[BACKGROUND] [NOISE] Good afternoon. Good afternoon. Good afternoon [BACKGROUND]. I saw this on the way in yesterday and couldn't pass it up. For the benefit of anybody who is watching this video at some future time, today is Tuesday, September 18, 2018, and it is pouring rain outside. A little evidence of what was Hurricane Florence now, probably depression in Florence, and the people in this room are soaking wet. Which tells me one of two things. Either you are the people who didn't have enough sense to stay home and watch the YouTube Capture videos later [LAUGHTER], or you are the people who absolutely had to have the answer to the percent yield problem that we were working on last time [LAUGHTER]. Here's the handout for today. If you have questions about the percent yield problem very [inaudible] [BACKGROUND]. It could be worse, it could be in Wilmington, North Carolina [BACKGROUND]. I have a cousin who lives there. He was sensible enough to get out, so he's helped. At the end of last class, we were talking about solutions. Today, we're going to get a little bit more quantitative with our discussion of solutions and the reactions that take place in them, and one concept we had touched on at the end of our last class was the concept of expressing the concentration of a solution using the term molarity. Whenever you see the big M like this, then what that means is the amount of solute in moles present in one liter of the solution that we're talking about. Molarity means moles per liter, whenever you see that M, that should be what you're thinking in terms of. What I want to do today is walk you through a few examples of things that appear in your lecture notes and your textbook about how to use the concept of molarity particularly as it pertains to chemical reactions. For those of you following along in the lecture notes, this problem appears on page 32. Actually what it says on page 32 is how would you make 250.0 milliliters of a 0.100 molar solution of silver nitrate? Well, hopefully, the basic idea is you get yourself a 250-milliliter flask, there are flasks called volumetric flask that you can fill to a particular line, and when you do, you will know that you have exactly 250 milliliters of whatever it is in that flask, and the point is you're going to weigh out some solid silver nitrate, put them in the volumetric flask and some water, swirl it around the dissolve and silver nitrate, then fill up the volumetric flask to the line knowing that you have exactly 250 milliliters in there, put a cap on and shake it up, there is your solution. But the question really comes down to you, how many grams of silver nitrate do you need to weigh out to do this? Well, bearing in mind that that M means moles per liter, how would you suggest we get started solving this problem [BACKGROUND]? Yeah. [inaudible]. Okay. The point is if molarity is moles per liter, amount of substance divided by volume, and if you want to get moles by itself, just multiply both sides by the volume. Volume times molarity is going to be equal to mole quantity of whatever the solute is. We know the volume, 250 milliliters, we know the molarity, 0.100, this should be straightforward. We throw in a factor of 1000 here to convert milliliters to liters, but the point is, if you set it up this way and multiply through, everything cancels out except moles, and so if you punch this up in the calculator, you should get 0.0250 moles of silver nitrate necessary to make this solution. Then since what we really want to know is how many grams of silver nitrate that corresponds to what do we have to do at this point to finish off the problem? Yes? Find the molecular weight. Find the molecular weight or formula weight of silver nitrate and then multiply through by that. You've got a periodic table and you look up the atomic weights of silver, nitrogen, and oxygen and figure this out. It turns out to be a 169.873 grams per mole of silver nitrate. Again, but there's our calculator to be three sig figs are allowed because we only have three sig figs and the concentration originally 4.25 grams of silver nitrate. I'm going to just hammer just a little bit [NOISE]. Hopefully, this is pretty straightforward, but does anybody have any questions about how this works? The basic recipe, in this case, is get a 250-milliliter volumetric flask, weigh out 4.25 grams of silver nitrate, put the silver nitrate in the volumetric flask, add some water, scrawl to dissolve, add more water until you have exactly 250 milliliters, put a cap on it, shake it up [BACKGROUND]. Anybody still writing? Yes. Okay. We'll wait. [NOISE] Everybody good? The next problem I want to show you appears on the same page of lecture notes, page 32 [BACKGROUND]. One concept we were talking about last time is the difference between a dilute solution as opposed to a concentrated solution. Not too surprisingly, perhaps if you want to take a concentrated solution of anything and make it a more dilute solution out of it, all you have to do is add the solvent, which is definitely going to be water. The point is though, as you go from the concentrated solution to the dilute solution, does the amount of solute present in that solution change? No. No, because the only thing you're doing is adding water or whatever your solvent is, the amount of solute remains the same. Bear in mind that when chemists talk about amounts of substances, what they're really referring to is mole quantities. In the course of a dilution process, the number of moles of solute present initially is going to be the same as the number of moles of solute present at the end of the process. But as we just saw on the previous slide, when we're talking about solutions, if we know the volumes and molarities, we can simply multiply the two together to get the number of moles of whatever it is we happen to be talking about. This equation, the initial times M initial, the product of the initial volume, and the initial molarity of your more concentrated solution is equal to V final M final, the volume and molarity of your more dilute solution is very useful in terms of figuring out what sort of recipe or what sort of proportions you need to use to make up a more dilute solution of just about anything. Here's the problem that appears in the lecture notes. Suppose you want to make 100.0 milliliters of a 2.90 molar aqueous solution of hydrochloric acid, HCl. You can purchase concentrated hydrochloric acid, which is 11.6 molar from a chemical company, and the question is, how many milliliters of the more concentrated acid does it take to make the 100 milliliters of the 2.90 molar solution which is what you want? Suggestions for how we might go about solving this problem [BACKGROUND]. Yes. You add [inaudible]. In other words, what this simply boils down to is, this is an equation in four variables. If you know three of them, solving for the fourth should be easy. We do know three of them. The only thing we don't know is what volume we want to use initially in the more concentrated acid. But we know the concentration of the more concentrated acid, we know what volume and molarity we want to have once we're done. [NOISE] We can take the equation as it is and enter the three quantities that we know. M initial 11.6 molar, V final, 100.0 milliliters, M final, 2.90 molar. Then we get V initial by itself, all we do is divide both sides by 11.6. [NOISE] Molarity cancels out. You're left with milliliters. I hope the answer, 25.0 milliliters at least makes intuitive sense. Because if you had to round 11.6 to the nearest whole number, what would it be? Twelve. Twelve. How about 2.9? [OVERLAPPING] Three. Three. That's about a factor of four difference in concentration. Therefore, it shouldn't be too surprising to find that the two volumes, initial and final, are also related by about a factor of four. You can always double-check yourself by estimating if you choose to do that. [NOISE] Start making sense? Wait a moment for people who are finishing writing and we'll move on to other things. [NOISE] Anybody need more time? Before we go on to the next problem, I want to call your attention to something that appears in your textbook for two reasons. One of your classmates emailed me with a question about something and I want to go ahead and answer that question for everybody just so you know, what's in and what's out for your exam coming up soon. The question that somebody sent to me was, in section 4.2 where it says, classifying chemical reactions. Do you need to know the so-called solubility rules in table 4.1? In other words, do you need to be able to simply look at the formula of an ionic compound and figure out whether it's going to dissolve in water or not? The answer is, no. For your first exam, we may revisit this topic a little bit later on in the semester. But for your first exam, you can ignore this table. I was glad that that person called my attention to that table because there's actually a typo in it. This is one thing I'm finding out. [NOISE] I hadn't really explore this textbook very much before this course began and as I mentioned before, this is my first time teaching this course. But I did find a typo in this table. [NOISE] This is the formula for the chromate ion as CrO_3 with a minus 2 charge. That's wrong. The formula for the chromate ion is CrO_4 with a minus 2 charge. But that gives me a chance to talk about something else I wanted to talk about in any way. I will go ahead and do that. [NOISE] This is a more conceptual question, but I don't think the arithmetic should become complicated. Suppose you had solid potassium chromate and you dissolve it in water. What happens when it dissolves? Well, what happens when anything dissolves depends on whether the compound is made of molecules, or made of ions, and at this point we've talked about those combat. When ionic compounds like this one dissolve in water, the individual ions separate from each other. As written, this is obviously not a balanced equation. What do we need to do to balance it? [NOISE] Say again. [inaudible]. [NOISE] Put 2 on the right in front of the potassium ion. Yeah. There are two potassiums in the formula. There should be two potassiums over here. Now just for argument sake, suppose we made up a solution of potassium chromate that was 1.00 molar potassium chromate. What would be the concentration of the potassium ions at the chromate ions in that solution? [NOISE] Chromate ions first. [NOISE] I'll rephrase. If we dissolve one mole of this stuff in enough water to make one liter of solution, we know that it's going to break apart and form ions. What's the concentration of the chromate ion going to be? Somebody said quietly. One mole. Say it again? One mole. One molar, yeah. In other words, the basic principles of stoichiometry apply here also. One mole, one mole, one molar, one molar. What's the concentration of potassium ions going to be? [OVERLAPPING] Two. Two molar. Formula says, two potassium's for every one mole of this. One mole of this dissolves, you get two moles of potassium ions. You're talking about just the total concentration of ions here. [NOISE] That would be three molar in total ions. The terminology that you used here, let me just say that solution is one molar and chromate ions, two molar in potassium ions, and three molar in total ions. We'll file that away for future reference because we will come back and revisit that concept probably later on in the semester also. Concept makes sense at least intuitively? Everybody comfortable? Let's move on. Because there's a section towards the end of chapter 4 in your textbook, the title of which is quantitative chemical analysis and earlier we were talking about combustion analysis problems. That's where your textbook puts the information on combustion problems. But they also put information here about titrations. How many of you have done a titration before? Some. Those of you who haven't, you will get that opportunity to do so in lab before the semester is over. Let's see. They show pictures here about how to read a burette, how to deal with hydration, things like that. I'll have more to say about that a little bit later on and just see if there's any other interesting pictures in here that can show you note that for the moment. But I do want to talk about a titration problem that appears in your lecture notes. Actually before we get to the titration problem, [NOISE] problem on page 33. In fact, let me just read the problem on page 33 to you. Because this is a pretty good summary of why I don't work in the laboratory anymore. A scientists knocks over a volumetric flask, that's me, containing 120 milliliters of a 2.9 molar solution of HCl, spilling the solution on the lab bench. Since HCl is an acid, a base must be used to neutralize it, but the only base available is a 1.97 molar solution of sodium carbonate. What volume in milliliters of the sodium carbonate solution is needed to neutralize the HCl solution given that the balanced equation that you are about to see on screen. Now the point is simply this. It's a stoichiometry problem. So far, in talking about stoichiometry problems, the problem is to get from grams of A, grams of B. Problem is, balanced chemical equations tell us about moles. We typically can't go directly from grams of A to grams of B, but we can do is go from moles of A to moles of B, and we know how to convert between gram quantities and mole quantities. Up here, divide by the molecular weight of A, over here, multiply by the molecular weight of B. Of course, to go from moles to moles we use the coefficients in the balanced equation. But now we are dealing with solutions. By the way, let me just point something out here, which I hope is obvious. What's the little g after CO_2 mean? Gas. [OVERLAPPING] Gas. Little l after H_2O? Liquid. Liquid. From time to time in your textbook and in the lecture notes, you will see little parenthetical notations like that after a chemical formula, to basically just call your attention to the fact that carbon dioxide is a gas, water is a liquid. When you actually carry out this reaction, you'll see fizzy bubbles of carbon dioxide gas being produced. What's the little aq mean here? Aqueous Aqueous solution. If I were reading this equation out loud, I would say two moles of aqueous hydrochloric acid and one mole of aqueous sodium carbonate react to form one mole of gaseous carbon dioxide, one mole of liquid water, and two moles of aqueous sodium chloride. Again from time to time, you'll see things like that just when they want to point out what the physical state of something happens to it. Main point though, it's a stoichiometry problem. [NOISE] It turns out, no surprise. We can solve this problem using basic stoichiometry techniques. Where step 1 is convert something into moles. Step 2 is convert moles of one thing to moles of something else using the coefficients of the balanced equation. Step 3 is to convert moles back to whatever it is we want to know which in this case is a volume of this solution in milliliters. How do we convert from volume of a solution to moles? [NOISE] We we're just doing it five minutes ago, look back in your books. Yes. [inaudible] molarity? Volume of A times molarity of A should give me moles of A. To go in the opposite direction, moles to volume, what do you do? [NOISE] If you multiply by molarity to do this, you divide by molarity to do that. Should hopefully at least make intuitive sense. But if not, you can always set up conversion factors. Step 1, [NOISE] what do we know? We know the volume of HCl that got spilled, and we know its concentration in molarity. I'm dividing by 1,000 here to convert milliliters to liters so that when I'm done with this part of the calculation, I'll have an answer in moles. I find that 0.290 moles of HCl got spilled. By the way, I trust you know, that in the event of an actual acid spill, you'd grab the bottle of base and start neutralizing, you don't grab the calculator and start calculating. But this is a stoichiometry problem, so that's what we're doing. Step 2. What's step 2? Yeah. R multiplied by the number of moles or divided by two moles of HCl. The point is step 2 involves using the mole ratio to convert moles of HCl into moles of sodium carbonate. The mole ratio is 2:1, 2 moles of HCl and 1 mole of sodium carbonate. You're correct, you divide by 2. 0.145 moles of sodium carbonate is going to be needed to neutralize the acid spilled. [NOISE] Finally, step 3. How do we convert that to a volume in milliliters? [NOISE] Yeah? Divide it by the molarity of sodium carbonate. We multiply by molarity in step 1, we're going to divide by molarity in step 3. 0.145 moles of sodium carbonate divided by the molarity, which is 1.97 moles per liter, and multiply by 1,000 to express the final answer in milliliters, 73.6 milliliters of the aqueous sodium carbonate solution is going to be necessary. [NOISE] Are there any questions about any aspect of this stoichiometry problem as it applies to this reaction that takes place in aqueous solution? Couple things to point out. In step 1, we divide it by 1,000, in step 3, we multiplied by 1,000. Obviously in the long run those two factors of 1,000 are going to cancel out. If you omit that part, you're still going to get the right answer. The only reason I did it this way was because I was getting things in moles. If I leave out this part, I get everything in millimoles. If millimoles are fine with you, then feel free to work in millimoles, you'll get the right answer. The other trap that people sometimes fall into with a problem like this one is to say, "Well, we have two molarities and a volume, why can't we solve this the same way we solved the previous problem where we have the equation in four variables and we know the three of them so we plug things in and get the fourth one? That equation was M_initial V_initial equals M_final V_final. Or to reconstitute that for this problem, M_A V_A from the acid equals M_B V_B from the base. Why doesn't that work in this particular case? What aspect of the problem does this equation not take into account? Yeah. Because there's three products. I'm sorry. Just because there's three products. There's three parts? No, three products on [inaudible] Well, there are three products, but that's not really the crucial point. Go ahead. I thought I saw a hand that way that come back down. Taking into account those two HCL [inaudible] Yeah. In other words, the reason I recommend that people go through the usual stoichiometric procedure here, you do have to take the mole ratio and the balance of the equation into account. This equation does not take mole ratios into account, which for a dilution problem is perfectly okay. The point is there are only two circumstances in which you would want to use this equation. One is for a dilution. The other, you can get away with it for chemical reactions in which the mole ratio between the two things you're looking at happens to be a 1:1 mole ratio. Other than that, we strongly recommend that you solve the problem using the typical stoichiometry matter. Step 1, convert something to moles. Step 2, convert moles of one thing to moles of something else. Step 3, convert moles of something else back to whatever it is you would like to. Questions about the concepts, the arithmetic, anything? I do want to get back to that titration problem I'm going to show you, but there's a few people still copying things down, so we'll wait another minute or so. [BACKGROUND] Anybody need more time? [NOISE] Quick piece of terminology before we begin the titration problem, which appears in your lecture notes. [NOISE] Again, this last section in chapter 4 in your textbook is [NOISE] talking about different chemical analysis. For something like this stoichiometric calculation, we're talking about gram quantities, [NOISE] that thing is referred to as gravimetric analysis. All gravimetric analysis means is that what you're actually measuring in a laboratory setting is gram quantities of the materials that you're working with. [NOISE] As opposed to something like a titration, [NOISE] which is called volumetric analysis. Because what you are directly measuring is volumes of solutions and converting them into mole quantities using the process that we just described a few minutes ago. The example [NOISE] of volumetric analysis that I want to show you, is an acid-base titration. Again, you'll be doing this later on in the semester in lab for those of you who haven't already done one. [NOISE] The typical setup for a volumetric analysis looks something like this. [NOISE] There's a better picture in your textbook of what a burette looks [NOISE] like. But the main point is a burette is a long piece of glass tubing with a valve at the bottom called the stopcock, that allows you to either slowly or quickly, dropwise [NOISE] or milliliter at a time, add one solution to another solution. For this particular acid-base titration, we have some unknown acid which we will call HX, sitting in our beaker. To it, we're adding a solution of sodium hydroxide that we have previously prepared and standardized so we know what its concentration is. I'll show you the details in a moment. Now I did have in mind to try to do a little demonstration for you today, but the weather didn't cooperate, so maybe I'll try to do it on Thursday. The point is the success of a titration [NOISE] like this, involves including one other thing. Who knows what that one [NOISE] the other thing is? Yeah. >> A color indicator? >> Yeah. In other words, the question is, how do you know when you're done? The idea in doing something like this is to combine the proper volumes of these two solutions so they exactly react with each other or neutralize each other. [NOISE] But the problem is if your sodium hydroxide [NOISE] solution looks like water and your acid solution looks like water, and when you get done mixing them together, you have a solution that looks like water, how do you know when you're done? The answer is you have to add something in there, typically we call an acid-base indicator, whose purpose is to change color [NOISE] when the solution goes from acidic to basic or vice versa. We'll see if we can put that together for you for Thursday. Anyway, here's the problem that appears in the lecture notes. What is the molecular weight [NOISE] of the acid HX, [NOISE] if a sample of 229 milligrams of HX can be exactly neutralized by titrating it with 29.70 milliliters of a 0.0965 molar aqueous solution of sodium hydroxide? [NOISE] As you probably guess, this is basically a stoichiometry problem. How do we get it's weight? [NOISE] That's step 1 of a typical stoichiometry problem. [NOISE] Yeah. >> Covert the molarity to moles. >> The point is, here's the molarity that we know the sodium hydroxide, and we also know the volume of the sodium hydroxide that we used. One of the nice things about using a burette like this, it allows you to make very precise measurements of volume. You can get this volume to four sig figs here. But the main point is we know that volume times molarity gives you moles. We're going to start here, and multiply by the molarity to get moles of sodium hydroxide. If we do that, [NOISE] 29.70 milliliters divided by 1,000 to get liters. Multiply by 0.0965 moles per liter. Shows us that we use [NOISE] 2.87 times 10^-3 moles of sodium hydroxide in this titration. That's step 1. What's step 2? [NOISE] Yeah. >> Use the mole ratio. >> Use the mole ratio. In this case, what's the mole ratio? One-to-one. The arithmetic for step 2 is pretty easy since it is a one-to-one ratio of acid to base. [NOISE] If 2.87 times 10^-3 moles of sodium hydroxide reacted. Then 2.87 times 10^-3 moles of HX must also have reacted. Now, normally [NOISE] just looking at our flowchart for a moment, we can go from moles [NOISE] to grams by multiplying by the molecular weight. [NOISE] However, we don't know the molecular weight. We're trying to find the molecular weight. Given [NOISE] the information that is currently on the slide, how would you solve this problem? How would you figure out what the molecular weight of HX is? What's the one piece of information we haven't used yet? [BACKGROUND] We haven't used the 229 milligrams. What do we do to find the molecular weight here? Yeah. We do the gram [inaudible] Molecular weight has units of grams per mole. We can easily convert milligrams to grams, here's moles, [NOISE] 229 milligrams divided by 2.87 times 10 to the negative third moles. Throw in a factor of 1000 here to convert milligrams to grams gives us a molecular weight for HX of 79.9 grams per mole. One of the reasons for doing something like this would be to possibly try to identify HX. Specifically, what element X might happen to be based on a knowledge of its molecular weight. Any speculation as to what element X might be? I can show you a periodic table, but I don't know how easy it is to see things on this particular periodic table from where you're sitting. Obviously, while you're doing your exam, you'll have a periodic table available to you that should be easier to read. How might we figure out what element X is? Would anybody care to speculate? I'm sorry? Can you just subtract them [inaudible] Okay. The point is if HX weighs is 79.9, hydrogen weighs 1, basically, so we would subtract that and whatever element X is would have to weigh about 78.9. Then you go to the periodic table and you try to find an element whose atomic weight is about 78.9. It turns out that the element bromine actually has an atomic weight of 79.9, so HX should weigh 80.9, but within the limits of experimental error, that's not too bad. The point is, it certainly makes more sense to suggest that X is bromine as opposed to chlorine or iodine or something whose atomic weight doesn't come anywhere near that. Concepts make sense. Arithmetic makes sense. Any questions about this problem? All right, that being the case, you may draw a line in your notes at this point. This is the end of what you are responsible for, for exam number 1. Which is two weeks and two days from right now. Now, let me caution you against a psychological trap that will be very easy to fall into. We do want you of course to be preparing for your first exam. We don't want you to be ignoring what we're going to be talking about between now and then because we're going to keep going. When we were originally trying to put together the schedule for this semester, we were hoping to get the first exam, not two weeks from now, but one week from now. Didn't work out that way. Part of the problem is simply scheduling classes or scheduling exams that take place outside normal class time. The trap for you guys this semester and again for the benefit of anybody watching the videos at some future time this is happening fall of 2018. Your first exam and your second exam are only three weeks apart, so be preparing for your first exam. Don't be ignoring what we're going to be talking about starting right now, which is fair game for your second exam. [NOISE] The title of Chapter 5 in your textbook, is Thermochemistry. Thermochemistry is a subsection of thermodynamics. We found out on the first day of class that we have quite a few engineering majors in this course. I'm going to ask your indulgence on something, especially if you're an engineering major. Put up with me as I talk about Chapter 5. Then once this course is over, delete memory file. [LAUGHTER] Because in your future engineering classes you're going to be talking to people who actually know things about thermodynamics and you should ignore me. But for the time being, I will give you what little I do understand about thermochemistry and thermodynamics. At the beginning of the semester, we gave you a definition of chemistry that appears in your lecture notes. Part of what it says in that definition of what the study of chemistry involves has to do with the relative amounts of matter and energy involved in chemical reactions. Well, at this point we've beaten relative amounts of matter to death, that's called stoichiometry. Let's talk about energy. What is energy? Can anybody define it, or give me an example? [BACKGROUND] Yeah. The capacity to do work The capacity to do work. Energy is something that an object has if it is able to do work. What's work? Don't tell me what's done by things that have energy. Yeah. The exertion of force. Okay. One definition of work, force times the distance. In other words, if I have my pen sitting here and I push it across the table, I have exerted a force on that pen, I've moved it through a certain distance, therefore, I have done work on the pen. Why I'm able to do work on the pen? Because I have energy. Where do I get my energy from? Because I had breakfast and I had lunch today, but that's another part of the store. Hopefully, this [inaudible] sounds reasonable, everybody okay? All right. What are some types of energy that you know of? Yeah. Potential Potential energy. Yeah. Kinetic energy. Kinetic energy. Let me know others, yes. Thermal? Thermal energy, also known as heat. Okay. Others. Yes. Radiation in the form of light. Radiation in the form of light. Okay, that's fine. This is why I don't make a good chemical engineer, I need one syllable to describe things that chemical engineers need about five syllables to describe. Yes. [OVERLAPPING] God, that's just going to recorded, didn't it? Okay. Mechanical? Mechanical energy, that's actually another term for kinetic energy, but okay. Yes. Potential energy. I think she said that. I'm sorry. Okay. Yeah. Chemical energy. We're going to get to talking about chemical energy sooner or later. That's actually another form of potential energy, but we'll see more about that shortly. Okay, these are all good answers, but let's explore what some of them mean. Kinetic energy, energy that something has if it's in motion. You might recognize this equation for kinetic energy, 1/2 mv squared, where m is the mass of the object in question, and v is the velocity with which it is moving. You can use that equation to calculate kinetic energy if you know an object's mass and its velocity. Potential energy is energy that in some sense or another is stored. Now here's an example, and there are many different kinds of potential energy, but I'll just show you one example here. Here's my pen sitting on the tabletop. Now I pick it up. Does it have more or less potential energy than it did a moment ago? [OVERLAPPING] More. More, because if I let it go, what's going to happen? [OVERLAPPING] It's going to fall. [NOISE] It falls. Some of that extra potential energy just got converted into kinetic energy as it falls down. The equation that is used to calculate that kind of potential energy. Potential energy equals MGH, where M is the mass of the object, G is the force of gravity acting on the object, and H is the height of the object above the table data. However, there are other forms of potential energy, so you can't always use this equation to calculate potential energy. Now we will, in the course of what we're talking about here, talk about chemical energy. Which as we said a few moments ago, is one form of potential energy, stored within compounds in the form of chemical bonds. Now we will say more later on in the course about what the different kinds of chemical bonds are, and how you can think about these things. For the time being, you can just think of chemical bonds as the glue that holds the atoms together to make molecules, or make compounds. But if we think of chemical energy as a form of potential energy, that will turn out to be a useful thing to do. [NOISE] While you're still in that, the Section 5.1 in your textbook is called energy basics, and that's what we're talking about now. It wouldn't surprise me if a lot of these concepts that we're going to mention to the rest of the day, do you sound familiar. If you had any exposure to this thing before at all. Anyone need more time and this line? [NOISE] Everybody ready? You may be aware, there are numerous different units that can be used to measure energy, but I'm looking for a couple of the more common ones. What's one unit that's commonly used to measure energy? Yes. >> Joules. >> Joules, what's another? >> Kilojoules. >> Kilojoules, joules, more or less the same thing factor [inaudible]. Somebody said? >> Calories. >> Calories. Now, let's point out a couple of things here. First of all, can anybody define what is meant by a calorie? Yes, please. In the back. >> Raising the energy of water [inaudible]. >> Basically, correct answer. One calorie, is the amount of heat or if you prefer thermal energy, that's fine, needed to raise the temperature of one gram of water, one milliliter water is the same thing as one gram of water, because density is one gram per milliliter. One calorie is the amount of heat energy needed to raise the temperature of one gram of water. Now the technical nitpicky definition is, from 14.5 degrees Celsius to 15.5 degrees Celsius. But what matters is not the absolute temperatures, what matters is the change in temperature. I hope it's obvious that that's a one-degree change in temperature. When most people think about calories, they think about how many calories of food they ate that day. This kind of calorie, is not to be confused with the nutritional calorie. Sometimes written with a capital C, sometimes written as kilocalories, abbreviated kcal. The point is calorie defined this way or written with a lowercase c. Calorie defined this way, written with a capital C. We will touch on nutritional calories as they come into play here, before we're done with chapter 5. The definition of the joul, goes back to that equation for kinetic energy we showed you a moment ago, 1/2 MV squared. If we measure the mass of the object in kilograms, and its velocity in meters per second, then we can calculate its kinetic energy in joules using that equation. If I take a two-kilogram object, and impart enough energy to it to give it a velocity of one meter per second. I have imparted one joule of kinetic energy to that object. Joules are the metric units of energy derived from other metric units. Since joules and calories are measuring the same thing, there has to be some relationship between them, and here's the relationship between them. The actual definition of the calorie [NOISE] is what fits this equation. One calorie is 4.184 joules. This number should be thought of as an exact number, which means that it has an infinite number of Sig Figs. Which means if you find yourself doing a calculation using that number, that's not going to be the limiting number of Sig Figs in the calculation. Have you already heard of joules and calories before? >> Yes [OVERLAPPING] >> Okay. We'll be using them as we discuss thermochemistry in Chapter five. [NOISE] We want more time with this one. All right. Going back to the concept of potential energy for a moment, there are basically two situations in which the potential energy of a system increases. We already showed you one of them. Pen sitting on the tabletop, then I pick up the pen. It has more potential energy up here than it does down there. Just in general, if you take two objects that are attracted to each other and you move them further apart, the potential energy increases. The two objects in this case are the pen and the planet. Move them farther apart , higher potential energy. Conversely, if you take two objects that repel each other and bring them closer together, the potential energy increases. It didn't happen to bring any bar magnets with me, but have you ever done that thing where you take two bar magnets and try to push the two north poles closer together? You feel revulsion? Yep. Or for that matter, if my fist represents two protons or two electrons, two protons positively charged, repel each other. Push them closer together, potential energy goes up. Make sense intuitively? Terminology that you may have heard before. Exothermic and endothermic. Some reactions are exothermic, that means that they give off heat. Some reactions are endothermic, that means they consume heat. Prefix exo means out, prefix endo means in. We talking about both kinds of reactions as we go. Now, just using the information that's on the screen and thinking about trying to either form or break a chemical bond between two atoms that are attracted to each other, let's take a fairly straightforward gaze like say, an ionic bond between a positive sodium ion and a negative chloride ion. If you bring together a sodium ion and a chloride ion and form that ionic bond, is that going to be an exothermic or endothermic process? You're dying to raise your hand, ain't you? If you've got the chance take a shot. Endothermic. Nop, sorry. [LAUGHTER] Give up tales. Let's think about it this way. The pan is the sodium ion. The planner is the chloride ion. Started, let them come closer together. Lower potential energy, right? Now in this case, that became kinetic energy. But when you deal with the ions, it's given off in the form of heat. Forming a bond is an exothermic process. By contrast, if you want to take two things that are attracted to each other and move them farther apart, you have to put energy in to do that. I have to do work on the pen to make that happen. That means I'm putting some of my energy into separating these things. Turns out that in general, forming a chemical bond is exothermic, breaking a chemical bond is endothermic. In general, if you want to separate atoms that are connected to each other somehow, you have to put energy in. If the atoms are attracted to each other in the first place and you want to let them move closer together and form a chemical bond, you get energy out. Terminology makes sense? Concepts make sense? We'll take a look at one more slide, I just want to make sure these concepts make sense to you. But go ahead and finish jotting this down first. Do we have this information, because you will need it for the next slide? Let's talk about a reaction called the thermite reaction. I used to do this as an in-class demonstration. I got tired of torching tabletops so I don't do it anymore. You can all stop. [LAUGHTER] You can look up a video and see what it's like. It's not the same. I know it's not the same. Yeah. I just found out what a bunch of little creatures you are. [LAUGHTER]. We'll summarize it this way. The reaction itself is very straightforward. Involves metallic aluminum reacting with ferric oxide to form aluminum oxide and iron. However, the reason it's called the thermite reaction is that this gives off a lot of heat, so much so that the iron that's formed is formed usually as a puddle of molt iron. One application of this reaction is for welding big pieces of metal together. Say, for example, you were putting down railroad track and you wanted to weld together two big pieces of metal for the rails. Put some of the thermite mixture at the juncture of the two pieces of metal, set off the reaction, puddle of molt iron quickly solidifies, melted up weld the two pieces of metal together. What I want you to do for right now, is just imagine this reaction as though it occurs in the following two steps. Remember, Dalton said basically all that happens during a chemical reaction is you rearrange the atoms to make new compounds. Step 1, we take apart the ferric oxide. Now we have iron atoms, oxygen atoms. Step 2, we take the oxygen atoms from step one and combine them with the aluminum to make aluminum oxide. Now, based on what we were saying on the previous slide, step 1, exothermic or endothermic? Endothermic. Endothermic. Because to go from the compound to the individual atoms, you have to break chemical bonds. Therefore, step 1 is going to be endothermic or energy consuming. How about step 2? Exothermic. Exothermic. Because now you're taking individual atoms, bringing them together making a compound. You're making bonds in that process. Therefore, step 2 must be exothermic. Is the overall reaction exothermic or endothermic? Bear in mind that we form molten iron in the process. I'll rephrase the question. Would you want to put your hand into the middle of this reaction? [OVERLAPPING] No. No. Why? Because it's too hot or too cold? [OVERLAPPING] Too hot. If it feels hot to your touch, is it giving off heat or consuming heat? [OVERLAPPING] Giving off heat. Giving off heat. We call that which of these two terms? [OVERLAPPING] Exothermic? Exothermic. If the overall reaction is exothermic, which step involves more energy, step 1 or step 2? [OVERLAPPING] Step 2. Step 2. The assumption, which turns out to be a valid assumption, is that whatever number we associate with this, whatever number we associate with that, if we take the sum there, we can figure out that this must be giving off more heat than this is consuming for the overall process to be exothermic. Makes sense? [NOISE] We will resume here on Thursday, plus I will try to put together that little indicator, demonstrate that for you. I'm not thermite. Sorry. [BACKGROUND] [NOISE] Are you not going to do this experiment with us? No, sorry. [inaudible] it. [BACKGROUND] Why [inaudible] combust during this experiment? Would you first like combustion during this experiment? Yes. Depends on how close you get to it All right. I don't need safety equipment to do the experiment, right? [BACKGROUND] You need safety equipment to do everything. [BACKGROUND] You can't tell just by looking at that. What you can tell is look up there saying, the iron is giving off [inaudible] the reaction must be giving off a lot of heat to make that happen. Then once you know there is [inaudible] then you should be able to say that the amount of heat involved in step 2 must be greater than the amount of heat involved in step 1. We only know it because it's [OVERLAPPING] [inaudible] Okay. We have a question. Yeah. What are your office hours? Wednesdays and Fridays, 9:15-11:00, and if that doesn't work for you, then send me an email [inaudible] [inaudible] Okay. So Monday to Friday? Wednesdays and Fridays. Wednesday and Friday. 9:15-11:00. All right. I'll [inaudible] Run by at 2:33. Feel free to drop by. All right. I'll be there on Friday. Thank you. Thank you. [BACKGROUND] [inaudible] Tell you what, right now all we've told you is the bond forming is exothermic but breaking is endothermic. Yeah. We'll be discussing those concepts in more detail on Thursday. Okay. I'm hoping by the end of Thursday this class will make more sense. Okay. Thank you. [BACKGROUND] [inaudible] Workshops are available. [NOISE] Here's what Dr. [inaudible] has assigned me today. It doesn't change any of the meeting times or locations or anything like that. But since [inaudible] perhaps, most of my students and they really into workshops do not complain. What he's done is to take his crew, which is required to do workshops and split them up. So then instead of having like 25 people in one room, it's going to be more like 12 and 13, which is good because that way if my guys do the second level workshop [OVERLAPPING] So [inaudible] to follow. You can just follow the instructions and do the work [inaudible] and see what [inaudible] [BACKGROUND] Thank you. Professor. Yeah. I have a question. [inaudible] that we [inaudible] 1, 2, 3. Okay. About the thermite reaction, I started chapter 5 [BACKGROUND] Well, it's obviously a very exothermic reaction. It's not an explosive reaction. It doesn't produce a gas. Now, if you use the thermite reaction to ignite your explosive, that's not [inaudible] but that involves something other than just thermal. [BACKGROUND] but can it cut through stuff? It would be easier to use something like an acetylene torch, but well, I'll put it this way; one way the thermite reaction has historically been done, which I would never do in this class maneuver science really, but in our people from embracing those two good. You're saying that thing like a little plaid flower pot? Yeah. Then we can generate a multiplier and it burns in industry, the bottom of the plaid flower pot and you have a dish of water underneath [inaudible] and then the heat breaks down the water into hydrogen and oxygen that makes the hydrogen [inaudible] I'm sure you can find some crazy videos [inaudible] [NOISE] [BACKGROUND] Check it out on the Internet and see [inaudible]. All right. Thank you. [BACKGROUND]
chem103-090-20180918-140001.mp4
From Dana Chatellier August 24, 2019
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