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Ch.15 - Chemical Equilibrium

Chapter 15, Problem 91

Carbon monoxide and chlorine gas react to form phosgene: CO(g) + Cl2(g) ⇌ COCl2(g) Kp = 3.10 at 700 K If a reaction mixture initially contains 215 torr of CO and 245 torr of Cl2, what is the mole fraction of COCl2 when equilibrium is reached?

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Hello everyone. So in this video we're taking a look at our ice box, but we're gonna go ahead and apply this icebox to pressures instead of our traditional moles that were usually used to, so let's go ahead and rewrite our equation that's given to us in the problem, which is right over here. So we have our N 204 And then we have our equilibrium arrows and two moles of RNO. two. So of course, to set up the icebox, we need the I. C N E. If you call our I sense for initial we have received for change and then we have our E for equilibrium before I start. Let's go ahead. And also remind ourselves that on the left side we're gonna go ahead and lose reactant since and we're gonna go ahead and gain products. Okay, something to keep in mind when we're doing our filling up this ice table. Okay, so for our initial N204, that's going to be 750 tours For RN. 0. 2, we will have none for initial again, because we are losing reactant, we're gonna go ahead and minus X. Because we don't know how much exactly. And as for our end to we're gonna go ahead and gain the product. So we'll have plus two X. Because we see that too there. Alright, now at equilibrium then there'll be 7 50 minus X. For the equilibrium on the right side is going to be zero minus or zero plus R. Two X. And that's just going to be two X. So now that we have this table then we can go ahead and calculate for our constant value R K p value. The equation for RKP value is going to be the pressure of RN. 0 2 to the second power divided by P L'art N 204. So in the problem they gave us that K. P value. We go ahead and actually use that then. So we have 47. equal into and from the ice table and go ahead and fill out the pressures. So on the numerator will be two X to the second power and then we have 7 50 minus X in the denominator. So simplifying this just a little we get set 47. B. Going to four X squared all over again. 7 50 minus X. We go ahead and scroll down a little bit so we have more room. So sort of go ahead to isolate or just simplifying this here. Then our gets -47.9 X. Equalling 24 X squared. We're gonna go ahead and put everything on one side so we'll get zero equal into four X squared plus 47.9 X minus R. 35,925. And as you can see here then we have something that looks sort of familiar. We can actually go ahead and use the quadratic formula. So if we use our quadratic formula which is available on a calculator, we'll get that the X. value will equal to 89. Tour. So the pressure of N. 02 because you go to two X. That we said. So plugging our X. And then so 89 0. We'll get that. The P. Of N. 02 is equal to 178. Tour. I'll go ahead and start this over here now. So we know that P is proportional to most. So we can simply use the pressures to calculate the more fraction then so mole fraction of R. N. 02 is equal to moles of N. 02 over the total moles Which of course always says it is proportional will be pressure of n. 0. 2 over the total pressure. So plugging in those values then we'll have 178 tour that will just calculate for on top on the bottom. It'll be 178 tour Plus Our 7 50 -89. Alright, so putting that into my calculator, my mole fraction will equal to 0.212. It doesn't need darker colour. So 212. That will be my mole faction. Alright, this is going to be my final answer for this problem. Thank you all so much for watching