Consider the following reaction where Kc = 5.10.10-" at 548 KNHCls) : `WHz(g) HCI?)Areaction mixture Was found to contain 334*10-* moles ofNH,Cls) 1.42 10- ,...

Consider the equilibrium $$ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) $$ At $745 \mathrm{~K}$ the equilibrium constant $K_{\mathrm{c}}=50.0 .$ Suppose that $0.75 \mathrm{~mol} \mathrm{HI}(\mathrm{g}), 0.025 \mathrm{~mol} \mathrm{H}_{2}(\mathrm{~g})$, and $0.025 \mathrm{~mol}$ $\mathrm{I}_{2}(\mathrm{~g})$ are placed into a sealed 20.0 - $\mathrm{L}$ flask and heated to $745 \mathrm{~K}$ (a) Is the system at equilibrium? (b) If not, in which direction must the reaction proceed to reach equilibrium? (c) Calculate the equilibrium concentrations of all three substances.

This question says that further reaction shown here, K C equals 0.513 that tells us of a reaction vessel initially contains. And to go for a concentration 0.5 Moeller, what are the equilibrium Concentrations of each of these components. So the two pieces of information we have for the initial concentrated into a four an equilibrium concert with respect to see. So if you want to find out the equilibrium concentrations um, uh, we need to soften equally when concentrations Um, so first thing we wanted d'oh is we want to at least get an equation for equivalent concentration. So we were right. The chains that happens here as negative x. Some concentration will be lost in 204 and some will be gained by the products. But the story geometric ratio between and No. Two and into a four is 2 to 1. So where is the reactant lose minus sechs products will gain two X. We can write these, um, equilibrium is as the initial with the change is the point. 05 minus X and zero plus two X, which is just two X. Now we have an equation to represent equilibrium concentrations, But we need to solve for X. Of course we do. Have Casey given us here, So let's plug in all of these patrol on arrow here. So we're gonna say that 0.5 13 equals construction N o to where say is to x, and that's gonna be squared because of this, um, term in front of n 02 Revived by the concentration of into a world said was point five minus sechs. And so, if you were to expand this out great is the quadratic formula. You would get four x squared plus 0.513 thanks, minus 0.2 five. That's six five equals zero. And if you were to solve this quadratic equation, which I won't do here, you'd get one possible exits positive. And one that's negative. The negative, of course. Can't be it because your concentration isn't gonna be negative. So it's gonna be the positive term. That positive term is X equals 0.3 eat five. Now that we have that, we can play it back into each of these equations. We can figure out that the concentration of N 204 equals 0.5 minus 0.3 eight five. And concentration, uh, two equals, um, two times 0.3 eight five. And these two values are 0.1 one is our 0.0 115 more and 0.7 seven Moeller. So it takes a lot of work to get here. But from just knowing the equilibrium, constant and one of the initial concentrations, we can show that at equilibrium the concentration of into a 4.115 moller and the concentration of 2.77

L. M. You're given this chemical equation here And they say that one more of a. And one will be are placed into a 0.4 liter container. And after equilibrium has established their point to more of C. Is present in that container. So what you want to do is you want to calculate the concentrations and set up an icebox. So we know that the initial concentrations of AMG is 2.5 because you just do one developed by 0.4 and you find that the Equilibrium concentration of C is 0.5. Since the final, you have 0.20 more of c. So you do 0.2 divided by 0.4, which is the container size, which will give you 0.5. So based on this you can basically find that ex the change in concentration would just be 0.5 and you just want to plug that into all you have. So this will be d equilibrium concentration would just be one and then um for D. N. A. It would just be uh To so you just subtract 0.5 and to find the K. You just plug it into the formula products. Overreacting. So since there's a coefficient to in front of you have to do one square, which is just one times 0.5 Over the concentration of reactant, which is two squared, which is just four. So you just do .5 divided by four, which will give you 0.125. So this will be your K. value based on the information given.

Must calculate the Librium constant for this reaction here, starting with 26.9 grams Seiko, let's convert two moles 28 points your one grams of CEO in one more. This will yield 0.9604 moles carbon dioxide. So for the majority of carbon monoxide in this will be the condition. Similarity 0.960 formals over the volume given its 5.19 leaders in this will work out 2.185 Moeller. We then have 2.34 grams of each to convertibles, 2.16 grams of each to for more beach to in this or go to 1.161 Moore's Beach, too. So for the polarity each to and again, this is an initial value. 1.161 Moles departed by 5.19 leaders in this whole workout, too. When 2 to 4 Moeller and we have 8.65 grams of C H 30 h over two moles, 32.4 grams per mole and this war co two 0.2700 moles. So for the movie already and this would be the military D in equilibrium. This will work up to 0.5 to 0 Moeller Moeller using these values that set apart his table. Its CEO guess ad to H two guests. An equilibrium with see Page 30 H. Gas values that we just saw for but 185 22 4 0.5 to 00 there. I mean, a ship this week here minus minus. Plus went +05 to 0 when 05 to use zero and one is your four. Let's work out 2.133 But 1 to 0 I see would be the military c three h about it for the polarity of CEO and each to square, including our values. 05 to 0 over 133 0.1 to 0 squared and working out or equilibrium constant. Three. Sig figs will find that it is equal to 27.2

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.