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List of types of equilibrium, the condition of a system in which all competing influences are balanced, in a wide variety of contexts.

Equilibrium may also refer to:

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  • Le Chatelier's principle | Chemical equilibrium | Chemistry | Khan Academy
  • Introduction to reaction quotient Qc | Chemical equilibrium | Chemistry | Khan Academy
  • Reactions in equilibrium | Chemical equilibrium | Chemistry | Khan Academy


Let's say we had the reaction molecule A plus molecule B is in dynamic equilibrium with molecules C plus D. Which just means that the rate of the forward reaction is going at the same rate as the backward reaction, or the reverse reaction. There will be some equilibrium concentrations of A, B, C, and D, and we can figure out what the equilibrium constant is if we want. And I'll say it again. I've said it like four times so far. The fact that the forward reaction rate is the same as the backward reaction rate doesn't mean that all of the concentrations are the same. The concentrations themselves of each of the molecules could be very different. They're just not changing anymore because the forward and backward rates are the same. Now, given this, given that we're at equilibrium now, what's going to happen if I add more A to the system? So remember, it was in equilibrium. The concentrations were constant. But now all of a sudden, I'm adding more A to the system. So now, the odds of an A and a B particle, even though I'm not adding any more B molecule to the system, the odds are slightly higher that an A and a B are going to collide in just the right way, so the forward reaction is going to be more likely. So if we add more A to the system, you're going to have more A's. They're going to bump with more B's, so the B's are actually going to go down a little bit, right? Because more B's are going to be consumed. And even more important, the C's and D's are going to definitely be increased. And the way it would really happen, you would add more A. Those A's would bump into some more B's, and so this forward reaction, all of a sudden, it's rate would go faster than this backward reaction. So the reaction would go in that direction. Then you would have more C's and D's and maybe some of those are more likely to bump and go back in this direction. Eventually, you would reach a new equilibrium. But the bottom line is you'll be left with more A, a little bit less B because you didn't add more B. So more B's going to be used to consume with those extra A's you just added. And then those are going to produce more C's and D's in equilibrium. And you can imagine, if you added more A and more B, let's say if you added more B as well, then the reaction is going to go in the forward direction even more. I don't think this is an amazing insight of the world. I think this is kind of obvious, that if you stress this reaction by adding more on this side, that naturally it's going to move in the direction that relieves the stress. So if you add more A, you're going to have more A's bumping with B's and go in that direction and maybe consume a little bit more B's. If you add more of both, the whole thing's going to go in that direction. Likewise-- let me rewrite the reaction. I'll do it in a different color. A plus B, dynamic equilibrium, C plus D. If I add more C-- I think you get the point here-- what's going to happen? Well, that's going to drive A and B up, and it's maybe going to consume a little bit extra D. And then if you added more C and D, then, of course, it's going to produce a lot more A and B. And this idea, it seems pretty common sense, but there's a fancy name for it, and it's called Le-- let me put a capital L-- Chatelier's principle. If you've watched enough of these videos, you know I have to be careful with my spelling. And all it says is that when you stress a reaction that's in equilibrium, the reaction will favor the side or one side of the reaction to relieve that stress. When they say stress the reaction, that's like adding more A, so the reaction's going to move towards the forward direction to relieve the stress-- the quote, unquote stress-- of that more A. I mean, that's not stress in its traditional way of thinking about stress, but that is a kind of stress. You're somehow changing it relative to it. It was nice and comfortable before in a nice, stable environment. So given Le Chatelier's principle, let's think of some other situations. Let's say if I had A plus B plus some heat, and that produces some C plus D. And maybe it produces some E as well. So if I were to add heat to this system, what would happen? So in order for the reaction to progress in the forward direction, you need heat. The more heat you have, the more likely you're going to progress in the forward direction. So Le Chatelier's principle will say we're stressing this reaction by adding heat, so the reaction will favor the direction that relieves that stressor. And so to relieve that stressor, you have more of this input, so you're going to consume more A. So the stable concentration of A once we reach equilibrium will go down. B will go down because they're going to be consumed more. The forward reaction is happening more. And then C, D, and E would go up. Now, if you did the opposite. Let me erase what I just did. Let's say instead of adding heat, you were to take away heat. So let's say you were to take away heat. Let me make sure my cursor's right. So if you took heat away from the reaction, what will be favored? Well, then you're going to be favoring it in the other direction because there'll be less heat here. I mean, all of this is together. There'll be less heat for this reaction to occur, so this rate will start dominating this rate over here, right? If you take away heat, the rate of this reaction will slow down, this one will be bigger, and so you'll have more movement of concentration in that direction, or the reverse reaction will be favored. Now, let's think of another stressor-- pressure. Now, imagine that we had-- we mentioned the Haber process before, and this is the reaction for the Haber process. Nitrogen gas plus 3 moles of hydrogen gas in equilibrium with 2 moles of ammonia gas. Now, what's going to happen if I apply pressure to this system? I'm going to apply pressure. So if you think about what happens with pressure, everything all of a sudden is getting squeezed, although the volume isn't necessarily decreasing, but something is somehow making all the molecules want to be or forcing them to be closer together. Now, when things are getting closer together, the stress of the pressure could be relieved if we end up with fewer molecules. Think about it this way. PV is equal to nRT. We learned this multiple times, right? And let's say we could write P is equal to nRT/V. Now, if we increase the pressure, how can we relieve that? Remember, Le Chatelier's principle says that whatever's going to happen is going to relieve the stressor. The reaction is going to go in the direction that it relieves it. Well, if we lower the number of molecules, then that will relieve the pressure, right? You'll have fewer things bouncing against each other. So if we lower the number of molecules where you can kind of view it-- I mean, I shouldn't have written it this way, because it's not quite an equation, but I want you to think of it that way. Let me erase this. This probably wasn't the best intuition. If I have a container-- nope, too shocking. If I-- nope, same thing. If I have a container and I'm applying pressure to it, and in one option I could have 2 molecules-- let's say I could have 4 molecules in some volume. And in another situation, let's say they get merged and I only have 2 molecules, right? In either of these, the reaction can go between these, these 4 could merge to make 2 molecules. Actually, let me use this example up here. Let's say this nitrogen molecule is this blue one here. Actually, let me do it in a more different color. This brown one right here, it can merge with 3 hydrogen. It could produce this. So this is another way of writing this reaction, maybe in a more visual way. Now, if I'm applying pressure, if I'm applying pressure to this system, so pressure I just imagine is kind of more force per area from every direction, which of these situations is more likely to relieve the situation? Well, the situation where we have fewer molecules bumping around because it's easier to kind of apply or I guess squeeze them together than when you have more molecules bumping around. I'm doing this very hand wavy, but I think it gives you the intuition. So if you apply pressure to the system, if pressure goes up, you're applying-- this doesn't mean the pressure goes down. This means pressure is applying to the system. But the pressure is going up, what side of the reaction is going to be favored? The reaction's going to be favoring the side of that has fewer molecules. And this side has 2 molecules, although they'll be bigger molecules obviously, because it's not like we're losing mass in one direction or the other, as opposed to this situation where we have 4 molecules, right? 1 mole of nitrogen gas and 3 moles of hydrogen. And just to bring this all back to the whole idea that we saw earlier with the kinetic equilibrium, let's just imagine a reaction like this. And to show that it works with Le Chatelier's principle is consistent with everything we've learned with equilibrium constants. So let's say we had the reaction 2 moles, or the coefficient of two, 2 A's in the gaseous form plus B in the gaseous form is in equilibrium with C in the gaseous form. And let's say initially where our first equilibrium, our concentration of A is 2 molar, or our molarity is 2, our concentration of B is 6 molar concentration, and then our concentration of C is 8 molar. So what's the equilibrium constant here? The equilibrium constant here is the product, concentration of C, that's 8 molar divided by 2 molar squared, because of this, 2 squared times 6. Which is equal to 8/24, which is equal to 1/3. Now, let's say we were to add more A, and I'm not going to say exactly how much. We could actually get quite complicated with the mathematics, but let's say after adding more A, we have a new concentration. Now, let's say our concentration of A is 3 molar. You might say, hey, Sal, didn't you add 1 molar? No. I actually added probably more than 1 molar. What happens is, whatever I added, that'll push the reaction towards the right direction, or towards the forward direction, and so some of it will get consumed and go in that direction, but whatever's left over is here. So I might have added more than 1 molar concentration to this system. But whatever was extra beyond the 1 was consumed, and I'm just left with this equilibrium concentration of 3. So I didn't necessarily add 1. I could've added more than that. And let's say that our new equilibrium for C is 12 molar, which is consistent with what we say. We should be producing-- if we add some A, then our concentration of C should go up, and the intuition is that the concentration of B should go down a little bit, because a little bit more B is going to be consumed, because it's going to be colliding with-- or it's more likely to collide with-- more A particles or A molecules. So let's see what B's new concentration is. So remember, the equilibrium constant stays constant. So our equilibrium constant is now going to be equal to the concentration of C, right? That was the reaction. So it's 12 molar-- whoops-- I don't have to write the units here-- divided by our new concentration of A, that's 3. But remember the reaction. The coefficient on A is 2. So it's 3 squared times the new concentration for B, right? There's no coefficient here so I don't have to worry about any exponents. And let's just solve this. So you get 1/3 is equal to 12 over 9B. So if we just cross-multiply, we get 9 times the concentration of B is equal to 3 times 12, which is 36. And so divide both sides by 9. The new concentration of B is 4, or 4 molar. So B is equal to 4 molar. So that makes sense. We added some more A to the reaction. So we started with 2 molar of A, 6 molar of B, 8 molar of C. When we added more A, at the end, we added a bunch. It went in that direction, maybe it went back and forth a little bit, but it stabilized at 3 molar of A, 12 molar of C, so C went up, so that definitely went up. And notice, our stable equilibrium concentration of B actually went down, and this is consistent with what we were saying, that the reaction moves in that direction, more C gets produced, a little bit of B gets consumed. So anyway, hopefully you're fairly comfortable now with the whole notion of stressing a reaction and Le Chatelier's principle.

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This page was last edited on 16 April 2020, at 16:55
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