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.
2.08a Le Chatelier's Principle Part B
Loading...
Do curso por University of Kentucky
Química Avançada
312 ratings
Na lição
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.
Conheça os instrutores
Dr. Allison SoultLecturer
Chemistry
Dr. Kim WoodrumSr. Lecturer
Chemistry
Now that we have a general understanding of Le Chatelier's principle,
the assist of an equilibrium will respond to some disturbance of an equilibrium
that tried to minimize that disturbance.
We looked at adding, and removing reactants and
products, now we're going to look at some other things that we can change that will
disturb the equilibrium of the system.
The first one we'll look at, is changing the volume of container.
The end result of this, is actually looking at a change in pressure.
But there are two ways to change the pressure of a container.
Adding another substance to the mixture, or changing the volume.
We're only worried about the gas-based substances when we look at change in
the volume of a container because those are the only species that are affected by
an increase, or decrease in pressure.
Solids and liquids will not change their properties, as we change the volume or
change the pressure.
When I look at increase in volume,
what we have is those gas molecules will expand to fill that volume,
and as a result, the pressure inside that container will decrease.
If I had a decrease in pressure, what is going to happen to the reaction,
is that I'm going to favor the side, that has an increased number of moles.
I need to compensate for that lost pressure,
by doing something that will cause an increase in the pressure.
It's just the opposite.
And so what I look at, is my reactive side has 3 moles of gas.
My product side has only 1 mole of gas.
And so, if I can increase the amount of reactants present, what I'm going to see
is, I'm going to compensate for that decrease in pressure by making more moles
of gas that will counteract that decrease with an increase of pressure.
So, when we have an increase in volume and we see a decrease in pressure.
The reaction will favor the side with more moles of gas.
For this example, it's favoring the left side.
However, it's going to depend on the specific parameters of your reaction.
The opposite is also true.
If I decrease the volume, I increase the pressure.
I've increase the number of molecules in able particular space and so
we have an increase in pressure, and
now the system needs to responds to minimize that increase of pressure.
And it's going to do that by decreasing the pressure, and
the way that happens is I go from 3 moles of gas to 1 mole of gas.
And so when we see a decrease in volume,
it's going to favor the side that has fewer moles of gas.
We do however, have to watch for reactions that having mixed phase.
We still can look at favoring the side with more moles of gas, or
the side with fewer moles of gas depending on what change we apply.
But we do need to be careful, when we look at a reaction.
In this reaction, two magnesium plus oxygen, yielding 2 MgO.
What we have here, is just 1 mole of gas on the left side of the reaction,
and we have no moles of gas on the right size of the reaction.
We can still apply our same argument but we have to look at just the moles of gas,
not the moles of every single substance present in the balanced reaction.
So here we look at an example, we had our methanol, our hydrogen, and
our carbon monoxide, we have a system at equilibrium.
We decrease the volume by lowering the piston, we're forcing these molecules into
a smaller space, so we're increasing the pressure.
And as a result, we need to do something to re-establish equilibrium,
that will lead to a decrease in the pressure.
Because we need to counteract that disturbance to the system.
And, the thing that will counteract that increase in pressure will be to take some
of our reactants, and
form more products because we're favoring the side that has fewer moles of gas.
If we increase our volume.
Again, we have our system at equilibrium, we've increased the volume, and
what we are doing is lowering the pressure of the system because we have the same
number of molecules but in a much larger space, so we'll see a lower pressure.
In order to establish equilibrium,
we need to have some response to that disturbance of increasing the volume that
will counteract that increase in that lower pressure.
So that we're going to is we want to increase the pressure, so
we need to take some of our products, and decompose them into CO and
H2 because that's going to lead to the formation of more moles of gas.
And therefore, we will counteract that increase of volume,
which lowered the pressure, by producing more moles of gas,
which results in an increase in pressure.
Now, if I change the pressure by the addition of an inert gas.
I'm still changing the total pressure of the system, but
what I find is that I'm not changing the partial pressure of any of my substances.
Therefore, changing the pressure through the addition of an inert gas
will not have any effect on the equilibrium of this system.
One thing we have to worry about is, what is an inert gas?
We frequently, see that helium and nitrogen are used in inert gases.
However, we do have to use nitrogen with caution as an inert gas
because it can either be a reactant, or a product in our overall reaction.
For example, I wouldn't want to try to use hydrogen as an inert gas for
this reaction because I already have hydrogen.
And what I'm really adding is one of my reactants, not adding an inert or
unreactive gas.
[SOUND] We do not see any shifting as the reaction towards formation of
more reactants or more products with the addition of an inert gas,
because the partial pressures of each gas don't change.
Just the total pressure does.
And remember, when I'm looking at equilibrium constant, K p values,
where p stands for pressure, I'm looking at the pressure of methanol,
pressure CO, pressure of H2 squared.
Since there's nothing in the equilibrium expression about the inert gas,
because it's not part of the overall reaction.
I don't need to worry about it at all.
And what I'm going to see is that,
my equilibrium is not disturbed by the addition of an inert gas.
[SOUND] So let's look at an example here.
If I have a reaction at equilibrium, we add our N2 molecules to this.
And what we see is that our N2 molecules are still present at the end.
But we also see there,
once our equilibrium is here that we're not really re-establishing equilibrium,
we've just taking the exact same thing we had initially, and
after we added the N2 the number of molecules of CH3OH remains the same.
The number of molecules of H2 and of CO all remain the same.
So we have no effect on the equilibrium through the addition of an inert gas.
[SOUND] Now we need to look at the effect of changing the temperature.
Before we can do that, we do have to figure out,
do we have an exothermic reaction, or an endothermic reaction?
Now when I look at the first reaction here,
I see a delta H value of minus 90.5 kilojules.
What that tells me is that, this is, in fact, an exothermic reaction.
The negative sign indicates that reheat is being lost from the system.
And so what we can actually do is treat heat like a product,
because heat is coming off of the system.
If I look at an endothermic reaction,
such as I have down here, what I see is my delta H value is positive.
That's how I know it's an endothermic reaction, but
what I can also figure out, is that heat must be behaving as
a reactant because I have to put heat in to get this to reaction to proceed.
[SOUND] So when we look at a change of temperature for an exothermic reaction and
an endothermic reaction, we have to further subdivide this down.
What if I increased the temperature of an exothermic reaction,
what if I decreased the temperature?
What if I increased of an endothermic reaction,
what if I decreased the temperature?
And we have to worry about all four of those scenarios.
So let's first look at what happens, when we take an exothermic reaction.
This is where we have heat as a product, so we write heat on the product side.
And what happens, when we increase the temperature?
Well, if I have some reaction, and
we'll just write a generic reaction here, A plus B going to C, plus heat.
Because heat is a product here.
If I increase the temperature, what I'm really doing is I'm adding heat.
And, what did we learn earlier, if I add a product,
what I'm going to see is that the reaction is going to shift away from that,
towards the formation of more reactants.
So for a decrease in temperature of an exothermic reaction, we end up with
favoring a reactant, so some of that C will decomposed into A and B.
And will counteract that increase in temperature by using up some of that
heat to draw this reaction to the left, to shift this reaction towards the left side.
The opposite is true, if I decrease the temperature.
If I lower the temperature, what I'm really doing is I'm removing heat.
[SOUND] And when I remove heat, the reaction is going
to need to respond in order to counteract that change.
So heat is a product, I remove heat.
Our reaction is going to ship towards the right, towards the formation of more C.
So in this case, it's favoring the products.
And this is true for all exothermic reactions.
If we increase the temperature we'll favor the reactants.
If we decrease the temperature we'll favor the product.
So that helps us with the exothermic reactions.
But now we needs to look at the endothermic reactions.
And in this case, we treat heat as a reactant.
So let's say, we have the reaction D plus E plus Heat.
Remember, I'm looking at an endothermic reaction, so heat is cons,
being consumed by the reaction.
Is treated as a reactant, and we go to forming f.
Now, what I could look at, is the same thing I looked at for exothermic.
Increase and decrease in temperature, how does the system respond?
Remember that I said for increasing the temperature, we're adding heat again.
[SOUND] And if I add heat to an endothermic processes.
What I'll see happening, is that I'm adding heat.
The reaction is going to shift, toward the formation of more products.
In this case, f.
And it does this because heat is now a reactant.
And if I'm adding heat, it's just like adding any other of my reactants.
So we see the formation of more product, of more ads.
We see the decrease in the amount of D and E, but we've minimized
that increase in heat that we experienced by the increase in temperature.
We can also look at the opposite.
We can decrease the temperature of the system.
And so when we decrease the temperature, what we're looking at is removing heat.
[SOUND] So we remove heat from the system, and so what's happening here then,
is losing heat, and so now the reaction is going to have to shift to the left to
compensate for some of that lost heat.
So when we have an endothermic reaction in which we decrease the temperature,
we favor the formation of more reactants
[SOUND] So let's look at an example of this.
Here we have our hydrogen, our carbon monoxide, and our methanol.
We add heat.
Notice that the number of reactants in the, on the products on the left
panel is equal to the number of reactants and products in the middle panel.
We're adding heat to this.
This is our disturbance to our equilibrium.
And we need to figure out,
what's going to happen relative to Le Chatelier's Principle.
I see that I'm dealing with an exothermic reaction because it has a negative value,
and whenever I see that I'm going to go ahead and
write in plus heat, just as a reminder that heat is behaving like a product.
So if I add heat, what's going to happen?
Well, the reaction is going to shift towards the left.
We need to compensate for some of that additional head that has been added
to the heat that is already present in the reaction from our delta H of reaction.
And so I see heat is being treated as a product,
so if I add product, I'm going to shift my reaction to the left.
And this is what we see here.
I've lost a methanol molecule, and I've increased the number of CO molecules and
the number of H2 molecules.
If I look at lowering the temperature, or removing heat from the system,
I can still start with my same system at Chemical Equilibrium.
I'm cooling the reaction vessel, so I'm disturbing the equilibrium.
Notice, that at this point, in panel two,
we still have all of the same number of each species, as we did in panel one.
But now we've removed heat by decreasing the temperature, so
the system has to respond to that change.
Our delta H, same reaction we were looking at before, is still exothermic.
And so since were it's exothermic, we treat heat as a product.
Now, when I look at this, decreasing the temperature,
I am actually removing heat by decreasing the temperature,
[SOUND] and as a result, am going to shift this reaction to the right.
And so what I see is I actually, get the formation of an additional methanol.
I've lost a carbon monoxide, and I've lost two
of my hydrogen molecules to form that new CH3OH molecule.
If we looked at endothermic reactions, what we'd see is exactly, the opposite.
Because in those reactions,
remember it's endothermic, heat is going into the system.
Therefore, heat is being consumed, and heat is treated as a reactant.
>> In this reaction,
we're looking at the temperature dependence of a complex ion equilibrium.
We have a hydrated cobalt salt,
and we're going to take advantage of Le Chatelier's Principle.
We're going to put it in a heated water bath, and put it in ice and
see how it changes colors.
[SOUND] So as we have here now, we have a test tube that was in the ice bath, and
we have a test tube that was in the hot water bath that looks purple, and
we have our control that was left at room temperature.
This is an endothermic reaction, so as we put heat into the system,
we drove this reaction to the right, and we get to the cobalt chloride ion.
As we put it in the ice bath, we drove the reaction more towards the left side, and
so then we have the hydrated cobalt ion at present.
>> So now we can go back through these disturbances that affect equilibrium.
We can change the concentration of a reactant and see how that affects it,
and that could be either increasing the amount of a reactant or
decreasing the amount of a reactant.
We can change the concentration of a product, again either increasing or
decreasing.
We could change the pressure.
We could do that in two ways, one of which is changing the volume of a container,
which will have an effect on the shift in equilibrium, or we can add it in our gas.
It will also increase the pressure, or
we can remove it in our gas that's already there, and we'll decrease the pressure.
But this has no effect on the equilibrium.
Because what we're looking at is the fact that we're only worried about the partial
pressures of each of the component gases.
And last, we can look at changing temperature, and
we have to look how it happens, if we increase or decrease temperature.
And what we are dealing with an endo or exothermic system.
[SOUND] Now, lets look at an example, PCI5 decomposes according
to the endothermic process shown, which one of the following
includes only changes that will favor the decomposition process?
When we look at Le Chatelier's Principle,
we see that there are many factors which affect which way the reaction will shift
in response to that disturbance in equilibrium.
Before we get started in looking at these changes,
we want to first look at the information given in the problem.
And we see that the reaction is endothermic,
which means delta H is greater than 0.
It's a positive value.
And that means,
energy is being put into the system which means heat is behaving as a reactant.
So, now, we have heat plus PCl5 in the gas phase
is in equilibrium with PCl3 in the gas phase, plus Cl2 also in the gas phase.
So, the first thing we want to look at, is what happens when I add or
remove one of my reactants or products.
For part A, I see I'm adding PCl5, and
if I add PCl5 it's going to drive the reaction to the right, because it's
going to respond to that disturbance in the equilibrium of too much PCl5.
By trying to use some of that, and it's going to favor the decomposition process.
The addition of Cl2 will do just the opposite.
When we add a product, we now have too much product,
the reaction is going to shift it, shift in order to accommodate that extra, and
actually will drive this reaction to the left.
The removal of Cl2 will drive the reaction at the right,
because now we're missing some Cl2, so we'll drive that to the right for
both C and D, and then for E if we add PCl3, we're adding a product.
We're going to drove the reaction to the left.
Just as in when we added the Cl2.
Now, we can look at the change in temperature.
We determined that the reaction was endothermic because it was given in
the sys, in the problem.
We see that heat is going to behaving as a reactant.
And as a result, we look at what will happen if I add or remove heat.
Well, if I decrease the temperature, I'm actually removing heat from the system,
and so it looks like I'm removing a reactant.
And when I remove a reactant, I drive the reaction to the left.
So for anytime I'm decreasing the temperature,
I'm going to drive the reaction towards the left side.
Conversely, if I increase the temperature, it's like adding heat to the system.
So if I add heat, I'm going to drive the reaction to the right.
Just as when I added PCl5,
my other reactant, I drive the reaction to the right.
Now, we have to look at how changes in volume affect the equilibrium.
One thing we need to note, is that when we decrease the volume, so
if we see a decrease in volume, we seen an increase in pressure.
And when that happens,
the system is going to respond to try to minimize that increase in pressure.
And one way you can do that, is by minimizing the number of moles of gas.
So, when I increase the pressure, I will see it will go to fewer moles of gas.
For this system, that means going towards the reactants or towards the left.
We have 1 mole of gas on the left side, and we have 2 moles of gas on the right.
When I increase the volume, I'm going to see that it's going to shift to the right.
Because, I'm favoring the more moles of gas.
Because, as I increase the volume, I decrease the pressure, and our system
is going to trying to compensate that by making more moles of gas.
So we're going to favor, in this case, the right side or
favor the decomposition process.
Decreasing volume is to the left, increasing to the right, and
decreasing to the left.
So when I look at my answer choices, what I'm looking for is, which of these sets of
changes are going to favor only the decomposition, or only going to the right?
And where I see that in, is answer choice d.
Removal of CO2 favors the products, increasing the temperature, and
increasing the volume.
So D is our best answer.
[SOUND]
Explore nosso catálogo
Registre-se gratuitamente e obtenha recomendações, atualizações e ofertas personalizadas.
O Coursera proporciona acesso universal à melhor educação do mundo fazendo parcerias com as melhores universidades e organizações para oferecer cursos on-line.
© 2018 Coursera Inc. Todos os direitos reservados.
