I have been taught that solids do not appear in the equilibrium expression or rate laws because they do not have a concentration. Also that they do
not affect the equilibrium or rate. But I've been thinking. I2(s) and H2(g) can react to form 2HI right? even though it would be slower than I2(g) +
H2. And increasing I2(s) would shift equilibrium right also?
Another problem is when you add MnO2 to H2O2. Adding more MnO2 makes the reaction go faster than if you added a little bit right?
So is there a way to incorporate solids into these expressions?
**********
For NaCl(s) <--> Na+ + Cl-, people say that adding more NaCl will not affect equilibrium so you don't put the solid in the expression. Well,
what about N2(g) -------> N2(aq)? Adding more N2 will not affect equilibrium when it becomes saturated either. Otherwise, according to the ideal
equilibrium constant, you can dissolve and INFINITE amount of N2 into water! Hmm.. this is confusing me.
[Edited on 5/4/2006 by guy]Nerro - 3-5-2006 at 16:44
I don't know about the example with the MnO2 and the H2O2 but in the case of Iodine I'm pretty sure that happens because Iodine quite readily sublimes
to form I2(g). So that's really always I2(g) + H2(g) -> 2HI(g)PainKilla - 3-5-2006 at 17:00
If I understood your question correctly then:
First off, MnO2 is a catalyst for 2H2O2 ---> 2H2O + O2, adding more will only increase surface area of MnO2, so the decomposition will procede
faster. It is not incorporated into the expression at all.
What do you mean, about adding N2 to water, all reactions are reversible, equilibrium is established at a certain point, so adding more N2 wouldn't do
anything as it wouldn't dissolve or anything, just bubble through the water... (I don't think nitrogen is significantly soluble in water.)
It's not that solids don't have a concentration, it's just that they are pure. The rate equations are based on concentration, so something pure can be
omitted basically. That's my understanding of it anyway.guy - 3-5-2006 at 20:15
Quote:
adding more will only increase surface area of MnO2, so the decomposition will procede faster
Isn't it the same thing when you add more liquid or gas, you increase reactant collisons?
In the N2 in water example, I was trying to say that if you increase the pressure of nitrogen then, theorectically more should dissolve and the Keq
will still be constant (the ratio is the same). The point is that when water is saturated with N2, increasing pressure will no longer affect
equilibrium, just like a solid dissolving in water. So is it because that insoluble solids saturate water a lower concentrations that adding more
solid will be ineffective?
But what I dont get is that reactions are supposed to happen when particles collide with each other, and of course gases can collide with solids too,
so it must affect it somehow. What's the difference between gaseous I2 and solid I2 reacting with H2? What is differnent between I2(s) ---> I2(g)
and I2(l) --> I2(g)? For the liquid iodine to gas, adding more liquid can affect the constant but adding more solid iodine wont do anything?
[Edited on 5/4/2006 by guy]
[Edited on 5/4/2006 by guy]
[Edited on 5/4/2006 by guy]guy - 3-5-2006 at 21:54
The Answer?
I remember now about something my chem teacher told me. She said, "when you get into higher level chemistry, nobody uses Molarity in equilibrium
expressions, its all activities." Well, I'm still not 100% sure what activities are but this site shows solubility equilibriums including the ACTIVITY of the SOLID in the expression. It seems that the activity of solids is very close to 1
so it is ignored. At least it makes more sense now. I hate it when they teach the simplified version without mentioning that it is not entirely
correct; it just confuses me! . For example did you know pH does not exactly
equal -log[H+]?Darkblade48 - 3-5-2006 at 22:01
Quote:
Originally posted by guy
I remember now about something my chem teacher told me. She said, "when you get into higher level chemistry, nobody uses Molarity in equilibrium
expressions, its all activities."
This is true, I learned how to calculate equilbrium type problems using activities in my second year of physical chemistry (University)
Quote:
Originally posted by guy
For example did you know pH does not exactly equal -log[H+]?
This is one of the other things that are "assumed" to be correct at the high school level
[Edited on 5-4-2006 by Darkblade48]Magpie - 4-5-2006 at 09:27
I offer my humble understanding:
If the solid acts only as a catalyst then increasing the surface area of the catalyst increases the reaction rate, both forwards and backwards,
equally, but does not affect the equilibrium constant.
The classic example I have often seen to explain the role of a solid is:
CaCO3(s) <---> CaO(s) + CO2(g) (constant T, P)
The amount of solid CaCO3 and CaO does not effect the equilibrium pressure of the CO2, as long as some of each are present. So the
equilibrium constant is often expressed as just Kp = pCO2.
In your example of I2(s) + H2(g) <----> 2HI (g) I would think that this would also be true once equililbrium is established.
Prior to equilibrium I would agree that the amount of solid I2 (or more properly the amount of surface area) would affect the reaction rate.
I think "activities" are another matter and not really germane to your question.
[Edited on 5-5-2006 by Magpie]guy - 4-5-2006 at 13:40
Yeah I kind of figured that it didnt really have to do with activities. The gas dissolving in water paradox is still confusing me. For example solids
dont appear in equilibrium for dissolving because it is saturated and adding more is ineffective. But what about gases, they also can be saturated in
water so there should be a point where the gas does not appear in the expression too.
So for the example
CaCO3(s) <---> CaO(s) + CO2 at constant T,P
adding solid will not do anything, but theorectically if this was gaseous CaCO3 or gaseous CaO it would make a difference? I don't see the
difference. Gas is kind of like atomized solids which just means it has a higher surface area.
[Edited on 5/4/2006 by guy]Magpie - 5-5-2006 at 09:15
I think you are right in your suppositions. If a substance (gas, liquid, or solid) is heterogeneous rather than being homogeneously
mixed then I don't think the amount of mass affects the equilibrium relationship.
I guess something goes from hetero to homo when it goes from a particle or glob (however small) to molecules. Again, just my supposition.
It's hard to find much explanation of this phenomenon. It's always puzzled me a bit also.guy - 5-5-2006 at 14:14
I have another question. Is it possible for a reaction to never reach equilibrium? Such as when you dissolve NaCl in water, the expression is Keq =
[Na+][Cl-], and say you didnt add enough salt, so when the molarities are multipled it is less than Keq. That doesn't make sense to me because
there's always a chance that the ions will come together and form a solid,even for a split second.Magpie - 5-5-2006 at 18:02
My understanding is that in reversible reactions both forward and backward reactions are always occuring. However, up until equilibrium is reached
the forward reaction is faster.
If you believe this, then yes I guess NaCl must be formed, however briefly or minutely.unionised - 7-5-2006 at 01:12
There are two things here.
The ammount of a solid doesn't affect the position of the equilibrium.
It certainly does affect the rate of reaction.guy - 7-5-2006 at 02:07
If I am using the collision theory, then solids should affect equilibrium to a VERY small extent. (Activity of solids is close to one).
Equilibrium expressions come from rate laws. So if solids affect the rate of reactions then it should affect equilibrium?
Using the salt dissolving in water example in previous replys, I think that in solids can affect equilibrium to a certain point until the impact
becomes negligable. Just like how a gas can dissolve in a water to a certain point where additional pressure will not affect solubility.neutrino - 7-5-2006 at 07:07
Quote:
Originally posted by unionised
The ammount of a solid doesn't affect the position of the equilibrium.
It certainly does affect the rate of reaction.
But the rate of the forward and reverse reactions affect the position of the equilibrium.
How can something then affect the forward rate but not the equilibrium? Unless it affects the reverse rate to the same degree...?
[Edited on 7-5-2006 by neutrino]unionised - 8-5-2006 at 12:50
I can prove that it does not influence the position of the equilibrium by using the law of conservation of energy. Imagine a reaction between a
solution and a solid that produces a gas, get 2 containers, a big one and a little one and connect them with a tube with a tap in it. Put lots of the
soild in the big container and little of it in the small one. Add the solution Open the tap and let the gas push a piston, close the tap and (here's
the biggie) if the position of the equilibrium changes the piston will move as more or less gas is produce with the new eqm. Open the tap again and
the piston moves back. Connect the piston to a generator (and the tap (which is presumed to be frictionless)).
Free energy which is forbidden- so the eqm can't depend on the amount of solid.
"How can something then affect the forward rate but not the equilibrium? Unless it affects the reverse rate to the same degree...?"
It does affect them to the same degree.
In an idealised case, I put in twice as much solid so there's twice as much area to react so the reaction goes twice as fast.
The important bit is that statement holds for the forward and reverse reactions.
[Edited on 8-5-2006 by unionised]guy - 8-5-2006 at 14:24
unionised, I dont 100% get your example. Maybe a picture can help?
If a 1g of iodine(s) where to be placed in a large evacuated container, it would form some I2(g), till it reaches, let's say, 0.2 atm. Let's pretend
the equilibrium pressure for I2(g) is 1.0 atm. Its impossible for 100% the solid to vaporize, so there will be a VERY small amount of solid left.
The system is now at equilibrium (rates of sublimation and deposition are equal). If more solid were to be added, then the pressure of the gas will
be able to reach to 1.0atm, and then after that additional solid will become ineffective. So how do you explain that?Magpie - 8-5-2006 at 18:15
Guy said:
Quote:
If a 1g of iodine(s) where to be placed in a large evacuated container, it would form some I2(g), till it reaches, let's say, 0.2 atm. Let's pretend
the equilibrium pressure for I2(g) is 1.0 atm. Its impossible for 100% the solid to vaporize, so there will be a VERY small amount of solid left.
I don't think this statement is true. If the final pressure is only 0.2 atm then 100% of the I2 solid will have vaporized.
This is the same as placing a small amount of water in a large evacuated vessel. If the final pressure is below the vapor pressure of water at the
ambient temperature then there will be no liquid water left.guy - 8-5-2006 at 18:32
Well I didn't really do the actual mass, but I mean a very small amount.
I think there is always a chance of the particles forming a solid as long as they collide, even if its only 0.001% chance. If it is not in
equilibrium does that mean the rate of the foward reaction is infinitely faster than the rate of the reverse?
[Edited on 5/9/2006 by guy]Magpie - 8-5-2006 at 18:47
Equilibrium is not reached until you see solid I2 in the vessel. At that point P = 1.0 atm (per your hypothetical example). Adding more I2(s) will
not affect this pressure of 1 atm.
This is just like a saturated salt solution. You have not reached equilibrium until you see solids in the solution. Adding more solids changes
nothing.unionised - 8-5-2006 at 21:54
If 2 molecules (or a few) colide together to form a solid then the local concentration is very high and you can't use the bulk vapour concentration as
an indicator of what should happen.
Also, the "solid" that forms is very much a temporary thing- it's not at equilibrium with the vapour - that's why it falls apart.
[Edited on 9-5-2006 by unionised]guy - 14-5-2006 at 12:13
If a system is not a equilibrium does that mean that there is still a flow of energy? And is it possible for a closed system never to reach
equilibrium?Magpie - 14-5-2006 at 14:13
There are different types of equilibria. They exist when there are no net changes taking place in the system.
There are phase equilibria, i.e., liquid-vapor, solid-vapor, liquid-solid, etc, as well as chemical reaction equilibria.
If a closed system is given enough time I do not see why it shouldn't come to equilibrium. IIRC the Gibb's free energy will have been minimized at
this point.
A closed vessel with only water vapor in it at T=20C and P=10mmHg is not going to change. It is at equilibrium.guy - 14-5-2006 at 19:59
Le Chatelier's Principle
I know what the Le Chatelier is, but what I don't know is the mechanics of it. I tried googling but had no success. How and why does the Le
Chatelier principle work. I don't want to hear "because it needs to keep the K constant", etc. I need an explanation that describes how it actually,
physically works (like collision, etc). Thanks.
[Edited on 5/15/2006 by guy]neutrino - 15-5-2006 at 13:43
Any equilibrium is composed of two reactions: a forward and a backward. Say you had the following equilibrium:
A + B <--> C + D
The forward reaction is
A + B --> C + D
The backward reaction is
C + D --> A + B
If you increase the concentration of A and / or B, there will be more frequent collisions between them and the forward reaction will speed up. The
reaction will thus shift to the right until the concentrations are in equilibrium again. The same applies to increasing C and / or D.
I hope that helps.guy - 15-5-2006 at 13:47
Yeah that is what I thought. But then that explanation would agree that increasing solid (increase surface area and thusly collision rate) would
affect equilibrium too.unionised - 16-5-2006 at 11:20
It still does and it still increases the rate of the forward and backward reaction so the effects cancel.guy - 16-5-2006 at 12:55
Why does it not cancel out for gases?unionised - 18-5-2006 at 11:16
Good question.
Let's start with a couple of relatively real examples so I can remember what I'm on about. H2+I2 <--> 2 HI
and CO2 <--> C +CO
Under the right conditions these 2 reactions are both reversible. Heated HI can decompose to the elements and heated CO can deompose to C and CO2. In
one case, all the materials are gases (the H +I reaction only really works if it's pretty hot) so what needs to happen is for 2 HI molecules to bump
into eachother (or (simultaneously) into the wall or into an H2 or an I2 or even another HI) with enough vigour to ensure that the activation barrier
is overcome and they will react to give H2 +I2. This can happen throughout the volume of the gas.
For CO decomposition the reaction can go by one of two paths. It can take place entrirely in the gas phase leaving CO2 and a sinlge atom of C as the
vapour.
On the other hand, it can take place at the surface of the solid carbon.
That gives 2 similar equations
2CO --> C(g) +CO2(g) or
2CO --> C(s) +CO2(g).
The difference in the energy here is the energy needed to vaporise an atom of carbon. Since carbon is a solid it is relatively difficult to vaporise,
so the second path, giving the much more stable solid, is much easier.
Practically all the reaction thus takes place at the surface of the solid. For the gas phase reaction the whole of the mixture is available for the
reaction to take place.
I think that's the basis of the difference.guy - 18-5-2006 at 22:05
Yeah I think you are on to something. It is true that rate IS proportional to surface area and NOT the mass of a solid. So therefore, equilibrium is
related to the surface area of a solid. So equilbrium is reached when there is enough surface area for rates to balance out. I am correct so far?
Now I think I'm close to the conclusion but I can't completely figure it out so maybe can come up with one for me? I can't figure out the rest of this conclusion: "Adding more solid increases mass
but...."
[Edited on 5/19/2006 by guy]daeron - 19-5-2006 at 05:04
the rxn rate isnt proportional but dependent on the surface area of the solid, the larger the surface are the faster the mass transfer. If the rxn is
mass transfer limited that increasing the surface area will increase the rxn rate, however if its limited by the kinetics than one doesnt need to
increase the surface area. another thing one can do is, if theres a case of a g-l-s system:
-increase the mixing power input(better g-l, l-s mass transfer)
-higher presssure and/or temperature, but P is of more importance here if you want to influence the mass transfer&equib.
-look into the gas holdup, solid holdup,etc
however a chem. equilibrium is rarely affected much by the increase of surface area, since that will only affect the rxn rate in one direction, and
since it is a system at chem. equilibrium one gets as a result faster rxn rate in the opposite direction, hence silight or no change in equilibrium.
so what im tryin to say is that in order to shift the equlibrium one must first look into the system conditions, the as will just speed up all the rxn rates in the chem equilibrium, and a equilibrium will be reached in a shorter period(i.e. the equilib.
conversion rate, concentrations and all)
. For example did you know pH does not exactly
equal -log[H+]?
