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Studying for a test? Prepare with these 2 lessons on Chemical equilibrium.

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# Keq intuition

Video transcript

Let's see if we can develop some intuition as to why the equilibrium constant equation looks the way it does. Just as review, this is it: equilibrium constant. It would be the concentration of our molecule Y raised to its coefficient power or, if we're thinking in moles, raised to the number of moles. If we think of these as kind of the mole ratios, or the molar ratios -- or we could just view them as the molecular ratios, either way --times the concentration of our molecule Z. Now, we're not doing some calculus here. d is just the number of moles we need of Z for every c moles of Y, b moles of X,and a moles of V. So it's Z to the d power divided by the concentration of V to the a power and X to the b power. So it's a nice, little, clean equation, but why does it look this way? And I actually made a video earlier today where I started exploring this with natural logs. And I think I got someplace, but that one started to break down. And I think I've come up with a much simpler reason why this looks this way. So I've deleted that video, and I think I've come up with a much more intuitive one that explains more of why this works and actually some of the other things we're going to learn about the equilibrium constants in future videos. So what makes a reaction happen? Or what does equilibrium mean? It means the rate at which the forward reaction is happening. So that means that the rate of this happening, of V plus X turning into Y plus Z -- I can't forget the coefficients-- is going to be equal to the reverse reaction, is equal to the rate of the reverse reaction. So our c moles of Y plus d moles of Z turning and going the other way, into the V and the X with certain ratios. It doesn't mean necessarily that the concentrations are the same, because we could have one where we end up heavily favoring the forward reaction. Where we end up with much higher concentrations of Y and Z, or we might heavily favor the backwards reactions where we have more V and X. But when we're in equilibrium, we're saying that our concentrations have reached a stability point, which implies that the rate of going in this direction is equal to the rate going into that direction. So let's just think a little bit about what drives these rates, what drives these rates of reactions. In order for this forward reaction to happen that I drew in purple, what needs to happen? We have to have a molecules of V roughly. And let's say in any volume of space, we have to have some V molecules, and preferably a V molecules, being in the vicinity of b X molecules. So there's got to be b of these X molecules, and they have to be in the right configuration and in the right place and kind of close enough in order for the reaction to happen. So the reaction is really going to be driven by, if you think about it, the probability of finding a V molecules and b molecules all within close enough confines that they can actually react. So you could say that the rate is going to be driven by -- maybe it's going to be proportional. Let's say it's just equal to-- let's say some constant that takes into account things like temperature and how the molecules are actually configured. Because it's not dependent just on them being there. You have to have worry about their kinetic energies. You have to worry about their shape, because some shapes are going to be more conducive to reaction than others. So let's just let that be taken into account with a K. And we're talking about the forward reaction, right? So in order for the forward reaction to happen, let's call that K plus for the forward reaction. We have to have a molecules of V there and b molecules of X. So what's the probability of having a molecules of X? Or what's a rough approximation of the probability? Well, the concentration. Let's think about this a second. When we write the concentration of the molecule V, which I think when I did this was the blue one right here, what is that given in? That is given in moles per liter. Moles is just a number, so this tells us, look, in any given volume, roughly how many of the molecules do you expect to find? That's what concentration is. So if I wanted to figure out the probability of finding a of these molecules, because that's how many I need, I need to multiply this by itself a times, because I need a of them. The probability of having just one molecule in just some small fraction, you would just use the concentration once. But you're going to use it a times, because you want a of those molecules there, right? You could look at it like what's the probability of having five heads? Well you would multiply the probability of one head five times. So the forward reaction probability is going to be the concentration of V to the a power, and, of course, that's not enough to have the reaction happen. You also need to have b of the X molecules there. So you have the concentration of X to the b power. And I want to make sure you understand this. My claim is that this is approximation -- or actually it's a pretty good way of calculating-- the probability. So let me write it this way. The rate is equal to some constant that takes into account the temperature and the molecular configurations times the probability of having a V molecules and b X molecules in a sufficiently small area all at the same time. And the best way to approximate that is with their concentration. Obviously, the higher the concentration, the higher the moles per liter, the more likely you're going to find that many of molecules in kind of that little small space that you care about, and the temperature and the configuration are going to matter more. But if you use the concentration as the probability of a -- let me switch colors. If the probability of having a V molecule in some volume -- if we assume that the solution is homogeneous, that the V molecules are roughly evenly distributed, it's going to be -- this isn't even an approximation. It's going to be the concentration of the V molecules times the volume under which we care about. If we want the probability of a, where a is a number, it could be five V molecules, a V's in some volume, it's the probability of finding this a times. So it's going to be equal to -- and this is just from the probability concepts that we learned in the whole probability playlist. So if you want to have five heads in a row, it's 1/2 to the fifth power. If you want to have V molecules there, five of them at the same time in some volume, or a of them, it's going to be V to the a power times the volume. If you also care about the probability so you want all of that, so a V's and b X's in some volume, then you're going to have to multiply all of them together. So it's going to be equal to the concentration of V to the a power times the concentration of X to the b power times the volume. So the probability of finding the right number of V particles and X particles in the right place in some volume is going to be proportional to exactly this. And we're saying that the reaction rate, the forward reaction rate, is also proportional to this thing. So that's where we get the forward reaction rate. So the rate forward is equal to the concentration of our V molecules to the a power times the concentration of our X molecules to the b power. Now, if we want to find the reverse rate, so this is the rate forward. If we want to find the rate of the reverse reaction, let's say that that's equal to some other constant -- let's call that K-minus-- the same exact logic holds. We're just going in this direction now. If we look at our original one, we're going in that direction. So for this reaction, we do the same thing. We literally just do different letters, so the reverse reaction is just going to be the concentration of the Y molecule to the c power, because we need c of them there roughly at the same time, times the concentration of the Z molecule to the d power. Now, just at the beginning of the video, we said that equilibrium is when these rates equal each other. I wrote it down right here. So if the reverse rate is equal to some constant times this, and the forward rate is equal to some constant times that, then we reach equilibrium when these two are equal to each other. Let me clear up somespace here. Let me clear this up, too. So when are they going to be equal to each other? When the forward rate -- the forward rate is this. That's our forward constant, which took into account a whole bunch of temperature and molecular structure and all of that-- times the concentration of our V molecule to the a power. You can kind of view that as what's the probability of finding in a certain volume -- and that certain volume can be factored into that K factor as well-- but what's the probability of finding V things, a V molecules in some volume. And it's the concentration of V to the a power times concentration of X to the b power -- that's the forward reaction-- and that has to equal the reverse reactions. So K-minus times the concentration of Y to the c power times the concentration of Z to the d power. Now, if we divide both sides by -- let me erase more space. Nope, not with that. All right. So let's divide both sides by K-minus and both sides by this, so you get K-plus over K-minus is equal to that, is equal to Y to the c times Z to the d. All of that over that-- V to the a times the concentration of X to the b. Let me put this in magenta just so you know that this was this K-minus right here. And then, these are just two arbitrary constants, so we could just replace them and call them the equilibrium constant. And we're there where we need to be. We're at the formula for the equilibrium constant. Now, I know this was really hand wavy, but I want you to at least get the sense that this doesn't come from out of the blue, and there is -- at least I think there is-- there's an intuition here. These are really calculating the probabilities of finding -- this is the forward reaction rate probabilities proportional to this. Because the more V concentration you have, the more likely you're able to find it. Although if you need more of those particles around, you're going to have to multiply that concentration by each other, because the probability's going to get lower. Because you need more of them together in order for the reaction to happen. Same thing for everything there. But all this is derived from is that the forward reaction should be equal to some constant times the reverse reaction. Or actually, their rates should be equal, but then when you actually calculate the probability, you'll have a constant in there. Anyway, hopefully, I didn't confuse you, but I just wanted to give you that this isn't just some random equation. It really does, I think, come from the reality that the higher the concentration you have, the more the probability you have of the actual molecules bumping into each other.