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# Small x approximation for small Kc

## Video transcript

we're going to talk about the small X approximation the small X approximation is an approximation method used to solve equilibrium problems maybe the most important thing to remember that I'll start out with is that the small X approximation doesn't always work it works best and you should generally try to use it only when your reaction is strongly product or reactant favorite so strongly product or reactant favorite so you might be asking at this point okay well how do you know when your your reaction is strongly favoring either products or reaction or reactants and the answer to that is it depends on the value of your equilibrium constant K so if something is strongly favoring product that means K is really big and by really big we usually mean or I usually mean it's on the order of 10 to the fourth or larger and we're talking about KC here where our equilibrium constant is in terms of molar concentration and similarly when we say something is reactant favored that means that KC is really small so we have mostly reactants around and not very many products and when KC is very small I mean that KC is less than or about equal to 10 to the minus 4 so these are the situations where the small X approximation works really well and then anywhere in between where KC is between 10 to the minus 4 and 10 to the 4 it actually doesn't work as well and you'll want to use a different method you'll either usually want to use the quadratic equation or you might want to use successive approximation which we're not covering in this video we'll be covering those things in separate videos so how does a small X approximation work the first step is you assume your reaction goes fully to the side you you know it's going to go based on K so assume reaction goes 100% to either products or reactants depending on whether K is really small or large our second step is to use this information from step one to make your ice table so initial concentration change in concentration and equilibrium concentration will then solve for the X in our ice table assuming that X is really small hence the name of the method the small X approximation and we'll go into a little bit more specifics about exactly what we mean by that when we do an example and then the last step which is really important really really really important is to check your answer and I usually do that by plugging equilibrium concentrations back into the K C or equilibrium constant expression to make sure the concentration we did get as our answers make sense and we get the value of K that we thought we should get so to make more sense of all of these things we are going to go through an example so we're going to go through an example of when we have a really small equilibrium constant so that means that kc is less than or equal to 10 to the minus 4 and that means since our equilibrium constant is really small that we have mostly reactants and not very much product in our reaction pot when we reach equilibrium so the example reaction that we're going to talk about today is iodine gas I 2 and it's going to be in equilibrium with - I - also gas and the K value for this particular reaction at the temperature we're interested in is five point six times ten to the minus twelve so that's a pretty small number and it's definitely smaller than ten to the minus four so we should be able to use the small X approximation we have a little bit more information about our reaction we know the initial concentration of i2 gas is 0.45 molar and we know that the initial concentration of I minus is zero so based on this we're going to go through the steps for using the small X approximation we're going to assume reaction goes 100% to either products or reactants here we know that we have mostly reactants if we start out with all reactant and no product we assume that it at equilibrium we're still going to have mostly reactant it's going to change a little bit so some of this is going to turn into I minus and so we'll say minus X for some small concentration of i2 that gets converted and that minus X will turn into two moles of I - but we don't expect X to be very big and that's the key to being able to use this approximation so if we add these up to get our equilibrium concentrations we get 0.45 molar minus X for I 2 and we get 2x for our I - concentration now we can use our KC expression to solve for X when we assume X is small so KC for this particular problem is the concentration of our product I - squared since we have a sodium at your coefficient of 2 there divided by the concentration of i2 and if we plug in our equilibrium concentrations from our ice table we get that this is 2 x squared divided by 0.45 molar - X so far we haven't made any approximations now we can make our approximation we're assuming X is small and specifically we're assuming that X is a lot smaller than 0.45 molar what that means is we can assume that 0.45 molar minus X is approximately equal to 0.45 molar so X doesn't really change the concentration of i2 at equilibrium very much so that simplifies our expression for KC quite a lot we have four x squared in the numerator from square and two and squaring X and in the denominator we no longer have X so this makes a much easier equation to solve and this is all equal to five point six times ten to the minus twelve awesome so now what we can do is we can multiply both sides by 0.45 and we can divide both sides by four and what that gives us is that x squared is equal to six point three times ten to the minus thirteen and if we take the square root of this we get that X is equal to seven point nine times ten to the minus seven molar so that is what we get for X so here is the second most important step besides making our approximation in the first place we need to make sure that our approximation makes sense we've made a change somewhere we've decided X is small and we've ignored it in our equation in one part of our equation so we have to make sure that our original assumption gave us a common sense answer so we see that X is seven point nine times ten to the minus seven we assume that was a lot smaller than 0.45 which looks to be about true it's about eight actually it's about six orders of magnitude sorry it's about six orders of magnitude smaller than point four five so so far so good but we can make sure that it really is right by checking our answer so we can plug in our X and calculate KC again so then we get KC is equal to two times our x value seven point nine times 10 to the minus 7 molar and all of this squared divided by 0.45 molar minus our x-value seven point nine times ten to the minus seven more if you multiply this all out what you get for KC or what I got for KC is five point six times ten to the minus twelve and that matches what we had urgently in our problem so this tells us that a our approximation was pretty good it gave us an x value that gave us concentrations that make sense and B which is another reason why we want to check our answer B it tells us that we didn't make any silly math errors which is really easy to do in this kind of problem
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