2.08 Le Chatelier's Principle Part A

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From the course by University of Kentucky

Advanced Chemistry

354 ratings

University of Kentucky

Advanced Chemistry

354 ratings

A chemistry course to cover selected topics covered in advanced high school chemistry courses, correlating to the standard topics as established by the American Chemical Society. Prerequisites: Students should have a background in basic chemistry including nomenclature, reactions, stoichiometry, molarity and thermochemistry.

From the lesson

Chemical Equilibrium

This unit introduces the concept of chemical equilibrium and how it applies to many chemical reactions. The quantitative aspects of equilibrium are explored thoroughly through discussions of the law of mass action as well as the relationship between equilibrium constants with respect to concentrations and pressures of substances. Much of the discussion explores how to solve problems to find either the value of the equilibrium constant or the concentrations of substances at equilibrium. ICE (initial-change-equilibrium) tables are introduced as a problem-solving tool and multiple examples of their use are included. From a qualitative standpoint, Le Châtelier’s principle is used to explain how various factors affect the equilibrium constant of a reaction along with the concentrations of all species.

2.01 Dynamic Equilibrium6:28

2.02 Law of Mass Action7:34

2.03 Law of Mass Action for Combined Reactions4:36

2.04 Relationship Between Kc and Kp6:51

2.04a Relationship Between Kc and Kp2:34

2.05 Calculating the Equilibrium Constant13:09

2.05a Finding Kc5:04

2.06 Reaction Quotient5:55

2.06a Predicting Reaction Progress with the Reaction Quotient1:55

2.07 Calculating Equilibrium Concentrations12:54

2.07a Finding Equilibrium Concentrations, part 14:59

2.07b Finding Equilibrium Concentrations, part 23:20

2.07c Finding Equilibrium Concentrations, part 34:22

2.08 Le Chatelier's Principle Part A9:55

2.08a Le Chatelier's Principle Part B19:50

2.08b How Changes Shift the Equilibrium5:18

Meet the Instructors

Dr. Allison Soult
Lecturer
Chemistry
Dr. Kim Woodrum
Sr. Lecturer
Chemistry

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.

Our objective in this unit is to understand

the concept of Le Chatelier's principle

and to be able to predict

how different changes will effect

the system at chemical equilibrium.

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.

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.

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.

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.

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

over the concentration of our reactants.

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.

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.

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.

So lets look at a drawing here of what is going on.

Here I have my reaction at equilibrium in the first panel.

I go to the second panel, I have added

additional carbon monoxide.

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.

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.

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.

If I decrease the amount of

CH_3OH present, if I remove some of that methanol

so now looking at my equilibrium constant

k equals products over reactants.

I have lost some of my methanol

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.

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