. Removing it will shift the EQ to the left.

Let's say no water is present. Can you even write down the reaction? No.
Thus, water must be present in Keq.

A different example: HNO3 = H+ + NO3- (this process normally takes place in water but we do not really need water to write down the reaction, thus Keq willl not have water in this case)
Click to expand...

This logic is demonstrably incorrect. Solids involved in dissociation reactions are clearly reactants yet are left out of the Keq expression.
 
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gettheleadout said:
This logic is demonstrably incorrect. Solids involved in dissociation reactions are clearly reactants yet are left out of the Keq expression.
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I agree that my illustration does not explain everything away. I guess my point is that we should not try to slap blindly a simple "rule" from a test prep book on everything.

Are you talking about Ksp? Notice that this particular constant even has a special name. Essentially, it is a simplification of Keq which we derived by assuming that the concentration of the solid is constant. In other words, Ksp is not Keq.

Let's look at Ka or Kb. Why do they also have special names? Because they are not actually Keq. They are Keq/[H2O]. (***Edit: Keq =Ka/[H2O]) We are allowed to do this because there is a huge excess of water and, thus, [H2O]=constant

Back to the original example, nothing is mentioned about water being in a huge excess. We could have a container filed with NO2(g) and a few drops of water.
 
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ChemTutor said:
I agree that my illustration does not explain everything away. I guess my point is that we should not try to slap blindly a simple "rule" from a test prep book on everything.

Are you talking about Ksp? Notice that this particular constant even has a special name. Essentially, it is a simplification of Keq which we derived by assuming that the concentration of the solid is constant. In other words, Ksp is not Keq.

Let's look at Ka or Kb. Why do they also have special names? Because they are not actually Keq. They are Keq/[H2O]. We are allowed to do this because there is a huge excess of water and, thus, [H2O]=constant,

Back to the original example, nothing is mentioned about water being in a huge excess. We could have a container filed with NO2(g) and a few drops of water.
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Fair enough. My issue with all of this is simply that 1) this "rule" about pure solids and liquids, even when they are reactants, has been consistently presented in the chemistry courses I TA for over the last couple years, and the MCAT is specifically intended to test introductory course-level science knowledge, and 2) this is a question from a prep book that itself gives guidelines that it presents as hard rules...and then presumes to test an exception without mention of it? What?
 
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SOlids are different because their concentration is a function of their density which doesn't change.
 
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ChemTutor said:
Let's look at Ka or Kb. Why do they also have special names? Because they are not actually Keq. They are Keq/[H2O]. We are allowed to do this because there is a huge excess of water and, thus, [H2O]=constant,
Click to expand...

I don't think this is true though. I was under the impression that Ka and Kb are just specific instances of Keq.

For example, with a generic acid dissociating in water,
HA + H2O <-> A- + H3O+
Ka = ([A-][H3O+])/HA]
 
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cheesier said:
SOlids are different because their concentration is a function of their density which doesn't change.
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Solids don't have a concentration... Pick any concentration measure, molarity, molality, whatever, and concentration is based on solute in solvent. Solutes are solvated, solids aren't.
 
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gettheleadout said:
Solids don't have a concentration... Pick any concentration measure, molarity, molality, whatever, and concentration is based on solute in solvent. Solutes are solvated, solids aren't.
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You are using a very limited definition of concentration. A mass percent calculation for a solid is a measure of concentration, for instance. Some questions might ask you to find the percent concentration of a substance. The reason solids are always left out of equilibrium calculations is because the ratio of components in a solid does not ever change. In terms of equilibrium expressions, you are correct that you wouldn't include the concentration because they are not solvated and there's no measure of molarity or anything like that, but solids definitely aren't concentrationless (they have specific gravity, mass percent, density, etc, all measure of concentration - notice that none of these metrics would change during the course of a reaction).

Think about a liquid nucleophile in an Sn2 reaction. Obviously addition or removal of it affects equilibrium. The snag seems to be regarding water as a liquid, but there's no reason that it would be treated any differently.
 
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Well we can argue about terms and things like this that are beyond the technical level of understanding needed for the MCAT but I still don't see how this answer can be considered consistent with what TBR says in their own chapter.
 
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Yeah it is semantics, but I think it's important to know in order to understand why solids and liquids are treated differently (concentration of a particular component of the liquid phase can change, whereas the solid phase is always fixed). I feel like on the actual thing it's probably safe to assume that changing a reactant species shifts equilibrium if it's a gas, liquid, or aqueous solution.
 
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brlin said:
I don't think this is true though. I was under the impression that Ka and Kb are just specific instances of Keq.

For example, with a generic acid dissociating in water,
HA + H2O <-> A- + H3O+
Ka = ([A-][H3O+])/HA]
Click to expand...

Yes, because [H2O] stays ~ constant and is very large compared to acid concentration ([H2O] =~56M)

Keq = ([A-][H3O+])/HA][H2O]
Keq =Ka/[H2O]
 
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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.
 
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So the gaseous NO2 isn't dissolved gas in solution?

Edit: Maybe I'm stupid but I kind of assumed it was. They would have to tell us right? Or would a dissolved gas be considered aqueous? I don't think I've ever had to deal with that before.

Also BRT, I take it this means we don't have to worry about weird technicalities where a pure liquid would actually be included in the Keq expression?
 
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BerkReviewTeach said:
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.
Click to expand...

Thank you BRT, that was very enlightening!

And thank you to everyone else who contributed. I still don't like it, but that answer finally makes sense to me.
 
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gettheleadout said:
So the gaseous NO2 isn't dissolved gas in solution?

Edit: Maybe I'm stupid but I kind of assumed it was. They would have to tell us right? Or would a dissolved gas be considered aqueous? I don't think I've ever had to deal with that before.

Also BRT, I take it this means we don't have to worry about weird technicalities where a pure liquid would actually be included in the Keq expression?
Click to expand...

This system would have to be a closed container with a gas space above the solution, so that the NO2 could dissolve into the solution as well as be released from the solution. It seems unfamiliar in terms of chemistry, but if you think about the respiration process, it's kind of what's happening there.

In physio, we consider the PCO2/[HCO3-] ratio in our equilibrium expression. When CO2 is released from the blood to the lungs, we get a complex equilibrium involving a solute and a gas.

HCO3-(aq) + H+(aq) <=> H2CO3(aq) <=> CO2(g) + H2O(g)

As we learn in physio, the pH increases from about 7.2 to about 7.4 in the process, so even though the CO2 is a gas and not in the solution itself, it still impacts the equilibrium in the blood.

I hope this helps, because this particular passage is tough because of the large amount of reactions going on simultaneously and the unfamiliar nature of some of them.

As to when a pure liquid can be included in the equilibrium expression, at the MCAT level, it would be best to think of it as when you reach the point where the concentration is low enough that it's hard to decide whether it's a pure liquid (in contact with itself mostly) versus a solute (surrounded by other molecules). This is where the activity concept in P chem comes into play, but that's beyond the scope of the MCAT (I assume).
 
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