Yeah. >> Good morning. I'm fighting a little bit of a sore throat today, so my voice gives out all together. >> Maybe glass will end all that early today. >> A stretch to say that my voice, you fell together. >> The rest of the lecture will be delivered an interpretive dance, we're recording, and it just doesn't translate well. >> At the very end, the last class we had got started talking about the subject matter of Chapter five. >> And today we're going to say a little bit more about what thermochemistry is. Matter of fact, we're going to spend a fair amount today introducing some terminology and making sure you're clear on what certain words mean. But hopefully before two days over, you get a better sense of why some of these concepts are important. >> Last time we were hopefully mostly refreshing your memory about some concepts that we presume you would see it at some point before this. Hopefully everybody has an idea of what energy is. >> Heard about. >> Things like kinetic energy and potential energy. Maybe the one new concept that we hit you with last time was the concept of chemical energy, which is just one form of potential energy that is stored within compounds, are within molecules in the form of the chemical bonds that hold the atoms together. And again, hopefully before today is over, we will have fleshed out the ideas of potential energy and chemical energy and a little bit more detail because these will be useful for the rest of our discussion. >> Here, Jeff are flat. We had also talked about jewels and calories, the conversion factor between the two for 0.184 joules. Same thing as one calorie. >> And it is worth noting that when most people think about calories in the context of food, that's not the kind of calories we're talking about here. >> Nutritional calories are kilocalories. >> Thousands of the little C calorie is defined as the amount of heat needed to raise the temperature of one gram of water by one degree Celsius. >> And a reminder that it's going to be temperature changes, they turn out to be important, but more about that later. >> So to begin today, let's talk a little bit more about potential energy. So right now I have a pen sitting on the desktop here, and I'm going to pick it up and hold it up about here somewhere. >> And then of course, when I open my hand and falls, where does Penn have more potential energy? >> Up here or down here? Realize if only for but somebody flip a coin, Yes, neither does have more potential energy in one place to the other. >> No. It's definitely too early for a trick question. Yes, that's right. Because when it's up here and then we let it go, Of course, it starts moving in the course of moving, some of its potential energy changes to kinetic energy. >> So now that it's sitting on the desktop again and as less potential energy than it had when it was up here. >> Because all of its potential energy that had up here got translated to kinetic energy as it made its way down at the table. There are two situations in which the potential energy of an object or a system increases. One of those is if you take two objects that attract each other and move them farther apart, in this case, the two objects that attract each other, or the pin and the earth by virtue of the force of gravity. >> So I take this band and move it farther away from the Earth, its potential energy goes up. >> Conversely, if you have two objects, the repel each other and move them closer together, potential energy goes up. One example that you may have experienced if you're working in two bar magnets and try to force the two north poles together. >> Are the two south poles together, you will feel a repulsive force between those two. >> If you are trying to fight that repulsive force and push those things closer together, you're increasing the potential energy. >> If you take your hands off the bar magnets repel each other, that's when she would decrease. >> Now the significance of this for our purposes, because we are going to be talking about chemical reactions and the energy changes that take place there. Well, let me just refresh your memory on the pen. When the pen is up here, it has a certain amount of potential energy. >> When it's down here, it has less potential energy. >> Why? >> Because some of that potential energy got translated into kinetic energy. >> Well, when chemical reactions take place, some of the potential energy stored within the molecules in the form of chemical energy, can at times be given off to the surroundings, generally in the form of heat. >> The term exothermic refers to a chemical reaction or any other process that gives off heat or gives off energy. >> Exo means out, therm means heat. >> So an exothermic process gives off heat or gives heat out to the surroundings. The opposite of that is an endothermic process. In no means in an endothermic processes consume energy or require you to put in energy or heat to make them happen. >> When a chemical bond forms, that is an exothermic process. Energy is given off when a chemical bond breaks. >> That's an endothermic process. >> Energy is conserved. >> This should in principle, make sense, but let me give you a specific example. And we, I talk too much about chemical bonds at this point, but here's why don't we have talked about ions like sodium ions and chloride ions. >> Okay? >> We also mentioned the fact that there's really no such thing as a molecule of Saul. >> Because what a salt crystal is, is a whole bunch of positive sodium ions, sodium cations, and a whole bunch of negative chloride ions or anions being attracted to each other. Suppose we were to now separate these two. Is that going to happen by itself or do we have to make that happen? You have to make that happen because in general, positive and negative charges attract each other. >> These do you want to be closer together? >> We could separate them if we want to. >> We have to put energy in to do that. >> This is breaking a chemical blah says down here, bringing a chemical bond is an endothermic zooming process because we have to put energy in to make that happen. >> The opposite of that, simply having these things nearby and allowing them to come closer together represents the formation of a chemical bond which is exothermic, gives off potential energy to the surroundings, as we'll see, typically in the form of heat, three-body. Okay, at the moment with the meaning of the word exothermic and endothermic. Here's the reaction that I used to do in class is a chemical demonstration. But I'd given up on that because I got tired of torching tabletops. Its reaction called the thermite reaction. And if the term therm is in there someplace, you know that there's going to be a fair amount of heat involved. It's a fairly simple reaction. It just involves the reaction of >> Elemental aluminum with ferric oxide, which is basically rest to form aluminum oxide and iron. >> But what does this little subscript l mean? >> Yep. >> Now, under normal circumstances, iron is not liquid. >> This reaction gives off so much, IIT is sufficiently exothermic that the iron that is form, forms a little puddle of molten iron. It's very useful for welding, especially if you're welding big pieces of metal together. Like say for example, you were a railroad engineer looking to put down a railroad track and you have to weld two big pieces of metal together around this. And one way to do it, but some of the thermite mixture at the junction of the two pieces of metal you want allow together set off the reaction. >> Little patentable iron forms, very quickly, solidifies and Wells the two pieces of metal. >> Yet what I want you to do for the moment is imagine this reaction as though it is occurring in the following two steps. >> Step one involves basically taking apart the ferric oxide. >> Specifically, we separate the iron and the oxygen into the individual atoms that they're made. From. >> There we take the oxygen atoms that we got from step one and combine. Those are the aluminum to form aluminum oxide. Now, bearing in mind what we said before about the sodium chloride situation, think about step one, exothermic or endothermic? >> Endothermic because what you have at step one is atoms that are attracted to each other, iron, oxygen. And you're separating, Well, like we indicated for the sodium chloride case, separating things like that, that attract each other requires extra energy into the system, which makes it an x, it makes it an endothermic process more. Or the Boy, I hope it's obvious that in step one, we're actually breaking the chemical bonds, the process separating the atoms. So since we break chemical bonds and step one, step one is endothermic or energy consuming. >> How about step two, exothermic working and the Mac exothermic is correct. >> In step two, what do you think? The exact opposite, we are forming chemical bonds between aluminum and oxygen. >> We're taking two things that are attracted to each other and moving them closer it again, that should cause a decrease in potential energy. And the potential energy decreases. >> The energy has to go somewhere. >> Where it goes is do the surroundings in the form of heat. So since we are forming chemical bonds and step two, step two tends to be exothermic or energy releasing. Since we form a little puddle of molten iron up here, is the whole reaction exothermic or endothermic? >> And exothermic. >> If the iron, the form, forms a solid, that everything would be normal. >> But this reaction gives off sufficient amounts of heat so that the higher the forms, forms as a puddle of molten iron. Therefore, the overall reaction must be exothermic. >> Now, the moment we haven't said anything about the amount of energy involved in either step one or step two. >> But if step one is endothermic and step two is exothermic, and the overall reaction is exothermic. Which step, which you think would involve more energy? Step one, step two, step two is correct. >> That makes the assumption that we can add and subtract the amounts of energy involved. >> But that turns out to be a perfectly valid assumption. >> Step two gives off more energy than step one can assumes, which is why the overall process is exothermic. >> More about that in a little bit. >> So the point is some chemical reactions can sue me. >> Some chemical reactions give offering. And the term that's used to describe the amount of heat released or consumed by a chemical reaction is referred to it as the heat of the reaction. In the course of our discussion, whenever you see q written here, Q represents He. >> The subscript r represents the chemical reaction. >> So q subscript R is simply the heat of the reaction. Another term that means fundamentally the same thing is related to a concept that you will see discussed in your textbook and your lecture notes called enthalpy and the symbol for enthalpy as a capital H. >> And the thinking is that maybe what that capital H came from a massively, he makes one wonder why they went to this term. >> But here's the reason why people would do this for a living. When to this term, enthalpy is the potential energy of a system at constant pressure and the significance of a constant pressure thing. And you can read more about this in your textbook, but suffice to say that there's a good many chemical reactions that take place at constant pressure. That is to say they take place over a very short period of time in the course of doing the thermite reaction, which in the demonstrations I used today, would they like maybe five or ten seconds? The atmospheric pressure in this room didn't change very much. >> Atmospheric pressure does tend to fluctuate slightly up and down, except when major >> And whether occur like there's a hurricane that might be thinking about coming up. >> These goes to in the next few days. >> At that point, we might see significant drops in atmospheric pressure of that happens. Yak, you're watching the weather reports can be interesting. But anyway, suffice to say this for right now, most chemical reactions take place sufficiently rapidly, the pressure remains constant. And if that's the case, that simply monitoring the change in potential energy of the system is simply another way of representing the amount of heat produced or consumed by the chemical reaction. This symbol, which is read as delta h subscript bar, is the same thing as Q subscript R. The thing that looks like a triangle is actually the capital Greek letter delta, which means change in scientific circles. So delta H subscript R is the change in enthalpy of a chemical reaction that's equal to q subscript R, the heat of that reaction. In other words, where does the heat of the reaction come from? A change in potential energy or a change in enthalpy as we go from reactants to products. It was arbitrarily decided at some point that Delta H subscript o would be a positive number for endothermic reactions at a negative number for exothermic reactions. So for example, the thermite reaction we were looking at a few minutes ago as the value of delta H, approximately negative 850 kilojoules per mole, the negative side, and this refers to the fact that that's an exothermic reaction. >> Still following this. Now, a few moments ago we asked you to take a look at the thermite reaction and imagine that it took place by way of that two-step process that we outlined. >> And there's a reason we can get away with that because what we really wanted to talk about was enthalpy. >> And enthalpy turns out to be an example of something called a state function. What we mean by a state function is simply any quantity whose current value is independent of any previous value that I've had and also independent on taken to get there. Another way of defining what a state function is, is to say that if x is a state function, then this statement will be true. Delta x. The change in x is simply the final value of x minus the initial value of x. >> Just to give you an example of something that's maybe a little bit more familiar that happens to be a state function. >> Let's consider temperature changes. Suppose we have either a pot of water that we're heating up on a stove or a beaker of water that we're eating up in the laboratory or something like that. And suppose we start out the water at a temperature of 32 degrees Celsius, and we warm it up until it's at 78 degrees Celsius. >> What is the change in temperature of the water? >> How will we figure it out? >> Yeah. >> Well, I have to figure it out as bracketing temperatures. If I do that, I should get 46 degrees Celsius as the temperature junk. And the point is it doesn't matter how I get from 32 to 78 is 32 is the beginning. And yet if I start at 32 degrees Celsius, warm it up to 95 degree Celsius, and then cool it back down to 78 degrees Celsius plus the change in temperature for the overall process. >> Somebody isn't the same thing because the final temperature is del 78 and the initial temperature was below 32. >> If I start at 32 degrees, cool it down to ten degrees, and then warm it up to 78 degrees. >> The change in temperature for the overall process, still boys experience doesn't matter how we get from 30 to 78 as long as they do so eventually. >> So temperature is a state function because d final minus the initial is your change in temperature regardless of the pathway you take to get them. >> Now to give you an example of something that we've talked about recently that is not bleached state function. >> It turns out that work is not a state function. Let me just illustrate that. >> And for the benefit of the viewing audience. >> Hello, my on-camera is the Chair on camera? That's important to demonstrate that work is not a state function. >> What I wanna do is propose to alternative pathways for dealing what I'm about to do, and that is to move this chair from where it is now. >> They. >> So here's pathway for accomplishing that. >> That's pathway here's badly. >> And be as Beth, Wendy, which pathway is more work? >> Be obviously, work is not a state function because work depends on the pathway taken, but divertor change does not. >> So temperature is a state function. >> So temperature turns out to be a state function and so does enthalpy. >> Turns out we want to find out delta H for a chemical reaction. >> We can find out the final and initial values of the enthalpy. >> Well, I have to do is subtract them. But if you think about it, what that means in the context of a chemical reaction, the initial state of a chemical reaction is simply the reactants, and the final state of the chemical reaction is simply the products. In the course of a chemical reaction, the reactants get converted into the products. So if we know the enthalpy of the products are and the enthalpy of the reactants aren't. That turns out, well we have to do is subtract. It. >> Turns out we can just about do more about that in a few minutes. >> But before we do a few more terms, we need to define. When people first started trying to measure things like this, they found out that they got different results depending to some extent on what the weather was like, like what the atmospheric pressure was that day when the temperature was that day, things of that nature. >> And so what they decided to do was to define a set of standard conditions under which they would make these measurements. >> The standard conditions, or a temperature of 25 degrees Celsius and an atmospheric pressure of one standard atmosphere. >> In the kind of units that we use in this country to talk about the weather. >> 77 degrees Fahrenheit, barometric pressure, 29.92 inches of mercury. They're probably pretty close to the standard conditions. >> Yesterday afternoon, it was a relatively nice sunny day. >> About three o'clock yesterday afternoon strictly is battling that. >> So the standard heat of reaction, or the standard enthalpy change of a reaction, is simply the heat of the reaction measured under these standard conditions >> Go look at this symbol right here. >> It looks kind of like the sample we showed you on the previous slide, but there's this little superscript 0, right? The little superscript 0 means that you are measuring under standard conditions. So whenever you see that little superscript 0, it means 25 degrees Celsius, one atmosphere. >> A thermochemical equation is simply an equation. >> Another chemical reaction that also includes the value of the standard heat of reaction. Section 5.3 in your textbook talks a little bit about thermochemical equations. >> We're trying to look that up right now. >> Yep, there it is, age 198 in your textbook for anybody following along their thoughts about what thermochemical equations. >> Ok, but here's a simple example of a thermochemical equation. >> This involves ion reacting with oxygen to form rust ferric oxide. >> And it gives you a value of delta H. >> Zeros of are the standard heat of reaction with history negative 196.5 kilocalories. Exothermic or endothermic. In the back, exothermic counted, you know, because of the negatives, they get a value of delta H exothermic reaction positive value built age, they're having range. Do the words make sense? >> Okay? >> Now there's one other thing I want you to notice about this particular thermochemical equation before I go to the next slide, notice two things. One is we are making one mole of this gum, and the other is that to do so, we are starting from the elements this compound is made from. >> And you'll see in a moment there's a reason I'm pointing out those things. >> The standard state of any substance is the most stable, normal physical form of that substance, be it solid, liquid, or gas, under the standard conditions. So the point is the standard conditions are sort of like a warm spring like day. The standard state of any substance is typically the state in which we normally experience. >> And going back to the previous slide for a moment. >> Higher is normally a silent. Oxygen is normally a gas. Rust, ferric oxide is normally a solid. So in this particular equation, all of the substances are in their standard states. And that is significant for defining this next term is, should be the last term we have to define today. The standard heat of formation of a compound rivers to the heat of reaction delta H zeros subscript r for the reaction in which one mole of that compound is formed from the elements that make it up with all of the suffixes involve being in their standard states. The symbol for standard heat of formation is delta H 0. The subscript f delta means change, h means ethyl being little circle on top means standard condition is 25 degrees Celsius. One, atmospheric pressure, f means formation of a compound from the elements that make it up. The subscript R down here, somebody refers to any chemical reaction, not necessarily one involving making a compound from its elements. >> But when you see that little subscript f down there, that means formation of a compound from the elements that make it up. >> And it turns out that we can look these things up. Again. >> Referring to your textbook, the title of Section 5.6 is standard enthalpies of formation. >> Nice a standard heats of formation. >> It's pretty much the same thing. >> But the real point is there's an appendix in the back of your textbook. It's appendix two, thermodynamic data at one atmosphere at 25 degrees Celsius. And the point is one of the things you can look up in this table is standard heats of formation for a whole bunch of different substances. >> In other words, people have gone ahead and made these measurements. So here is the point behind all of this. >> To me, I guess I'm not going to be able to do it. >> The point behind all of this is something called Hess's Law of each summation. >> Now, omega, when I asked, thinking about the thermite reaction as step one, step two. >> Step one is endothermic. >> Step do is exothermic. >> Whole thing. >> It's exothermic I ask which step involve more energy. Suddenly said to that's direct. >> And I said that that makes an assumption that you can simply add and subtract the amounts of heat involved. >> Turns out that's a good assumption. >> And that assumption is summarize what's called Hess's Law of Heat summation. What Hess's Law says is, if a reaction can be imagined to occur in a series of steps, then delta H for the overall reaction is just the sum of the values of delta H for the individual steps. In other words, one generic way of writing Hess's law, delta H for the reaction equals delta H for step one plus delta H for step two, was delta H for step three, etcetera. Everybody establish a guy. >> By the way, the keyword here is imagined because remember, and enthalpy is a state function, the pathway we take doesn't make any difference. We can imagine the reaction to take place by any pathway that's convenient. >> And here's the significance of that. >> Here is one pathway that we could imagine reaction taking place by way back way that we were talking about Dalton's atomic theory. >> And I used a legos analogy. >> Did read, describe what goes on during a chemical reaction. >> It take the Legos apart and build something else that's kind of like when we did a few moments ago that the thermite reaction, he said, all right, we're going to suggest this two-step pathway. Step one, we take apart the one compound, big individual atoms out of it. >> And then step two, we take some of those atoms along with the atoms of the other elements, put them back together, they can do combat in general, we can imagine any chemical reaction to go from reactants to products by first breaking down the reactants into the elements the made from, and then second, reassembling those elements to form the products. Does the reaction actually do that? >> Maybe, maybe not. >> We don't care because as long as we can imagine that to go by that pathway, if we can define the enthalpy of each step, then Hess's Law says we just add those two and we get delta H for the overall process. >> And it turns out that since we're going through the elements here, we can define delta H for each step in terms of the standard heats of formation that I was just showing you, you can look up in the tables in the back of your book. >> Because by definition, since we are starting from the elements in step two and assembling those to form the products that delta H for that step must simply be the standard heat of formation of the products. >> You look those numbers up for step one. >> If we were doing My elements to the reactance. That would simply be the standard heat of formation of the reactants board up going in the opposite direction. However, we have a way around that problem too. Let me ask you to just speculate for a moment. Suppose we have some re-entry, didn't matter what it is x being converted into y. And suppose the value of delta H for that reaction, a 123 kilojoules. What would you guess would be the value of delta H for the reverse process by which we can vary y into x. >> Say, same thing, except for one thing going to be positive instead of negative. >> In other words, the original reaction is exothermic. >> It gives off a 123 kilojoules to the surroundings. >> You want to do the reverse process, you're going to have to put that energy back again. So the reverse reaction is endothermic positive a 123 kilojoules. >> But the point is, if you reverse reaction, you simply change the sign of delta H. >> The amount of energy is the same. So it turns out that we actually can represent step one of our pathway using standard heats of formation. Because if we were going from the elements to the reactants, it would just be the standard heat of formation of the elements. But we're not going in the opposite direction. >> File that just makes it the negative of the standard heat of formation of the round. >> So Hess's Law says that we can ask, simply add these two things together to get delta H for the overall reaction. >> And since this is positive and this is negative, it's usually written this way. Products minus reactants. >> You simply look up the standard heats of formation in the tables in the back of your book. And add and subtract products minus reactants. >> So the point is, Hess's Law is discussed in your textbook in Chapter five, Section 5.5, where it's introduced. What I want to do for now is just to give you a chance to take a look at a couple of examples of using Hess's law this way. >> So the ways that use that second equation, the one that appears down at the bottom of the slide. >> You need to be able to look up standard heats of formation of your textbook. And of course, you see a problem like this on your exam coming up in two weeks from today, you will of course, be given standard heats of formation. >> We don't expect you to know those off the top of your head, with one notable exception that we'll talk about shortly. >> But first, let's take a look at this example. >> By the way, for those of you following along in the lecture notes, the examples I'm about to show you up here on page 43. >> And here's the first problem that appears on page 43. This is very much like one of the first reactions we look down, although maybe a little bit different because most of the time we think about combining hydrochloric acid and sodium hydroxide, or mixing solutions of those things together to form a salt and water. And in this case, we're talking about taking hydrogen chloride gas, and solid sodium hydroxide and combining them to form solid sodium chloride, liquid water. And the question is, what is the standard heat of reaction for this reaction? Given that the following standard heats of formation, and I would say this 200 tables in your textbook when there actually, because when you look at the tables in your textbook, you'll see that everything there is listed in kilojoules per mole. I got these from another reference source that listed everything in kilocalories per mole. However, it doesn't really matter if you're working in calories or joules as long as you're consistent. If you ever have to convert from calories to joules or vice versa. >> The conversion factor is 4.184 joules is the same thing as one camera. >> So given this information, along with this formulation of Hess's law from the previous slide. >> And we solve this problem, yep, killings by mining products, combining the reactants, and then just subtract it. >> You can do it. Now looking out there, the reaction, let me just move, everybody can see it. >> Which ones are the reactance, which ones are the products I will get is y hat, because otherwise you will get this peg. >> And the products are the things on the right side. >> In this case, the products are sodium fluoride and one of the reactants or the things on the left side, HCl and sodium hydroxide. >> So when we say products minus reactants, we mean right side, left side. >> But yes, you're absolutely allowed to combine these two, combine these two and then subtract this minus that naturally. >> And I did it slightly different, but you're allowed to do that. What I did was to take the negative 98.23, which is the standard heat of formation of sodium chloride. >> To that, add the negative 68.32, which is the standard heat of formation of liquid water. Subtract from that the 22.06, which is the standard heat of formation of HCl, and subtract from that the negative 101.99, which is the standard information of sodium hydroxide. >> But it comes down to the same thing. >> So essentially all you're doing here is adding, subtracting these numbers. Since you know each of them out to the hundreds column, you will allow the hundreds column in your answer. >> Patch this up on a calculator. >> The calculator, lil as usual, lied to you about sig figs. >> Most of the diamond calculators lie to you about sig figs. They give you too many sig figs. >> In this case, the calculator will give you enough patch this up on the calculator will just say negative 42.5, but you're allowed the hundreds column, so you should write this as negative 42.50 and of course, kilocalories per mole. But in terms of how to solve problems like this, it's just adding and subtracting numbers. >> You just have to look up the numbers in the tables in the back of your book or the tables that you'll be provided with. >> Everything makes sense. >> Simple enough. >> Now I'm just going to point this out because I know I'm going to be asked this question and sooner or later, so for the benefit of anybody watching at home, and just to illustrate the point you're just making a few moments ago, let's see who is awake enough to do a fairly straightforward math problem. >> What is eight plus six, minus seven, minus five in the back too, because eight plus six is 14 minus 77 minus five is five. >> What is the quantity eight plus six minus the quantity seven plus five? >> Yeah, also two, because eight plus six is 147 plus five is 1214 minus 12 is two. >> The second way is your weight. The first way is my way. >> Doesn't matter which way you get the same answer. But the reason I point this out is because what I did was to do this plus this, minus this, minus this. >> And I got my answer. >> Most people seem to refer to add these two, then add these two, then subtract. >> That's fine, you get the same answer. >> So it doesn't make any difference when they do it your way. >> Am I going to get the same answer? >> That way we don't expect you to know things like this on the dominant. >> So you will definitely be given means if you take a look at the corresponding problem that you'll see on the practice exam, you should see them. >> However, let's do one more problem because I want to point out a couple of other things about the standard heat of formation values. >> So here's the other problem for paid 43 electrodes. >> This is the chemical. >> It takes place when somebody lights a Bunsen burner in the laboratory, the gas that burns is called methane, with the formula CH4, which reacts with oxygen to form carbon dioxide and water. What is delta H for this reaction? >> Given these values for the standard heats of formation. >> Now couple things to point out here. >> First of all, why is the value for Walker on this slide not the same thing as the value for water from the previous line? >> Yes. >> In the previous example, we were talking about liquid water. In this example we're talking about water vapor. There will be times in the course of doing problems, they've been consulting the tables in the back of your textbook that you might see some compounds listed more than once. And that's because sometimes they're listed more than once because of their physical states. So what you're looking up numbers in the tables in your book, just make sure you're looking up the number that is not just the right compound with the right physical state of that compound for whatever problem you happen to be somebody. >> So that's one thing to pay attention to. The other thing here, alright, you'll notice again he's standard heats of formation for methane, CO2, H2O, not oxygen. The one thing we will ask you to know is the standard heat of formation for oxygen and for that matter, any other element. >> And if you think about what the definition of the standard heat of formation is, that should be obvious. >> So to refresh your memory, the standard heat of formation is the amount of energy involved when one mole of a compound is formed from the elements that make it up. >> By that definition, what must be the standard heat of formation of any element? >> It guesses there tend to be 0. >> And if you think about it, this should make sense because >> Since the definition of standard heat of formation is make the substance in question from the elements that make it up. Well, the only element that makes up oxygen is oxygen. >> How much energy is involved converting O2 and devote to none? What you will see when you look at that table of standard heats of formation in your textbook, is that there are a number of substances listed there for which the standard heat of formation is 0. And what all of those suffixes have in common is that they're all elements in their standard states. So to better answer the question that someone has asked me a few minutes ago. >> Yes, you will get standard heats of formation for compounds because they don't expect you to know those OWN. >> But the standard heat of formation for any element is 0. >> That should be easy enough to remember. >> Ok, so how do we solve this problem given the information that's on this line way back there. >> Okay? Almost basically correct. >> But there is one thing we have to take into account in this case that we didn't have to worry about in the previous case. In the previous example, let me just put the balanced equation back. All of the coefficients worldwide, but in this example, they're not in standard heats of formation are given in terms of kilo calories per kilo, joules per mole. >> So the answer you gave me add up negative 94.05, negative 7.8 zeros, subtract negative 17.89. >> Subtract theorem is almost right, except in this case, you have to multiply by the appropriate coefficients to convert kilocalories per mole. >> And it just kilocalories. >> In other words, you have to multiply the number for water by two. >> Technically f multiply this number by two also, but that number is zeros. >> So here's outweighs out two times the value for water, which is negative 7.80, plus technically one times the value for CO2. They give IT 4.0, five to minus, technically one times the value for methane, negative 17.89 minus technically do dimes 0. >> This one out to the hundreds column C, or allow the answer out to the hundreds column Punch it up on the calculator. Take the five sig figs that your lab. I hope it will not come as a big surprise that the combustion of methane is exothermic, right? >> The whole reason you light a Bunsen burners, you want to heat something else up. So you're doing this reaction that gives off heat. >> So you need somebody else. >> Negative 191.76 kilocalories is the standard enthalpy of this reaction, standard entropy change of this reaction nations. >> But again, it's really just adding and subtracting numbers. Everybody comfortable with the arithmetic? Okay, there are some similar problems to this at the end of chapter five in your textbook, but I don't recommend that you do for additional facts. Let me just briefly say where we're going from here. >> It will turn everybody was, we are going to take more of a look at Hess's law next time, because there are other ways to formulate that. >> But the real question for next time it's going to be, where do these numbers come from? The title of chapter five is thermochemistry. Thermo chemistry is the study of the energy changes that accompany chemical reactions. At this point, what we've seen is that where those energy changes come from is the breaking and forming of chemical bonds. Now the problem we have here is that we can't measure energy directly, but we have something else we can measure directly that we could use given the appropriate conversion factors to say more about entered in that sense, it's sort of like the situation with stoichiometry. Stoichiometry is all about Bowles. Well, we can't measure bowls directly. >> So what do we do? >> We measure grams divide by molecular weights to convert the moles. Or if it's a solution, we multiply by the volume of the solution by its molarity to get moles. In this case, we can't measure energy changes directly, but what we can measure directly, our temperature changes. The study of the energy changes that accompany chemical reactions. By making measurements of the temperature changes that accompany skeptical reactions is full calorimetry. Before too much longer in the laboratory, you're gonna be dealing some calorimetry experiments. >> And fundamentally, there are two types of devices that you can use to do this. >> The constant volume or rather Or the constant pressure or a coffee cup calorimeter. And I'm just going to show you a schematic diagram of each. This is basically when a bomb calorimeter looks like. Don't be put off by the use of the word bomb. It does not explode in the course of the experiment. >> The bomb is simply the vessel inside of which the chemical reaction takes place. >> Notice that it's immersed in a water bath that has a thermometer dipping down into it. So we could measure temperature changes. Then it will bomb calorimeter. It's probably about and $15 thousand. You guys have been doing chemistry for not quite a month, displaying were not about to turn you loose on the $15 thousand piece of apparatus when you guys are gonna get to do instead with works actually better than you might think that he's the $50 apparatus and $14.50 is a thermometer. So rule number one for the calorimeter. Your lab don't break the thermometer. >> But the main point is something as simple as a set of styrofoam cups nested together actually provides the reasonable amount of insulation to be able to make the kinds of measurements that people make in doing experiments like this that's coming up a couple of weeks from now in the lab. >> So what we'll do next time is more about experiments like this and how you can get information of all these things. >> And in addition to that, a little bit more about Hess's law. So I'll stop there for today and we'll resume the conversation Thursday. >> You see my voice? >> Yes. I mean, obviously once you tell me what they really should do some problems reworks, let's face it. If you're just reading stuff to study for these tests is not necessarily, hey, give me the same event something. The question, well, what is IR is normally a solid. That's the only reason I was a little bit here, because this reaction gives off a lot of heat, which tells me I am just OK. Let's think about it this way. We were just talking about like obviously you don't want your hand and that's why this is too hot or too cold. Wind if it feels hot because it's giving off heat, animal liar, either visits getting off, either, actually the reaction is exothermic. >> It happens to be a solid, but when is the amount of heat given off here wasn't enough? >> Another day you might have someone who does what I tell them to question number. Ok? So that by giving you the controller's name and he needs, of course, but there was a person in this class who did that. Well, I have a feeling that this person, because that person waited until the day before the exam starts. That person study where there are resources available to them that would be helpful. >> Person isn't going to get started doing before that. >> We really can predict good things for that person. So obviously one factor to study every day, do a little bit at a time of a bear to the overwhelming amount of stuff to do that age. And you know, again, you are your own example. You're contributing towards making a good, successful team. Is interested in general. We'll paste has been my academic background. Okay? >> Everything in the universe is made of chemicals, including apparel, some substances that are used to make it grow big, exam, but others are synthetic drugs. >> So I think it's useful to know about the properties of those things so that if you're going to mix them together, and to me that's an idea that comes into play. That design was used to a particular colors. Dies or chemicals in there to die fast. So yeah, there's a couple of examples of darpa had a textile design. Obviously, design involves a fair amount of creativity events crucial, but just in terms of being implemented. Design materialism, useful, I think as it could be or in relation with something. Does the name Stephanie will sound familiar to you? Probably every police officer in America should be giving thanks to knowledge is the former. Every plant scientist, recently passed away, who invented caviar horse Jaguars, a very sturdy resistant material because that's the stuff. Bullet-proof jackets, the police web design jackets obviously had to figure out a way to incorporate vocabulary material into voting. A police officer, I guess I'll just mention one other thing besides a few years ago, 100 years in material design project for us, obligation or something you want to make egress out of class. And I said, okay, and point you in the direction of a person in our department who works on Glass Explorer takes here facing all the brass instruments. Obviously, he's the person I know more about glass than anything else. But what is in glass using material? One day she just wanted to address those class completely class. But the point is you design a very nice-looking outfit or percentage of last, last resting place. Dilations, interest. Aig is obviously, people can be creative, but you are limited to the elements to them about what we know about university design attitude as well. So it's good to have some sense of what's possible. Different substances combine, combine properties in terms of what it is you're thinking about. Of course, you do have one of the interesting chemical research that is being created together. Okay. Mm-hm. Yeah. Yeah. Yeah.
Recording from Oct 4, 2016
From Dana Chatellier March 03, 2020
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