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### Topic: Writing Kp and Kc for heterogeneous equilibrium  (Read 7732 times)

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##### Writing Kp and Kc for heterogeneous equilibrium
« on: June 07, 2013, 07:36:56 AM »
Kp uses the partial pressure of the products (right hand side) and the reactants (left hand side). So for solids and liquids their partial pressures are constant at a certain temperature. But what about gases? At a certain temperature won't the partial pressure also be constant? By using PV=nRT P=nRT/V so V, R, n and T are constants so why is the partial pressure of the gas only used?

For Kc it uses the concentrations. In my book that stated that the concentration of solids and liquids as concentration=amount (in mol)/volume. Since the number of moles is calculated as mass/molar mass, concentration=mass/(molar massXvolume)=density/molar mass? If say A(s)+B(l) C(s)+D(g) then for A and B, won't the concentration be number of moles of A/volume of A and B? So if I had 1 moles of A and B each and 2 of each won't the concentration be different because now my volume is not just of A but includes B as well? Also, what about gases and aqueous solutes why are their concentrations considered to be not fixed? Shouldn't they be fixed as well?

Lastly, even if the solids and liquids partial pressure or concentration are fixed why can't they be in the equation? Or is this because it would give unnecessary information that would make it harder?

Thanks for the help

#### Corribus

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##### Re: Writing Kp and Kc for heterogeneous equilibrium
« Reply #1 on: June 07, 2013, 09:23:02 AM »
Kp uses the partial pressure of the products (right hand side) and the reactants (left hand side). So for solids and liquids their partial pressures are constant at a certain temperature. But what about gases? At a certain temperature won't the partial pressure also be constant? By using PV=nRT P=nRT/V so V, R, n and T are constants so why is the partial pressure of the gas only used?
How is n constant?  The amount of gas consumed or produced in the reaction is changing.

Quote
Lastly, even if the solids and liquids partial pressure or concentration are fixed why can't they be in the equation? Or is this because it would give unnecessary information that would make it harder?
It is because equilibrium constants are really supposed to deal with activities, not concentrations.  Activities are thermodynamic potentials, and by definition are really only meaningful when compared to some reference value.  For activities, the reference point usually chosen is the standard state of the substance, at which we define the activity to be equal to 1.  Pure liquids and solids are in their standard states, therefore the activity is 1, therefore they don't impact the equilibrium constant.

(You can be pedantic and put them into the equilibrium expression if you want, but this would only serve to confuse students who don't understand really that the concentrations are approximations of the activities.  And in any case they don't impact the value of the equilibrium constant, so what's the point?)
« Last Edit: June 07, 2013, 09:51:18 AM by Corribus »
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##### Re: Writing Kp and Kc for heterogeneous equilibrium
« Reply #2 on: June 07, 2013, 12:10:21 PM »
Pressure generally means gaseous pressure and partial pressures are thus gaseous partial pressures; solids and liquids don't have partial pressures at all. They don't contribute to the gaseous pressure (except in the sense that they may restrict the volume available for the gases, but what I'm saying is that they have no partial pressure of their own).

If solids and liquids do have a "partial pressure" it is not the same kind as we generally mean when we talk about pressure calculations in chemistry.

All these conclusions can be found in one of my recent discussions with Corribus, I can provide a link if necessary.

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##### Re: Writing Kp and Kc for heterogeneous equilibrium
« Reply #3 on: June 08, 2013, 12:53:50 AM »
Kp uses the partial pressure of the products (right hand side) and the reactants (left hand side). So for solids and liquids their partial pressures are constant at a certain temperature. But what about gases? At a certain temperature won't the partial pressure also be constant? By using PV=nRT P=nRT/V so V, R, n and T are constants so why is the partial pressure of the gas only used?
How is n constant?  The amount of gas consumed or produced in the reaction is changing.

Quote
Lastly, even if the solids and liquids partial pressure or concentration are fixed why can't they be in the equation? Or is this because it would give unnecessary information that would make it harder?
It is because equilibrium constants are really supposed to deal with activities, not concentrations.  Activities are thermodynamic potentials, and by definition are really only meaningful when compared to some reference value.  For activities, the reference point usually chosen is the standard state of the substance, at which we define the activity to be equal to 1.  Pure liquids and solids are in their standard states, therefore the activity is 1, therefore they don't impact the equilibrium constant.

(You can be pedantic and put them into the equilibrium expression if you want, but this would only serve to confuse students who don't understand really that the concentrations are approximations of the activities.  And in any case they don't impact the value of the equilibrium constant, so what's the point?)

Oh but isn't n fixed at equilibrium? That's why I'm having some trouble understanding this. At equilibrium everything is already constant. So I don't get why only gases or aqueous stuff are included. Won't they also be fixed?

Thanks

#### Corribus

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##### Re: Writing Kp and Kc for heterogeneous equilibrium
« Reply #4 on: June 08, 2013, 02:05:06 AM »
The relative activities (remember: potential energy) of the products and reactants determine where the equilibrium point is.  All of these values are relative to something else, so you have to pick a point of reference.  The point of reference is by convention given a value of 1, and we call this the standard state.  The standard state is not dependent on temperature - that is, the standard state for a pure substance can be different at different temperatures.  Pure liquids and solids are considered standard states, and so have an activity of 1.  Pure gasses are in their standard states at unit pressure.  Technically a pressure criterion should probably be defined for "pure liquids and solids" as well, but these states of matter are far less compressible than gasses, and so the difference in activity between a liquid at one pressure and a liquid at another pressure is going to be very (very) small.  So we can approximate it by saying that any pure liquid (or solid) is in a standard state regardless of pressure.  Also note that some solids have different crystalline phases, even if the substance is pure (example: carbon vs. diamond.  One must be considered the standard state, the other will have an activity expressed relative to it.  Doesn't matter which; consistency is the important thing.)  The pressure is important for gasses because gasses are compressible and so pressure-volume work is significant, so activity of a gas changes substantially as a function of pressure.  This is why you must define a gas standard state as being at a specific pressure, and why we must use pressure as the activity (it is expressed relative to the unit pressure), even for "pure gasses".  For simplicity, the pressure that is chosen is unit pressure.  Note that all this assume ideal gas behavior.  For real gasses, fugacity is used as activity, as it takes into account the deviations due to intermolecular forces.

http://en.wikipedia.org/wiki/Standard_state

This is my old understanding of it, anyway.  Most people don't deal often with activities, myself included, so I'm not used to thinking about them.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

#### curiouscat

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##### Re: Writing Kp and Kc for heterogeneous equilibrium
« Reply #5 on: June 08, 2013, 02:23:31 AM »
Technically a pressure criterion should probably be defined for "pure liquids and solids" as well,

Isn't that the Poynting Factor? But yes, mostly it is close to 1.

#### Corribus

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##### Re: Writing Kp and Kc for heterogeneous equilibrium
« Reply #6 on: June 08, 2013, 02:29:30 AM »
Honestly, I'm not familiar with this term, but I imagine someone has come up with a factor to describe it  If a Mr. Poynting* was the man, so be it.

* Or Mrs. Poynting.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

#### curiouscat

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##### Re: Writing Kp and Kc for heterogeneous equilibrium
« Reply #7 on: June 08, 2013, 03:02:05 AM »
Honestly, I'm not familiar with this term, but I imagine someone has come up with a factor to describe it  If a Mr. Poynting* was the man, so be it.

* Or Mrs. Poynting.

"For a pure fluid in vapor-liquid equilibrium, the vapor phase fugacity is equal to the liquid phase fugacity. The exponential term represents the Poynting correction factor and is usually near 1.0 unless pressures are very high. "

#### Corribus

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##### Re: Writing Kp and Kc for heterogeneous equilibrium
« Reply #8 on: June 08, 2013, 10:21:56 AM »
Well how 'bout that.  I'm not sure whether to be proud that I was able to predict something like that should exist or piqued that I didn't know it already did.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

#### curiouscat

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##### Re: Writing Kp and Kc for heterogeneous equilibrium
« Reply #9 on: June 08, 2013, 10:25:32 AM »
Well how 'bout that.  I'm not sure whether to be proud that I was able to predict something like that should exist or piqued that I didn't know it already did.

As a Chemical Engineer I remember getting test problems on it. Only reason I remember. Never needed it in practice.

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##### Re: Writing Kp and Kc for heterogeneous equilibrium
« Reply #10 on: June 08, 2013, 10:34:34 AM »
Can I just check my understanding - due to the vapour equilibrium a pure liquid forms (i.e. liquid to gaseous state equilibrium), the liquid has some effect on the pressure of the system, in that the gaseous phase form of the substance exerts its own partial pressure which you'd have to add up with that of all the other gases to find the total pressure. However the pure liquid itself, in liquid form, doesn't have an effect on the total pressure (in the sense of gaseous total pressure as we usually mean) but rather the gaseous form of the liquid, produced by the vapour-liquid equilibrium, does.

#### Corribus

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##### Re: Writing Kp and Kc for heterogeneous equilibrium
« Reply #11 on: June 08, 2013, 10:50:16 AM »
As a Chemical Engineer I remember getting test problems on it. Only reason I remember. Never needed it in practice.
I can see why this would be more important for a ChemE to know.  "Regular chemists" don't often deal with condensed phase systems under high pressures.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

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##### Re: Writing Kp and Kc for heterogeneous equilibrium
« Reply #12 on: June 08, 2013, 10:52:08 AM »
Can I just check my understanding - due to the vapour equilibrium a pure liquid forms (i.e. liquid to gaseous state equilibrium), the liquid has some effect on the pressure of the system, in that the gaseous phase form of the substance exerts its own partial pressure which you'd have to add up with that of all the other gases to find the total pressure. However the pure liquid itself, in liquid form, doesn't have an effect on the total pressure (in the sense of gaseous total pressure as we usually mean) but rather the gaseous form of the liquid, produced by the vapour-liquid equilibrium, does.

I think so. In my syllabus we learned about Raoult's law and the pressure exerted by the liquid state is the mole fraction of the liquid in the solutiom multiplied by its pure vapour pressure of the gas at that temperature. But in that case, if I were to have a gas in the closed container shouldn't some of it also be a liquid? Because by Raoult's law if there's are partial pressure of the gas then shouldn't it also have some of the gaseous substance at liquid state as well?

But still, once at a reaction's equilibrium shouldn't the concentrations of the aqueous and gaseous products or reactants be fixed?

#### curiouscat

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##### Re: Writing Kp and Kc for heterogeneous equilibrium
« Reply #13 on: June 08, 2013, 11:35:33 AM »
But still, once at a reaction's equilibrium shouldn't the concentrations of the aqueous and gaseous products or reactants be fixed?

If by fixed you mean unchanging with time then yes, it is indeed.