- [Tutor] For the reaction of iron two oxide
plus carbon monoxide goes to solid iron and carbon dioxide.
The equilibrium constant Kp is equal to 0.26 at 1000 Kelvin.
Our goal is to find the equilibrium partial pressures
of our two gasses, carbon monoxide and carbon dioxide.
If the initial partial pressures
are 0.80 atmospheres for carbon monoxide
and 0.40 atmospheres for carbon dioxide,
we can use the reaction quotient Q,
to predict which direction that reaction will go
to reach equilibrium.
So let's calculate Qp
and Qp
is equal to,
first we think about our products
and we leave solids out of equilibrium expressions,
and therefore we also leave it out of our expression for Qp.
So we're gonna leave out iron.
We're going to include carbon dioxide since it's a gas.
So we're gonna write the partial pressure of carbon dioxide,
and since we have a coefficient of one
in front of carbon dioxide,
it's the partial pressure raise to the first power
divided by, next we look at our reactants,
and we have a solid, so we're gonna leave that out.
So we have another gas, carbon monoxide.
So this will be the partial pressure of carbon monoxide
raised to the first power,
since there's also a coefficient of one.
Next, we plug in our partial pressures
at this moment in time.
The partial pressure of carbon dioxide
is
0.40,
and the partial pressure of carbon monoxide
is
0.80
atmospheres.
So we plug those into our expression for Qp
and 0.40
divided by 0.80
is equal to
0.50.
So Qp
at this moment in time is equal to 0.50.
Since QP is not equal to Kp at this moment in time,
the reaction is not at equilibrium.
So Qp is equal to 0.50 and Kp is equal to 0.26.
So Qp
is greater
than Kp.
And when Qp is greater than Kp,
there are too many products and not enough reactants.
Therefore the net reaction is going to move to the left.
Next let's fill out our I.C.E table for this reaction.
I stands for the initial partial pressure in atmospheres.
And so the initial partial pressure of carbon monoxide
was 0.80 atmospheres.
And the initial partial pressure of carbon dioxide
is 0.40 atmospheres.
C stands for change.
And E stands for the equilibrium partial pressure.
Calculating Qp allowed us to realize
that the net reaction moves to the left.
And if the net reaction moves to the left,
we're going to lose some carbon dioxide
and we're going to gain some carbon monoxide.
So first let's think about carbon dioxide.
We're gonna lose some of it,
but we don't know how much and therefore
that's gonna be represented by X.
So we're gonna write minus X here for carbon dioxide.
And since the coefficient is a one
in front of carbon dioxide,
and it's also one in front of carbon monoxide,
if we lose X for carbon dioxide,
we're going to gain X for carbon monoxide.
Therefore the equilibrium partial pressure
for carbon monoxide
would be
0.80
plus X.
And the equilibrium partial pressure for carbon dioxide
would be
0.40
minus X.
Our next step is to write an equilibrium constant expression
for this reaction.
So Kp is equal to the partial pressure of carbon dioxide
divided by the partial pressure of carbon monoxide.
So the expressions for Kp and Qp look the same,
but the difference is for Kp,
it would be the equilibrium partial pressures only.
Whereas for Qp,
it's the partial pressures at any moment in time.
And since for Kp,
we're talking about the equilibrium partial pressures,
we can take those directly from our I.C.E table
and plug them in.
So we can plug in the equilibrium partial pressure of CO2
and the equilibrium partial pressure of carbon monoxide.
Here we can see our two equilibrium partial pressures
plugged into our Kp expression
and also the equilibrium constant Kp
is equal to 0.26 for this reaction,
so that's plugged in as well.
Our next step is to solve for X.
So we multiply both sides by 0.80 plus X,
and we get this and then we do some more algebra
and we get down to 1.26 X is equal to 0.192.
So
0.192
divided by
1.26
is equal to
0.15.
So X
is equal to
0.
1
5.
Now that we know that X is equal to 0.15,
we can go back to our I.C.E table
and solve for the equilibrium partial pressures.
So for carbon monoxide,
the equilibrium partial pressure is 0.80 plus X.
So that's equal to
0.80
plus
0.
1
5,
which is equal to
0.
9
5
atmospheres.
So that's the equilibrium partial pressure
of carbon monoxide.
For carbon dioxide,
the equilibrium partial pressure is 0.40 minus X.
So
0.40
minus
0.15
is equal to
0.
2
5
atmospheres.
So that's the equilibrium partial pressure
for carbon dioxide.
Finally,
we can use the reaction quotient Qp
to make sure that these two answers,
for equilibrium partial pressures are correct.
So we can write that Qp is equal to
the partial pressure of CO2
divided by the partial pressure of CO.
And we can plug in those,
those equilibrium partial pressures.
So this would be
0.25 atmospheres,
was the equilibrium partial pressure of carbon dioxide
and 0.95 was the equilibrium partial pressure
of carbon monoxide.
And 0.25 divided by 0.95
is equal to
0.
2
6.
And
since Kp
is also equal to
0.26
at this moment in time,
Qp
is equal
to Kp
and the reaction is at equilibrium.
Therefore we know we have the correct
equilibrium partial pressures.