Le Chatelier's principle (video) | Khan Academy
To define chemical equilibrium we need to ask some important questions about reactions: Does a reaction always run its full course so that all the reactants are. The equilibrium constant, Keq, is defined at equilibrium conditions and [B]eq. In a chemical reaction, chemical equilibrium is the state in which both reactants and products are present in concentrations which have no further tendency to.Equilibrium: Crash Course Chemistry #28
How will the total pressure change as the system approaches equilibrium at constant temperature? So, our carbon monoxide is reacting with our hydrogen in a one to two ratio to give us methanol. And this reaction is reversible.
We also know the equilibrium constant for this reaction is And we know that the temperature is staying constant. So, we are going to break this problem up into two parts. In part one, we're gonna try to figure out, using the reaction quotient, whether our system is at equilibrium or not. So, for this reaction, our reaction quotient 'Q' is the product concentration. Divided by the concentration of our hydrogen gas; to the second power, because of that stoichiometrical efficient.
And then also in the denominator, we have our carbon monoxide concentration. We can calculate 'Q' by plugging in the concentrations of these, at this particular moment in time.
And we can calculate the concentrations using the volume of the vessel, which is one liter and the mole quantities. We know that concentration is just moles divided by volume. And since we're dividing everything by one, the initial concentrations will be the same as the number of moles.
So, if we write that out, for carbon monoxide, the initial concentration is 1. For hydrogen, it's 1. Now, if you did the opposite. Let me erase what I just did. Let's say instead of adding heat, you were to take away heat. So let's say you were to take away heat. Let me make sure my cursor's right.
So if you took heat away from the reaction, what will be favored? Well, then you're going to be favoring it in the other direction because there'll be less heat here. I mean, all of this is together. There'll be less heat for this reaction to occur, so this rate will start dominating this rate over here, right?
If you take away heat, the rate of this reaction will slow down, this one will be bigger, and so you'll have more movement of concentration in that direction, or the reverse reaction will be favored. Now, let's think of another stressor-- pressure. Now, imagine that we had-- we mentioned the Haber process before, and this is the reaction for the Haber process.
Nitrogen gas plus 3 moles of hydrogen gas in equilibrium with 2 moles of ammonia gas. Now, what's going to happen if I apply pressure to this system? I'm going to apply pressure. So if you think about what happens with pressure, everything all of a sudden is getting squeezed, although the volume isn't necessarily decreasing, but something is somehow making all the molecules want to be or forcing them to be closer together.
Now, when things are getting closer together, the stress of the pressure could be relieved if we end up with fewer molecules. Think about it this way.
PV is equal to nRT. We learned this multiple times, right? Now, if we increase the pressure, how can we relieve that? Remember, Le Chatelier's principle says that whatever's going to happen is going to relieve the stressor.
The reaction is going to go in the direction that it relieves it. Well, if we lower the number of molecules, then that will relieve the pressure, right?
You'll have fewer things bouncing against each other. So if we lower the number of molecules where you can kind of view it-- I mean, I shouldn't have written it this way, because it's not quite an equation, but I want you to think of it that way. Let me erase this. This probably wasn't the best intuition. If I have a container-- nope, too shocking. If I-- nope, same thing. If I have a container and I'm applying pressure to it, and in one option I could have 2 molecules-- let's say I could have 4 molecules in some volume.
And in another situation, let's say they get merged and I only have 2 molecules, right? In either of these, the reaction can go between these, these 4 could merge to make 2 molecules. Actually, let me use this example up here.
Let's say this nitrogen molecule is this blue one here. Actually, let me do it in a more different color. This brown one right here, it can merge with 3 hydrogen. It could produce this. So this is another way of writing this reaction, maybe in a more visual way. Now, if I'm applying pressure, if I'm applying pressure to this system, so pressure I just imagine is kind of more force per area from every direction, which of these situations is more likely to relieve the situation?
Well, the situation where we have fewer molecules bumping around because it's easier to kind of apply or I guess squeeze them together than when you have more molecules bumping around.
I'm doing this very hand wavy, but I think it gives you the intuition.
Dynamic equilibrium - Chemistry LibreTexts
So if you apply pressure to the system, if pressure goes up, you're applying-- this doesn't mean the pressure goes down. This means pressure is applying to the system. But the pressure is going up, what side of the reaction is going to be favored? The reaction's going to be favoring the side of that has fewer molecules.
Static equilibrium does not necessarily mean there are no moles on the reactant side after the reaction has occurred. This is only an example of static equilibrium. Equilibrium shifts towards one side or the other depending on concentration, temperature, pressure, and volume.
Le Chatelier's principle is not the same as dynamic equilibrium; they are similar but distinct concepts. Le Chatelier's principle describes how equilibrium can change. Dynamic and static equilibrium describe how equilibrium behaves. A reversible reaction is a reaction that can proceed in the direction of products to reactants.
A reaction is at dynamic equilibrium if the rate of the forward reaction is equal to the rate of the reverse reaction. It is at static equilibrium if the reaction has occurred and the forward and reaction rates are both equal to 0. The rate of the reaction is given by: