[BACKGROUND] Good afternoon. Today is October 31st, 2019. Happy Halloween to everyone. Some people came dressed for the occasion. This actually is my Halloween costume. There's a reason I say that. I have two cousins who were both born on Halloween. You think that's a good time? It's not. [LAUGHTER] Try going to a nice restaurant to celebrate your birthday and being the only one who's not in costume. My cousin John, [NOISE] lives in North Carolina, he's a wild and crazy person and probably would have enjoyed my job, which involves blowing things up and making a mess as you've seen. My cousin Carol is not only a Halloween kid but a breast cancer survivor. [NOISE] October of course, I should say was because it's the last day of it breast cancer awareness month. Carol wrote a book called It Is What It Is about her experience with breast cancer. There's a picture of around the back surrounded by her two daughters. Good news is that it's been 12 years and she remains cancer-free. I stole the quote here. [APPLAUSE] I'm going to tell her about this video, so she'll hear that. [LAUGHTER] I stole this thought for the day from the last [NOISE] page of the book. It says, "Why not go out on a limb? Isn't that where the fruit is?" Happy birthday today to Carol and John [NOISE] and many more. Today's handout, there's another aside, but it follows up on something we were talking about previously, namely osmotic pressure. You can read this article. Basically it has to do with what's called osmotic energy and I thought this might be of interest, especially to the engineering majors in the audience. Because basically what engineers live for is trying to harness new sources of energy. Some people are thinking that osmosis can be harnessed to generate power and the article talks in some detail about that. Since we were just talking about those things, I thought that might be interesting for you to have and you can take a look at whenever you would like. I want to go back to where we were at the end of last class. Earlier, we had shown you a diagram that looks like this [NOISE] to describe how the concentration of a reactant changes as the reaction unfolds. The point is even though the slope of this line changes over time, if we only study the initial phases of the reaction where that line remains relatively straight, the slope is relatively constant. That is, we can study what's called the initial rate of the reaction. The method of initial rates is what we can use to figure out the rate law or rate equation for a particular chemical reaction. Section 12.3, in your textbook is titled Rate Laws. There are several problems in Section 12.3 that talk about essentially the method of initial rates and give you more practice with [NOISE] the kind of thing we were doing last time. But I want to just go over this problem briefly again, just in case it wasn't making sense last time. The idea is one factor that influences the rate of a reaction is the concentration of the reactants. If there's more than one reactant, you can change the concentration of one reactant at a time, measure the effect of that has on the initial rate of the reaction, and therefore come to some conclusions about what the exponents are in the rate equation. Last time we showed you a whole bunch of data, all of which appears on page 121 in your lecture notes for those of you following along. What I asked you to do last time was find me a couple of pairs of experiments here where one concentration stays the same and the other one changes. If you compare line 2 with line 3, we see that the concentration of HCl remains constant, but the concentration of C_5H_10 changes and here's the impact of that has on the rate. If we compare lines 4 and 5, the concentration of C_5H_10 is pretty much constant, but concentration of HCl changes and here's the impact on the rate. What we're able to show last time and if you compare lines 2 and 3, here's the ratio of the two concentrations of C_5H_10 the change. Here's the ratio of the two rates. The point is this comes out to be three, this comes out to be three, but the concentration is raised to the a exponent. The exponents are usually small whole numbers. In this case, since 3^1 is 3 then exponent a has a value of one. For lines 4 and 5, here's the two concentrations of HCl and here's the two rates, the concentration of C_5H_10 remains unchanged. This turns out to be 3.11. This turns out to be 1.77^b. It turns out that 1.77 squared is pretty close to 3.11. So exponent b turns out to be two, and then we can write the rate law or the rate equation looking like this. Then to determine the value of the rate constant k, we can just choose any one experiment. The data shown here comes from experiment number 1 on the previous slide. Let things in. We do get some funny units from time to time for the rate constant, but you'll see later on today why those units matter. You would be able to see why different reactions, depending on what the rate equation looks like, actually have different units for the rate constant. Suffice to say when we set things up like this, moles per liter divided by moles per liter cancels out, nothing else does. Seconds to the minus 1 and moles per liter squared in the denominator, which is moles per liter to the minus 2 power, show up as the units in this particular rate constant. Now would anybody like to go over any of that in more detail as we left off things last time at the end of class when people were probably thinking more about food than about chemistry? This makes sense to everyone. [NOISE] Now the point is, there are other problems like this in your textbook and we would like you to explore those as you prepare for your third exam, which is now three weeks from today. Now the thing I want to call your attention to you for the moment are the exponents again. The exponent of C_5H_10 is 1, the exponent of HCl is 2. The reason for calling your attention to that [NOISE] is the next concept I want to talk about is the order of a reaction. [NOISE] The order of the reaction is very straightforward if you know what the rate equation looks like. But you're not going to know what the rate equation looks like until you go into the laboratory and do a series of experiments using the method of initial rates and figure it out that way. But once you know what the rate equation looks like, then the order of the reaction is simply the sum of the exponents in the rate equation. For example, [NOISE] if the rate equation looks something like this, rate equals rate constant k times the concentration of some reactant a^1, we don't normally bother writing in the one, but hopefully, you know that means raised to the first power. Well, there's a grand total of one exponent there and it's one, and therefore, the sum of the exponents is one. Therefore, this is called a first-order reaction. For a second-order reaction, there are two possible rate equations. Either the rate is equal to the rate constant k times the concentration of one reactant squared, or rate constant k times the concentration of reactant A times the concentration of reactant B. Both of those raised to the first power, 1 plus 1 is 2. Either of those would be a second-order rate equation. From the example we were looking at at the end of last class that we just give a quick and dirty review of, the rate equation came out looking like this. Rate equals the rate constant k times the concentration of C_5 H_10 to the first power times the concentration of HCl to the second power. Since 1 plus 2 is 3, that means the overall order of this reaction is a third-order reaction. Occasionally, when people want to refer to one particular reactant and its impact on the rate, they might refer to the exponent of one reactant, C_5 H_10 as an exponent of one. Some people would say that this reaction is first-order with respect to C_5 H_10 concentration of HCl as an exponent of two. Some people would say that this is a second-order reaction with respect to HCl. But overall, since 1 plus 2 is 3, this is a third-order reaction. Those interesting units on the rate constant that we were showing you on the previous slide. I'll just show you that again, real quick. Units like that are typical for a third-order reaction. [NOISE] Would anybody like more time with this slide? Now, even though I don't set foot in the laboratories associated with this course very often, and if any feeling I know when you folks are just going into lab, what the first question out everybody's mouth is for the TA. How long is today's lab going to take? Well, that's what we're talking about, we're talking about reaction kinetics. How long does the reaction take? The point is, if you know what the rate equation looks like and if you know what the concentrations of the reactants are going to be, you can typically figure that out. However, it's difficult to do that just from the rate equation. You need a slightly different form of the rate equation. What I call in the lecture notes, the concentration versus time equations, are what your textbook calls in Section 12.4 the integrated rate laws. To keep life relatively simple we're going to talk just about first-order and second-order reactions as we discuss this. Suppose you're going into the laboratory to do the reaction that you know is a first-order reaction with a rate equation looking something like this. Let's say you know the value of the rate constant plus [MUSIC] [NOISE] Wait we have to get the footage. [LAUGHTER] [MUSIC] Over there. [APPLAUSE] [LAUGHTER] I had nothing to do with that and nobody told me that was going to happen. [LAUGHTER] How do you follow that? [LAUGHTER] Don't know, but we're going to try. [BACKGROUND] Back to the Monday. We want to know how long the reaction's going to take. We know it's first-order, we know what the rate constant is. But when you look at this rate equation, there's nothing there that suggests time. The point is if you want to know how long the reaction is going to take you're asking a question like, "How many seconds, how many minutes, how many hours is this going to take?" We need an equation that has time in it somewhere. We can convert this into an equation that looks like that using calculus. [NOISE] How many of you have taken or are taking calculus? Fair number. If I said integral calculus, how many hands would still go up? [NOISE]. Fair number. Calculus is not a prerequisite or corequisite for Chem 103 so we are not going to presume that you know how to do calculus in this course. For those who do, and especially those who raised their hands for integral calculus, you might want to see if you can figure out where this equation comes from. For those who haven't had calculus yet, you just going to have to take my word for it that using integral calculus, you can convert this equation into this equation. What does the ln in this equation mean? Natural logarithm. You may not have to know about calculus for this course, but you do have to know about logarithms. We haven't done much with logarithms so far this semester, but we're about to. I'll bring that other slide back in just a moment. The most important thing for right now in dealing with logarithms is knowing which of two possible buttons on your calculator to push. Because there are two kinds of logarithms. One kind called common or base 10 logarithms is what happens when you push the L-O-G key on your calculator. The definition of a common or base 10 logarithm is the power to which the number 10 must be raised to obtain the number that you're talking about. For example, what is the log or logarithm of 1,000? Three. Three. Because 10 to the third power is three, so what power do you have to raise 10 to, to get this number? The answer is three. What's the log of a million? Six. Six, 10 to the sixth power? One million. What's the logarithm of 10. [NOISE]. One. One. What's the logarithm of 1? Zero. Zero. All that makes sense, so far? Yes. Okay. What's the logarithm of 2? Well, if you don't know off head, that's fine. That's what your calculator is for. However, I would hope it would make sense that since two is the number of between 10 and one, then the logarithm of 2 would be a number between the logarithm of 10, which is 1, and the logarithm of 1 which is 0. If you do this on your calculator, you find that the logarithm of 2 is approximately 0.3. [NOISE] The point is, you can use the log key on your calculator to take the logarithm of any positive number. That's base 10 logarithms. Natural logarithms, which is what the LN key on your calculator is all about. Different kind of logarithm, although similar in some ways, the main difference is the base number, for common logarithms, the base number is 10. For natural logarithms, the base number is e, and e is an irrational number with a value of approximately 2.718. In this case, the natural logarithm is simply the exponent to which e must be raised to get the number in question. Whenever I don't have this thing squared up, it's just going to [inaudible] Now, if you don't happen to have LN key on your calculator, all is not lost as long as you have an LOG key on your calculator. Because the conversion factor between base 10 logarithms, LOG and natural logarithms LN is simply 2.303. If you multiply the LOG logarithm of any number by 2.303, you get the natural logarithm of that number. Hopefully everybody has heard of logarithms before. Hopefully everybody understands the difference between LOG and LN when you touch those buttons on your calculator. In a moment, will show you how you can do a problem like the one we were just alluding to for a first-order reaction using logarithms. But before we do, let people finish with this slide. [BACKGROUND] [inaudible] [NOISE] [NOISE] [NOISE]. Everybody, okay, here? Yes. All right, actually, before I show you a problem, I'm going to go back to that previous slide when everybody's finished with that. We are going to show you an example of both a first-order reaction and a second-order reaction. The point is, the first-order concentration versus time equation looks like this. For a second-order reaction, if you start with a second-order rate equation looking like this. Once again, integrate this using integral calculus, you can get a [NOISE] second-order concentration versus time equation that looks like this. There are no logarithms involved here, but we do have reciprocals of the initial concentration A with the subscript i and the final concentration A with a subscript f. In the first-order concentration versus time equation, rate constant k times the amount of time t that has elapsed is equal to the natural logarithm of [NOISE] the ratio of the initial concentration to the final concentration. Finish with this slide, and then we'll look at some problems. [BACKGROUND] All right, everybody okay here with what we're doing? Yes. [BACKGROUND] Let's look at a first-order reaction and actually, this problem is in the lecture notes on page 123. Here's the reaction in question. Here is its rate equation and rate constant. Rate constant k times the concentration of C_4H_9Br to the first power, rate constant k has a value of 0.010 seconds to the minus one power. Notice that those are different units for the rate constant we showed you earlier. The reason for the different units is this is a first-order reaction, the other was a third order reaction. You might wonder why there's no concentration of hydroxide in here because it's obviously one of the reactants. For the moment, we'll just say that there is no concentration of hydroxide term there, we'll discuss more about why that's the case later on. Main point is there's only one exponent here. It's a one, therefore, that's a first-order reaction. The problem is asking, suppose the initial concentration of C_4H_9Br is point 0.100 mole. How long will it take, that is what is the value of t, the time for 99.0 Percent of the C_4H_9Br to be consumed? In other words, what we're going to say is we're going to set this reaction, let it run, and we're going to call it done, when 99 percent of the reactant has disappeared. Here's the first-order concentration versus time equation from the previous slide. The rate constant k is given, t, the time is what we're trying to find. The initial concentration is given. How do we figure out, what the final concentration is? All right, I'll give you a hint. If 99 percent of the reactant is consumed, what percentage remained? One percent. One. [NOISE] What's one percent of the initial concentration? How would you find that number? Divide by a 100. Divided by a 100 of course. If we divide this by a 100, that means the final concentration is 0.00100 molar. Now that we know the initial concentration, the final concentration, and the rate constant, we can just enter all of that into the first-order concentration versus time equation. Here's what happens if we do all that. 0.010 seconds to the minus 1, that's the rate constant k times t, is what we're trying to find, equals natural logarithm of the initial concentration, 0.100 molar divided by the final concentration, 0.00100 molar. Dividing this out just gives you a 100. The natural logarithm of a 100 turns out to be 4.606. To get t by itself, we just divide 4.606 by 0.010. Notice that seconds to the minus 1 in the denominator becomes seconds to the plus 1 in the numerator. The answer comes out in seconds. To the two sig figs are allowed here because you only know the rate constant to two sig figs. We find that the reaction takes 4.6 times 10^2 seconds. Four hundred and sixty seconds is a little bit less than eight minutes. Set up the reaction, go down the hallway, get yourself a cup of coffee, come back, reaction is properly done. Any questions about the arithmetic or any of the concepts involved? [NOISE] All right, that's a first-order example. When people are done with this slide, will show you a second-order example also on page 123 in the lecture notes. [BACKGROUND] [NOISE] Everybody, how are they in here? Let's look at a second-order example., Now as I go to the second-order slide, just take another look at this reaction up here. C_4H_9Br plus hydroxide ion gives C_4H_9OH and bromide ion. Because the second-order example I'm going to show you looks a lot like that. In this case it's CH_3Br reacting with hydroxide ion, to form CH_3OH plus a bromide ion. But in this example, it turns out that the rate equation is a second-order reaction. First-order with respect to both CH_3Br and hydroxide ion. The reason I wanted to point that out is you simply cannot just look at the balanced equation of a chemical reaction and decide what the order of the reaction is, decide what the rate equation looks like. In this case, it works out that the exponent's here, one and one are the same as the coefficients in the balanced equation. In the previous example that wasn't true. In the example we saw at the end of class last time, that wasn't true. There are for that and we'll talk more about those later on. Anyway, here's the problem on page 123. If the rate constant K for this reaction is 0.0214 moles per liter to the minus one seconds to the minus one, these are typical units for a second-order rate constant. If the initial concentration of both reactants is 0.100 molar, suppose we set this reaction up, go have lunch, come back in an hour. What's the concentration of CH_3Br going to be? Well, here's the second-order concentration versus time equation. What do we know? We know the rate constant k. We know the initial concentration so we can put that in here. We're trying to find the final concentration, and we know time. But what units must time be in? Look at the units in the rate constant. In other words, we just put in one for time for one hour. That's going to be wrong. What units do we need time to be in? Seconds. Seconds. Because we have seconds to the minus one in the rate constant. How many seconds are there in an hour? I'm hearing no response. How many seconds are there in a minute? Sixty. Sixty. How many minutes are there in an hour? Sixty. Sixty. Thirty-six hundred, 60 times 60. Some of you probably count all 3600 while I'm talking over here. That's okay. The point is, we know everything but the final concentration. But at this point, this arithmetic. We multiply these two guys together, seconds times seconds to the minus one cancels out. The left side here is going to have units of moles per liter to the minus one power. 1 divided by 0.1 is 10, but that also has units of moles per liter to the minus one power. I finally figure out what the final concentration is. We're going to add 10 to both sides, which that gives us 87 on the left side, then just take the reciprocal of both sides. 1 over 87 turns out to be 0.0115. Therefore, as it turns out, the final concentration of CH_3Br is 0.0115 molar. Compare that to the initial concentration which was 0.1 mole. Approximately what percentage of our reactant has been consumed after one hour. This is what we started with. This is what's left after one hour. Roughly what percentage was consumed? [inaudible] Say again? Eighty eight or something. Yeah, ighty 88, 89 percent, something like that. Yeah, that's about right. In other words, it took one hour for this reaction to consume not quite 90 percent of the reactant. Compared to the previous example, the first-order case, where 99 percent of the reactant is consumed after less than eight minutes. Which of those two reactions is happening faster? The first-order or second-order? First order. First-order. There's another way to see that a little bit more easily than going through all of this. If you just want an approximate idea of how long the reaction is going to take, I'll show you a shortcut as soon as everybody is done with this slide. Would anybody like more time here? Yes. Here's the shortcut. Actually this is talked about in your textbook on page 686 under the heading of the Half-life of a Reaction. Let me show you this diagram, which'll giving you the basic idea behind half-life. Now this particular reaction happen to involve a gas as the reactant. Instead of measuring concentration, they measured pressure. But the point is, we start out with a pressure of, in this case, I'm a 150 torr. After about 13,000 seconds, we're down to 75 torr, which is half of the 150. In other words, 13,000 seconds is the time that it took to get down to half of what the initial concentration was. If we wait another 13,000 seconds, brings it up to about 26,000 seconds. We're down to half of that. Half of 75 is 37-and-a-half and so on. The point is that the definition of half-life is simply the time it takes for the concentration of the reactant to decrease to half of its initial value. You can figure out half-life using the concentration versus time equations , because by definition, the final concentration when you're talking about half-life, is simply half of the initial concentration, that is a initial divided by 2. For example, for a first-order reaction, here's the first-order concentration versus time equation. But if we enter as our final concentration, the initial concentration divided by 2, the term in parentheses simply reduces down to two, and the natural logarithm of two turns out to be 0.693. The time involved, and I should point out, this symbol, T with a subscript one-half is simply the symbol for half-life. It's a symbol. It doesn't mean divide time by one-half, it doesn't mean multiply time by one-half, it doesn't mean do any arithmetic operation. It's just a symbol. Half-life. For a first order reaction, the half-life is 0.693 divided by the rate constant. But since the rate constant is constant, that means the half-life for a first-order reaction is constant and will remain constant throughout the lifetime of the reaction. In the example we just showed you, the fact that at least through two half-lifes, the half-life seems to be remaining constant at about 13,000 seconds. Suggests that this is a first-order reaction. If we carry out the same arithmetic for a second-order reaction. Here's the second-order concentration versus time equation. But again, if we enter A initial over 2 for A final and do the arithmetic, [NOISE] it turns out the half-life for a second-order reaction is 1 over the rate constant times the initial concentration. But notice the difference here. This includes a concentration term and this does not. This is constant throughout the lifetime of the reaction. But since the concentration of the reactant changes as a function of time, second-order half-lives are not constant because the concentration will keep on changing. But the point is these equations are somewhat easier to deal with than the concentration versus time equations. If all you want is a rough idea of how long the reaction is going to take, we'll show you an example, actually, two examples, going back to those same reactions we were just looking at. You can come to the same conclusion about which one is happening faster. But we'll let people finish with this slide first. [BACKGROUND] Anybody need more time here? [BACKGROUND] Everybody okay? Going back to our two previous examples. Here's our first-order reaction from before. K for that first-order reaction was 0.010 seconds to the minus 1 power. What's the half-life of this reaction? Well, it's a first-order reaction. First-order half-life equation is simply 0.693 divided by K. If we do that and again, the seconds to the minus 1 in the denominator becomes seconds in the numerator. The answer comes out in seconds. For the two sig figs we're allowed, 69 seconds. In other words, after a little more than a minute, this reaction is half done. By contrast, here's our second-order reaction from last time. Here's the rate constant and the initial concentration of the reactants. What's the half-life for this one? Well, for a second-order reaction, here's the half-life equation. Plug in K and the initial concentration. Moles per liter times moles per liter to the minus 1 cancels out. Seconds to the minus 1 in the denominator becomes seconds in the numerator, the answer comes out in seconds. [NOISE] The arithmetic comes out to be 467 seconds. Again, a little bit less than eight minutes. In other words, the second-order reaction is half done in the time that it took the first-order reaction to be 99 percent done. Which reaction is happening faster? [BACKGROUND] The first-order reaction. Another way to think about that is to say whichever one has, the shorter half-life is the one that's going faster. Do the concepts make sense? As a general guideline, especially when talking about a first-order reaction. Suppose you want to let the reaction go until it's like 99.9 percent done. That's 10 half-lives. After 690 seconds up here, that reaction would be 99.9 percent done. But that only works for first-order reactions because they have constant half-lives. [BACKGROUND] [NOISE]. Would anybody like more time here? Section at 12.5 in your textbook is called collision theory. It gives you a little bit of an idea of what collision theory is all about. I want to talk about baby elephants again. Remember the baby elephants problem? We figured out last time that we need male elephants and female elephants, so we have to open the gauges and let them mix together. Even so, just like not every collision between molecules results in a successful reaction, not every collision between elephants results in baby elephants. Obviously to make baby elephants [inaudible] [NOISE] male elephants to collide with female elephants. But there are two things that must be true about that collision. First of all, the collision must have the correct orientation. If your elephants bump into each other head-on, you're just going to get elephants with headaches and then no baby elephants. [LAUGHTER] This diagram is trying to show a reaction between a chlorine atom, which is this green thing over here, and a molecule of a compound called nitrosyl chloride, NOCL. The desired products are a molecule of NO, nitric oxide, and a molecule of chlorine. That reaction will work if things collide this way so that the two chlorine atoms come in contact with each other. But it won't work if they collide with a different orientation so that the chlorine atom here never gets close to this fluorine atom here. They just bounce off each other, that's the end of it. If a collision between molecules has the wrong orientation, the reaction won't work. [NOISE] The other is that if you want to make baby elephants, you need for the collision between elements to be sufficiently vigorous. That is, it needs to have a certain minimum energy. The movement of molecules and their kinetic energy is related to their temperature as we indicated earlier. We're going to talk about the impact of temperature on reaction rates. [NOISE] As a general rule, if you want to get a reaction to go faster, warm it up. If you want to get the reaction to go about two or three times as fast as it's already going, increase the temperature by about 10 degrees. But when you look at a rate equation like this one, you don't see any terms in there that have anything to do with temperature. How can we use this equation to say something about temperature? Well, it turns out that the rate constant k is only constant at a particular temperature. K, the rate constant depends on a number of other factors. One of them is E, the base number for natural logarithms 2.718. Another is the temperature in kelvins. [NOISE] Another is R. Now this is R like PV equals nRT. However, when you multiply pressure times volume, it turns out you can express that in energy units and say we can rewrite R here using energy units as 8.314 joules per mole per kelvin. Another factor that comes into play is this thing, E with a subscript a. That is something called the energy of activation for a reaction. We'll talk more about the energy of activation when people finish with this slide. [NOISE] Question? Is that E subscript [NOISE] n? This part right here? Yeah. Capital E with a subscript small a. You'll see that again on the next slide. [BACKGROUND] [NOISE] [BACKGROUND] Would anybody like more [NOISE] time here? [NOISE] [BACKGROUND] The definition of the energy of activation [BACKGROUND] is simply the minimum amount of energy that a collision between molecules needs to have for a reaction to take place. Let me show you a diagram that illustrates this concept. [NOISE] We can get most of this on-screen. Let me just show you the x-axis for a moment, and we'll get you back to this slide. This is a graph of potential energy on the y-axis. When they say reaction pathway on the x-axis, that means the reactants are over here, the products are over here. You could think of the x-axis as simply top, [BACKGROUND] but what's really trying to say is how far along have we gotten from reactants to products. Now, here's the idea. [NOISE] At relatively low temperatures, molecules are moving around slowly like this and they bump into each other. Well, the kinetic energy gets translated to potential energy because at the moment they collide with each other, they're not moving relative to each other anymore, their potential energy increases. But if they're moving relatively slowly, only goes up a little bit. An analogy here. Think of this as a hill, and you've got this boulder on this side of the hill and you want to roll this boulder over to the other side of the hill. Well, if you roll it over halfway up the hill [BACKGROUND] and then let go, what's [BACKGROUND] going to happen? Fall back. Fall right back to where it was. Nothing accomplished. But raise the temperature. Now the molecules move around real fast and when they bang into each other, that's going to be a big increase in potential energy. If a more vigorous collision gets you up to the top of the hill, then I got to do is get the rock one more shove and it goes out to form products. There's a way to find the energy of activation using the Arrhenius equation that we just gave you a few moments ago. By the way, this fellow Arrhenius, I think we mentioned him earlier this semester Svante Arrhenius was a Swedish scientist who were the first ones to win the Nobel Prize for chemistry. He basically convinced people that there were such things as ions, but he dabbled in a lot of other aspects of chemistry as well, including kinetics. If we take the Arrhenius equation from the previous slide, and take the natural logarithm of both sides, we can rewrite the Arrhenius equation looking like this. Natural logarithm of the rate constant k equals, the negative the energy the activation divided by R times 1 over the temperature plus the natural logarithm of A. A is a constant in the Arrhenius equation that includes a number of other factors, including the geometry of the molecules and things like that. We're not going to worry about what A is in this course. However, take a look at this equation. What's the equation of a straight line in general? Y equals mx plus b [OVERLAPPING]. Y equals? Mx plus b. Mx plus b. The reason for rewriting the Arrhenius equation in this form is now it has the form of the equation of a straight line, y equals mx plus b. If we let the natural logarithm of the rate constant be y, and 1 over the Kelvin temperature be x. Plot that on a graph. Typically what's done here is to measure the rate constant at various different temperatures, and then make a graph that looks something like this one. If we graph the natural logarithm of k on the y-axis and 1 over the Kelvin temperature on the x-axis. We draw the best straight line we can through our data points. We get hopefully a straight line with a slope equal to the negative of the energy of activation divided by R, we gave you the value of R previously. What I would like to do is walk you through an example of how to do that. However, the hour is getting late. Many of you are already falling asleep. This is when I needed the dancing clouds of ninjas to wake people up [LAUGHTER]. But failing that, I think we will stop here for today and we will walk you through a problem like this next time. [BACKGROUND] Have a happy and safe Halloween [NOISE]. We will see you again in November. [OVERLAPPING] [NOISE] The only thing we did before was to review what we've done at the end of class last Thursday. I will show you that. [OVERLAPPING] The very first thing we did today was this. [inaudible] We show them this slide which has a whole bunch of data. [OVERLAPPING] Then [inaudible] use that data to figure out exactly what the range of what it looks like and the equation looks like that. That's all we did before this [inaudible]. [OVERLAPPING] A couple of other people have turned in exams for regrading already. If you have an answer that you think is marked wrong to be marked right, you want to get credit for it just circle the number of that question, and then turn your exam back in grade. You can do it anytime. [inaudible]. As long as it's circled, that's all I care. [inaudible]. Well, the point is it's a multiple-choice test. The only thing I can see is that if you had the thing, right, and it was marked wrong then you can submit that for regrading. I'm not sure what else [OVERLAPPING]. Check again what's close to the answer D, and if you think you got more points coming let know? Well it depends with what it is if it has anything to do with the logistics or running of the lab, that's a better question for Dr. [inaudible] I can answer any chemistry questions related to the lab. I think I mentioned it before. But I forget I mentioned [inaudible] I still don't know the answer. Then I would go talk to Dr [inaudible]. Her office is over in Brown Labs she probably still there actually. If you want to do that now you can do that. [BACKGROUND] [NOISE]
chem103-080-20191031-140000.mp4
From Dana Chatellier October 31, 2019
513 plays
513
0 comments
0
You unliked the media.
Video Created by UD Capture Classroom Recording in Smith 120 on 2019-10-31 14:00:00.
…Read more
Less…
- Tags
- Appears In
Link to Media Page
Loading
Add a comment