字幕列表 影片播放 列印英文字幕 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.