This question success to consider this generic reaction where a goes to B and C and asked us to find the equilibrium concentrations of A B and C for three different values of K, C or equilibrium, constantly respect to see where the concentration and asked us to assume that the initial concentration A is one Moller and that there's no product. And, uh, yeah, so let's begin. We can express, um, the equal in concentrations with equations. So let's say that a loses some amount. The change with negative we're gonna lose a and regenerate being C so cane, some not being C. The magnitude of this change depends on the story geometry, this reaction. There is no term in front of these values, so that's a 1 to 1 to one ratio of A to B to C. So if this is negative X, this is positive X and this is positive X and then equilibrium is different between the initial and the change that B one minus X and then gear plus x backs. Your post sex lex Great. And here's the equilibrium. Constant equation. Concentration of beetles, constraints of see no exponents was expected of one that I haven't shown because thes don't have terms in front divided by the concentration of the Greek reactions, which is also has no expletive because we have an expert of one, because there's no term in front of a here so we can unplug in each of these short little equations we generated for the equilibrium, Constance, this would be equal to X times X over one of minus X. Great. And, um, Casey is what's changing in each of these parts. So in a K C is equal to one being seems equal two point 01 and then see Casey is equal to, um, one time some negative fifth. And so we're gonna have to find the equal, even concentrations for these three chemicals under these different equilibrium constants. And just intuitively, since Casey's the concentration of products of a reactant CE as, um, this number gets smaller, the amount of reactant should be increasing. The denominator should be getting bigger because this fraction is getting smaller. We expect that this should go up as we go down here. I'm not gonna work these out, of course, to take a lot of time, but only thing you're changing his Casey here. So if you plug in a one here, you can then get the quadratic formula from this by rearranging some terms. And then you can solve the quadratic for X, and we can plug X back into each of these. So I'll say that for, um, this 1st 1 x equals 0.62 for this next one again. Same exact thing plugged in, point a one. Rearrange your terms, get the quadratic So the a quadratic get X equals 0.0 95 And then for this one, you get X equals 0.3 two and the stress plenty. And Casey here solving for X by solving the quadratic equation. Now let's plug these exes all back into here. I'm going to kind of block this off, and I'm gonna put concentrations for a right here and move down the line. And so if you do one minus 10.62 this would be a 0.38 Well, we're one minus. Pointed on five is about 0.9. Moeller rounding a little bit and one minus. This comes out of your out of just about toe. One does really change like we expected. As Casey gets smaller, this fraction gets smaller. Tom in Arabic. It's bigger. So the concentration of the rectus increase. That's exactly what we see. And then each of these terms is just X was just 0.62 more. This is 0.95 boy, as you expect, if reactant CE are increasing concentration as we decrease Casey, the products are going down. And this is, um, three 0.2 inside raised scientific notation times 10 to the negative third. Well, where and same over here to tend to the negative. Third more so the these values here, the equilibrium concentrations for A B and C given different equilibrium, Constance.