If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

# Le Chatelier's principle

## Video transcript

let's say we had the reaction molecule a plus molecule B is in dynamic equilibrium with molecule C plus D which just means that the rate of the forward reaction is going at the same rate as the as the backward reaction of the reverse reaction there will be some equilibrium concentrations of a b c and d and we could figure out what the equilibrium constant is if we want and I'll say it again I send 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 more a so remember it was in equilibrium it was at equilibrium the concentrations were constant but now all of a sudden I'm adding more aid to the system so now I'm going to 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 that are slightly higher that an A and a B are going to collide in just the right way so we're going to if the forward reaction is going to be more likely so if we add more a to the system it's going to you're going to have more A's they're going to bump with more B's so the bees are actually going to go down a little bit right because more bees are going to be consumed and even more important the C's and DS are going to definitely be increased and the way it would really happen you would add more a those a's would would bump into some more bees and so this forward reaction all of a sudden its rate would go faster than this backward reaction so the reaction will go in that direction then you'd have more C's and DS and maybe some of those are more likely to bump and go back in the distraction and eventually you would reach another 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 the more B is going to be used to consumed with those extra aids you just added and then those are going to produce more C's and DS in equilibrium and you can imagine if you added more a and more be let's say if you added more be as well more be then both of these are going to then the the reaction is going to go into the in the forward direction even more even more I don't think this is you know that amazing insight of the world I think this is kind of obvious that if if you stress this reaction by adding more on this side that naturally it's going to move in the direction that releases relieves the stress so if you add more a you're going to able have mores bumping with B's and go in that direction maybe consume a little bit more B's if you add more both the whole thing is going to go in that direction likewise if you let me rewrite their reaction do a different color a plus B dynamic equilibrium C plus D if I add let's do dark 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 going to be going to consume a little bit of extra D and then if you added more C and D then of course it's going to produce a lot more a lot more a and B and this idea this isn't this it's you know it seems pretty common sense but there's a fancy name for it it's called la should let me put a capital L shitali A's principle should tell Li A's if you've watched enough of these videos you don't have to be I have to be careful with my spelling principal 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 of the reaction to relieve that stress when they say stress the reaction that's like adding more a so the reaction is 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 and it's in this 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 list should tell yeas principle let's think of some other situations let's say if I had let's say a plus B plus some heat and that produces some I don't know some C plus D and you know maybe produces some a as well so if I were to add heat to this system what would happen so if I were to add Heat 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 let's tell the telly as principal will say we're stressing this reaction by adding heat so the reaction will go will favor the direction that relieves that that stress or and so to relieve that structure you have more of this input so you're going to consume more of more a so the the stable concentration of a once we reach equilibrium will go down B will go down because they're being 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 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 is 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 then you need you'll you there'll be less heat here I mean all 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 the reaction for the Haber process nitrogen gas plus three moles of hydrogen gas in equilibrium with two moles of ammonia gas now what's going to happen if I apply pressure to this system I'm going to apply pressure pressure so if you think about what happens to pressure you know everything all of a sudden is getting squeezed although it's not necessary the volume isn't necessarily decreasing but something is somehow making all the molecules want wanna-be or forcing them to be closer together now when things are getting closer together the the stress the stress of the pressure could be relieved as if if we we end up with fewer molecules think about it this way think about it this way PV is equal to NRT we learned this multiple times right and let's say let's say we could write P is equal to NRT over V now if we if we increase the pressure right if we increase the pressure how can we we were relieved that remember what we tally as principle says that whatever is going to happen is going to relieve the stress or the reaction is going to go in the direction that it relieves it well if we lower if we lower the number of molecules and 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 this is you know it's not 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 if if I'm let me let me erase this this probably wasn't the best intuition if I'm if I have a container nope too shocking if I know same thing I have a container and I'm applying pressure to it and in one option.i I could have two molecules let's say I could have let's say I have four molecules in some volume and another situation let's say I get merged and I only have two molecules right and either of these the reaction can go between these four could merge to make two molecules and let actually use this example up let's say this nitrogen molecule is this blue one here actually me do it more different color this brown one right here it can move the three hydrogen it could produce this so this is 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 you know 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 it squeezed them together then when you have more molecules bumping around this is you know 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 if pressure goes up you're applying this isn't this doesn't mean 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 is going to be favoring the side that has fewer molecules and this side has two molecules although they'll be bigger molecules obviously because we're talking like we're losing mass in one direction or the other as opposed to this situation where we have 4 molecules right one mole of nitrogen gas and then three moles of hydrogen and just to bring this all back to the whole idea that we saw earlier with the with the kinetic equilibrium is let's just imagine a reaction like this and to show that it works with that this letelier's principle is consistent with everything we've learned with with equilibrium constants so let's say we have the reaction two moles or or the coefficient of two two 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 we're in our first equilibrium is our concentration of a is two molar or our molarity is two our concentration of B is six molar concentration and then our concentration of C is eight Miller so what's the equilibrium constant here the equilibrium constant here is the product concentration of C that's eight molar divided by two molar squared because of this two squared 2 squared times six which is equal to 8 over 4 times 8 over 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 then you add 1 molar no I probably actually added probably more than one molar what happens is I add whatever I added that will push the reaction towards the right direction or to the forward direction and so some of it will get consumed and go in that direction but the whatever is left over is here so I might have added more than one molar concentration to this to this system whatever was more 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 have added more than that and let's say that our our new equilibrium for C is 12 molar right which is consistent with what we say we should be producing if we add some a if we're adding a then our concentration of C should go up and the intuition is is that the concentration of B should go down a little bit because a little bit more B is going to be consumed 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 the the equilibrium constant stays constant so our equilibrium constant is now going to be equal to the concentration of C all right that was the reaction so it's 12 molar whoops 12 molar I don't have to write the unit's 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 concentration 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 9 B 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 to a 2 molar of a 6 molar of B 8 molar of C when we added more a at the end and you know we added a bunch of when in that direction maybe the one 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 some more 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 the shitali a's principle
AP® is a registered trademark of the College Board, which has not reviewed this resource.