# Can I use Le Châtelier's Principle to prove Henry's Law?

by Bipolarity
Tags: chatelier, henry, principle, prove
 P: 760 So suppose you have ammonia gas suspended above a column of water. Because both ammonia and water are polar, some of the ammonia will immediately dissolve into the water, forming a solubility equilibrium. $$NH_{3}(g) \rightleftharpoons NH_{3}(aq)$$ Henry's Law states that: $$S_{g}=kP_{g}$$ However, if I use the equilibrium constant expression, I get: $$K_{sp} = \frac{RT[NH_{3}(aq)]}{P_{NH_{3}(g)}}$$ According to this expression, if I increase the pressure by a factor of λ (or decrease the volume by λ), then Q will decrease by λ, so equilibrium will shift to favor the production of aqueous ammonia, however I am not so sure that this will be proportionate to the pressure increase. I see a necessary contradiction between Henry's Law and Le Châtelier's Principle. OF course I have probably made some type of mistake, but I can't find it... I would appreciate any help. THanks! BiP
 Admin P: 21,706 Ignoring other things I don't understand in your post, both equations state exactly the same - ratio of pressure and concentration equals some constant.
P: 760
 Quote by Borek Ignoring other things I don't understand in your post, both equations state exactly the same - ratio of pressure and concentration equals some constant.
Ah I see it now thanks! I was confusing myself needlessly.

BiP

P: 76

## Can I use Le Châtelier's Principle to prove Henry's Law?

There seems to be something fishy going on here, $k_{H}$ and $K_{sp}$ do not have the same units, to begin with.
P: 760
 Quote by mjpam There seems to be something fishy going on here, $k_{H}$ and $K_{sp}$ do not have the same units, to begin with.
It shouldn't matter what their units are. As long as they are constants, everything is fine. In fact, via a little substitution, it is clear that $$K_{sp} = kRT$$

I was confused myself at first, but it seems crystal clear now.

BiP
P: 76
 Quote by Bipolarity It shouldn't matter what their units are. As long as they are constants, everything is fine. In fact, via a little substitution, it is clear that $$K_{sp} = kRT$$ I was confused myself at first, but it seems crystal clear now. BiP
It does matter what the units are, because it is meaningless to equate two values the units of which are different.
P: 760
 Quote by mjpam It does matter what the units are, because it is meaningless to equate two values the units of which are different.
When did we equate them? We simply proved that the ratio of aqueous ammonia and gaseous ammonia is some constant, as Borek correctly pointed out.

BiP
P: 76
 Quote by Bipolarity When did we equate them? We simply proved that the ratio of aqueous ammonia and gaseous ammonia is some constant, as Borek correctly pointed out. BiP
Must have been this:

 Quote by Borek Ignoring other things I don't understand in your post, both equations state exactly the same - ratio of pressure and concentration equals some constant.
The equilibrium coefficient is a dimensionless quantity, whereas the Henry's Law constant has dimensions of pressure over concentration.
P: 760
 Quote by mjpam Must have been this: The equilibrium coefficient is a dimensionless quantity, whereas the Henry's Law constant has dimensions of pressure over concentration.
I don't see your point. The units is completely irrelevant to this thread. All that matters is that Henry's Law is correctly predicted by Le Châtelier's Principle. Besides, my question has already been resolved.

BiP
P: 76
 Quote by Bipolarity I don't see your point. The units is completely irrelevant to this thread. All that matters is that Henry's Law is correctly predicted by Le Châtelier's Principle. Besides, my question has already been resolved. BiP
The equation equating the equilibrium coefficient $K_{sp}$and the Henry's Law constant $k_{H}$ is incorrect. The right-hand side and the left-hand side do not have units.
P: 760
 Quote by mjpam The equation equating the equilibrium coefficient $K_{sp}$and the Henry's Law constant $k_{H}$ is incorrect. The right-hand side and the left-hand side do not have units.
I never made any equation equating the two things...
P: 76
 Quote by Bipolarity I never made any equation equating the two things...
What's this then?

 Quote by Bipolarity It shouldn't matter what their units are. As long as they are constants, everything is fine. In fact, via a little substitution, it is clear that $$K_{sp} = kRT$$ I was confused myself at first, but it seems crystal clear now. BiP
P: 760
 Quote by mjpam What's this then?
This was not from my original post. I assumed you have a problem with the equations in my original post, because if you did not, you wouldn't have posted in the first place.

In In any case, you can derive this easily from the two equations in my original post. I do not see why you are making a fuss of this, as Borek cleared it all up in one sentence.

Also, the $K_{sp}$ is not a unitless quantity. Its units are not expressed, simply because they depend on the reaction, but for a particular reaction, they have units. It is meaningless to write them out, since we are concerned only with the concentrations/pressures. In this case, the units are consistent with $kRT$, but again, the units are irrelevant! That is the beauty of equilibrium constants.

BiP
P: 76
 Quote by Bipolarity This was not from my original post. I assumed you have a problem with the equations in my original post, because if you did not, you wouldn't have posted in the first place. In In any case, you can derive this easily from the two equations in my original post. I do not see why you are making a fuss of this, as Borek cleared it all up in one sentence. Also, the $K_{sp}$ is not a unitless quantity. Its units are not expressed, simply because they depend on the reaction, but for a particular reaction, they have units. It is meaningless to write them out, since we are concerned only with the concentrations/pressures. In this case, the units are consistent with $kRT$, but again, the units are irrelevant! That is the beauty of equilibrium constants. BiP
The units are not irrelevant. If you are going to equate an expression with an equilibrium constant it has to be unitless or the equation is meaningless. Also, $K_{sp}$ is unitless, the units of the various pressures or concentration are standardized against some standard pressure or concentration, usually 1 M or 1 atm.
 Admin P: 21,706 I told you there were things in your post I didn't understand. Units of equilibrium constant are tricky. In thermodynamics K is unitless, as it is defined with unitless activities. But we often ignore activity coefficients and use concentrations or pressures in K, then it has units.
P: 760
 Quote by Borek I told you there were things in your post I didn't understand. Units of equilibrium constant are tricky. In thermodynamics K is unitless, as it is defined with unitless activities. But we often ignore activity coefficients and use concentrations or pressures in K, then it has units.
Hmm, if that is true, then my equation relating the two may be incorrect. It doesn't change anything though.

BiP
 P: 76 Actually, equilibrium constants are always unitless, regardless of whether one uses activities, concentrations, or pressures to calculate them. The units of concentration or pressure are, in a sense, "suppressed" because all thermodynamic calculations are done with reference to a standard state, which has the same units as the quantities used and divide each of the quantities in the equation. Most general chemistry text books (or at least the ones I used in high school and college) don't mention the unit "suppression" and tend the gloss over the whole topic of the units of the equilibrium constant, but the mathematics used in thermodynamics strongly implies the equilibrium constant is unitless, as it is often the argument of logarithms or set equal to unitless quantities.
P: 760
 Quote by mjpam Actually, equilibrium constants are always unitless, regardless of whether one uses activities, concentrations, or pressures to calculate them. The units of concentration or pressure are, in a sense, "suppressed" because all thermodynamic calculations are done with reference to a standard state, which has the same units as the quantities used and divide each of the quantities in the equation. Most general chemistry text books (or at least the ones I used in high school and college) don't mention the unit "suppression" and tend the gloss over the whole topic of the units of the equilibrium constant, but the mathematics used in thermodynamics strongly implies the equilibrium constant is unitless, as it is often the argument of logarithms or set equal to unitless quantities.
Ah I see, that makes sense now because my textbook too does not talk much about the units, but in the chapter on thermodynamics it uses the natural log of K and relates it to the Gibbs free energy.

But what answer does this realization offer to the title of this thread?

BiP

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