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Can I use Le Châtelier's Principle to prove Henry's Law? 
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#1
Feb2112, 12:32 PM

P: 783

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.
[tex] NH_{3}(g) \rightleftharpoons NH_{3}(aq) [/tex] Henry's Law states that: [tex] S_{g}=kP_{g}[/tex] However, if I use the equilibrium constant expression, I get: [tex] K_{sp} = \frac{RT[NH_{3}(aq)]}{P_{NH_{3}(g)}} [/tex] 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 


#2
Feb2112, 01:47 PM

Admin
P: 23,530

Ignoring other things I don't understand in your post, both equations state exactly the same  ratio of pressure and concentration equals some constant.



#3
Feb2112, 03:51 PM

P: 783

BiP 


#4
Feb2112, 06:49 PM

P: 76

Can I use Le Châtelier's Principle to prove Henry's Law?
There seems to be something fishy going on here, [itex]k_{H}[/itex] and [itex]K_{sp}[/itex] do not have the same units, to begin with.



#5
Feb2112, 07:33 PM

P: 783

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


#6
Feb2112, 07:53 PM

P: 76




#7
Feb2112, 08:02 PM

P: 783

BiP 


#8
Feb2112, 08:32 PM

P: 76




#9
Feb2112, 09:19 PM

P: 783

BiP 


#10
Feb2112, 09:45 PM

P: 76




#11
Feb2112, 10:13 PM

P: 783




#12
Feb2112, 10:29 PM

P: 76




#13
Feb2112, 10:42 PM

P: 783

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 [itex]K_{sp}[/itex] 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 [itex]kRT[/itex], but again, the units are irrelevant! That is the beauty of equilibrium constants. BiP 


#14
Feb2112, 10:51 PM

P: 76




#15
Feb2212, 02:41 AM

Admin
P: 23,530

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. 


#16
Feb2212, 08:33 AM

P: 783

BiP 


#17
Feb2212, 12:32 PM

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. 


#18
Feb2212, 12:40 PM

P: 783

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


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