The Keq is [C]4 / [A]2 **3 (where the numbers in this line are exponents) (I can’t make B be capital, there.)
I cannot answer Mr. Smartypants Student as to why, chemically or mathematically, the coefficients of the equations become exponents. I know this was found sort of by accident while running experiments, but if nature is running this way consistently, isn’t there more of a reason than “That’s the way God made it”? I’m comfortable with pi being a constant concerning circles because it simply is, but this seems to have a better explanation than that. But I don’t think anyone ever told me. I just did it.
It is helpful to understand the concept of a “dynamic equilibrium.” When a chemical system is in equilibrium, this doesn’t mean that it has stopped reacting. Molecules know nothing of the overall concentration of a system; on the molecular level, reactions are constantly occurring. “Equilibrium” just means that the reagent concentrations are such that the forward and reverse reaction rates are the same.
Suppose you have a chemical reaction in which, microscopically, W + 2X <-> Y + Z. By “microscopically” I mean that the forward reaction occurs in one step, requiring one atom (or molecule) of W to collide with two of X to produce one of Y and one of Z.
The probability of a forward reaction happening is therefore proportional to the concentration [W] of W and to the square of the concentration of X: for the reaction to occur we need one W and two X all present in a small region of space. So the forward reaction happens with a rate r[W][sup]2[/sup], for some rate constant r. Similarly the reverse reaction happens with a rate s[Y][Z]. At equilibrium, we want these to be the same:
r[W][sup]2[/sup] = s[Y][Z]
r/s = ([Y][Z]) / ([W][sup]2[/sup]) .
Thus at equilibrium this ratio is a constant (since it is the ratio of two constants).
This is basically a derivation of the law of mass action; the Wikipedia page gives a reasonable starting point, but any good intro college chem text should give a good summary too. I’ve only shown what happens in a simple one-step reaction. Most chemical reactions involve some intermediate state, often with radicals or other unstable states, so you have to imagine a series of these simple reactions, all finding their own dynamic equilibria, and combine all of these equilibria to get the overall equilibrium constant.
To expand a bit more on that: We’ve got some small region of space that the reaction takes place in. For it to proceed, what do we need? Well, we need a W in that region: The probability of that is proportional to the concentration of W. We need an X there: The probability of that is proportional to the concentration of X. And we need another X in there: The probability of that second X is also proportional to the concentration of X. Since we need all three of those, the probabilty of having all three (and thus the reaction rate) is proportional to the product: [W][X]
Thanks, people. I told my B.S. Physics brother someone would know this on this board. He was skeptical.
Are we assuming that the molecular size is essentially immaterial here, the way we assume for gases? I can’t see an adjustment for molecular size.
Also, I’ve been looking at Ksp, and an old question occurred to me again. I understand that if you put in the solute, it will create the dissolved ions in a proportion that will generate that same Ksp at the given temperature. Somehow I’m having a problem with the common ion effect.
If we add in .1 mole OH- ion in a liter of water before adding Fe(OH)2(aq) of course that will impact the dissolving. However, I keep thinking that the Fe(OH)2 would make some more OH- by dissolving, and then the actual concentration of OH- would increase, making the true concentration more than .1M, whereas in the actual math, we leave it as .1M.
I’m sure there’s a decent explanation. Is it that we’re always using the beginning concentrations to get started? In that case, the molarities of the three different aqueous substances are never shown in their true amounts at the same time. It’s more of “if we put in this amount of A, according to the Ksp, we will get this amount of B, allowing that it will use up some of the A.” So the amount of the A put in will not remain the same, but that’s not part of the answer; that’s the given portion of the question.
As a chemist, I feel obligated to point out that reactions requiring 3 molecules are pretty damn rare - as used an example, “W + X + X <> Y+Z”
Imagine a billiards table. Now imagine all the balls on the table moving in random directions at random speeds. They can collide with the wall, or each other.
Collisions between two balls at once will be fairly common. A collision among three balls, simultaneously - very rare. The intermediates in a chemical reaction — “X+Y <> XY <> Z”, where XY is the intermediate - these persist for time scales that are incredibly short. I’m talking nano-pico-femto seconds.
So how often do three balls make contact on that time scale? Very, very, very infrequently. Many orders of magnitude less frequently than two balls make contact.
On the other hand, there are lots and lots of sequential reactions. Example:
X + Y <> Z
Z + X <> A
Z + Y <> B
If you take a decent course on kinetics you learn how to apply the math to figure out what the rate constants for the chemical reaction:
Yes, you’re right that the Fe(OH)[sub]2[/sub] is producing some more OH[sup]-[/sup], the actual [OH[sup]-[/sup]] is (.1+a). But if Fe(OH)[sub]2[/sub] dissolved well, you’d need to add a LOT of Fe(OH)[sub]2[/sub] for a to be significant; since it dissolves quite badly, the solution saturates before “a” becomes significant… say, big enough to be detectable by a first-rate pHmeter (detection of Fe[sup]2+[/sup] in water: add NaOH, if you get a brown cloud you have Fe[sup]2+[/sup]). What happens is that many teachers “skip” in their explanation the part about “at this kind of amount or with such a value for the equilibrium constant, we can assume that any variation in pH caused by the solute is insignificant.”
If you want to do the complete exercise, you need to add a second equation (generally the water constant). Many people get tripped by exercises which do need this second equation but don’t specifically state so; any chemistry student should remember that 10[sup]-14[/sup].
Oh, I know that. But for a pedagogical example I wanted something more generic than 2X or W+X for the left-hand side, to show why the exponents work like they do. I mentioned intermediate states briefly but didn’t want to confuse the main issue. It probably would have made more sense to write the reaction as Y+Z <-> W + 2X, since the equilibrium state for this reaction is likely to be far to the right.
The dependence of the reaction rates on all parameters except for the concentrations is just rolled into the reaction rate constants (r and s, in my example above); I probably should have mentioned that. These rate constants obviously depend on a lot of other things. Temperature is obvious; rates for fundamental reactions often grow with temperature T by a factor of something like e[sup]E/(kT)[/sup] (the Arrhenius rate law), since particles at a higher temperature have a higher kinetic energy and are able to more easily overcome any barrier potential E between reactant state and product state (k here is Boltzmann’s constant). The speeds of the species involved are inversely proportional to their masses, so the rates depend on the molecular weights as well. If one of the reacting species is a complicated molecule, it may have to be in a particular orientation (relative to the other reactants) or configuration for the reaction to proceed.
Well, after bothering my engineer dad, a doctor, his chemistry major wife, and a paid tutor, I’m still stuck on another issue. Here is the gist of my email exchange (sorry, I can’t get the subscripts on this board):
Me: Zumdahl’s Intro Chem, 4th Ed: [SIZE=“4”]CH4(g) + 2O2 CO2(g) + 2H2O(l)[/SIZE] Tell whether the equilibrium will shift to the right, left, or will not be affected.
Ch. 16 35a. Any liquid water present is removed from the system. Solutions manual: No change (water is in the liquid state)
If any present water is removed, then the reaction will have to use up more reactants to replace that water, right? Won’t that be a shift to the right, as the new static molarities of the reactant gasses will go down?
Tutor: No, that’s not right. The amount of water you have on the right side does not affect the equilibrium.
Pure liquids and solids are not part of the equilibrium expression, your book is correct as far as that goes. All that matters is that there be some amount of the solid or liquid present to keep equilibrium. If ALL of the water is removed in the question you’re asking, then there would no longer be an equilibrium.
Think of another type of equilibrium system… water vapor in a closed room with an open bucket of water. The liquid water is in equilibrium w/ the water vapor in the room. If you have a drop of water or an entire swimming pool filled with water will not change the concentration of water vapor in the air. I know that seems strange, but that’s the way it works. All that matters is that some liquid water be present for equilibrium.
Me: I still don’t get it. You said as long as there is some liquid water, there will be equilibrium. But the book says any liquid water is removed. If that’s the case, then it could be said that there’s not an equilibrium for a moment, but then wouldn’t the left side create more product, meaning that the left side was more used up, meaning that the equation shifted to the right?
The reaction is occurring among the molecules in the gas phase. So only the water-vapor concentration matters. It sounds like you understand this part.
I don’t particularly care for the (l) notation here, though this may be because I am misunderstanding what it means. There are some reactions where gas-phase reactants combine to give a product that is essentially completely condensed-phase, such as the famous
HCl(g) + NH[sub]3[/sub] --> NH[sub]4[/sub]Cl(s) ;
for this reaction the notation (s) makes perfect sense to me. But anywhere near room temperature, the water product in your methane-oxidation reaction will either be pure vapor or in a liquid-vapor equilibrium with a nontrivial vapor pressure, so there is some H[sub]2[/sub]O(g) around as well. (Any real chemists who can comment on this usage of (l)?)
So what happens if some or all of the liquid water is removed from the system? Let’s imagine, just so that we can have something particular to talk about, that we have a closed constant-pressure reaction cylinder (with a movable piston at the top, keeping the vessel at 1atm), with a vacuum port on the bottom so that we can remove water. To start with, there’s a puddle of liquid water on the bottom, with some gas-phase quantities of all four species (CH[sub]4[/sub], O[sub]2[/sub], CO[sub]2[/sub], H[sub]2[/sub]O) in equilibrium above the water.
First, suppose we remove some of the water, but not all of it. I specified a constant-pressure vessel, so the gas volume has stayed constant. The partial pressures have not changed, so the main reaction and the liquid surface
H[sub]2[/sub]O(l) <–> H[sub]2[/sub]O(g)
are still in equilibrium.
Now if we remove all of the water, the gas-phase partial pressures are still unchanged, so this equilibrium is still unchanged. The only question is whether some of the gas will condense into a liquid. But the current water vapor pressure in the cylinder is right at the dew-point: the gas has exactly as much water vapor as it’s able to support without condensation, since otherwise some of this vapor would already have condensed out. So there’s no excess water condensing out, and no change to the equilibrium.
Now let’s get a little trickier. Let’s put the reaction in a constant-volume flask instead of a constant-pressure vessel. If we remove all of the water, then this increases the gas volume, and hence decreases its pressure. Just changing the pressure for a gas-phase reaction can cause a change in the equilibrium position, but in this case it won’t (why?). So again in this case there is no change in the equilibrium.
Finally, though, suppose remove some of the water from the constant-volume flask, but not all of it. Again this doesn’t directly affect the equilibrium of the gas-phase reaction. But now the water has a lower vapor pressure, which means that the liquid surface
H[sub]2[/sub]O(l) <–> H[sub]2[/sub]O(g)
is no longer in equilibrium; more liquid will evaporate, increasing the product concentrations, and the reaction will shift, but to the left!
PS: Quote this reply to see how to get subscripts.
PPS: Zumdahl was my chem professor; he was a great lecturer.[/namedrop]