Transcription
In this video, I'll show you how to solve the Alex problem called "Solving Combustion Thermochemistry Problems." There are three different versions of this Alex problem, and in this particular video, I'm going to be showing you how to solve the problem that has to do with burning a substance in a calorimeter. If you are working on the version of the Alex problem that has to do with calculating the number of canisters you need for a wilderness expedition, then look for my video called "Solving Combustion Thermochemistry Problems Part Two." If you are working on this version of the Alex problem, then look for my video Part Three.
So, for this particular problem, we have a substance that is being burned or combusted in a constant pressure calorimeter. The very first thing that I want you to do is write and balance a combustion equation for this particular substance. Whatever substance you have assigned to you—I have C4H6—in the combustion reaction, that substance is going to be reacting with O2, so that's your other reactant. The products of that reaction are going to be CO2 and H2O. You're just going to be entering in whatever molecule you have right here, and then you need to balance this equation.
So, I'm going to balance mine real quick, and let's see. I've got eight carbons, I've got 12 hydrogens, and 22 oxygens (16 plus 6 is 22). It is super important that you get this equation written and correctly balanced. If your equation is not balanced correctly, you will get this problem wrong.
This problem is telling us that I have 5 grams of this particular substance. I'm just going to make a note of that right here: five grams of this unknown. You can see here that I went ahead and looked up the molecular weight of this C4H6. I looked it up online, and I'm just going to tell you that we are going to need to have this number converted into moles at some point in the problem-solving process. So, let's just go ahead and do that gram-to-mole conversion right now, just so that we're ready for it when we need it.
So, we have five grams, and to do that conversion, we want grams down on the bottom and moles up on top. We're using the molecular weight as our conversion factor: one mole is 54.091 grams. Those gram units are going to cancel, and that is going to be five divided by 54.091, and that's 0.09244 moles. I'm just going to add that right here; we're going to need this number of moles later, and it'll be handy to have that calculation done when we need it.
Okay, so this problem is asking, "What is this even asking us?" This problem is asking us to calculate the standard heat or standard enthalpy of formation of this particular C4H6 compound. So, it's asking us to calculate the standard enthalpy of formation of C4H6. Don't be tempted to look the value up in a data table because one of the things that Alex has done is created some funky data here to make these numbers not match up perfectly with what's in a data table. So again, don't be tempted to just go to a data table and look it up because that's not going to work for this particular problem.
The enthalpy of formation—we're not going to be able to calculate it directly, but I'm going to show you the equation that we're going to use to calculate it. One of the things that you have learned is that you can calculate the standard enthalpy of a reaction by using Hess's law and the enthalpies of formations of the products and the reactants: products minus reactants. Hess's law enthalpy of formation is going to look something like this: we're going to take the enthalpies of formations of all of our products, and from that, we're going to subtract the enthalpy of formation of all of our reactants. This is just kind of a fundamental thermodynamics concept that you have probably learned, but it just doesn't seem like it makes sense or doesn't seem like it necessarily belongs in this problem where we're talking about calorimetry.
So, let's take this enthalpy of reaction and expand it and apply it to this particular equation. The enthalpy of the reaction using Hess's law is going to be the enthalpy of formation of the products, and for me, that is—and remember, they're being multiplied by their stoichiometric coefficients—so for me, that is eight times the enthalpy of formation of CO2. For you, it's not necessarily going to be eight; it's going to be whatever your stoichiometric coefficient is. Then to that, I'm going to add six times the enthalpy of formation of water, and again, for you, it might not necessarily be six; you need to look at what your stoichiometric coefficient is. This right here is why this equation needs to be balanced.
Then those are the products, and from that, we're going to subtract the enthalpies of formations of our reactants right here. So, I have 2 times the enthalpy of formation of C4H6, and this is what I'm actually trying to solve in this problem. I also have 11 times the enthalpy of formation of oxygen, so I need to scoot this over and make room for that: plus 11 times the enthalpy of formation of oxygen.
Now, these enthalpy of formations—this right here, this is actually what we're trying to solve for in this problem. This is what Alex wants us to solve for. So, the enthalpy of formation of our other substances, the CO2, the enthalpy of formation of the H2O, and the enthalpy of the formation of the O2, these are things that we're going to look up in Alex's data tab. I already looked them up; I've got them copied right here. The enthalpy of reaction—we are going to calculate this from the calorimetry data in the problem.
The enthalpy of reaction is going to come from all this calorimetry data, and we're going to calculate the enthalpy of reaction using the calorimeter constant pressure calorimeter equation, which is Q = smΔT. This is for the calorimeter for the water in the calorimeter. I'm just going to call it "cal." It's important that you know that we're doing this particular calculation for the water in the calorimeter because we need to use the specific heat for the water, the mass of the water, and the temperature change of the water.
The specific heat of water is a constant: 4.184 J/g°C. That's a constant that I have used so many times I just have it memorized because I've used it so much. But this number you can look up if you don't have it memorized. 4.184—you might see it with units of Kelvin instead of degrees Celsius; it's the same thing mathematically. The mass—there, I need the mass of the water. So, I don't want to use this five grams; that's the mass of my compound X. I want to use the mass of the water, which is 20 kilograms. I need that mass to be in units of grams, not kilograms, so I'm going to take that 20 and then I'm going to multiply it by a thousand so that it is in units of grams, not kilograms.
Last but not least, I need the temperature change of the water. The problem is telling me that the water temperature rises by 2.701 degrees Celsius, so that's nice; we don't have to do that calculation. And we take a look at how our units are going to cancel out here: the grams will cancel, the degrees Celsius are going to cancel, and we are going to be left with units of joules.
Let's carry this calculation out: 4.184 times 20 times 10^3 times 2.701, and this is 226,019.68 joules. Now, as you probably know, these numbers are typically expressed in kilojoules, not joules, so I'm just going to divide by a thousand. I don't need the calculator to help me with that, but I want to have that number saved in my calculator, and I'm really holding on to a lot of significant figures here. That's okay.
Also, you probably know that these numbers are typically—well, actually, I'm not going to say it that way. So, this is the number of kilojoules that are released for the combustion of 5 grams of C4H6 specifically. These numbers, as they are tabulated, they're typically tabulated in kilojoules per mole. So, this number that we calculated is not specific to a mole of C4H6; this number is for 5 grams or 0.09244 moles.
To kind of make that number specific and match up with the way these numbers are expressed, what we're going to do is take the number of moles—this is where we need it. We're going to take that kilojoule, and we're going to divide it by the number of moles that we actually have. This is going to give us a heat or enthalpy in units of kilojoules per mole.
0.09244 gives me a kilojoules per mole heat of 2445.04 kilojoules per mole. And this, again, I'm going to be careful; this is the heat of the water. So, this is the Q for the water, and we're not really concerned about the heat of the water; we're actually really concerned about the heat of the C4H6. So, this number right here is not equal to—it’s directly equal to the enthalpy of reaction, but they are definitely related. All of the heat that is absorbed by the water, this 2445.04 kilojoules per mole, that's being absorbed by the water; that is all energy that is being released by C4H6 in this combustion process.
They are exactly the same in terms of value or magnitude; they are different in terms of sign because this reaction—the temperature is rising. As the temperature is going up, that means that this is an exothermic process, and that means that the reaction has a negative value of ΔH. We calculated a positive value of Q for the water because the water is absorbing heat. So, for water, Q is positive; the water is absorbing heat because the reaction is losing heat, and that means for the reaction, ΔH is negative.
So, what I'm going to do is kind of squeeze this in: Q is 2445, ΔH for the reaction is negative 2445.04 kilojoules per mole. So now where we're at is that we know this number right here, we have the ability to look these numbers up, and we're ready to start the process of solving for this ΔH of formation.
So, I'm going to get this data out of the way, and this next thing that we're going to do is going to get a little bit funky. So, this is for the ΔH of reaction for this reaction right here. We are specifically concerned about the ΔH for C4H6. There are two C4H6 molecules involved in this reaction, so what we need to do for every two of these C4H6 molecules that burn, we're going to basically perform this reaction twice.
So, we're going to take this stoichiometric coefficient—whatever the stoichiometric coefficient is for your compound X—and you are going to multiply it by your ΔH value, and that is an uncomfortable step in this process. So, my ΔH is actually negative 4890.08 kilojoules per mole. All these numbers are like a tongue twister.
So again, what we've done here, just to recap, the way we're using these moles: we calculated the absolute amount of heat that was being generated, we divided it by the number of moles of this particular substance—the exact number of moles of the substance—to make it specific for kilojoules per mole. Then what we're doing is saying that this number right here is based on the quantity of C4H6. When this reaction runs, it uses two C4H6 molecules every time it runs, so we're going to take this ΔH value that we calculated, and we're going to multiply it by 2 to get the exact amount of heat that is being evolved every time this particular reaction is running.
Okay, so now we are ready. I'm going to move this stuff out of the way; we're ready to start plugging some of these numbers in. Oh, I realized I erased my data, and I haven't used it yet. I got to bring my data back. There we go.
Okay, so my ΔH of reaction is negative 4890.1 (I'm just going to say 0.1 and I'm going to leave the units off to try to save space, but my units are kilojoules per mole) and that is equal to 8 times the enthalpy of formation for CO2, which is negative 393.5 kilojoules per mole. Get my brackets on there: six times the enthalpy of formation of water, which is negative 241.8 kilojoules per mole, minus 2 times the enthalpy of formation of C4H6 (my unknown), plus 11 times the enthalpy of formation of oxygen, which is just zero. So, I'm just dropping that whole term right out of this equation; it's complicated enough as it is.
Instead of writing ΔH for C4H6, I'm just going to write X because that is the variable that we're trying to solve for. Now what we're left with is some algebra: negative 4890.1, and then I'm going to do all this math on the right-hand side. I've got 8 times negative 393.5—be really careful with the signs in this problem because I think that if you're going to make a mistake, that's definitely one of the most common mistakes that you're going to make.
Take my answer plus 6 times negative 241.8—just taking my time, making sure that I get all the signs correct—that is negative 4598.8 minus 2X. I keep doing my algebra: negative 4890.1 minus that answer gives me negative 291.3 equals negative 2X. Divide both sides by negative 2, and this gives me an X value, which is the enthalpy of formation of C4H6, of 145.65 kilojoules per mole. Alex wants this to three significant figures, so that's going to be 146 kilojoules per mole.
This is a really frustrating problem to solve because there are so many different things that you have to calculate, and there are so many places when I'm working on this problem I've been getting the problem wrong—really close to correct, like maybe I calculate 146 and the answer is 147, and that's coming from maybe me rounding a little too quickly when I shouldn't be, which is one of the reasons why I held on to so many significant figures for such a long time.
My recommendation to you when you're working on this problem—obviously, because it is such a beast—that you make sure you're giving yourself plenty of time to work on it, but holding on to as many significant figures as you can all the way up until the very end. Be super, super careful with your signs, and this step right here, which is really, really hard, and the process here—I'll write this down—of multiplying it by the stoichiometric coefficient of your unknown substance in order to get the correct ΔH value, making sure that your ΔH value has the correct sign. There are so many places that you could screw this problem up. Be really patient with yourself while you're working on it.