WEBVTT chemistry/general-chemistry/ow 00:00:00.000 --> 00:00:02.100 Hi, welcome back to Educator.com. 00:00:02.100 --> 00:00:07.900 Today's lecture from general chemistry is going to be on chemical kinetics. 00:00:07.900 --> 00:00:10.300 Here is our brief overview of the lesson today. 00:00:10.300 --> 00:00:13.300 As always we will go ahead and start off with our introduction. 00:00:13.300 --> 00:00:17.200 Then we are going to get into what is really at the heart of chemical kinetics. 00:00:17.200 --> 00:00:21.600 That is the rate of chemical reactions; that is how fast does a reaction occur? 00:00:21.600 --> 00:00:28.900 We are then going to see what are the mathematical equations which actually attempt to quantify the rate of an equation. 00:00:28.900 --> 00:00:31.600 That is what we call a rate law. 00:00:31.600 --> 00:00:37.500 Our objective is to then try to derive these equations for any given reaction. 00:00:37.500 --> 00:00:40.300 There is two sets of experiments that you can do to perform this. 00:00:40.300 --> 00:00:42.800 Number one is what we call the method of initial rates. 00:00:42.800 --> 00:00:45.600 Number two is what we call the integrated rate laws. 00:00:45.600 --> 00:00:50.800 We will then jump into how a reaction proceeds, the step by step process which is 00:00:50.800 --> 00:00:58.700 called a reaction mechanism, followed by the effect that temperature has on reaction rates. 00:00:58.700 --> 00:01:02.500 We will go ahead and wrap up the lesson with a brief overview of 00:01:02.500 --> 00:01:10.500 catalysis followed by our summary and a pair of sample problems. 00:01:10.500 --> 00:01:20.800 Chemical kinetics basically is the area of chemistry that examines the factors which can influence the rate of a reaction. 00:01:20.800 --> 00:01:24.200 What do we mean by factors that can influence the rate? 00:01:24.200 --> 00:01:35.100 There are several--temperature, pressure, the reactant concentration, the addition of a catalyst, and mechanical force. 00:01:35.100 --> 00:01:44.000 We are going to see that basically as temperature goes up, the rate goes up. 00:01:44.000 --> 00:01:52.600 We are going to see that in general as pressure goes up, the rate goes up. 00:01:52.600 --> 00:02:00.100 We are going to see that as the concentration of reactant goes up, the rate goes up. 00:02:00.100 --> 00:02:03.100 When we add a catalyst, we are going to define what that exactly means. 00:02:03.100 --> 00:02:06.100 But this also increases the rate. 00:02:06.100 --> 00:02:09.400 Finally mechanical force, this is what you probably do in lab. 00:02:09.400 --> 00:02:16.900 All we mean by mechanical force is something as simple as stirring or shaking. 00:02:16.900 --> 00:02:21.700 Of course this will increase the rate of the reaction. 00:02:21.700 --> 00:02:29.800 This is what we mean by factors which can influence the rate of a reaction. 00:02:29.800 --> 00:02:34.600 Let's take a look though how a typical reaction progresses through time. 00:02:34.600 --> 00:02:37.700 Consider the reaction A plus 2B goes to C. 00:02:37.700 --> 00:02:45.200 This is telling me that for every one of A, two Bs are required. 00:02:45.200 --> 00:02:47.500 One C is going to be produced. 00:02:47.500 --> 00:02:55.900 Remember we are only looking here at the forward direction only. 00:02:55.900 --> 00:03:10.500 When we go in the forward direction, we say that reactants are consumed which means concentration of reactant go down. 00:03:10.500 --> 00:03:20.800 The products are made which means that the concentration of product goes up. 00:03:20.800 --> 00:03:25.900 It looks like because twice the amount of B is going to be consumed as A, 00:03:25.900 --> 00:03:41.300 the rate of consumption of B is going to be double that of A. 00:03:41.300 --> 00:03:45.300 It is very typical for us in chemical kinetics to graph this. 00:03:45.300 --> 00:03:56.900 We can go ahead and graph the concentration of the reactant with respect to time and the concentration of product too. 00:03:56.900 --> 00:04:00.500 At time zero, I have no product; I only have reactants. 00:04:00.500 --> 00:04:06.900 Down here right at the origin, this is the concentration of C initial. 00:04:06.900 --> 00:04:11.300 The way that concentrations change with respect to time is not linear. 00:04:11.300 --> 00:04:19.900 But instead it is going to be characterized by a simple curve just like that. 00:04:19.900 --> 00:04:24.300 You notice that at some time t, the concentration of C stops changing. 00:04:24.300 --> 00:04:28.800 We are going to plateau; it is not going to increase forever. 00:04:28.800 --> 00:04:32.100 How about the concentration of A and B? 00:04:32.100 --> 00:04:38.200 The concentration of A let's say is here and the concentration of B. 00:04:38.200 --> 00:04:40.600 Let's say we had equal amounts. 00:04:40.600 --> 00:04:47.000 Let me go ahead and draw now a curve that represents the concentration of A changing with respect to time. 00:04:47.000 --> 00:04:53.600 It is going to be a mirror image of the concentration of C changing because they are both 1:1 ratio. 00:04:53.600 --> 00:04:56.500 But how is the concentration of B changing? 00:04:56.500 --> 00:05:01.800 The concentration of B, it is going to drop at double the rate. 00:05:01.800 --> 00:05:15.400 The concentration of B is going to drop much much much... quicker than the concentration of A. 00:05:15.400 --> 00:05:24.800 You notice that for all of the reactions, there reaches a point where the curves plateau 00:05:24.800 --> 00:05:36.700 which means that the concentration of reactants and the concentration of products no longer changes. 00:05:36.700 --> 00:05:40.500 We are going to talk about this region in another lecture. 00:05:40.500 --> 00:05:45.800 This is what we call the equilibrium region. 00:05:45.800 --> 00:05:51.000 In chemical kinetics, what we are interested in is really the start of the reaction, the early part 00:05:51.000 --> 00:06:00.600 of the reaction where we actually can monitor change in the concentration of reactants and products. 00:06:00.600 --> 00:06:03.800 That is the representation graphically; how about mathematically? 00:06:03.800 --> 00:06:08.000 It turns out that the rate of change is going to be equal to the following. 00:06:08.000 --> 00:06:21.700 The change in the concentration of A, Δ[A]Δt, this is going to be equal to 00:06:21.700 --> 00:06:32.400 1/2 Δ[B]Δt which is equal to the change in concentration of C Δ[C]Δt. 00:06:32.400 --> 00:06:40.800 We have to put a negative sign in front of the reactants such that the overall number is 00:06:40.800 --> 00:06:47.700 going to be positive because Δ[A]Δt is going to be slope of the curve. 00:06:47.700 --> 00:06:50.800 This is going to be slope of the curve you see. 00:06:50.800 --> 00:06:55.700 This is going to be the slope of that curve too. 00:06:55.700 --> 00:07:09.500 This gives us a general equation for relating the rates of change of each reactant and product concentration with respect to time. 00:07:09.500 --> 00:07:14.200 Now that we have gone through a brief introduction, let's go now into 00:07:14.200 --> 00:07:18.700 what is at the core of chemical kinetics--that is the rate law. 00:07:18.700 --> 00:07:27.200 A rate law is going to be a mathematical equation which relates the rate of reaction to the concentration of the reactants. 00:07:27.200 --> 00:07:38.100 Basically the rate law tells us that the rate of a reaction is proportional to the concentration of reactants typically. 00:07:38.100 --> 00:07:41.300 When we write out the rate law, all rate laws have the following equation 00:07:41.300 --> 00:07:47.300 Of the form rate which is going to be usually in molarity per second... 00:07:47.300 --> 00:07:52.400 is equal to some constant k times the concentration of A raised to 00:07:52.400 --> 00:07:58.000 some power times the concentration of the B raised to some power, etc. 00:07:58.000 --> 00:08:04.500 Let's go ahead and define what each of these mean. 00:08:04.500 --> 00:08:11.400 k is what we call the rate constant. 00:08:11.400 --> 00:08:28.600 The rate constant is going to be unique to a reaction at a certain temperature. 00:08:28.600 --> 00:08:39.500 In addition, we are going to see that the rate constant, the units of k vary. 00:08:39.500 --> 00:08:42.300 We are going to see that very soon. 00:08:42.300 --> 00:08:49.200 x and y are what we call rate orders. 00:08:49.200 --> 00:08:57.000 Once again x and y are what we call rate orders. 00:08:57.000 --> 00:09:13.700 If x is equal to 1, we say reaction is first order with respect to the concentration of A. 00:09:13.700 --> 00:09:24.400 If x is equal to 2, we say the reaction is second order with respect to the concentration of A. 00:09:24.400 --> 00:09:36.700 In addition, if the sum of the rate orders x plus y is equal to 0, we say zero order overall. 00:09:36.700 --> 00:09:45.200 If the sum of the orders is equal to 1, we say first order overall. 00:09:45.200 --> 00:09:52.000 If x plus y is equal to 2, we say second order overall. 00:09:52.000 --> 00:09:58.300 This is just some terminology that we want to introduce and clarify. 00:09:58.300 --> 00:10:01.500 Let's go ahead and get back to the units of k. 00:10:01.500 --> 00:10:04.400 The units of k will vary. 00:10:04.400 --> 00:10:19.000 For zero order, the units of k is just going to be reciprocal seconds. 00:10:19.000 --> 00:10:27.300 For first order, the units of k is going to... excuse me. 00:10:27.300 --> 00:10:29.400 For zero order.. my apologies. 00:10:29.400 --> 00:10:36.100 For zero order, the units of k is going to be molarity per second. 00:10:36.100 --> 00:10:40.400 For first order, the units of k is just reciprocal seconds. 00:10:40.400 --> 00:10:48.700 For second order overall, the units of k is going to be inverse molarity inverse second. 00:10:48.700 --> 00:10:57.600 You can easily plug that back into the equation for k and see that the units will cancel. 00:10:57.600 --> 00:10:59.600 This is the general rate law. 00:10:59.600 --> 00:11:02.500 Rate is equal to some constant k times the concentration of A raised to 00:11:02.500 --> 00:11:06.200 the x power times the concentration of B raised to the y power. 00:11:06.200 --> 00:11:13.200 But let's go ahead now and see what the significance of the rate orders are. 00:11:13.200 --> 00:11:17.900 Consider the following reaction: A plus 2B goes to C. 00:11:17.900 --> 00:11:26.600 The rate law is given to us to be k times A squared times the concentration of B. 00:11:26.600 --> 00:11:31.800 Let's go ahead and study what happens if we change the concentration of one of these reactants. 00:11:31.800 --> 00:11:49.700 If the concentration of A doubles holding B constant, then we see that the rate is going to increase by 4. 00:11:49.700 --> 00:11:54.000 It is going to quadruple. 00:11:54.000 --> 00:12:07.800 If the concentration of A triples holding B constant, we see that the rate is going to increase by a factor of 9. 00:12:07.800 --> 00:12:19.400 What if the concentration of B doubles holding concentration of A constant? 00:12:19.400 --> 00:12:25.800 If that happens, the rate is just going to increase by 2. 00:12:25.800 --> 00:12:36.500 If the concentration of B triples holding A constant, we see that the rate is going to increase by a factor of 3. 00:12:36.500 --> 00:12:39.700 You see what the significance of the rate order is. 00:12:39.700 --> 00:12:58.200 The rate order is really proportional to the sensitivity of a reaction rate on 00:12:58.200 --> 00:13:07.700 the concentration of a specific reactant molecule, on the concentration of a specific reactant. 00:13:07.700 --> 00:13:15.200 Pretty much we see that the larger the value of these rate orders, the more sensitive 00:13:15.200 --> 00:13:22.300 your reaction is to a change in concentration of a specific reactant. 00:13:22.300 --> 00:13:31.000 Basically what the kinetics deals with is focusing on this rate law and solving for x, y, and k. 00:13:31.000 --> 00:13:40.000 In other words, our goal is to solve for the rate constant and rate orders. 00:13:40.000 --> 00:13:42.300 We can do this in one of two ways. 00:13:42.300 --> 00:13:48.500 The first experimental way to derive a rate law is what we call the method of initial rates. 00:13:48.500 --> 00:13:54.500 In the method of initial rates, you basically do what we just did. 00:13:54.500 --> 00:14:16.800 You change one reactant concentration holding all else constant and seeing how the rate varies. 00:14:16.800 --> 00:14:19.600 Let's go ahead and take a look at the following. 00:14:19.600 --> 00:14:25.000 You are usually given some table of data. 00:14:25.000 --> 00:14:33.400 The table of data is going to list different concentrations, different molarities of each reactant and the rate that was measured. 00:14:33.400 --> 00:14:41.600 For example, in experiment number one, this concentration of iodide was found. 00:14:41.600 --> 00:14:46.500 This concentration of thiosulfate was found; this is the initial rate. 00:14:46.500 --> 00:14:53.900 In experiment number two, we see that the concentration of thiosulfate was fixed. 00:14:53.900 --> 00:15:01.400 The concentration of iodide was tripled; we see that the rate also tripled. 00:15:01.400 --> 00:15:13.300 What does that mean?--basically as the concentration of iodide tripled, the rate tripled. 00:15:13.300 --> 00:15:17.200 This is a 1:1 correspondence. 00:15:17.200 --> 00:15:27.400 Anytime you have this 1:1 correspondence, it is a rate order of 1; rate order of 1. 00:15:27.400 --> 00:15:33.000 Let's now see what the rate order is for S2O8^2-, thiosulfate. 00:15:33.000 --> 00:15:41.100 For thiosulfate, we are going to look at experiments two and three. 00:15:41.100 --> 00:15:53.200 You see that as the concentration of thiosulfate tripled... I'm sorry. 00:15:53.200 --> 00:15:56.000 We have to do experiments one and three, not two and three. 00:15:56.000 --> 00:16:05.700 Here as the concentration of thiosulfate tripled, iodide was held constant. 00:16:05.700 --> 00:16:08.600 What happened to the rate?--the rate tripled. 00:16:08.600 --> 00:16:15.600 As S2O8^2- tripled, the rate tripled holding everything else constant. 00:16:15.600 --> 00:16:26.000 This was also a 1:1 correspondence; the rate order is also 1. 00:16:26.000 --> 00:16:28.900 Therefore we have the following rate law. 00:16:28.900 --> 00:16:34.600 The rate of this reaction is equal to some constant k times the concentration of I^- 00:16:34.600 --> 00:16:42.400 raised to the first power and the concentration of S2O8^2- raised to the first power. 00:16:42.400 --> 00:16:51.600 The next thing now that we have the rate orders, we can go ahead and now solve for k. 00:16:51.600 --> 00:16:59.600 We can solve for k by simply plugging in any experiment--one, two, or three--directly into the equation. 00:16:59.600 --> 00:17:08.800 Solve for k using any experiment number. 00:17:08.800 --> 00:17:17.300 For example, let's use experiment one; the rate was 0.044 molarity per second. 00:17:17.300 --> 00:17:21.500 That is going to be equal to the rate constant k times the concentration of I^- which is 00:17:21.500 --> 00:17:30.500 0.125 molar times the concentration of thiosulfate, 0.150 molar, all raised to the first power. 00:17:30.500 --> 00:17:43.500 Then you can just use your algebra to go ahead and solve for the value of k for this second order overall reaction. 00:17:43.500 --> 00:17:50.100 Here we are going to get units of inverse molarity inverse seconds. 00:17:50.100 --> 00:17:58.100 That is how we use the method of initial rates--very straightforward experiment to determine the rate law. 00:17:58.100 --> 00:18:05.000 The second way of determining a rate law is to use what we call integrated rate laws. 00:18:05.000 --> 00:18:15.100 Integrated rate laws, they quantify the relationship between the reactant concentration and time; and time. 00:18:15.100 --> 00:18:20.100 Basically these are all derived mathematically. 00:18:20.100 --> 00:18:23.900 You should always ask your instructor if you need to know how to derive it or not. 00:18:23.900 --> 00:18:28.300 For zero order rate law, the rate is equal to the rate constant k. 00:18:28.300 --> 00:18:31.200 The integrated rate law is the following. 00:18:31.200 --> 00:18:36.500 The concentration of A at any given time is equal to the initial concentration of A minus kt. 00:18:36.500 --> 00:18:44.700 For the first order overall, the integrated rate law is natural log of A0 minus natural log of A equal to kt. 00:18:44.700 --> 00:18:53.500 Finally the second order integrated rate law is 1 over A minus 1 over A0 is equal to kt. 00:18:53.500 --> 00:18:58.800 The nice thing about these equations is that they are all linear. 00:18:58.800 --> 00:19:06.400 If several concentrations are determined at different times, you can get the rate constant graphically. 00:19:06.400 --> 00:19:16.400 For a zero order overall reaction, we are basically going to graph the concentration of A as a function of t. 00:19:16.400 --> 00:19:22.400 That is going to give us a nice straight line with slope equal to ?k. 00:19:22.400 --> 00:19:33.700 For first order overall, we can go ahead and graph the natural log of the concentration of A versus t. 00:19:33.700 --> 00:19:39.600 We are also going to get a straight line whose slope is equal to ?k. 00:19:39.600 --> 00:19:49.000 For second order overall, we are going to plot 1 over the concentration of A which is equal to time. 00:19:49.000 --> 00:19:53.800 Here we are going to get a nice straight line with a positive slope equal to k. 00:19:53.800 --> 00:19:58.300 Once again this is a graphical determination of the rate constant anytime 00:19:58.300 --> 00:20:09.800 you have data from several different time intervals and measured reactant concentration. 00:20:09.800 --> 00:20:13.800 Another thing we like to talk about too is the half-life. 00:20:13.800 --> 00:20:21.600 For zero order, the half-life is equal to the initial concentration of A over 2k. 00:20:21.600 --> 00:20:27.700 For first order, the half-life is equal to the natural log of 2 over k. 00:20:27.700 --> 00:20:34.800 For second order, the half-life is equal to 1 over k times the concentration of A0. 00:20:34.800 --> 00:20:40.800 Once again you should ask your instructor to make sure if you have to know the derivation or not. 00:20:40.800 --> 00:21:01.500 Basically the half-life is very important because it tells us the time required to reach 1/2 of the initial concentration of A. 00:21:01.500 --> 00:21:07.300 Again that is what we call integrated rate laws. 00:21:07.300 --> 00:21:10.900 The next thing we are going to look at is what we call a reaction mechanism. 00:21:10.900 --> 00:21:18.500 A reaction mechanism basically represents the step by step reactions which when combined give you the overall net reaction. 00:21:18.500 --> 00:21:24.700 For example, let's say we had the following given, step one. 00:21:24.700 --> 00:21:33.200 Step one was 2NO gas going on to form N2O2 gas. 00:21:33.200 --> 00:21:39.400 This tells us that the reaction is occurring both in the forward and reverse directions. 00:21:39.400 --> 00:21:44.800 This is what we call the reversible reaction; this is usually very very fast. 00:21:44.800 --> 00:21:57.700 Step number two is going to be O2 gas plus N2O2 gas going on to form 2NO2 gas. 00:21:57.700 --> 00:22:03.400 You are told that this reaction is very very slow. 00:22:03.400 --> 00:22:16.100 The overall reaction is going to be 2NO gas plus O2 gas going on to form 2NO2 gas. 00:22:16.100 --> 00:22:22.400 You notice that N2O2 gets cancelled out. 00:22:22.400 --> 00:22:31.400 Any item in a chemical reaction mechanism that gets cancelled is what we call an intermediate. 00:22:31.400 --> 00:22:49.800 That is it is both formed and consumed during the course of a chemical reaction; formed and consumed during a reaction. 00:22:49.800 --> 00:22:56.500 The reason why we care about what the slow step is is because of the following. 00:22:56.500 --> 00:23:02.200 The slow step, which is in this case step two, is like the weakest link in your chain. 00:23:02.200 --> 00:23:06.700 It is like the slowest person on your track and field relay team. 00:23:06.700 --> 00:23:12.600 The slow step determines, it limits how fast a reaction can go. 00:23:12.600 --> 00:23:19.100 We call this the rate limiting step. 00:23:19.100 --> 00:23:28.500 The rate limiting step is equal to the rate of the overall reaction. 00:23:28.500 --> 00:23:32.400 If we were to write out the rate law for this, we would get the following. 00:23:32.400 --> 00:23:36.900 The rate is equal to the rate constant k2... 00:23:36.900 --> 00:23:38.800 This is k2, of the second step. 00:23:38.800 --> 00:23:42.100 Times the concentration of O2 raised to the first power 00:23:42.100 --> 00:23:47.400 times the concentration of N2O2 raised to the first power. 00:23:47.400 --> 00:23:58.000 We can always use the coefficients as the orders if the reaction you are 00:23:58.000 --> 00:24:03.700 looking at is a part of the mechanism, what we call an elementary reaction. 00:24:03.700 --> 00:24:11.300 Coefficient is equal to the rate orders for elementary steps only. 00:24:11.300 --> 00:24:14.600 Otherwise you would have to go through integrated rate laws or method of initial rates 00:24:14.600 --> 00:24:18.400 to go through the whole process again to find what x and y are. 00:24:18.400 --> 00:24:20.600 When we look at this, we have a problem. 00:24:20.600 --> 00:24:23.200 We have N2O2 appearing in this rate law. 00:24:23.200 --> 00:24:25.500 This is the intermediate. 00:24:25.500 --> 00:24:28.500 You can never have an intermediate appearing in the rate law. 00:24:28.500 --> 00:24:32.700 We have to do something about this; what we do is the following. 00:24:32.700 --> 00:24:40.700 We use what is called a steady state approximation. 00:24:40.700 --> 00:24:52.300 You utilize the fast equilibrium step where the k1 times the concentration of NO squared 00:24:52.300 --> 00:25:02.000 equals to k-1 times the concentration of N2O2 where k1 represents the forward. 00:25:02.000 --> 00:25:04.800 k-1 represents the reverse. 00:25:04.800 --> 00:25:07.400 What we do then is we solve for the intermediate. 00:25:07.400 --> 00:25:18.000 Concentration of N2O2 is equal to k1 over k-1 times the concentration of NO2 squared. 00:25:18.000 --> 00:25:21.900 We then plug this back into our experimental rate law. 00:25:21.900 --> 00:25:25.800 Rate is equal to k2 times the concentration of O2 00:25:25.800 --> 00:25:34.100 times k1 over k-1 times the concentration of NO squared. 00:25:34.100 --> 00:25:39.000 You can collect all of the constants together and just call that what we call kobserved. 00:25:39.000 --> 00:25:46.600 We get left with O2 times the concentration of NO squared. 00:25:46.600 --> 00:25:54.800 Here we have our final rate law that is going to be the rate law for the overall reaction. 00:25:54.800 --> 00:25:58.800 It is just by coincidence here that the rate orders are the coefficients. 00:25:58.800 --> 00:25:59.800 That is not usually the case. 00:25:59.800 --> 00:26:11.500 But that is how we solve for it where kobserved is equal to k2k1 over k-1. 00:26:11.500 --> 00:26:17.100 Once again this is how you deal with reaction mechanism problems. 00:26:17.100 --> 00:26:19.300 Let's now move on to the next topic. 00:26:19.300 --> 00:26:23.200 This is the relationship between temperature and reaction rate. 00:26:23.200 --> 00:26:28.100 Basically in general, as temperature increases, so does the reaction rate; just think about this. 00:26:28.100 --> 00:26:33.700 You know the tea bag is going to brew faster in warm water than cold water. 00:26:33.700 --> 00:26:37.200 You can see that visually happening right before you. 00:26:37.200 --> 00:26:42.000 In order for a reaction to occur, reactant molecules must do two things. 00:26:42.000 --> 00:26:46.500 They must collide; they must collide with sufficient energy. 00:26:46.500 --> 00:26:52.800 They must collide in the proper orientation; sufficient energy and proper orientation. 00:26:52.800 --> 00:26:56.200 Basically we can look at a sample here. 00:26:56.200 --> 00:27:08.000 Let the y-axis be fraction of sample; let the x-axis be temperature. 00:27:08.000 --> 00:27:19.100 At any given temperature, I am going to have a bell curve distribution of molecules just like that. 00:27:19.100 --> 00:27:21.500 Let's call this T1. 00:27:21.500 --> 00:27:29.500 What happens to T2?--what happens when we have a hotter temperature? 00:27:29.500 --> 00:27:32.000 When I have a hotter temperature, my bell curve is going to shift just like that. 00:27:32.000 --> 00:27:38.100 I call this T2; T2 is greater than T1. 00:27:38.100 --> 00:27:43.100 Let's say that in order for the sufficient energy, in order for the 00:27:43.100 --> 00:27:48.700 reaction to occur, let's go ahead and make that as a dotted line. 00:27:48.700 --> 00:27:57.500 I am going to call this dotted line EA; EA is equal to activation energy. 00:27:57.500 --> 00:28:11.400 What this is, it is the minimum energy required for the reaction to proceed, for collisions to occur. 00:28:11.400 --> 00:28:24.600 Basically anything below EA, anything less than EA, you get zero collisions and no reaction. 00:28:24.600 --> 00:28:31.700 Anything greater than or equal to EA, you get collisions; therefore a reaction will occur. 00:28:31.700 --> 00:28:36.700 Basically you see that as you go from T1 to T2, 00:28:36.700 --> 00:28:43.000 the fractional molecules with an energy greater than EA significantly increases. 00:28:43.000 --> 00:28:59.200 At T2, larger percent of molecules with an energy greater than or equal to EA. 00:28:59.200 --> 00:29:09.400 This is why as temperature goes up, so does the rate of a reaction in general. 00:29:09.400 --> 00:29:13.800 We have a nice equation which can actually quantify this. 00:29:13.800 --> 00:29:16.000 This is called the Arrhenius equation. 00:29:16.000 --> 00:29:26.600 The natural log of k1 over k2 equals to EA over R times 1 over T2 minus 00:29:26.600 --> 00:29:39.600 1 over T1 where R is our universal gas constant in terms of energy, 8.314 joules per K times mole. 00:29:39.600 --> 00:29:48.800 Temperatures T1 and T2 are kelvin temperatures; what this basically says is the following. 00:29:48.800 --> 00:29:53.700 If I do a series of reactions at different temperatures and I calculate the 00:29:53.700 --> 00:30:00.000 rate constant, I can then do graphical determination of the activation energy. 00:30:00.000 --> 00:30:09.200 If I graph natural log of k1 over k2 as a function of 1 over temperature, 00:30:09.200 --> 00:30:15.600 I am going to get a nice straight line with slope equal to ?EA over R. 00:30:15.600 --> 00:30:28.700 Again the Arrhenius equation is very useful because it gives us graphical approximation of the activation energy for a reaction. 00:30:28.700 --> 00:30:33.400 Again this is the relationship between rate and temperature. 00:30:33.400 --> 00:30:36.700 Finally the last factor we are going to study is a catalyst. 00:30:36.700 --> 00:30:39.700 Basically a catalyst's job is to do the following. 00:30:39.700 --> 00:30:47.400 A catalyst assists reactant molecules to be in the proper orientation for proper collision to occur. 00:30:47.400 --> 00:30:52.900 What that does is that the activation energy is lower. 00:30:52.900 --> 00:31:07.500 If this reaction represents without a catalyst, the activation energy is going be basically right here. 00:31:07.500 --> 00:31:13.000 This is the energy that you must overcome for the reaction to go form products. 00:31:13.000 --> 00:31:18.500 I can then proceed and draw another curve where this is a catalyst now. 00:31:18.500 --> 00:31:24.000 With the catalyst, you see that the activation energy EA is much lower. 00:31:24.000 --> 00:31:32.600 Activation energy catalyst is going to be always much less than the activation energy no catalyst. 00:31:32.600 --> 00:31:40.400 Because of that, with the lower activation energy, that means a faster reaction. 00:31:40.400 --> 00:31:47.100 A catalyst again speeds up a reaction by lowering the activation energy. 00:31:47.100 --> 00:31:52.800 We see that of course the nice about catalysts is that they can be reused over and 00:31:52.800 --> 00:32:00.800 over again because during the course of a reaction, they are recovered; they are recovered. 00:32:00.800 --> 00:32:03.500 That is catalysis. 00:32:03.500 --> 00:32:07.400 Let's now get into our summary before we jump into our sample problems. 00:32:07.400 --> 00:32:11.700 Kinetic studies the factors that can influence the rate of a chemical reaction. 00:32:11.700 --> 00:32:15.700 We saw that we can have two main experiments to help us determine 00:32:15.700 --> 00:32:20.000 the rate law--the method of initial rates and the integrated rate law. 00:32:20.000 --> 00:32:26.200 We found that in a reaction mechanism, that the slowest step dictates the overall reaction. 00:32:26.200 --> 00:32:32.300 That is what we call the rate limiting step. 00:32:32.300 --> 00:32:43.400 Finally we introduced the Arrhenius equation which gives a mathematical relationship between the reaction rate and the temperature. 00:32:43.400 --> 00:32:54.800 The nice thing about this equation again, this gave us graphical estimate of EA. 00:32:54.800 --> 00:32:57.500 Let's now get into a pair of sample problems. 00:32:57.500 --> 00:33:00.400 A certain first order reaction has a half-life of twenty minutes. 00:33:00.400 --> 00:33:04.200 Calculate the rate constant; that is part A. 00:33:04.200 --> 00:33:08.700 Part B, how much time is required for this reaction to be 75 percent complete? 00:33:08.700 --> 00:33:12.200 Let's go ahead; you are told that the reaction is first order. 00:33:12.200 --> 00:33:21.100 For a first order reaction, T1/2 is equal to the natural log of 2 over the rate constant k. 00:33:21.100 --> 00:33:27.100 You are told that the rate constant is 20.0 minutes. 00:33:27.100 --> 00:33:34.300 Here the rate constant is simply going to be equal to the natural log of 2 divided by 20.0 minutes. 00:33:34.300 --> 00:33:43.700 That gives us our answer in units of reciprocal minutes; that is part A. 00:33:43.700 --> 00:33:45.700 Let's go ahead and do part B now. 00:33:45.700 --> 00:33:49.100 How much time is required for this reaction to be 75 percent complete? 00:33:49.100 --> 00:33:56.600 As soon as you see the word time, you should immediately, immediately, think integrated rate law because 00:33:56.600 --> 00:34:01.200 method of initial rates does not have time in it; only integrated rate laws does. 00:34:01.200 --> 00:34:07.500 For the first order integrated rate law, it is the natural log of the initial concentration 00:34:07.500 --> 00:34:16.000 of A minus the natural log of the concentration of A is equal to kt. 00:34:16.000 --> 00:34:20.300 We know what k already is because we already solved for that in part A. 00:34:20.300 --> 00:34:22.500 We are good to go on that. 00:34:22.500 --> 00:34:25.000 The question is asking for how much time is required. 00:34:25.000 --> 00:34:28.000 This is what we are trying to solve for. 00:34:28.000 --> 00:34:34.300 All that matters is is what is the identity of A0 and what is the identity of A? 00:34:34.300 --> 00:34:39.300 You are told that how much time is required for this reaction to be 75 percent complete? 00:34:39.300 --> 00:34:43.500 Let's say A0 is going to be 100. 00:34:43.500 --> 00:34:49.700 If the reaction is 75 percent complete, that means only 25 percent of A is remaining. 00:34:49.700 --> 00:34:52.500 A is going to be 25. 00:34:52.500 --> 00:34:57.300 Again we now have enough information to solve for t. 00:34:57.300 --> 00:35:05.000 We are going to get our final answer of t in units of minutes. 00:35:05.000 --> 00:35:07.100 How do you know if you have done something wrong? 00:35:07.100 --> 00:35:19.400 You should always check your answer because if you get a ?t, again that just doesn't make physical sense. 00:35:19.400 --> 00:35:21.400 You know you have done something mathematically wrong. 00:35:21.400 --> 00:35:25.200 Always check your answer for a negative time. 00:35:25.200 --> 00:35:27.100 That is sample problem number one. 00:35:27.100 --> 00:35:29.800 Let's go ahead and move on to sample problem number two. 00:35:29.800 --> 00:35:32.000 Consider the following reaction. 00:35:32.000 --> 00:35:36.700 2N2O5 gas goes on to form 4NO2 gas plus O2 gas. 00:35:36.700 --> 00:35:45.200 Here we are given several rate constants that were found at several temperatures. 00:35:45.200 --> 00:35:49.700 The only equation that we know that deals with this is the Arrhenius equation. 00:35:49.700 --> 00:35:54.000 The natural log of k1 over k2 is equal to 00:35:54.000 --> 00:36:00.900 EA over R times 1 over T2 minus 1 over T 1. 00:36:00.900 --> 00:36:05.100 We know that we are going to be using this equation quite easily. 00:36:05.100 --> 00:36:10.100 Here we can calculate EA from there. 00:36:10.100 --> 00:36:14.500 The nice thing about this is because this is going to give us a nice straight line, 00:36:14.500 --> 00:36:33.300 we can use any pair of k and T data points to go ahead and solve for EA. 00:36:33.300 --> 00:36:39.000 Once again this is going to be using the Arrhenius equation to solve for EA. 00:36:39.000 --> 00:36:46.000 Our units of EA is going to be in kilojoules per mole. 00:36:46.000 --> 00:36:48.600 Next one is what is the order of the reaction. 00:36:48.600 --> 00:36:53.500 This is kind of a trick question; this is something I have asked students before. 00:36:53.500 --> 00:36:56.700 This is something I have seen asked by other instructors before. 00:36:56.700 --> 00:36:59.200 The order of the reaction, you don't have to do any work for that. 00:36:59.200 --> 00:37:05.900 The reason is because they already give you the units for k. 00:37:05.900 --> 00:37:14.500 The units of the rate constant tell us the rate order. 00:37:14.500 --> 00:37:27.900 It is only first order where the units of k is reciprocal time; first order overall. 00:37:27.900 --> 00:37:30.900 Again just watch out for that when you do problems. 00:37:30.900 --> 00:37:36.800 Again the units of k tell us a great deal of information without doing any work. 00:37:36.800 --> 00:37:42.200 That is our lecture from general chemistry concerning kinetics. 00:37:42.200 --> 00:37:43.300 I want to thank you for your time. 00:37:43.300 --> 00:37:45.000 I will see you next time on Educator.com.