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in this video we're going to be talking about using the small X approximation to solve equilibrium problems when KC is large and by large I mean that KC is greater than or approximately equal to 10 to the fourth so we have another video before this that talks about using the small X approximation for when KC is small and I would say that's probably a simpler what place to start if this is the first time you're seeing it and in this video we'll be talking about the opposite situation as we talked about in the previous video the steps for solving this kind of problem are fourfold so there are four steps the first step is to assume that the reaction goes 100% in the favorite direction so the favorite direction when K is really large is that means we're going all the way to products we're assuming we have almost no starting materials left the second step is to set up an ice table and then to solve for X in our ice table assuming that X is small hence small X approximation and then the last and possibly the most important step is to check our answer so make sure the size of X is actually small compared to what we said it was smaller than and also to make sure that it gives us the right answer when we plug it back in to calculate KC in this video we're going to go through an example problem the example is the reaction of an O gas reacting with chlorine gas to make 2 n o CL gas and this particular reaction has a K value that is equal to 6 point 2 5 times 10 to the 4th so that is indeed on the order of 10 to the 4th so we should be able to use a small X approximation we're going to do a two ways so we're going to set up I stable and we know the initial concentrations are 2.0 molar for n o and 2.0 molar for CL 2 and we have no nocl at the beginning so if we are setting up a nice table as we normally did we would say okay well we're going to make some amount of nocl so we'll say minus 2x for our our n o since for every one CL 2 we use up we'll use up to n o which means we'll have minus X for CL 2 and we'll make 2 X molar concentration of our product and then the end once we reach equilibrium we'll have 2.0 molar minus 2x and we just get that by adding the initial concentration with the change in the same way we get 2.0 molar minus X for chlorine gas and for our product we just get 2x at equilibrium let's look back at our original steps that we talked about at the very beginning of this video we just made a nice table so that's step 2 did we assume that the reaction goes to 100% in our favorite direction it turns out we didn't we actually skipped step 1 to set up our ice table in what seems like a pretty natural way and it turns out that's not good we sip so we'll see what happens when we skip step 1 and keep going so if we keep going we can solve for X assuming X is small and we'll get for our equilibrium expression we get KC is equal to 2x squared divided by 2.0 molar minus X which is our cl2 concentration and multiply that by 2.0 molar minus 2x and that's all squared since we have that stoichiometric coefficient of 2 in front just to point out we already started with a balanced reaction that's pretty important or else our KC expression would be wrong so make sure everything is balanced before you get started so we've written our expression for KC and now we are going to erroneously assume we're just going to be like oh I don't care I'm just going to assume X is small I don't care if that's a good assumption at all so if X is small what we're really saying is X is a lot smaller than two molar here and here so we're also saying 2x is a lot smaller than two molar so we're saying this is approximately equal to 2x squared divided by 2.0 molar since X is a lot smaller than two molar we'll just ignore it entirely and X is a lot smaller than 2x or sorry X is a lot smaller than 2.0 molar even if you multiply X by two so this is going to be 2.0 molar squared and so if we multiply this out we get that this is equal to x squared over this two squared is canceled out by this two so we actually get x squared over two is equal to arc AC which is six point two five times ten to the fourth so that's what we get for x squared so if you multiply both sides by two we get that x squared is equal to one point two five times ten to the fifth which means X is equal to the square root of that which is 354 so this is where well okay so we've gone through steps two and three and now we're going to check our answer we're going to see if our answer makes sense so we're saying that the change in concentration here is 354 that clearly doesn't make sense because it actually gives us a negative concentration here so we're getting negative concentrations for n o and cl2 so that's bad so that doesn't make sense the other thing is that this already tells us that our assumption was really bad we assumed to solve our equation for KC we assume that K is a lot smaller than 2 or sorry we assume that X is a lot smaller than 2 but then we got that X is 350 which is clearly not smaller than 2 so it looks like skipping step 1 was bad so let's give this another try try number 2 what was step one step one says assume the reaction goes 100% in the favored direction and since KC is really large for this particular problem we're saying it's going all the way to products what this basically means is that we're going to set up our ice table again except that this time our initial concentrations are going to be assuming that our reaction already wins all the way to products and we can figure that out we can figure out those initial concentrations using stoichiometry so in the beginning we started out with two molar n oh and two molar CL 2 and we know that the N o and CL to react in a two to one ratio so since we have the same amount of both of these are limiting reactant will be the N oh and so that will get used up completely when we get to equilibrium so that means if we assume that it goes all the way to products we'll have 0 molar n Oh left and we'll make 2.0 molar of our product and since we had our CL to an excess we'll use 1 molar of it to react with 2 molar of our n oh and we'll have 1 molar left over so this is the special step where we actually followed step 1 and assumed we got 100% of product and the reason why we made this assumption was because we know that K is really large so we know that at equilibrium we should have all product or mostly product so now let's go through the other steps we assume that it's going to all product but we're not quite at equilibrium so if we're going to equilibrium that means we assume that a little we'll see a little bit of n-no at equilibrium so that will actually be plus 2x and we'll expect to see a little bit more cl2 at equilibrium so that'll be plus X because of the stoichiometry and that would give us minus 2x for the nocl we expect a little of this to get used up to go in the reverse reaction so then if we add everything together from the initial and change then we get 2x for our NO concentration we get X for a cl2 concentration and we get 2.0 minus 2x for our product concentration so far so good so now we're going to set up our KC expression just like we did before so KC is still equal to 6.25 times 10 to the 4 and this is equal to the concentration of nocl squared so that is 2.0 minus 2x squared divided by 2x squared which is our concentration of no.2 X oh whoops I made a mistake this is actually 1.0 plus X sorry so this should be X plus 1 point or molar so that is our full expression for KC using are using our equilibrium concentrations and we have made zero approximation so far but now we're going to assume that we can assume now we're going to assume that X is small we're going to assume that assume so if X is small what we really mean is that X is a lot smaller than one molar and we're also saying it's a lot smaller than two molar so then we're saying that our numerator is approximately equal to 2.0 molar squared because we're saying that this is small I'm going to make this an approximately equal sign the 2x squared stays the same since we're assuming X is a lot smaller than 1.0 molar this just becomes 1.0 molar so now if we multiply this all out we get that four divided by four x squared is equal to KC and the fours cancel out so x squared is equal to one over K C or six point two five times ten to the four and then if we take the square root of both sides here we get that X is equal to let's see we get X is equal to four point O times 10 to the minus 3 molar so this is where common sense is needed to make sure that well everything work this time around so first of all we can ask ourselves okay is this X actually a lot smaller than the numbers we said it was smaller than so now we're comparing X to one which it's about three orders of magnitude smaller so that's good and we can also compare 2x - 2 and again it's about three orders of magnitude smaller so so far so good but our final test will be to plug it back in our KC expression so if we plug in our value of x we get that KC is equal to 2.0 molar minus 2 times 4.0 times 10 to the minus 3 molar all squared divided by 2 times 4.0 times 10 to the minus 3 molar and that's also squared 4 and that's the enno concentration at equilibrium and then the last part is our cl2 concentration which is 4.0 times 10 to the minus 3 molar plus 1.0 molar and so we multiply that all out what you get is AI KC is equal to six point two three times ten to the fourth and so we can compare that to the value of KC we started out our problem with which is let's see it is six point two five times ten to the four this is actually pretty good we did make an approximation so answer isn't exactly right but it's pretty close and so if we wanted to get it even closer there are other methods we could use but four more for most purposes this is actually this tells us that our approximation was good so we can see that when KC is really large what we need to do is assume that we have on 100% product when we're setting up our ice table and that will help us safely assume that X is small

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