- [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.