2015 AP Chemistry free response 2a (part 1 of 2)

- [Voiceover] Ethene, C2H4, molar mass of 28.1 grams per mole, may be prepared by the dehydration of ethanol, C2H5OH, molar mass 46.1 grams per mole, using a solid catalyst. A setup for the lab synthesis is shown in the diagram above. The equation for the dehydration reaction is given below. So, we have the ethanol, and then in the presence of a catalyst, we're going to yield, after our reaction, some ethene and some water, and then we have some metrics on the actual reaction. A student added a 0.200 gram sample of ethanol, this is ethanol, C2H5OH, to a test tube, using the setup shown above. So this is the glass wool with ethanol, so the ethanol is right over here. There's a solid catalyst right over there. The student heated the test tube gently with a Bunsen burner until all of the ethanol evaporated, and gas generation stopped. So, here you see the heating, all of the ethanol here evaporates, the gas generation stops. When the reaction stopped, the volume of gas collected was 0.0854 liters at 0.822 atmospheres and 305 Kelvin. The vapor pressure of water at 305 Kelvin is 35.7 torr. So, what you have is, in the presence of the catalyst, you have the ethanol, it's going to react, it's a dehydration, it's a dehydration reaction right over here, so you're going to produce ethene and water, and so, you cool it down, and so the ethene gets captured at the top, it's going to be in a gaseous state. The water, it's going to be water vapor, but then, of course, once you've cooled it down, you're going to have liquid, you're going to have liquid water here as well. So, let's try to answer their questions. Calculate the number of moles of ethene, one, that are actually produced in the experiment and measured in the gas collection tube, and two, that would be produced if the dehydration reaction went to completion. Part b, calculate the percent yield of ethene in the experiment. All right, so let's tackle part one. Let's tackle part one first, and so, somehow we have to figure out the, we have to figure out the actual moles produced in the experiment and measured in the gas collection tube. So, what do they tell us here? They said until all of the ethanol evaporated and gas generation stopped. When the reaction stopped, the volume of gas, so this volume of gas, so this is our volume of gas right over here, that's where our ethene is, our volume of gas was 0.0854 liters, so we underline that, at 0.822 atmospheres, so that's the total pressure there. The temperature is 305 Kelvin, and then they give us the vapor pressure of water is 35.7, at 305 Kelvin, is 35.7 torr. The vapor pressure of water, you could view that as a partial pressure of water, and the reason why that's useful is the partial pressure of water plus the partial pressure of the ethene is going to add up to the total pressure here. And so, we can use that information to figure out the partial pressure of the ethene. So, the partial pressure of the ethene. And if we know the partial pressure of the ethene, well, then we can use the Ideal Gas Law to figure out how many moles of ethene, how many moles of ethene are actually right over here. How can we do that? Well, just a little reminder of the Ideal Gas Law. We have pressure, and if we're talking about one particular thing, here we'd be talking about the partial pressure of ethene, times volume, they give us the volume right over here, is equal to the number of moles times the gas constant times the temperature. Well, we have the pressure, we have-- or, we'll be able to figure it out, we have the volume, we have the gas constant. You don't have to memorize these things. Right over here, I copied and pasted what they give you at the front of the actual AP exam, so that you don't have to memorize these constants and things like that. So, we have the gas constant, and then we have it in different units, depending on which one we want to use, and then, of course, the temperature, they give it to us. It is 305 Kelvin. So, if we want to figure out the number of moles, right over here, you can divide both sides by RT, and so you could say n is equal to, is equal to R, sorry, n is equal to, we divide both sides, if we divide both sides by RT, we're going to get PV, and I'll say the partial pressure of ethene, the partial pressure of ethene, times the volume, times the volume, over, over R times T over R times T. And so, let's figure out everything over here. So, what is going to be the partial pressure of the ethene? Well, the pressure of, the partial pressure of the ethene is going to be equal to the total pressure, the total pressure minus the partial pressure, or you could say the vapor pressure, of the water, the partial pressure of, the partial pressure of the water. So, this is going to be equal to, our total pressure is 0.822 atmospheres, 0.822 atmospheres. Now, what's the pressure of, what's the partial pressure of the water? They tell us the vapor pressure of water at 305 Kelvin is 35.7 torr. Well, they gave us the total pressure in atmospheres, so we have to convert from, we have to convert from torrs to atmospheres if we want to stay in atmospheres. So I'll do it over here. So, the pressure, the pressure of, the partial pressure of the water is equal to 35.7 torr, which is a unit of pressure. But if we want to convert it to atmospheres, once again, you don't have to memorize these things. They give it to you on the test. They tell us, one atmosphere is 760 torr. So, we wanna get rid of the torr. We wanna get rid of the torr, and we want it in terms of atmospheres, so one atmosphere is 760 torr, so this is going to be equal to 35.7 divided by 760. So it's this is the total pressure, minus the partial pressure of the water. Let's see, the partial pressure of the water, let me figure this out. It is going to be 35.7 torr times one atmosphere for every 760 torr, so I'll just divide by 760, divided by 760, is equal to, and I have three significant digits here. So, 0.0470. 0.0470. So this is going to be 0.0470 atmospheres. So, let me write that down. So, the total pressure minus the partial pressure of the water, 0.0470 atmospheres, is going to be equal to, is going to be equal to-- So, let me just take my-- Well, I could just make that a negative, and then add it to the total pressure, so, plus, plus .822 or 0.822 is going to be equal to... So, the partial pressure of our ethene, three significant digits here, is going to be 775, 0.775. So, 0.775. And so now, we can substitute that in. We know everything to figure out the number of moles. N is going to be equal to 0.775. The units here are atmospheres; we can always check to make sure that we get that right, atmospheres. Times our volume. Well, they tell us our volume. It's 0.0854 liters, 0.0854 liters, and then we divide that by RT. Now, which version of R do we use? Well, we're dealing with atmospheres, moles, and Kelvin, so we could use, we could use this right over here. If we had converted everything to torr, we would use this constant. So, our R is going to be 0.08206. They're giving us more significant digits. Doesn't hurt to use them, but later we have to just realize that we're gonna convert to only three significant digits, because that's the minimum, that's what we have. These are three, three, this is more than three signifi-- This is four significant digits. But let's just make sure the units work out. So, this is going to be-- So this is the units here are liters times atmosphere, and so those will cancel with these, divided by, we have to remember this is in, this is in the denominator right over here, divided by moles times Kelvin, and we multiply that times the temperature, which is 305 Kelvin, 305 Kelvin. And we'll see, we could see that, well, this atmospheres is going to cancel with that atmospheres, that liters is going to cancel with that liters, this Kelvin is going to cancel with that Kelvin, and so, in the denominator, you're dividing by moles, so that's gonna be the same thing as just having moles as your units. And so, this is all going to be equal to, we deserve a little bit of a, we deserve a little bit of a drumroll here. This is equal to-- So, 0.775 atmospheres times .0854 liters divided by .08206, so that gets us that, and then we're going to divide by 305. We still have that in the denominator, divided by 305, is equal to, and if we do three significant digits, it's 0.00264. So this is approximately, I could say, 0.00264 264 moles. Did I write that right? I have a bad memory. 0.00264, yeah, I rounded down to 264 moles. And there you have it, that's part one. The number of moles of ethene that are actually produced in the experiment and measured in the gas collection tube.