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  • Hi. It's Mr. Andersen and this chemistry essentials video 64. It's on chemical equilibrium. And

  • I wanted to start with a demonstration. What I'm doing is moving water from one jar to

  • another. And so these will represent reactants and products. And you can see that as I move

  • it from the left to the right side I'm using a larger glass to move that water from the

  • reactant side to the product side. But as I try to move it back to the product side

  • I can't quite fill that all the way up. Now this would take about seven minutes to do

  • so I'm going to speed it up about 1000 times. And so you can see that the reactants are

  • being converted to products. Some of those products back to reactants. And you might

  • say the left side is going to empty out. But I can't quite get all the water out. So what

  • happens is it reaches an equilibrium state. Where even though I'm moving water back and

  • forth, the amount of water that I'm moving back and forth is equal. So it's reached equilibrium.

  • And so in a reversible reaction reactants are converted into products, products into

  • reactants and over time it will reach an equilibrium state. And what's going on is that reactants

  • and products are moving back and forth at an equal rate. And so we can measure that

  • constant as the concentration of products divided by the concentration of the reactants.

  • And we learned in the last video that before we reach K we could also calculate Q, which

  • is going to be the same thing before we've actually reached equilibrium. And so what

  • we can do is we can compare our Q and K values and we can make predications about what's

  • going to happen in this reaction. And so an example that I'll use throughout this is the

  • Haber Process. That's that process by which we can make ammonia. So we're taking nitrogen

  • and hydrogen gas, converting that together and we're making ammonia. And so it's a reversible

  • reaction. So if we were to look at the concentrations over time in this process you would get a

  • graph that looks like this. And so we have way more reactants at the beginning then we

  • have products. In fact for products we have zero at the beginning. So eventually they're

  • going to reach a point where they stay the same over time. And so that would be our equilibrium

  • point. And so now I'm going to show you a different graph where we're going to look

  • at, instead of the concentration of reactants and products, the rate of the reaction, the

  • rate forward and the rate backwards. And we would get a graph that looks like that. Now

  • why do those two things merge? Remember going back to that demonstration, the amount that

  • I was moving from the left to the right in the forward direction equals the amount that

  • I'm moving back in the reverse direction. And so we would have reached equilibrium at

  • this point. And so if I were to show you how I calculate K and Q values, let's start real

  • simple. Let's say we're just looking at the total number of reactants and products, I

  • could figure this out at equilibrium. How do I calculate that? I simply take the ratio

  • of products to reactants. How many products do I have right here? We'll say that that's

  • 2. What are reactants? That's 4. And so what's my K value going to be? It's going to be 0.5.

  • Now let's move it further back in equilibrium. What's it going to be? Same thing. What's

  • it going to be here? Same thing. It's going to be 0.5. In other words that K or equilibrium

  • constant remains the same as long as we're at equilibrium. But let's look at what it

  • looked like before that. So let's calculate our Q values, which is before we reach equilibrium.

  • Let's move it right back here. What is my products at this point? My products at this

  • point is around 0.5. What are my reactants? It's going to be around 5. And so I'm going

  • to have a value of around 0.1. Or if I move it right here what's my Q value going to be?

  • It's going to be 0 because I have 0 products. And so what happens if we ever have a Q value

  • lower than our K value? That means it's going to move towards the right. In other words,

  • since our Q value is less than K the reaction is going to move to the right. In other words

  • we're going to move from reactants to more products. If we're to look at a reaction like

  • this, let me show you our K values again, they're going to be exactly the same, because

  • we have the same number of products and the same reactants. Could you calculate the Q

  • value right here? Well reactants and products are equal. And so our Q value would be 1.

  • What about right here? It's going to be 5. And so in this case our Q values is actually

  • greater than our K value. So now it's going to move to the left. We're going to move from

  • products to more reactants. And so again, if you're given values of Q and K you should

  • immediately be able to figure out do you go to the right or do we go to the left. And

  • sometimes this is confusing. So just put it on a number line. If I put the K and the Q

  • values on a number line, K is where I want to go. Q is where I'm at. Where am I going

  • to go? I have to move to the left. So I have to move from products to reactants. Now let's

  • actually look at the chemicals that are going on in this reaction. Let's look at the gases

  • that are going on in this Haber process. And this would be a typical equilibrium. So in

  • this case we're starting with a larger amount of hydrogen gas and no ammonia. And over time

  • it's going to reach an equilibrium state where the concentration of all of these stay the

  • same. Now we could start with different amounts of each of those. It's also going to reach

  • that same kind of an equilibrium state over time. So once the concentrations remain constant

  • over time, we've reached equilibrium. And so when I was showing you those first calculations

  • I was kind of hiding the difficulty in solving these problems from you by making it very

  • very simple. We just divided the products by the reactants. Now when you solve an actual

  • K value you're going to have to figure what is that equilibrium content based on the equation

  • or based on the chemicals that are in that reversible equation. And so this is what that

  • general form looks like to calculate K. And so if we have a reaction where these are our

  • two reactants over here where the small a and the small b represent the moles of that

  • chemical. This is going to be the form that we write it in. And so if we were to start

  • with the products, remember the products were on top, we're going to have the concentration

  • of C which is going to be this chemical right here and it's raised to the power of that

  • mole, which is going to be raised to the power of c. This is going to be the concentration

  • of D raised to the power of small d. And so let me show you an example of that. It might

  • be a little bit easier. So let's say this is our equation right here. How could I write

  • the equation for my equilibrium constant? Well using this as a model I'm going to put

  • my products on the top and my reactants on the bottom. And so it's going to look like

  • this. And so this is going to be my nitrogen gas, right here. So the concentration of that

  • multiplied times the concentration of water raised to the second power. Why is it raised

  • to the second power? Because our mole value is going to be 2 right out in front. Now let's

  • look on the reactant side. We're going to have our NO raised to the 2 power, concentration

  • of, times the hydrogen raised to the 2 as well. Now if I were to give you a problem

  • like this, could you solve that for the Haber process? Could you figure out our equilibrium

  • constant? You may want to pause the video and give this a try. But this is the answer

  • right here. It's going to look like that. So this is a little bit different than the

  • last one. Since we only had one product we're going to only have one thing on the top of

  • our K. And so that's going to be the ammonia raised to the second power. Our hydrogen is

  • raised to the third power. And then our nitrogen is raised to the first power essentially.

  • And so let me show you how to calculate K. These are the molar values at equilibrium.

  • Could you figure out K? Well first we start by writing out the reaction. So it's going

  • to be ammonia raised to the second power. Nitrogen times hydrogen raised to the third

  • power. And then you're simply going to plug those values in. So we're going to put 0.157,

  • which is our concentration of the ammonia in the top. We multiply this out and I'm going

  • to get a value of 0.602. And so that's going to have 3 significant digits. So that's calculating

  • K. You should be able to do that. Now let's look at an ice table. An ice table is going

  • to show you, remember, the initial concentration, that's where the i comes from, the change

  • and then the equilibrium concentrations at the end. And so if you were given this equation

  • right up here of a reversible reaction, you should be able to calculate the Q value. The

  • Q value is going to be right here. Now how did I write this out? Again it's water raised

  • to the second power because we have two moles right here times nitrogen. This is going to

  • be real simple because both of these values are going to be 0. So if I were to plug those

  • values in I get 0. And I get a Q value equal to 0. So let's say that that's our initial

  • state. Could you calculate our equilibrium state if I give you those molarities at that

  • point? Well, same kind of a thing. K value is going to be equal to this. If I plug in

  • those values here, and then multiply it out, I get a value of around 650. Two significant

  • digits we should have in this one right here. Okay. So once you get your Q and your K, you

  • should know immediately which way this reaction is going to go. Is it going to go toward the

  • right or is it going to go to the left? Well this pretty easy. Since our Q value is less

  • than our K value it's going to move towards the right. We're going to have to convert

  • our reactants into products. We could actually figure out the rest of our table. We could

  • figure out our change values right here. If this is 1 and this 0.062 we're actually subtracting

  • 0.038 from that. So I could calculate all my values across right here. My change values.

  • And what you'll see is there's an interesting pattern right there. If we were to look at

  • all of these values they are essentially the same. We're either subtracting the same amount

  • or we're adding the same amount. Why are those the same? It's because the molarities in the

  • front are going to be the same as well. And so what we've really discovered here is stoichiometry.

  • So if we look at this one we're adding the nitrogen gas. Why is there only half as much

  • as this? It's because we only have 1 mole of that. And so stoichiometry is really going

  • to guide this second row in an ice table. So this would be one thing you might have

  • to do. Calculate the Q value. Calculate the K value. And then figure out what's in the

  • middle. But this is a common problem as well. Let's say I only give you my initial state.

  • I only give you my Q value. And my K value. And then you have to figure out the rest of

  • the table. Well how would you solve that? First things first. You can see that our Q

  • value is going to be less than our K value. So is it going to move to the right or to

  • the left? It's going to move, we know it's going to move to the right. And so I can actually

  • fill these is with variables. And so what I am going to do is I'm going to subtract

  • a given amount. We'll call that x. Now why am I putting a 2x here? If we go back and

  • look at our equation at the top, there's 2 moles of this. And only 1 mole of these other

  • 2 gases. And so I know the ratio between these is going to be a 2 to 1 to 1. Why are these

  • negative and this positive? It's because we're subtracting the amount of reactants that we

  • have. Since that equation is moving from the left to the right we know that this is a negative

  • value right here. Okay. What do I do next? I could just figure this out. So I'm going

  • to put my 1 minus x. How am I doing that? 1 minus x here and I've got to figure out

  • my equilibrium point. Again I don't know what x is, but now since I know what my K value

  • is, I know that is equal to 16 and I only have 1 variable, I should be able to solve

  • for that. So how would you set that up? First we're going to write the equation for the

  • equilibrium constant. So how do I do that? Here's my concentration of HI raised to the

  • second power. Because we have two moles up here. And now I'm going to just plug those

  • values in. So what's going to go in here? It's simply going to be 2x. So I'm going to

  • put it in like that. And then I'm going to put my concentrations on the bottom. Since

  • those are both the same thing we could just square it. Now you might freak out at this

  • point because this is a 16 equal to, oh man this is going to get ugly. I'm going to have

  • to use the quadratic equation. Lot's of times it's much simpler than that. You can see that

  • we're squaring the value on the top. And we're squaring the value on the bottom. And so what

  • I could do is I could take the square root of both sides. And if I do that I get this.

  • I can solve for x as 0.66. And now I just put those x values back in my ice table before.

  • So this would be my negative x values right here. And this is going to be my 2x value.

  • And so just knowing my equilibrium constant I could figure out the rest of this table,

  • what's going on. So can you read graphs like this to figure out, based on the concentration

  • of reactants and products, if we've reached equilibrium? And if not, which way do we have

  • to go? And finally, can you use data to predict the direction of the reaction? Left to right?

  • Right to left? Or have we reached K? Could you calculate K? And then could you determine

  • the conditions at equilibrium if I give you K? I hope so. And I hope that was helpful.

Hi. It's Mr. Andersen and this chemistry essentials video 64. It's on chemical equilibrium. And

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平衡 (Equilibrium)

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    李掌櫃 に公開 2021 年 01 月 14 日
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