0:00
In this module, we will explore
Le Chatelier's principle and how it effects
the system at chemical equilibrium.
Because there are many examples to look at
we will divide this into two parts. Because there are many examples to look at
we will divide this into two parts.
0:23
Before we can start looking at the specific details
we need to understand the general concept of Le Chatelier's principle.
What it tells us is, that when a chemical
system, or chemical reaction What it tells us is, that when a chemical
system, or chemical reaction
is at equilibrium is disturbed
the system shifts in a direction to minimize the disturbance.
Well we see this in our everyday lives if you have ever been on an elevator.
If there is one person on a elevator
and second person gets on
everyone rearranges
in order to spread themselves out.
A third person gets on and we all move again
a fourth, and a fifth
until we get to the point where we just cannot move anymore.
but we try to spread ourselves out in the elevator.
We rarely ever see an elevator door open
and find four people standing really close
together on one side of the elevator, and the other side being empty.
As each person get on the elevator
we are disturbing the equilibrium
and the system, which is the people already in the elevator
have to respond, or shift and the system, which is the people already in the elevator
have to respond, or shift
in order to minimize that disturbance.
1:20
So how to we related Le Chatelier's principle back
to the concepts we already have explored with equilibrium?
Well if we look at a
room full of people.
We see that all the people are in the room
now this systems is not at equilibrium We see that all the people are in the room
now this systems is not at equilibrium
it has got to go to equilibrium. now this systems is not at equilibrium
it has got to go to equilibrium.
And so to get to equilibrium
what we see, is that 3 people leave the room
and when those 3 people are outside of the room
our system is now at equilibrium.
Remember equilibrium is products over reactants.
Or in this case we can look at it
as the ratio of the number of people outside the room
compared to the number of people inside the room
so here we have a system at equilibrium
we have 3 people outside the room
and rest of the people inside the room.
At time goes on
we have different people outside the room
here our blue people go back into the room
and the purple people have left.
We are still at equilibrium.
Because what we are seeing is the number of people
outside the room is still 3
and the rest of the people are inside the room
and so our ratio remains the same.
2:27
Now we have a system
at equilibrium again
and we disturb that equilibrium in some way.
We are not going to worry about what that disturbance is.
Say we add some more people. We are not going to worry about what that disturbance is.
Say we add some more people.
We change something about the ratio
when we disturb that system at equilibrium. We change something about the ratio
when we disturb that system at equilibrium.
What we see is that a new equilibrium is established.
So now this new equilibrium
has 4 people outside of the room and the rest of the people inside.
But the system had to respond to that
disturbance to the equilibrium.
Maybe we turned up the heat, maybe the music changed
maybe they ran out of food, something changed
To disturb that system which was at equilibrium
and it had to respond and at this point
equilibrium was restored. and it had to respond and at this point
equilibrium was restored.
There are many things that can effect a system at equilibrium.
All of these disturbances can happen to
reaction or to some system.
And the system will respond in such a way to minimize
this disturbance.
So we can look at a variety of topics and we will actually go through
examples of each of these, so we can see how they apply to
actual chemical reaction.
We can change the concentration of our reactant
or product.
We can change the pressure of the system
and we can change the pressure in two different ways.
One by changing the volume of the container
or the addition of an inert gas.
Remember that inert just means nonreactive.
And what we will find is that
that inert gas actually does not have any effect on the equilibrium.
We can also change the temperature
and we have to worry about is this an exothermic
process, where heat is being released
or is this an endothermic process
where heat is being consumed.
Because changes in temperature will affect
those reactions in different ways. Because changes in temperature will affect
those reactions in different ways.
4:10
To lets look at an example here where we are looking
at a reaction, we have our balanced chemical equation
Carbon Monoxide reacting with Hydrogen
to produce Methanol.
And what we are going to look at is we are going to change
the concentration of a reactant.
And so what happens if we
increase the concentration of a reactant.
4:28
We are going to see the formation of more product.
It is going to favor the formation of products.
Because if I add additional
carbon monoxide for example.
Now I have disturbed my equilibrium
we know that k
equals concentrations of our products
4:50
And if I add reactants I now have a
bigger number on the problem
That I would need to have in order to balance
and have the correct ratio with the concentration with the products.
What needs to happen is that some of the
excess CO that what added needs to be
consumed and formed into products.
As a result, we will also a
reduction in the amount of H_2 present because it is
needed to react with the CO. reduction in the amount of H_2 present because it is
needed to react with the CO.
5:13
We are going to see and increase
in the amount of CH_3OH formed
because we need to form products we need to
increase the concentration of products
and we need to decrease
concentration of reactants which we have also done.
And what this will allow the system to do is
to return to equilibrium.
Where we maintain that constant value for the equilibrium constant. to return to equilibrium.
Where we maintain that constant value for the equilibrium constant.
5:37
So now what happens if I decrease the amount of a reactant?
Lets say I remove some carbon monoxide.
I am not worried about the logistics or the
experimental parameters or the setup I need to remove it. I am not worried about the logistics or the
experimental parameters or the setup I need to remove it.
but the amount of CO available decreases.
What happens is the concentration
of our reactants decrease
and now we have disturbed this ratio.
We are no longer a system at equilibrium.
In order to restore the system to equilibrium
it is going to have to compensate
for that lost carbon monoxide
and as a result the reaction for that lost carbon monoxide
and as a result the reaction
is going to shift to the left for favor the formation of more reactants.
Some of our CH_3OH will decompose.
We will increase the amount of H_2 a little bit.
We will increase the amount of CO a little bit.
Until we get to the point
where the concentration of the products over the concentration of our reactants
is equal to k.
Now at this point, we are going to have a different amount of
CO a different amount of H_2 Now at this point, we are going to have a different amount of
CO a different amount of H_2
and different amount of Methanol, than we originally had.
But the important thing is that we return to that ratio
which is the equilibrium constant k.
6:52
Now the reaction needs to return to equilibrium.
In order to do this, it needs to consume Now the reaction needs to return to equilibrium.
In order to do this, it needs to consume
one of our carbon monoxides
and two of our hydrogens.
7:02
And when it does that, it will form
a molecule of CH_3OH
so if we look over here after equilibrium has been re-established.
We see we have increased the amount of CH_3OH present.
We have gone down to having 3 carbon monoxides
present, and we have also lost two of our hydrogens
because they were consumed.
When that carbon monoxide reacted.
So we are now back at equilibrium.
Though have a different number of molecules present
we are going to maintain that same ratio.
7:33
Now we can look at the same reaction and see what
happens when we change the concentration of a product.
Since there is only one product, we will be looking at methanol.
If I increase the amount of methanol present.
I see that the reaction now has too much methanol If I increase the amount of methanol present.
I see that the reaction now has too much methanol
I need to get rid of some of that methanol I see that the reaction now has too much methanol
I need to get rid of some of that methanol
and so as a result we are going to see the decomposition into
carbon monoxide and hydrogen.
So we are going to lose some of out CH_3OH
in order to get back to that equilibrium ratio.
So if I increase the concentration of methanol
I see that I am going to
favor the formation of more reactants.
8:23
and I need to make up for that loss.
So I need to produce some more methanol to make up some of what I have lost.
But in the process I am also going to
decrease the amount of reactants present But in the process I am also going to
decrease the amount of reactants present
and as a result I will get back to product, reactant ratio decrease the amount of reactants present
and as a result I will get back to product, reactant ratio
that is equal to the equilibrium constant k.
And so if we are decreasing the amount of products
I am going to favor the formation of more product.
I need to compensate for that lost amount of CH_3OH.
Now these are just two examples of what Le Chatelier's principle is telling us.
It is all about responding to some disturbance to the equilibrium.
In these two scenarios we looked at increasing
and decreasing the amounts of reactants and products.
We can also look at
our system at equilibrium
looking at our same reaction that we looked at before
now what we are looking at is we are adding more
methanol so here we have now what we are looking at is we are adding more
methanol so here we have
added more methanol to the system
and what we see happening is that when an
equilibrium is re-established
actually lose one of those methanol molecules
and actually form more
carbon monoxide
and I formed some more
hydrogen molecules, so these are my molecules that form
And what happens, is that now
I form more of my reactants
I am re-establishing that equilibrium
by the addition of methanol.
We can look at a similar idea for
the removal of methanol.
We would just see that the formation of our
products is favored
instead of the formation of reactants
that's favored when we add methanol.
Dr. Allison SoultLecturer
Dr. Kim WoodrumSr. Lecturer
