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Phosphorus pentachloride decomposes according to the chemical equationPCI,(g) = PCI;(g) + Cl,(g)Kc = 1.80 at 250 '*CA 0.2600 mol sample of PCl, (g) is injected...

Question

Phosphorus pentachloride decomposes according to the chemical equationPCI,(g) = PCI;(g) + Cl,(g)Kc = 1.80 at 250 '*CA 0.2600 mol sample of PCl, (g) is injected into an empty 3.35 L reaction vessel held at 250 "C. Calculate the concentrations of PCl, (g) and PCI; (g) equilibrium.[PCI,]0.002535[PCI;]

Phosphorus pentachloride decomposes according to the chemical equation PCI,(g) = PCI;(g) + Cl,(g) Kc = 1.80 at 250 '*C A 0.2600 mol sample of PCl, (g) is injected into an empty 3.35 L reaction vessel held at 250 "C. Calculate the concentrations of PCl, (g) and PCI; (g) equilibrium. [PCI,] 0.002535 [PCI;]



Answers

Phosphorus pentachloride decomposes at high temperatures. $$ \mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) $$ An equilibrium mixture at some temperature consists of $3.120 \mathrm{~g} \mathrm{PCl}_{5}, 3.845 \mathrm{~g} \mathrm{PCl}_{3},$ and $1.787 \mathrm{~g} \mathrm{Cl}_{2}$ in a sealed 1.00-L flask. (a) If you add $1.418 \mathrm{~g} \mathrm{Cl}_{2}$ without changing the volume, how will the equilibrium be affected? (b) Calculate the concentrations of all three substances when equilibrium is reestablished.

The problem 17.75. You have this chemical equilibrium here and they're asking you to calculate the partial pressures of all the species present. Actor. You have six months or pcr high place. And a four liter container at 252 degrees Celsius. So since we're looking for partial pressure we have to use the K. P. And perfectly they give you the K. P. As 1.92. So we just plug this into plugging all our information into the icebox. So Um since we're using partial pressures we have to first convert all the information we have here to find the partial pressure of p. c. 0. 5. So to do that, you use the formula um M times are times T where M. Is the concentration which you find by doing six modes divided by four leaders. And then you multiply that by R which is 0.08-1 leaders. Times 80 M over Calvin. And you multiply that by the temperature. And to convert south years into Calvin. You just add 273. So you will get 525 K. And when you calculate that you find that the partial pressure is approximately 64.65. And you just plug that into your icebox here 64.65. Make sure when you do your intermediate calculation to use the ungrounded form. But I'm just gonna write 64.65. Just to make it easier to see. And these will just be zero and the you subtract X. From them Reactant & Add X. two Products. And you will get your equilibrium expression is 64 26 5 minus tex. And these will just be X. Um So you plug in the equilibrium expression to your K. P. Formula which is K. P equals the concentration of the products which is P. C. Of A. Or P. C. 03. Actually these will be partial pressures PCO three times the partial pressure of PCO two over the partial pressure of Pc P. c. 0. 5. And this will be an equilibrium. So you put this in so you get 64.65. Sorry you get X squared over 64 .65 -1. And this will equal the K. P. Which they give you is 1.92. And you just um simplify this proportion and you get that um X squared plus 1.92 X minus 124.1352 equals zero. And the easiest way to solve this is just to plug this into the quadratic formula which is X equals negative B negative 1.92 because of minus square root of B squared -4 A. C. Which is negative 124.1352 and all that over to a. Which is just what and when you do that you get that X equals 10.2 to 3 80 M. And also a negative 12.143. And you just get rid of that because you can't have a negative partial pressure and you push plug this x into your icebox here and you get that you're the partial pressure. Okay? Um P. C. 05. The equilibrium would be around 54 point for a. T. M. And then the partial pressure of PC three, which also equals the partial pressure of cl two, Is around 10 point to ATM. So this will be your answer here.

Okay, so I bring all the information given us for this problem in black. So the first thing that we should do is we should calculate decay for this reaction at equilibrium. So in order to calculate K, we need the concentration of all of you reacted to the products. So you must do that first, right? Concentration is concentration is most divided by leaders, right? We have leaders, but we don't have most. Let's calculate most. So we see that mo's of PCL five is equal to 3.12 gram. Just how much we have divided by the molar mass. Right? So I've figured out the molar mass off all of these, and I've ruined them and read here. So we see that PCL five is to wait grams from Oh, all right, So we do that, we get point 015 most and Alice do most of PCL three. Tomoe of PCL three is equal to 3.845 grams, divided by the molar mass, which is 1 37 and we get plant 0 to 8 and then we go to most of of C L to get 1.787 divided by 71 which is 710.25 Most of those. So we figure out the most now we just got to find the concentration and we So we divide all of these most by the 10 leaders. Right here. We have concentration of PCL fa. Let me do you listen in different color, actually, you have concentration of p C. L five is equal to 0.15 moles divided by 10 leaders. We get playing 015 I m so m is my clarity. It's most of out of my leaders now we have concentration of PCL three is 0.28 most wanted by 10 liters Get put 028 now for C l two. We have 20.25 moles divided by 10 years and we get 100.25 and right, So we figure out all the concentration at equilibrium so we can use these concentrations are equally well to figure out Ok, right, so okay is equal to the concentration of the products over the concentration of the reactant. Right? So we have PCL agree times it by CEO too, over concentration of PCL five. All right, so we modified this the concentration on the same side. But we divide by the other side. So we plug in the concentrations that we found from the low. You see that PCL three is point 0 to 8. Really? That c l two is 20.25 and we see that P C 05 is 0.15 All right, put that into the calculator. Would get K two b go two point. 0 4/7 Now this K we can use to find the answer. So they tell us that we're adding 1.418 grams of seal, too into this equilibrium. So how would the eagle even be affected? Right. Well, if we were to add, we were increased CEO, too. Then the equilibrium was shift to the left. These will have more co two. So this more co two Well, uh, react with PCL dory to form or PCL five. Right? So the equilibrium was shift to the left. So, uh, they said that we add 1.418 grams of C l two and they want us to find the equilibrium concentration after we add it. So what we can do is figure out how many. What is the concentration of 1.418 grams of co? Two were much the same thing that we did in green and blue here. All right, so we have long points for 18 grams. We know that one mole of CO two is 71 grams Pramono, I'm gonna put ad right, and we get the answer as 0.0, to move. Now, let's calculate concentration. Well, we know that concentration. We have add concentration seal to to go to a 0.2 moles over the volume, which is 10 years, and we get 100.2 Right? So what is the total number of C l two? We have Well, we have, actually, I don't need this. Okay, so we add this concentration of C 02 to the already existing concentration. Right? So we know that at equilibrium making me a second I feel like So now we do do This is okay, So I c E o k. So we have 00.2 plus the concentration that we have right here just 0.25 and we get plenty 00 before five. So that is the concentration of CO two in our reaction. All right, so we have total concentration of co two. So we go up, we have we have points. Oh, for five for C E 03 We have 30.28 And for PCL five, we have 50.15 Right. So these are initial concentrations now. We gotta find the change in concentration. Well, we know that as your reaction proceeds, there will be so we know that the reaction shifts toe left. This means that Mawr PCL five will be forms we put plus X. We would from minus expert bolted these since all of these are 1 to 1 ratio in most. Right, So we have 111 Now we write the equilibrium expressions, we have 0.15 plus x, the plane. 0 to 8 minus X. And we have 0.0 for five minus x. This is a five here. Okay, so now we have decay that we calculated, right? And then we have these so we can use eso weaken, stop for accident. So we have from what we have 0.47 which is K. We have the same equally room expression, right? Well, we just have different values for concentration. Yes, we have t 0.0 to H minus X times it by 0.0 for five point. It's x 0.15 plus x eso. When we saw for X, we should get X to be equal 2.0, for eight. Now we use the X and the equilibrium expressions right here to find the equilibrium. Concentration. Right. So we have Let's see, for concentration of PCL five is equal to point a 15 plus X right. We know that X is equal 2.0 for eight. All right, so we get 0.198 as the concentration for PCL five PCL three We get 0.0 to H minus X. We know exes minus X. We know access 0.0 for eight, and we get 0.2 threes and then for CEO too. We have 0.0, 45 minutes X 450.0 for five minus 50.0! Four, eight Oh, it's playing 00 for eight for all of these. But the answers are correct. I 80.0 for eight, and we get point. Oh! 0402 Okay, Yeah. Okay, so these these are answers to this question.

For this question. Let's sell for Mill Aridjis. First we of three point Well, let's do this all at once. Um, polarity of PC of five. Initial. There's going to be 3.1 to 0 grams and then the molar mass. It's going to be too old, 8.5 grams per more, divided by one leader, someone leader flask. And this would give me went 0150 Moeller for the PCO five initial PCL three initial is going to be 3.845 We eat 45 grams on the molar mass. Just going to be 1 37.5 grams. Promote divided by one leader. This would give me point zero to eat zero Moeller and the clarity of seal to initial is 1.77 ground Spolar mouth 71.0 grams per mole. When a leader 787 but about 71.0 gives me 0.0 to 5 to Moller. So and then we're told we're gonna add 1.418 grams of seal too. So seal, too added, is 1.418 grams 71 grams per mole. Also in the one leader. Fosca, listen, Volume that's not changing, but 71. And this would be 0.200 Moeller. So let's set up a rights table here and we'll get PCL five in equilibrium with PCL three and seal too. The initial uh, we just calculated we've got 0.150 0.28 and went 0 to 80 in 0.0 to 5 to. And so we're gonna add 0.200 Moeller. So how the equilibrium be affected too, Right here. Uh, the addition of, um Siegel to causes the equilibrium to shift left. So we're going to see a shift towards the left. The change here would be plus minus and minus. And we're assuming that actually, these are present. Initially, we're told at equilibrium. So let's before we feel it filling the ice table. That's all for my k value here, which would be PCL three seal too, over PCL five, which would be equal to 0.0 to 80 went 0 to 5 to and 0.150 slit self. Right, Librium constant. And we get point 0470 We're gonna use that and will now solve X and calculate the equilibrium constant or the equilibrium values for everything once equilibrium has been reestablished. So here's our expression. Going head in, solving for X we will get a quadratic that you may go ahead and verify. Uh, I'm gonna go ahead and solved, so point we use the same equilibrium. Constant. We're gonna have 0.0 0.0 to 80 minus x and 0.0 45 to minus x. And we have 0.150 glass axe. So and find that the two routes we get our 0.1 or 0.115 and when 00 for 86 and we will have to will reject this route here is that will yield us negative values. And let's go ahead and solve. So at reestablished PCO five at the reestablish equilibrium is 0150 plus explain 00486 and this will be equal to zero point 0199 Moeller PCL three. At equilibrium reestablished equilibrium is 0.0 to 80 minus 800.486 And this gives me 0.231 Moeller and C 02 at the reestablish equilibrium is 0.45 to minus X and we get 0.403 Moeller. So here are the three mil Aridjis once equilibrium has been reestablished.

So the equilibrium reaction is PCL three for cl two is in equilibrium with PCL five and they're all gases. So our Casey is the concentration of the PCL five provided by the concentration of the PCL three Times The Concentration of the Cl two. So the concentration of our of our PCL three, he's gonna be moles over leaders to a point 0148 moles And we're in four L. So that gives us .003 70 more. The concentration of our PCL five is a .0126 moles By that by four L. So .00315 Moller. And then our concentration of our chlorine, This .08 70 moles in four liters. So 0- 18 Mueller. So we'll go ahead and plug those into our K. Expression or Casey expression. So that's going to be our PCL five, which is 00 315. Try to buy our PCL three our chlorine. So that's going to give us a K. C. Of 39 0.1


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