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A reaction is represented by this equation: $2 \mathrm{W}(a q) \rightleftharpoons \mathrm{X}(a q)+2 \mathrm{Y}(a q) \quad K_{c}=5 \times 10^{-4}$(a) Write the mathematical expression for the equilibrium constant.(b) Using concentrations of ?1 M, make up two sets of concentrations that describe a mixture of W, X, and Y at equilibrium.

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Chemistry 102

Chapter 13

Fundamental Equilibrium Concepts

Chemical Equilibrium

Aqueous Equilibria

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this question says the reaction is represented by this equation I've written here to w goes toe X and too high until his equilibrium constant for this equation is five times sending that go forth. And Part eight asks us to write a mathematical expression for the equilibrium constant. So we know that K C equals the concentrations of products of the reactant ce. So right that concentration of X times concentration of why this would be squared because there's a to here. All that is divided by the concentration of reactant CE, which is W that is also squared because it has the two in front. That's part A and B says using concentrations of less than or equal to one Moeller make up two sets of concentrations that describe a mixture of W, X and y at equilibrium. Basically, we have to set this equal to five times 10 negative forth, and we have to pick values of W, X and y that make this true and so the way you a near infinite number of ways you can do this. Uh, I've done it by basically picking a w value that's less than your equal toe one Moller and saying that X and Y are the same. So you could say that this is like X cubed, so x times why square to be the same thing is X cubed if X and wire the same. So what? That looks like it is five times 10 to negative floor equals. Um, I'm just gonna write this as some value X cubed over w squared. And for the first w I just chose one. Moeller squared. So, um, if you were to solve this you divide by so once, where does one divide five times doesn't make it 1/4 Buy one. You get one again and then you take the cube root of five times 10 to the negative four. You see that X equals zero 0.8 more, so that is equal to the concentration of X, which is also equal to the concentration of why and my hypothetical. And I said that one Moeller when the concentration of w Okay, this is just how I did it through again many ways to do this. But I said w equal to one. And then I made excellent. Why the same. So I could just simply take the cube root of that final answer and get X and y you plugged in 0.8 for X and Y here and one for W. Here. This would be true. Let's do another scenario. Square five times 10 negative four for if you can tell equals X cubed. And this time I used a 0.5 Moeller for W that is squared. If you solve for X X is now equal to 0.1 26 Moller And again, that's equal to the concentration. I should rewrite this real quick to apologize. I'm gonna rewrite this as that equals a concentration of X and the concentration. Why not the product of the to make that a little bit more clear. So this equals the concentration of X, which equals concentration. Why? Because I made them the same and I made the concentration 0.5 more for W. This is just one example of how you might do this problem to find sets of concentrations for W x and y that make this equilibrium constant true

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