[blaisem]

Hi, I am an undergraduate tutor at my university.  In General Chemistry, for an equilibrium expression such as:

K=(C^c)(D^d)(A^a)^-1(B^b )^-1

for a reaction aA + bB    cC + dD

we are taught to ignore the phase of any component in the reaction that is a solid or liquid.

So, today, someone asked me what to do if a reaction contains both gas phase and aqueous phase molecules.  Since an equilibrium of pressures and an equilibrium of Molarities can be interrelated by the ideal gas law, the closeness of their relationship made me unsure how to approach that problem. 

Surely, due to the different phases there must be some consideration to take into play.  Since one is solvated and the other is not, I thought maybe the concentrations would have to be expressed in ppm, but this consideration is not necessary when using the ideal gas law to convert from concentration equilibrium to pressure equilibrium.  Then, I realized I had no way to convert an aqueous concentration into a pressure.  I know these kinds of reactions exist because gases often react with other states, such as air contaminating a reaction, but I had always assumed the gas remains in the gas phase as it interacts rather than entering an aqueous phase.  Now I don't know.

Finally, I managed to further confuse myself when I realized that I had no idea why liquids and solids weren't considered in the first place in an expression for the equilibrium.  With all that said, my questions are the following:

1. Why are solid and liquid phase compounds not considered in an expression for an equilibrium?

2. How do you develop an expression for the equilibrium of a reaction that includes both a gas state and an aqueous state molecule as reactants?  Can they be equally well represented as an equilibrium of pressures as well as an equilibrium of concentrations?  What are the implications of such a reaction; in other words, how are these reactions taking place on a molecular level?  Are the gas phase molecules briefly considered "solvated" or do they condense for the extent of their interaction with an aqueous compound?

I would really appreciate any advice on this, even if it is simple redirection to the appropriate source.  I'm not here to burn anyone's time on a question if I should have found a way to answer it myself.  Thank you!

Blaise

[Vidya]

equilibrium is related to concentration of the solutes and gases
now when we have some solid then we can not express its concentration as moles / L because it is not in the solution .Similarly if pure liquids are there then again it can not be expressed as in terms of moles /L of the soln.Hence it is not included in equilibrium constant and they are considered to be constant as they are in pure state or you can say that they are taken as unity. 
If gas and aqueous phases are present together then we can calculate Kc by taking all of them in terms of concentrations.

[blaisem]

Why can't I take the density of a liquid, convert mass to moles, and divide it by the volume of the solution?

I'm not seeing the difference between a gas and a liquid in this case.  If a gas is capable of being in the solution, why not a liquid?  Both a gas and a liquid can also be expressed as moles per liter.

Even many solids, to some degree, have a certain amount of solubility and so are in solution.  We may not be able to measure it as moles / L because we do not know the moles of the solid in the solution; however, the same argument can be made for a gas, where some of the gas interacts with the solution while the rest is present as a vapor pressure above the solution.  Why, therefore, can a gas be expressed as moles per liter, included in Kc, but a solid can't be?

Sorry, it just strikes as there must be more to it than the reasons given for a liquid and solid, since these same arguments for why they cannot be included could also be applied to a gas, which apparently is somehow an exception.  Is anyone able to clarify this?  Thank you.

[fledarmus]

It seems that the question you are asking is what difference phase makes in reaction rates.

The concentration dependence of the reaction rate is actually a statistical likelihood of a molecule of one reactant bumping into a molecule of the other reactant, and is based on an even dispersion of molecules of the two reactants throughout the volume of the reaction. For gases and for solutions, this is a reasonable assumption - in gases, the molecules are far enough apart that the attractive forces between molecules are not sufficient to separate one type of molecule from the other - the statistical likelihood is that, if there is a 50% mixture of gases, the odds are 50% that the next molecule one gas runs into would be a molecule of the other gas. In a solution, the same thing occurs - all of the molecules of each component of the solution are separated by solvolysis, and the odds of a molecule of one solute bumping into a molecule of the other solute depend only on their concentrations.

If one of the molecules is present as a solid, this is no longer the case. Only the tiny proportion of molecules on the surface of the solid could possibly *Ignore me, I am impatient* into a molecule of a different molecule. Any molecules within the bulk of the solid can only *Ignore me, I am impatient* into each other, which doesn't yield a reaction. Consequently, the effective concentration of a solid in the reaction is much smaller than the amount of solid would indicate.

For a liquid, the same situation applies, if by liquid you mean that it has separated from the solvent. There is a difference between a compound present as a liquid and a compound present in solution. For example, if you are trying to run a reaction on toluene using water as a solvent, toluene would be present as a liquid, not in solution in the water. This means the only molecules of toluene which would have a possibly of reacting with anything dissolved in the water would be those molecules which were right on the surface between water and toluene. The effective concentration of toluene in the reaction is very small compared to the amount of toluene present in the reaction vessel.

If you are running a reaction on methanol in water, it is inappropriate to show methanol as a liquid in the reaction diagram; if anything it should be shown as (aq), meaning an aqueous solution. In this case, all of the molecules are evenly dispersed throughout the solution, and the concentration of the methanol in the solution does become part of the rate equation.

I hope that helps.

[blaisem]

It seems that the question you are asking is what difference phase makes in reaction rates.

The concentration dependence of the reaction rate is actually a statistical likelihood of a molecule of one reactant bumping into a molecule of the other reactant, and is based on an even dispersion of molecules of the two reactants throughout the volume of the reaction. For gases and for solutions, this is a reasonable assumption - in gases, the molecules are far enough apart that the attractive forces between molecules are not sufficient to separate one type of molecule from the other - the statistical likelihood is that, if there is a 50% mixture of gases, the odds are 50% that the next molecule one gas runs into would be a molecule of the other gas. In a solution, the same thing occurs - all of the molecules of each component of the solution are separated by solvolysis, and the odds of a molecule of one solute bumping into a molecule of the other solute depend only on their concentrations.

If one of the molecules is present as a solid, this is no longer the case. Only the tiny proportion of molecules on the surface of the solid could possibly *Ignore me, I am impatient* into a molecule of a different molecule. Any molecules within the bulk of the solid can only *Ignore me, I am impatient* into each other, which doesn't yield a reaction. Consequently, the effective concentration of a solid in the reaction is much smaller than the amount of solid would indicate.

For a liquid, the same situation applies, if by liquid you mean that it has separated from the solvent. There is a difference between a compound present as a liquid and a compound present in solution. For example, if you are trying to run a reaction on toluene using water as a solvent, toluene would be present as a liquid, not in solution in the water. This means the only molecules of toluene which would have a possibly of reacting with anything dissolved in the water would be those molecules which were right on the surface between water and toluene. The effective concentration of toluene in the reaction is very small compared to the amount of toluene present in the reaction vessel.

If you are running a reaction on methanol in water, it is inappropriate to show methanol as a liquid in the reaction diagram; if anything it should be shown as (aq), meaning an aqueous solution. In this case, all of the molecules are evenly dispersed throughout the solution, and the concentration of the methanol in the solution does become part of the rate equation.

I hope that helps.

Ah, I see.  This is very interesting.  Thank you for your input.

So, when we describe a reaction equilibrium via an equilibrium constant, solids and liquids are disregarded because their "effective concentration," or their ability to react in the system, is very small compared to other reactants.

Does this mean that the equilibrium expression isn't entirely accurate? Because however small a non-solvated molecule's ability to react is, there is still some potential for it to occur.

[fledarmus]

How large would you expect the error to be? (Here is where you get to play with math :-)

If you have equimolar amounts of two compounds in a reaction flask and both are totally solvated, you would expect that the odds of two different molecules colliding would be the same as the odds of two identical molecules coliding. If you have equimolar amounts of two compounds in a reaction flask, but one is a solid which has settled to the bottom, what are the odds that any molecule of the compound in solution will be next to a molecule of the undissolved material?

Assuming that you can get perfect mixing of your heterogeneous solution and every single particle of your undissolved solid is equidistant from every other particle, how small would the particles have to be to make even a 0.1% statistical contribution to collisions with the dissolved component?

The same calculations can be made for molecules in the gas phase.

You would get a larger source of error if the solid or gaseous component is not completely eliminated from the solvent. It is possible that even if most of your insoluble compound has precipitated, there is still a small amount left in solution. In that case, for total accuracy, you would have to include the equilibrium between solid and solvated compound.

[blaisem]

Ok, so I was thinking about your problem, Fledarmus.

If you have equimolar amounts of two compounds in a reaction flask, but one is a solid which has settled to the bottom, what are the odds that any molecule of the compound in solution will be next to a molecule of the undissolved material?

So, I attempted to do a basic model of this.  One difficulty was: How do I represent the probability of a collision?  What I decided to do was use the surface area of a sphere to represent the odds of one molecule's probability to be collided with.  I assumed both different molecules (my solvated and non-solvated molecules, respectively), were of equal radius, and the radius is 1.  I also assumed the solid has zero solubility in the solution.  These assumptions may turn out to be unnecessary, but we'll see.

So, I am dealing with molecules that are perfect spheres with a radius 1.  If I have a million "solid" molecules, the sum of the surface area of each molecule is:

[4*PI*r^2]*1000000 = 12.57 x 10^6 sq. units.

This number would correspond to their square area if the solid were completely solvated.  Then, for equal amounts of two reactants, there would be a 50% chance of collision.

Now, if the solid has zero solubility, it exists only as a solid.  I took two extremes.  If the particles are evenly divided on the floor of the glass, with spacing between each particle, then we could say that only the top half of the molecules are available to reaction.  In this case, their surface area is simply divided by two.  Their "ability" to collide or react with another molecule is halved, so there is a 25% chance of collision.

In the other extreme, the molecules all aggregate together at the bottom of the glass.  In this case, I decided to assume the form of a hemisphere on the bottom.  On an atomic scale, the edge of this hemisphere is jagged and not smooth.  This would increase the effective surface area, but given that orientation also plays a role in a reaction, this may increase or decrease reactivity.  I decided to exclude jaggedness.  I'm not even sure yet how to model it if I had the information necessary.

A million perfect spheres of radius 1 have a volume of

4/3*PI*1^3*1000000 = 4.19 x 10^6 cu. units.

This is the volume of the hemisphere.  The equation for the volume of a hemisphere is one half times the equation for the volume of a sphere.  ie:

(4/3)/2*PI*r^3=2/3*PI*r^3

The radius of this hemisphere is therefore:

[(4.19 x 10^6 cu. units)*(3/2)/(PI)]^(1/3)= 126 units.

The surface area of a hemisphere of radius 126 units is:

2*PI*(126)^2=9.97 x 10^4 sq. units.

Taking the ratio of this surface area, representing the availability for collision, with the maximum surface area available when solvated, I get:

(9.97 x 10^4)/(12.57 x 10^6) *100% = 0.8 %.

So, an unsolvated solid phase molecule is 99.2% less available than a fully solvated one.  But this final calculation is a little meaningless because in a solid there are lattice energies which I think should raise the activation energy of a reaction.

A lot of this is dependent on the radius of the molecules involved.  This could likely change the results by an order of magnitude or more.  Also orientation requirements could have a massive effect.  Furthermore, the collision velocity is halved, since only one molecule is moving and the solid phase molecule is at rest.  If the lattice energy raises the activation energy, but the molecules have half the velocity (quarter the kinetic energy), the reaction is probably even less likely to occur.  I think the energy required for the reaction and orientation make up the greatest amount of error in the model assumed, both of which should decrease the likelihood of a reaction occuring.  The model is more or less an "at best" scenario.  "At best," the K value is altered by 0.8%, which is really small.  I guess we can conclude that the contribution really is pretty much nothing.

EDIT

Ok, I made a mistake.  It turns out that, the more molecules you have, the greater this difference becomes.  I took one mole of particles, 6.022 x 10^23, performed the same calculation, and the 0.8% became 9.4 x 10^-7 %.  One million particles was just too small, unless you are dealing with a femtomolar concentration.  Considering lattice energies and kinetic energy of the collision, I think really that the result is even more infinitesimal than I had thought.  Even considering a solubility % of the solid is probably unnecessary.  The difference would still likely remain very small.

Assuming that you can get perfect mixing of your heterogeneous solution and every single particle of your undissolved solid is equidistant from every other particle, how small would the particles have to be to make even a 0.1% statistical contribution to collisions with the dissolved component?

I'm not really sure how to model this.  My understanding was that if all of the particles are equidistant from one another in a solution, then, that is the definition of being solvated?

[blaisem]

Nevermind I figured my question out when I thought about it some more.