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The density of trifluoroacetic acid vapor was determined at $118.1^{\circ} \mathrm{C}$ and 468.5 torr, and found to be 2.784 $\mathrm{g} / \mathrm{L}$ . Calculate $K_{c}$ for the association of the acid.

26.7

Chemistry 102

Chapter 13

Fundamental Equilibrium Concepts

Chemical Equilibrium

Aqueous Equilibria

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all right, this question says the density of try Floro acetic acid vapor was determined at 118.1 degree Celsius and 468.5 tour and found to be 2.784 grams per leader. That asked us to calculate Casey for the association of the asset. It's up here in the top, right is what that looks like. You have to individual try floor acetic acid molecules that coordinate together to form this associated for him. And so throughout this, I would refer to This is one and two, for example, for your pressure from one and pressure from two. That's what these are, and I'm also refer to them is D. The dissociated form and 80 associated for this is a really challenging problem. But I've worked through all the steps, so hopefully it'll make sense the end. So the way we can get Casey is first to find KP because we're given at least one pressure. They're probably easier to find the partial pressure. I'm sorry, the equilibrium constant with respect to pressure first KP and then solve for Casey. So the reaction that equation will be using is K C equals K p, divided by Artie raised to the Delta end where Delta End is the change in the number of moles in the reaction. As you can see here, the products have only one mole of gas. They're calling this coordinated form one molecule but there are two in the reactant ce. So the change of moles will be number in the products one minus the number of most reactions to which equals negative one. So if this were a negative one, you can rewrite. This is K P equals I'm sorry, k c equals K P R t. Okay, so let's tuck that away. We're gonna use that in a moment. So we have to use a number of equations here and kind of work thumb for our benefit. The first is the average density of this solution. If we had some mixture of the dissociated and associative forms, would be the mass of the two added together divided by the volume. That would be the density of the solution. You could rewrite this as the mass of one of those two forms divided by the volume plus the mass of the other form divided by the volume. And again, we don't know how much of either one there is. So we don't know what those masses are quite yet, but there's one more equation we can use, and you might have seen this. This is a slightly different way of writing the ideal gas law. The mass of the gas divided by the volume equals the Moler mass of that gas times the pressure divided by Artie. And this is just by rewriting what and in the PV equals and Artie equation, if you write and as more mass over Norm maciver more mass, you could get that. So this is just a re written form of the ideal gas law. You notice em over. V is what we can use for density here. And so what I've done is I've just taken this m over V, which equals np over Artie and plugged it in for each of these two forms. So we still have density equals the moler mass of gas number one, which is the disassociated trifle or acetic acid times the partial pressure from gas number one this again the dissociated divided by Artie, plus the molar mass of gas to. That's the associative form times the partial pressure of gas to divided by Artie. So again, all I've done is taken this rewritten ideal gas constant are ideal gas law and plugged it into this density equation. And if we now plug or rearrange our terms so that we can just get one of the pressures, we know that the total pressure of the system is the pressure from each of the gases. And so I've rearranged that just by subtracting p two from both sides. So the partial pressure of one of the gases equals the total pressure minus the pressure of the other gas. Straightforward. And also, we know from this equation, they gave us that the associated form should have twice the molar mass, the disassociated form. When you bring those two molecules together, you've doubled the molar mass. So the more massive gas to the associative form equals two times the molar mass of the US disassociated form. Okay, so we're gonna use these two equations. P one equals, um, total pressure minus B two and M two equals two times and one. We're gonna plug those into this big equation, so density equals I pulled out one over r t here em one and I plugged in P t minus p too for P one plus. And I've plugged in to M one for M two times p too. Okay, All I did was plug in terms there. Now I could pull out an M one from both of these. So I have m one over r t. And I have a P T minus p too. Plus to pee too. Said minus one p two plus two p two's gives me with plus one pito. So this is just a simplified version of that. And then I've rearranged it just by, um, uh, dividing by or multiplying by arty over m to get this fracture on the other side. So di rt over m one and then I've just subtracted Petey from both sides. So all this is is re arranging to get p two by itself. So the partial pressure of the second gas that's the associated form equals the density times the ideal gas constant and temperature divided by the molar mass of the first gas, the disassociated form minus the total pressure. All right, so it's just a little bit of outbreak rearranging, and I copied this over here. So again, the partial pressure of the second gas or the associated form equals the density that it gave us times. The ideal gas. Constant times temperature divided by the molar mass of the disassociated form minus the total pressure. I'm just plugging in. What they gave us density was 2.784 grams per leader. Ideal gas constant. Now we have to do it and using the ideal guys constant with units tour because everything will be in tour is 62.36 Leader tour Permal Calvin and the temperature gave us was 118 Celsius converted to Kelvin 391 divided by. And this is the molar mass try floor of acetic acid. The disassociated form. Okay, and it's attracting P two. I'm sorry, Petey. The total pressure, which it told us, was 468.5. With all that said done, you would find that P two equals 126.9 tour and P two, like I said, is the associative form on the right side, the products here. So the partial pressure of the associated form. There's 126.9 tour. Now It's easy to find the partial pressure of the dis associative form because we know that the total pressure minus the pressure of one of the two gives us the pressure of the other one. So if we do 468.5, the total minus what we just solved for the partial pressure of the associative form minus 126.9 tour will leave us with 341.6 tour, which is the partial pressure of the disassociated form over here on the left side. Okay, then we can write our KP equation. K P equals the partial pressure of the products, which is the associated form divided by the partial pressure of the reactions, which is the disassociated form. And that's squared because there are two of them on the left side here. So we just plug in their two numbers 1 26.9 divided by 341.6 squared gives us a K p of 0.1 Now from the beginning, if you remember flashing back real quick, we rearranged Casey by plugging in Delta End and we can say K c equals K p r t So you were in that here. Casey equals K p r t. Now we just plug in all of the terms we have. We know KP. We just saw forthis point. Oh, no. One. We know the ideal gas constant with tour units of 62.36 and we know the temperature was 3 91 Kelvin. If you saw that, you'll find that k C equals 24.38

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