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Hi. It's Mr. Andersen and this chemistry essentials video 64. It's on chemical equilibrium. And
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I wanted to start with a demonstration. What I'm doing is moving water from one jar to
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another. And so these will represent reactants and products. And you can see that as I move
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it from the left to the right side I'm using a larger glass to move that water from the
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reactant side to the product side. But as I try to move it back to the product side
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I can't quite fill that all the way up. Now this would take about seven minutes to do
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so I'm going to speed it up about 1000 times. And so you can see that the reactants are
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being converted to products. Some of those products back to reactants. And you might
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say the left side is going to empty out. But I can't quite get all the water out. So what
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happens is it reaches an equilibrium state. Where even though I'm moving water back and
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forth, the amount of water that I'm moving back and forth is equal. So it's reached equilibrium.
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And so in a reversible reaction reactants are converted into products, products into
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reactants and over time it will reach an equilibrium state. And what's going on is that reactants
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and products are moving back and forth at an equal rate. And so we can measure that
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constant as the concentration of products divided by the concentration of the reactants.
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And we learned in the last video that before we reach K we could also calculate Q, which
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is going to be the same thing before we've actually reached equilibrium. And so what
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we can do is we can compare our Q and K values and we can make predications about what's
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going to happen in this reaction. And so an example that I'll use throughout this is the
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Haber Process. That's that process by which we can make ammonia. So we're taking nitrogen
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and hydrogen gas, converting that together and we're making ammonia. And so it's a reversible
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reaction. So if we were to look at the concentrations over time in this process you would get a
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graph that looks like this. And so we have way more reactants at the beginning then we
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have products. In fact for products we have zero at the beginning. So eventually they're
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going to reach a point where they stay the same over time. And so that would be our equilibrium
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point. And so now I'm going to show you a different graph where we're going to look
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at, instead of the concentration of reactants and products, the rate of the reaction, the
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rate forward and the rate backwards. And we would get a graph that looks like that. Now
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why do those two things merge? Remember going back to that demonstration, the amount that
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I was moving from the left to the right in the forward direction equals the amount that
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I'm moving back in the reverse direction. And so we would have reached equilibrium at
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this point. And so if I were to show you how I calculate K and Q values, let's start real
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simple. Let's say we're just looking at the total number of reactants and products, I
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could figure this out at equilibrium. How do I calculate that? I simply take the ratio
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of products to reactants. How many products do I have right here? We'll say that that's
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2. What are reactants? That's 4. And so what's my K value going to be? It's going to be 0.5.
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Now let's move it further back in equilibrium. What's it going to be? Same thing. What's
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it going to be here? Same thing. It's going to be 0.5. In other words that K or equilibrium
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constant remains the same as long as we're at equilibrium. But let's look at what it
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looked like before that. So let's calculate our Q values, which is before we reach equilibrium.
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Let's move it right back here. What is my products at this point? My products at this
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point is around 0.5. What are my reactants? It's going to be around 5. And so I'm going
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to have a value of around 0.1. Or if I move it right here what's my Q value going to be?
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It's going to be 0 because I have 0 products. And so what happens if we ever have a Q value
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lower than our K value? That means it's going to move towards the right. In other words,
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since our Q value is less than K the reaction is going to move to the right. In other words
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we're going to move from reactants to more products. If we're to look at a reaction like
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this, let me show you our K values again, they're going to be exactly the same, because
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we have the same number of products and the same reactants. Could you calculate the Q
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value right here? Well reactants and products are equal. And so our Q value would be 1.
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What about right here? It's going to be 5. And so in this case our Q values is actually
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greater than our K value. So now it's going to move to the left. We're going to move from
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products to more reactants. And so again, if you're given values of Q and K you should
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immediately be able to figure out do you go to the right or do we go to the left. And
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sometimes this is confusing. So just put it on a number line. If I put the K and the Q
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values on a number line, K is where I want to go. Q is where I'm at. Where am I going
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to go? I have to move to the left. So I have to move from products to reactants. Now let's
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actually look at the chemicals that are going on in this reaction. Let's look at the gases
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that are going on in this Haber process. And this would be a typical equilibrium. So in
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this case we're starting with a larger amount of hydrogen gas and no ammonia. And over time
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it's going to reach an equilibrium state where the concentration of all of these stay the
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same. Now we could start with different amounts of each of those. It's also going to reach
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that same kind of an equilibrium state over time. So once the concentrations remain constant
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over time, we've reached equilibrium. And so when I was showing you those first calculations
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I was kind of hiding the difficulty in solving these problems from you by making it very
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very simple. We just divided the products by the reactants. Now when you solve an actual
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K value you're going to have to figure what is that equilibrium content based on the equation
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or based on the chemicals that are in that reversible equation. And so this is what that
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general form looks like to calculate K. And so if we have a reaction where these are our
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two reactants over here where the small a and the small b represent the moles of that
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chemical. This is going to be the form that we write it in. And so if we were to start
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with the products, remember the products were on top, we're going to have the concentration
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of C which is going to be this chemical right here and it's raised to the power of that
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mole, which is going to be raised to the power of c. This is going to be the concentration
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of D raised to the power of small d. And so let me show you an example of that. It might
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be a little bit easier. So let's say this is our equation right here. How could I write
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the equation for my equilibrium constant? Well using this as a model I'm going to put
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my products on the top and my reactants on the bottom. And so it's going to look like
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this. And so this is going to be my nitrogen gas, right here. So the concentration of that
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multiplied times the concentration of water raised to the second power. Why is it raised
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to the second power? Because our mole value is going to be 2 right out in front. Now let's
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look on the reactant side. We're going to have our NO raised to the 2 power, concentration
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of, times the hydrogen raised to the 2 as well. Now if I were to give you a problem
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like this, could you solve that for the Haber process? Could you figure out our equilibrium
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constant? You may want to pause the video and give this a try. But this is the answer
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right here. It's going to look like that. So this is a little bit different than the
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last one. Since we only had one product we're going to only have one thing on the top of
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our K. And so that's going to be the ammonia raised to the second power. Our hydrogen is
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raised to the third power. And then our nitrogen is raised to the first power essentially.
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And so let me show you how to calculate K. These are the molar values at equilibrium.
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Could you figure out K? Well first we start by writing out the reaction. So it's going
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to be ammonia raised to the second power. Nitrogen times hydrogen raised to the third
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power. And then you're simply going to plug those values in. So we're going to put 0.157,
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which is our concentration of the ammonia in the top. We multiply this out and I'm going
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to get a value of 0.602. And so that's going to have 3 significant digits. So that's calculating
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K. You should be able to do that. Now let's look at an ice table. An ice table is going
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to show you, remember, the initial concentration, that's where the i comes from, the change
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and then the equilibrium concentrations at the end. And so if you were given this equation
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right up here of a reversible reaction, you should be able to calculate the Q value. The
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Q value is going to be right here. Now how did I write this out? Again it's water raised
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to the second power because we have two moles right here times nitrogen. This is going to
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be real simple because both of these values are going to be 0. So if I were to plug those
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values in I get 0. And I get a Q value equal to 0. So let's say that that's our initial
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state. Could you calculate our equilibrium state if I give you those molarities at that
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point? Well, same kind of a thing. K value is going to be equal to this. If I plug in
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those values here, and then multiply it out, I get a value of around 650. Two significant
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digits we should have in this one right here. Okay. So once you get your Q and your K, you
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should know immediately which way this reaction is going to go. Is it going to go toward the
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right or is it going to go to the left? Well this pretty easy. Since our Q value is less
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than our K value it's going to move towards the right. We're going to have to convert
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our reactants into products. We could actually figure out the rest of our table. We could
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figure out our change values right here. If this is 1 and this 0.062 we're actually subtracting
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0.038 from that. So I could calculate all my values across right here. My change values.
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And what you'll see is there's an interesting pattern right there. If we were to look at
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all of these values they are essentially the same. We're either subtracting the same amount
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or we're adding the same amount. Why are those the same? It's because the molarities in the
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front are going to be the same as well. And so what we've really discovered here is stoichiometry.
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So if we look at this one we're adding the nitrogen gas. Why is there only half as much
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as this? It's because we only have 1 mole of that. And so stoichiometry is really going
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to guide this second row in an ice table. So this would be one thing you might have
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to do. Calculate the Q value. Calculate the K value. And then figure out what's in the
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middle. But this is a common problem as well. Let's say I only give you my initial state.
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I only give you my Q value. And my K value. And then you have to figure out the rest of
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the table. Well how would you solve that? First things first. You can see that our Q
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value is going to be less than our K value. So is it going to move to the right or to
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the left? It's going to move, we know it's going to move to the right. And so I can actually
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fill these is with variables. And so what I am going to do is I'm going to subtract
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a given amount. We'll call that x. Now why am I putting a 2x here? If we go back and
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look at our equation at the top, there's 2 moles of this. And only 1 mole of these other
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2 gases. And so I know the ratio between these is going to be a 2 to 1 to 1. Why are these
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negative and this positive? It's because we're subtracting the amount of reactants that we
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have. Since that equation is moving from the left to the right we know that this is a negative
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value right here. Okay. What do I do next? I could just figure this out. So I'm going
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to put my 1 minus x. How am I doing that? 1 minus x here and I've got to figure out
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my equilibrium point. Again I don't know what x is, but now since I know what my K value
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is, I know that is equal to 16 and I only have 1 variable, I should be able to solve
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for that. So how would you set that up? First we're going to write the equation for the
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equilibrium constant. So how do I do that? Here's my concentration of HI raised to the
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second power. Because we have two moles up here. And now I'm going to just plug those
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values in. So what's going to go in here? It's simply going to be 2x. So I'm going to
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put it in like that. And then I'm going to put my concentrations on the bottom. Since
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those are both the same thing we could just square it. Now you might freak out at this
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point because this is a 16 equal to, oh man this is going to get ugly. I'm going to have
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to use the quadratic equation. Lot's of times it's much simpler than that. You can see that
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we're squaring the value on the top. And we're squaring the value on the bottom. And so what
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I could do is I could take the square root of both sides. And if I do that I get this.
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I can solve for x as 0.66. And now I just put those x values back in my ice table before.
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So this would be my negative x values right here. And this is going to be my 2x value.
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And so just knowing my equilibrium constant I could figure out the rest of this table,
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what's going on. So can you read graphs like this to figure out, based on the concentration
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of reactants and products, if we've reached equilibrium? And if not, which way do we have
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to go? And finally, can you use data to predict the direction of the reaction? Left to right?
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Right to left? Or have we reached K? Could you calculate K? And then could you determine
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the conditions at equilibrium if I give you K? I hope so. And I hope that was helpful.