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Problem 125 Hard Difficulty

In the vapor phase, acetic acid molecules associate to a certain extent to form dimers:
$$2 \mathrm{CH}_{3} \operatorname{COOH}(g) \rightleftharpoons\left(\mathrm{CH}_{3} \mathrm{COOH}\right)_{2}(g)$$ At $51^{\circ} \mathrm{C}$ the pressure of a certain acetic acid vapor system is 0.0342 atm in a 360 -mL flask. The vapor is condensed and neutralized with $13.8 \mathrm{mL}$ of $0.0568 M$ NaOH. (a) Calculate the degree of dissociation $(\alpha)$ of the dimer under these conditions: $$\left(\mathrm{CH}_{3} \mathrm{COOH}\right)_{2} \rightleftharpoons 2 \mathrm{CH}_{3} \mathrm{COOH}$$ (Hint: See Problem 14.1 17 for general procedure.) (b) Calculate the equilibrium constant $K_{P}$ for the reaction in (a).

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Video Transcript

Okay, here we are ready for a super duper challenge problem. And this particular problem, um, is going to just really not your socks off here. So stay with me on this one and will consider if we can get it figured out or not. Okay, The problem. Um, in the vapor phase, acetic acid molecules represented right here, associate. And they form a dime er and a dime. Er is we have to die for two molecules hooked together, connected and were given the following information at 51 degrees Celsius, the pressure of acetic acid vapor system. The pressure is 0.300 point 03 42 atmospheres. And we have a flask with the volume of 360 milliliters, which, of course, we'd have to change leaders. Okay, so there's some information temperature, pressure and volume, and the vapors condensed and neutralized. So the vapor is condensed a neutralized with neutralized with 13.8 mill leaders. Ah, 0.568 Moller sodium hydroxide. And we have two things to dio. We're going to number one, find the degree of find the degree of dissociation. That's the first thing we're going to Dio. And the second thing is to calculate the equilibrium constant In this case, it's gonna give B k p for the reaction. So those other things that were going to dio Okay, so we'll be coming back and forth to these will have to go back and forth a little bit because this one, like it said, is pretty challenging. Now, within the problem, if you're following along in the same text, we have another problem that's referenced, um, for how to set this up for the general procedure. So I'm gonna be stealing some information about that. But let's make a plan. Our first plan. Our first step is to find malls, um, of our any o h. That one super easy. And then we're going Teoh our second thing. We're going to make the assumption that, um, if if all the acid were dime er all dimmer acid find bowls of acid, that's also going to be pretty easy. Um and then we're going to If all the acid is dimmer, find P and we use that doing do using PV nerd and then we're going Teoh use a magical equation and more to follow on that, um, and that will give us to find partial pressure we're going to find, Actually, what going to find here is what we we're told we were looking for was our, um let me see. Here was our degree of dissociation. I didn't spell it right dissociation. As I mentioned before, this is a tough problem, and we're still only on the first part. So we're gonna work on this part right now. Make sure that this is still going and let's begin. Okay. So first again, we're going to find bowls of N A. O. H. And to do this, it's simply enough. We're going to take the, um, polarity times the volume and leaders. So we're taking our mill arat E, which is zero point. What was it? 056? Eight times are leaders. And I think we had 13 point 8 13 eight leaders. So this is going to give me my moles. And we've got seven point 838 will worry about rounding later times. 10 to the minus fourth bulls. And that's in a O h. Step one. Okay, I'll put a little one there in case you want to go back and look. And Step two if all the acid were dime. Er, um let's figure out how many moles of the diamond acid we would have. And for this one, um, that's going to be the moles of any ohh. Which we just calculated in one over two. Because we're using a mole ratio here. So let's substitute, um, for this we would have I'll switch colors for this calculation. 7.838 times, 10 to the minus, fourth bowls divided by two. And we're going to get 3.919 times 10 to the minus fourth malls of and this is for the acid as dimmer. Okay, now the pressure as we've mentioned our third step, we're gonna use PV nerd to solve for pressure so the pressure is equal to end. We saw four moles that we're going to use our molds that we just calculated. I'm gonna go start this down here so that I've got the whole line and is 3.919 times 10 for minus fourth moles. And now we're going to use the atmosphere. The are the universal gas car. The gas constant for atmosphere 0.8 to 1 l A t m overcame all. And then we're gonna take that times the Kelvin temperature, which is our original temperature 51 plus 2 73 is 3 24 Kelvin Divide that by believe it was 0.360 But I'm going to check that. Yep. And solving for P doing our math on this one, we're going to get 0.0 2896 atmospheres. Okay, so now we're going to do our next part of this, and we've got our actual pressure. Are actual given pressure for the problem was given, as let me c 0.342 atmospheres. So that means not all of the asset existed as in the dime. Er, and this is the part where it gets a little funny looking, and I'm just going to give you our little magical equation here. Um, and this I got from another problem that we were referenced with in the resource is that we had available. So the total moles of acetic acid will be one minus a plus to a which, of course, equals one. That's not a I can't read my own writing here one minus Alfa Alfa. So it's gonna be one plus, Alpa. Okay. And then using partial pressures for the last piece of this part before we move on, the pressure that we observed is equal to the pressure that we calculated times one plus Alfa. So we're gonna have zero point. Remember, this number was 0342 atmospheres. That's gonna equal what we just solved for 0.0 2896 atmospheres, times one plus over. Do our magic math here. This will equal 0.181 And this is the degree of dissociation. Okay, so this is all part one. Now. I don't think anybody have room for will. Come back and reference this later. Um, for part two were asked to find KP. And how are we going to do this? Um, the equilibrium constant for this one is going to be KP equals. And then we've got p squared overpay, And this was for R. C. When you write down the formulas for these c h three c o age to that was for this one. And the reason this one squared is that was the one with the coefficient. Okay, so this is what we're finding. And, um, when I was talking about magical math for this part, here's we're going to do some little magical math on this and you're going to have to just trust me on this. I was very confused and I found this equation and I got the right answer. So I'm assuming it's the right way to do this. So if I take, I can substitute into this to ELTA and I'm gonna have to square this whole quantity over one plus a or Alfa Square. This times my observed pressure squared and this will be over one minus Alfa over one plus Alfa Times P observed. When do a little bit of rearranging on this little bit of combining and we will end up with four Alfa squared, he observed. I'm just going to write o B s on that one now and one minus Alfa squared so we can substitute our value in from our previous page, which was this value. So L fa equals 0.181 and we'll substitute that value in, and our final answer will be 4.63 times. Tend to the miners third, and that's our cape. That's our equilibrium. Constant, for this is the super tough problem. I probably couldn't have done it if I hadn't looked a few things up. So good luck if you try to do it. Thanks for listening.

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Raymond Chang, Kenneth A. Goldsby

Chemistry

Stephanie Castillo

Nadia Lara

Stephen Pulliam

Morgan Sizemore

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Chemistry

Stephanie Castillo

Nadia Lara

Stephen Pulliam

Morgan Sizemore

Raymond Chang, Kenneth A. GoldsbyChemistry

Stephanie Castillo

Nadia Lara

Stephen Pulliam

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Raymond Chang, Kenneth A. Goldsby

Chemistry