- [Instructor] An equilibrium constant has one value
for a particular reaction at a certain temperature.
For example, for this reaction,
we have oxalic acid turning into two H plus ions
and the oxalate anion.
The equilibrium constant, K C, for this reaction
is equal to 3.8 times 10 to the negative six
at 25 degrees Celsius.
However, the value of the equilibrium constant changes
if we change how we write the balanced equation.
For example, let's reverse everything.
So, instead of having oxalic acid on the reactant side,
let's have it on the product side.
And instead of having H plus ions in the product side
and oxalate on the product side,
let's put them on the reactant side.
Since we've reversed the reaction,
there's a new equilibrium constant.
So we're going to write K C for the reverse reaction,
is equal to,
and to figure out the value
for the new equilibrium constant,
since we reversed everything,
we take the inverse of the original equilibrium constant.
So to find the new K C value,
we take one over the original equilibrium constant
of 3.8 times 10 to the negative six.
And this is equal to 2.6 times 10 to the fifth,
also at 25 degrees Celsius.
Let's look at another example of how
changing how we write the equation
changes the value for the equilibrium constant.
And we start by looking at
the ionization of hydrofluoric acid,
turning into an H plus ion and a fluoride anion.
The K C value for this reaction is equal to
6.8 times 10 to the negative fourth
at 25 degrees Celsius.
This time we're going to multiply everything through
by a factor of two.
So it would be two H F
turning into two H plus, plus two F minus.
Our goal is to find the value
for the new equilibrium constant K C.
And since we multiplied everything through
by a factor of two,
we're going to take the old equilibrium constant
and we're going to raise it to the second power.
So this is K C is equal to
6.8 times 10 to the negative fourth.
And we're going to square that.
So this is equal to 4.6 times 10 to the negative seventh
at 25 degrees Celsius.
So since we multiplied through by a factor of two,
we raised the equilibrium constant to the second power.
If we had multiplied through by a factor of three,
we would have raised the old equilibrium constant
to the third power.
Let's use the two reactions we've just looked at here.
So we have oxalic acid turning into H plus and oxalate.
And the other one we looked at was hydrofluoric acid
ionizing to turn into H plus and F minus.
Let's use those two reactions
to calculate the equilibrium constant
for a third reaction.
And this third reaction shows oxalate
reacting with hydrofluoric acid
to form oxalic acid and the fluoride anion.
Our first step is to get these first two reactions
to look more like the third reaction.
For example, we have oxalic acid on the reactant side,
in the first reaction here,
and oxalic acid is on the product side
for the third reaction.
Also notice, oxalate is on the product side
for the first and over here for the third,
it's on the reactant side.
So we need to reverse the first reaction
to make it look more like the third one.
Here we have that reaction reversed.
And since we reversed the reaction, we would have to take
the inverse of the equilibrium constant,
which we did in the earlier example.
So let's go ahead and cross out this first one here,
because we're not going to need it anymore.
And let's move on to our next reaction
and try to make it look more like the third one.
So here we have hydrofluoric acid
turning into H plus and F minus.
And for our third reaction,
we have hydrofluoric acid on the reactant side,
but notice there is a coefficient of two
in the balanced equation,
and we have only a one in how we have it written.
So we need to multiply everything through
by a factor of two
to give us two H F and two H plus and two F minus.
And since we multiplied everything through
by a factor of two,
we would need to square the equilibrium constant
for the original reaction.
And once again, this is from our previous example.
So we already know that equilibrium constant value.
Next, let's add these two reactions together
and make sure they give us the third reaction.
So we have all of our reactants on one side
and we have all of our products on the other side here.
And we look for things that are the same
on both the reactants and the products side.
Well, there's two H plus on the left
and there's two H plus on the right,
so that would cancel out.
And now, notice that we would have
oxalate plus two H F going to oxalic acid plus two F minus.
So adding our two reactions together
does give us our third reaction.
Once we've confirmed that adding our two reactions together
gives us our third reaction,
we can use the equilibrium constants
for the first two reactions
to figure out the equilibrium constant for the third.
So the equilibrium constant
for this first reaction we're gonna call K one,
for the second reaction we're gonna call K two,
and the equilibrium constant
that we're trying to find for the third reaction
we'll call K C.
To find the equilibrium constant for the third reaction K C,
we need to multiply the equilibrium constants
for the first two reactions together.
So K C is equal to K one times K two.
We've already calculated the value for our K one
from an earlier example,
it was 2.6 times 10 to the fifth,
and we also calculated K two,
and it was 4.6 times 10 to the negative seven.
So to find K C,
the equilibrium constant for the third reaction,
we simply multiply those two together
and we get that K C is equal to point one two
at 25 degrees Celsius.
So whenever you add reactions together
to get a new reaction,
to find the equilibrium constant for the new reaction,
you simply need to multiply the equilibrium constants
of the reactions that you added together.