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Current time:0:00Total duration:9:40

- [Voiceover] We're gonna 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 favored. So strongly product or reactant favored. You might be asking at this point, okay, well, how do you know when your reaction is strongly favoring either products or reactants? 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. We're talking about Kc here where our equilibrium constant is interms 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 it 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 -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 -4 and 10 to the 4, it actually doesn't work as well, and you'll wanna use a different method. You'll either usually wanna
use the quadratic equation or you might wanna use
successive approximation, which we're not covering in this video. We'll be covering those
things in separate videos. So how does the small
x approximation work? The first step is you assume
your reaction goes fully to the side you know
it's gonna 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. We'll 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. Then, the last step,
which is really important, really, really, really important is to check your answer. I usually do that by plugging equilibrium concentrations back into the Kc or equilibrium constant expression to make sure the
concentration that 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. We're gonna go through an example of when we have a really small equilibrium constant. That means that kc is less than or equal to 10 to the -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 gonna talk about today is iodine gas, I2, and it's going to be an equilibrium with 2I-, also gas. The K value for this particular reaction at the temperature we're interested in is 5.6 times 10 to the -12. So that's a pretty small number,
and it's definitely smaller than 10 to the -4. 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- is zero. So based on this, we're
gonna go through the steps for using the small x approximation. We're gonna 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 at equilibrium, we're still gonna have mostly reactant. It's gonna change a little bit. So some of this is gonna turn into I-. So we'll say -x for some
small concentration of I2 that gets converted. And that -x will turn into 2 moles of I-, but we don't expect x to very big, and that's the key to being
able to use this approximation. If we add these up to get our
equilibrium concentrations, we get 0.45 molar minus x for I2, 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. Kc, for this particular problem, is the concentration of
our product, I- squared since we have a metric
coefficient of two there divided by the concentration of I2. And if we plug in our
equilibrium concentrations from our ICE table, we get
this 2x squared divided by 0.45 molar minus 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 4x squared in the numerator from squaring 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
5.6 times 10 to the -12. Awesome. So now what we can do is we
can multiply both sides by .45, and we can divide both sides by four, and what that gives us is
that x squared is equal to 6.3 times 10 to the -13, and if we take the square root of this, we get that x is equal to 7.9 times 10 to the -7 molar. So that is what we get for x. 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. We see that x is 7.9 times 10 to the -7. We assumed that was a
lot smaller than .45, which looks to be about true. It's about six orders of
magnitudes smaller than .45. So, so far, so good. We can make sure that it really is right by checking our answer. We can plug in our x
and calculate Kc again. So then, we get that Kc is equal to two times our x value, 7.9 times 10 to the -7 molar, and all of this squared divided by 0.45 molar minus our x value, 7.9 times 10 to the -7 molar. If you multiply this all out, what you get for Kc, or what I got for Kc is 5.6 times 10 to the -12. That matches what we had
originally 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 wanna check our answer. B, it tells is that we didn't
make any silly math errors, which is really easy to do
in this kind of problem.