1 00:00:00 --> 00:00:01 2 00:00:01 --> 00:00:02 The following content is provided under a Creative 3 00:00:02 --> 00:00:03 Commons license. 4 00:00:03 --> 00:00:07 Your support will help MIT OpenCourseWare continue to 5 00:00:07 --> 00:00:10 offer high quality educational resources for free. 6 00:00:10 --> 00:00:13 To make a donation, or view additional materials from 7 00:00:13 --> 00:00:17 hundreds of MIT courses, visit MIT OpenCourseWare 8 00:00:17 --> 00:00:20 at ocw.mit.edu. 9 00:00:20 --> 00:00:21 PROFESSOR: So, any questions from last time? 10 00:00:21 --> 00:00:27 We started doing reaction mechanisms. 11 00:00:27 --> 00:00:33 We did first order parallel reactions. 12 00:00:33 --> 00:00:38 So one of the things that we'll be stressing is that the 13 00:00:38 --> 00:00:43 mechanisms that we write down are ways to try to understand a 14 00:00:43 --> 00:00:46 complicated chemical process. 15 00:00:46 --> 00:00:53 And you take data, you can find intermediates sometimes. 16 00:00:53 --> 00:00:56 And you get rate laws from the data. 17 00:00:56 --> 00:00:58 You infer rate laws. 18 00:00:58 --> 00:01:02 And from these rate laws, you think up of a reaction 19 00:01:02 --> 00:01:09 mechanism and you make sure that what you measure is 20 00:01:09 --> 00:01:13 consistent with your mechanism that you make up to explain the 21 00:01:13 --> 00:01:15 complicated chemical process. 22 00:01:15 --> 00:01:18 And one really important thing to remember is that when you 23 00:01:18 --> 00:01:22 come up with a mechanism and you've got a bunch of data 24 00:01:22 --> 00:01:25 that supports it, it's great. 25 00:01:25 --> 00:01:29 But it doesn't mean that you've solve the problem. 26 00:01:29 --> 00:01:33 It doesn't mean that you've proven that the way that the 27 00:01:33 --> 00:01:36 chemistry proceeds is really the way it proceeds. 28 00:01:36 --> 00:01:41 Because there's always the possibility that there's an 29 00:01:41 --> 00:01:43 intermediate somewhere in there that you haven't 30 00:01:43 --> 00:01:45 been able to measure. 31 00:01:45 --> 00:01:49 And a lot of people got tripped up over the course of the last 32 00:01:49 --> 00:01:55 century, of making up mechanisms, getting a lot of 33 00:01:55 --> 00:01:56 data that supports the mechanism and missing 34 00:01:56 --> 00:02:01 something really important. 35 00:02:01 --> 00:02:05 So it's a common mistake, that people think they've proven a 36 00:02:05 --> 00:02:09 particular chemical mechanism just because they have 37 00:02:09 --> 00:02:12 data that supports it. 38 00:02:12 --> 00:02:15 It's up to a certain point that you know what you have. 39 00:02:15 --> 00:02:20 And we'll see examples of that today, probably. 40 00:02:20 --> 00:02:23 And then certainly next time when we start making 41 00:02:23 --> 00:02:25 approximations in our kinetics. 42 00:02:25 --> 00:02:28 And see where things can really go wrong if you 43 00:02:28 --> 00:02:30 don't have enough data. 44 00:02:30 --> 00:02:33 If you don't try to fish out really, really, tiny 45 00:02:33 --> 00:02:37 intermediates somewhere along the way. 46 00:02:37 --> 00:02:40 So last time we did then parallel first order reactions. 47 00:02:40 --> 00:02:46 And today we're going to march through and do parallel 48 00:02:46 --> 00:02:47 first and second order. 49 00:02:47 --> 00:02:50 So one is first order, the other one is second order. 50 00:02:50 --> 00:02:52 Things are going to get more complicated. 51 00:02:52 --> 00:02:54 And what you're going to get is a flavor of how 52 00:02:54 --> 00:02:55 to solve the problems. 53 00:02:55 --> 00:02:57 Setting up the problems. 54 00:02:57 --> 00:02:59 We're not going to do a lot of the algebra here on the board, 55 00:02:59 --> 00:03:02 because otherwise we would spend the next three 56 00:03:02 --> 00:03:04 weeks doing algebra. 57 00:03:04 --> 00:03:06 And that would be no fun at all. 58 00:03:06 --> 00:03:08 That doesn't mean that you're not going to be doing any 59 00:03:08 --> 00:03:10 algebra on the homework. 60 00:03:10 --> 00:03:15 Because part of it is figuring how to do it. 61 00:03:15 --> 00:03:22 So, we have a reaction where A goes to B plus C. 62 00:03:22 --> 00:03:24 And the first thing we did was, the first mechanism we wrote 63 00:03:24 --> 00:03:28 was a parallel mechanism where we had A goes 64 00:03:28 --> 00:03:32 B and A goes to C. 65 00:03:32 --> 00:03:34 That's the mechanism. 66 00:03:34 --> 00:03:42 And one other way to write it is A goes to B or C, in a way 67 00:03:42 --> 00:03:49 that is more, sort of describes the branching process, that 68 00:03:49 --> 00:03:53 when we talk about a branching ratio, the ratio of amount of B 69 00:03:53 --> 00:03:56 to C, this looks like the branching out of two 70 00:03:56 --> 00:03:59 different paths. 71 00:03:59 --> 00:04:03 And this time we're going to do where this is first order 72 00:04:03 --> 00:04:07 and this is second order. 73 00:04:07 --> 00:04:13 With the rate k1 and a rate k2. 74 00:04:13 --> 00:04:15 And the way that you do all these problems, you have to 75 00:04:15 --> 00:04:18 be very systematic about it. 76 00:04:18 --> 00:04:20 The first thing you do is you have to write down 77 00:04:20 --> 00:04:22 all your rate laws. 78 00:04:22 --> 00:04:25 And make sure that you include everything. 79 00:04:25 --> 00:04:27 So you start by writing the rate law for the 80 00:04:27 --> 00:04:28 destruction of A. 81 00:04:28 --> 00:04:32 And then the rate law for the appearance of B and for C. 82 00:04:32 --> 00:04:33 So for a, we have dA/dt. 83 00:04:33 --> 00:04:36 And again, I'm dropping the brackets around the A, because 84 00:04:36 --> 00:04:41 I don't want to carry around all these extra symbols. 85 00:04:41 --> 00:04:45 So we destroy A by making B. 86 00:04:45 --> 00:04:47 So that's going to be first order. 87 00:04:47 --> 00:04:50 And then we also destroy A by making C, and 88 00:04:50 --> 00:04:50 that's second order. 89 00:04:50 --> 00:04:55 So we have two paths out of A. 90 00:04:55 --> 00:04:57 To get rid of A. 91 00:04:57 --> 00:05:02 Then we write down the appearance of B, dB/dt. 92 00:05:02 --> 00:05:09 It's only through one path, first order in A, and then the 93 00:05:09 --> 00:05:16 appearance of C, only one path which is second order. 94 00:05:16 --> 00:05:18 And these are all our differential equations. 95 00:05:18 --> 00:05:24 And now the trick is, how do you solve these three 96 00:05:24 --> 00:05:27 differential equations in a way that you get meaningful 97 00:05:27 --> 00:05:32 information out. 98 00:05:32 --> 00:05:36 Well, the first thing to do is to, so these 99 00:05:36 --> 00:05:40 two here depend on A. 100 00:05:40 --> 00:05:42 So the appearance of B depends on A, the appearance 101 00:05:42 --> 00:05:43 of C depends on A. 102 00:05:43 --> 00:05:45 But this one here only depends on A by itself. 103 00:05:45 --> 00:05:48 So this is the first one you're going to start out with. 104 00:05:48 --> 00:05:50 Because you can put all of the A's on one side and 105 00:05:50 --> 00:05:53 all the time on the other side and integrate. 106 00:05:53 --> 00:06:02 So then you solve, and you can rewrite this as minus dA/dt is 107 00:06:02 --> 00:06:14 equal to k1 times A times one plus k2 over k1 times A. 108 00:06:14 --> 00:06:16 And then you put all the A's on one side, all the 109 00:06:16 --> 00:06:18 t's on the other side. 110 00:06:18 --> 00:06:28 And integrate from A0 to A minus dA over A times one 111 00:06:28 --> 00:06:33 plus k2 over k1 times A. 112 00:06:33 --> 00:06:39 That is equal to k1 from zero to t dt. 113 00:06:39 --> 00:06:41 And then you have to solve this integral here. 114 00:06:41 --> 00:06:44 And there's a reason why I factored out the A here. 115 00:06:44 --> 00:06:47 It's because this is of the form you can use to use the 116 00:06:47 --> 00:06:51 trick of partial fractions, which we mentioned this before. 117 00:06:51 --> 00:07:01 So you use partial fractions to solve this. 118 00:07:01 --> 00:07:04 And I'm not going to do it on the board. 119 00:07:04 --> 00:07:08 You turn the crank, you basically write one over A 120 00:07:08 --> 00:07:12 times one plus k2 over k1 A is equal to n1 of A plus n2 over 121 00:07:12 --> 00:07:14 this part here, and you solve for n1 and n2 and you plug it 122 00:07:14 --> 00:07:18 back in, and you redo your integral, et cetera. 123 00:07:18 --> 00:07:20 And you get your answer. 124 00:07:20 --> 00:07:23 Which I'm going to write down because I'm going to use it 125 00:07:23 --> 00:07:36 later. k1 A0 over e to the k1 times t, k1 plus k2 A0. 126 00:07:36 --> 00:07:38 So one of the things we immediately see is even though 127 00:07:38 --> 00:07:42 this mechanism isn't very complicated, just two paths, 128 00:07:42 --> 00:07:45 one first order, one second order, the solutions start to 129 00:07:45 --> 00:07:53 get not so simple pretty quickly. 130 00:07:53 --> 00:07:55 So it depends, there's an exponential on the bottom here. 131 00:07:55 --> 00:07:58 Depends on the initial concentrations and both rate 132 00:07:58 --> 00:08:02 constants are in there. 133 00:08:02 --> 00:08:03 The first thing you want to do is you want to 134 00:08:03 --> 00:08:09 look at limiting cases. 135 00:08:09 --> 00:08:12 Just like before. 136 00:08:12 --> 00:08:17 We did, and the first limiting case we're going to look at, so 137 00:08:17 --> 00:08:20 interesting to look at is this guy right here. k1 138 00:08:20 --> 00:08:22 and k2 A here. 139 00:08:22 --> 00:08:24 If one is bigger than the other, then things 140 00:08:24 --> 00:08:27 will cancel out. 141 00:08:27 --> 00:08:30 And we can get some intuition. 142 00:08:30 --> 00:08:34 So the first limiting case we're going to look at is where 143 00:08:34 --> 00:08:44 k2 A0, this term right here, is much smaller than k1. 144 00:08:44 --> 00:08:47 Now, we want to make sure that in these limiting cases that 145 00:08:47 --> 00:08:48 when I write something like this, that the units 146 00:08:48 --> 00:08:50 actually make sense. 147 00:08:50 --> 00:08:52 That I'm not saying, three apples are less 148 00:08:52 --> 00:08:55 than four oranges. 149 00:08:55 --> 00:09:00 So k1, it's first order, so the units are one over second. k2, 150 00:09:00 --> 00:09:02 it's a second order rate constant, so the units are 151 00:09:02 --> 00:09:07 one over second molar, times moles per liter. 152 00:09:07 --> 00:09:09 So the moles per liters cancel out and I have one over our 153 00:09:09 --> 00:09:12 second is less than one over second, so things are great. 154 00:09:12 --> 00:09:15 It I'm making the right kind of approximation here. 155 00:09:15 --> 00:09:17 What else is this approximation saying? 156 00:09:17 --> 00:09:23 This is saying that the rate into B, so this one is the 157 00:09:23 --> 00:09:28 faster one. k1, the rate into B, is the rate into, is k1, is 158 00:09:28 --> 00:09:35 faster than k2 times A, which is basically the rate into C. 159 00:09:35 --> 00:09:41 So this is saying that the rate into B, to form B, 160 00:09:41 --> 00:09:49 is faster than into C. 161 00:09:49 --> 00:09:53 So we can write down a sketch. 162 00:09:53 --> 00:09:57 We can sketch what we expect this approximation to be, then. 163 00:09:57 --> 00:10:04 And we know it's going to look something like this. 164 00:10:04 --> 00:10:10 We know that A is going to come down, exponentially. 165 00:10:10 --> 00:10:12 Without any structure to it. 166 00:10:12 --> 00:10:14 It's going to go either into B or into C, in some 167 00:10:14 --> 00:10:15 branching fractions. 168 00:10:15 --> 00:10:19 And we know that the rate into B is faster than into C, so B 169 00:10:19 --> 00:10:22 is going to come up like this. 170 00:10:22 --> 00:10:27 And C is going to come up, but slower. 171 00:10:27 --> 00:10:34 And there's going to be a ratio between these two guys. 172 00:10:34 --> 00:10:37 So we know the slope here is going to be slower 173 00:10:37 --> 00:10:38 then the slope for B. 174 00:10:38 --> 00:10:41 And we can check that out, within our approximation. 175 00:10:41 --> 00:10:47 We can write, at t equals zero, find out what these 176 00:10:47 --> 00:10:48 initial slopes are. 177 00:10:48 --> 00:10:50 For the creation of B and C. 178 00:10:50 --> 00:10:58 So dB/dt is the slope of the creation of B at the beginning. 179 00:10:58 --> 00:11:02 At t equals zero. 180 00:11:02 --> 00:11:05 And that's just the rate law. dB/dt at t 181 00:11:05 --> 00:11:11 equals zero is k1 A0. 182 00:11:11 --> 00:11:21 And dC/dt, the initial slope, t equals zero, is k1 A0 squared. 183 00:11:21 --> 00:11:28 Or k2 A0 squared, rather. 184 00:11:28 --> 00:11:36 So I'm going to write it as k2 A0 squared, like this. 185 00:11:36 --> 00:11:40 We said k1 is much bigger than k2 times A0. 186 00:11:40 --> 00:11:41 There's the same A0 here. 187 00:11:41 --> 00:11:48 So we see, just by writing this down, that our intuition that 188 00:11:48 --> 00:11:50 because the rate into B is much faster than into A, that 189 00:11:50 --> 00:11:51 it should look like this. 190 00:11:51 --> 00:11:54 Well it's borne out just like this very simple sort of 191 00:11:54 --> 00:11:59 looking at the initial rate where the slope here is much 192 00:11:59 --> 00:12:04 smaller than the slope here. 193 00:12:04 --> 00:12:10 And then we can take the approximation in here. 194 00:12:10 --> 00:12:13 And see what it implies. 195 00:12:13 --> 00:12:18 So let me do that here, let me just use my green chalk here. 196 00:12:18 --> 00:12:21 So k2 A0 is much less than k1. 197 00:12:21 --> 00:12:27 So this is going to go away. k2 A0 is much less than k1. 198 00:12:27 --> 00:12:29 Well, this is building up from zero. 199 00:12:29 --> 00:12:31 So this is always bigger than one here. 200 00:12:31 --> 00:12:33 So I can get rid of this as well. 201 00:12:33 --> 00:12:34 Because there's the k1 sitting here. 202 00:12:34 --> 00:12:39 And k1 is much bigger than k2 A0. 203 00:12:39 --> 00:12:44 Once I got rid of all these two guys, then the k1's cancel out. 204 00:12:44 --> 00:12:45 And this is looking very nice. 205 00:12:45 --> 00:12:48 Because now I can take that e to the k1 t 206 00:12:48 --> 00:12:50 and put it upstairs. 207 00:12:50 --> 00:12:56 A is equal to approximately A0 e to the minus k1 times 208 00:12:56 --> 00:13:00 time in this approximation. 209 00:13:00 --> 00:13:03 It looks like it's first order coming down, with 210 00:13:03 --> 00:13:05 a rate constant k1. 211 00:13:05 --> 00:13:11 It looks like basically the branching into C is nonexistent 212 00:13:11 --> 00:13:14 as far as, at least in this approximation, as 213 00:13:14 --> 00:13:14 far A is concerned. 214 00:13:14 --> 00:13:19 It's going to look like this thing here. 215 00:13:19 --> 00:13:25 It's going to look like first order A goes to B with 216 00:13:25 --> 00:13:26 rate constant k1. 217 00:13:26 --> 00:13:28 So if you didn't know, if you didn't know that there was 218 00:13:28 --> 00:13:32 another substance, C, that was being formed along the way, if 219 00:13:32 --> 00:13:34 you didn't measure it, if you didn't do the analytical 220 00:13:34 --> 00:13:39 chemistry and sort of fish it out, and you just looked at the 221 00:13:39 --> 00:13:43 two major components, A and B, in this case here, you'd 222 00:13:43 --> 00:13:45 measure a rate out of A, it would look like 223 00:13:45 --> 00:13:48 it's first order. 224 00:13:48 --> 00:13:50 It's great. 225 00:13:50 --> 00:13:51 Got my mechanism. 226 00:13:51 --> 00:13:54 A goes to B, period. 227 00:13:54 --> 00:13:57 And you would know there was a minor component that was being 228 00:13:57 --> 00:14:02 formed at the same time. 229 00:14:02 --> 00:14:15 OK, let's do another approximation. 230 00:14:15 --> 00:14:25 Let's do the other case, where k2 A0 is much bigger than k1. 231 00:14:25 --> 00:14:26 And, in this case here, you have to be a little 232 00:14:26 --> 00:14:27 bit more careful. 233 00:14:27 --> 00:14:31 You have to add that you're going to look at 234 00:14:31 --> 00:14:32 this at early times. 235 00:14:32 --> 00:14:36 And we're going to see why early times is important here. 236 00:14:36 --> 00:14:38 In just five minutes. 237 00:14:38 --> 00:14:41 And early times, and how you define early times. 238 00:14:41 --> 00:14:44 What does it mean for things to be close to time equals zero? 239 00:14:44 --> 00:14:48 Well, we have to have a reference time scale. 240 00:14:48 --> 00:14:51 The reactions, I'm going at a certain rate. 241 00:14:51 --> 00:14:54 And one second may be very slow. 242 00:14:54 --> 00:14:55 Or it may be very fast. 243 00:14:55 --> 00:14:57 Depending on the rate of the reaction. 244 00:14:57 --> 00:15:03 So k1 here defines the time scale of the problem. 245 00:15:03 --> 00:15:06 It's one over second, unit of one over second. 246 00:15:06 --> 00:15:10 And early times means that the times that I'm looking at 247 00:15:10 --> 00:15:15 compare to this rate, it's very small. 248 00:15:15 --> 00:15:19 So k1 t is less than one. 249 00:15:19 --> 00:15:23 That means that the time, during the time period that I'm 250 00:15:23 --> 00:15:26 looking at, hardly anything has happened to the 251 00:15:26 --> 00:15:30 branching of A to B. 252 00:15:30 --> 00:15:32 That that's what I mean by early time. 253 00:15:32 --> 00:15:38 So B is hardly being built up yet. k1 is our reference time. 254 00:15:38 --> 00:15:40 k1 is units of one over second. 255 00:15:40 --> 00:15:42 This has units of seconds. 256 00:15:42 --> 00:15:45 So it's all working out fine. 257 00:15:45 --> 00:15:47 In other words, I would have no idea how to define early times. 258 00:15:47 --> 00:15:50 I could say, it's in a year, it could be, if you're looking at 259 00:15:50 --> 00:15:53 plutonium decay, it's a 100,000 year half-life. 260 00:15:53 --> 00:15:55 A year would be an early time. 261 00:15:55 --> 00:15:58 So it would be really fast, compared to the process. 262 00:15:58 --> 00:16:02 But if you're looking at a reaction that takes a few 263 00:16:02 --> 00:16:04 nanoseconds than a year would be very, very long. 264 00:16:04 --> 00:16:06 So you really need your reference time 265 00:16:06 --> 00:16:11 somewhere in there. 266 00:16:11 --> 00:16:13 So this approximation means that essentially, no 267 00:16:13 --> 00:16:16 B is being created. 268 00:16:16 --> 00:16:19 While we're looking at the process. 269 00:16:19 --> 00:16:24 And so we might then very well expect that if we look at A and 270 00:16:24 --> 00:16:27 C, that the answer is going to be that we're going to see 271 00:16:27 --> 00:16:29 something that's second order. 272 00:16:29 --> 00:16:35 So we expect that the form of A, the way to write it is 273 00:16:35 --> 00:16:39 going to be one over A equals k1 t plus A0. 274 00:16:39 --> 00:16:44 So let's go ahead and take our complete solution and write it 275 00:16:44 --> 00:16:46 in the way that we might expect the answer to be. 276 00:16:46 --> 00:16:49 In terms of one over A rather than A here. 277 00:16:49 --> 00:16:56 So let's just invert it. e to the k1 times time, k1 plus k2 278 00:16:56 --> 00:17:07 A0, minus k2 A0 over k1 A0. 279 00:17:07 --> 00:17:11 This should be a minus sign here. 280 00:17:11 --> 00:17:14 There should be a minus sign here. 281 00:17:14 --> 00:17:16 It didn't matter for our approximation because we got 282 00:17:16 --> 00:17:24 rid of it anyway, but it would be a proper sign here is minus. 283 00:17:24 --> 00:17:27 And so now we need to make our approximation and 284 00:17:27 --> 00:17:29 figure out what this here. 285 00:17:29 --> 00:17:33 So if we take the approximation that k1 t is small, less than 286 00:17:33 --> 00:17:37 one, then what that should trigger in your mind if that if 287 00:17:37 --> 00:17:39 you have e to the something that small you should 288 00:17:39 --> 00:17:39 use a Taylor series. 289 00:17:39 --> 00:17:42 And expand this into a Taylor series. 290 00:17:42 --> 00:17:45 So we're going to see this multiple times. 291 00:17:45 --> 00:17:51 So we take e to the k1 t and expand it out as one plus 292 00:17:51 --> 00:17:55 k1 t plus et cetera. 293 00:17:55 --> 00:17:58 We're going to keep the first order terms in there. 294 00:17:58 --> 00:18:05 So we're going to plug this approximation into here. 295 00:18:05 --> 00:18:07 That's this approximation there. 296 00:18:07 --> 00:18:11 And then we're going to expand it out. 297 00:18:11 --> 00:18:20 So plug it in here, and, keeping it to first order in 298 00:18:20 --> 00:18:24 time, and multiplying it out, and I'm not going to do the 299 00:18:24 --> 00:18:27 intermediate steps, I'm just going to give you the 300 00:18:27 --> 00:18:30 stuff after doing the multiplications, we 301 00:18:30 --> 00:18:32 get k1 t squared. 302 00:18:32 --> 00:18:37 So that's basically k1 times one plus k1 t, and then you 303 00:18:37 --> 00:18:45 have plus k1 k2 A0 times t. 304 00:18:45 --> 00:18:50 That's this term here. k1 k2 A0 times t times that. 305 00:18:50 --> 00:18:54 And the k2 A0 times one gets subtracted out from 306 00:18:54 --> 00:18:56 this minus k2 A0 here. 307 00:18:56 --> 00:18:57 So that's all we have on top. 308 00:18:57 --> 00:19:00 And then there are terms of higher order in time. 309 00:19:00 --> 00:19:01 But we're not going to worry about this. 310 00:19:01 --> 00:19:05 We're going to stick to first order in time. 311 00:19:05 --> 00:19:08 Divided by k1 A0. 312 00:19:08 --> 00:19:11 And now you do your cancellations. 313 00:19:11 --> 00:19:15 We have our approximation at k1 times time is 314 00:19:15 --> 00:19:16 much less than one. 315 00:19:16 --> 00:19:20 So here we have k1 times k1 times time. 316 00:19:20 --> 00:19:23 So this term here means it's much less than this term here, 317 00:19:23 --> 00:19:24 because of our approximations. 318 00:19:24 --> 00:19:28 So we can get rid of this term here. 319 00:19:28 --> 00:19:29 Get rid of that. 320 00:19:29 --> 00:19:35 There's no reason for us to get rid of this one here. 321 00:19:35 --> 00:19:40 Because we've got k2 A0 there, which is much larger than k1. 322 00:19:40 --> 00:19:44 And so when you expand this out, all the 323 00:19:44 --> 00:19:45 k1's cancel out here. 324 00:19:45 --> 00:19:49 So so k1 cancels out that k1, we have one plus one. 325 00:19:49 --> 00:19:51 So this is basically one over A0. 326 00:19:51 --> 00:19:56 And then we have plus k2 A0 time over A0, so this ends 327 00:19:56 --> 00:20:07 up being equal to one over A0 plus k2 times time. 328 00:20:07 --> 00:20:09 Exactly what we expected. 329 00:20:09 --> 00:20:13 That if you put in the right approximation in the math, it 330 00:20:13 --> 00:20:16 behaves as your intuition would tell you. 331 00:20:16 --> 00:20:20 Which is that the branching into B nonexistent. 332 00:20:20 --> 00:20:23 And that you're looking at everything going to C. 333 00:20:23 --> 00:20:27 It looks second order in C. 334 00:20:27 --> 00:20:30 But in this case here, it's only early times. 335 00:20:30 --> 00:20:31 So if you keep going, what happens? 336 00:20:31 --> 00:20:35 If you keep going in time, beyond the time scale of this 337 00:20:35 --> 00:20:44 first order rate constant, and you wait long enough, then the 338 00:20:44 --> 00:20:46 amount of A keeps decreasing. 339 00:20:46 --> 00:20:50 Keeps getting smaller and smaller. 340 00:20:50 --> 00:20:53 So if you start the clock again, a little bit later, 341 00:20:53 --> 00:20:58 where A is much smaller, then k2 times the amount of A you 342 00:20:58 --> 00:21:04 have there, is no longer going to be much greater than k1. 343 00:21:04 --> 00:21:10 At some point along the way, k2 times A is going to 344 00:21:10 --> 00:21:13 be much less than k1. 345 00:21:13 --> 00:21:16 So at some point along the way, it's not going to be this 346 00:21:16 --> 00:21:17 approximation any more. 347 00:21:17 --> 00:21:19 It's going to be the previous one that we did. 348 00:21:19 --> 00:21:22 The one up there. 349 00:21:22 --> 00:21:24 So at early times we start with something that may 350 00:21:24 --> 00:21:27 look second order for A. 351 00:21:27 --> 00:21:31 Time goes on, time goes on, time goes on, A gets depleted. 352 00:21:31 --> 00:21:35 A gets depleted, and the rate k2 times A, which is sitting 353 00:21:35 --> 00:21:40 right here, this part here becomes smaller and smaller. 354 00:21:40 --> 00:21:43 And pretty soon this part wins. 355 00:21:43 --> 00:21:46 And it becomes first order. 356 00:21:46 --> 00:21:52 So what you'd expect to see, then, is if you were to plot as 357 00:21:52 --> 00:22:01 a function of A, you expect to see something that starts out 358 00:22:01 --> 00:22:04 as first order and then as A gets depleted, switches 359 00:22:04 --> 00:22:07 over to second order. 360 00:22:07 --> 00:22:09 Something that's second order here. 361 00:22:09 --> 00:22:11 And first order. 362 00:22:11 --> 00:22:12 This is the concentration of A. 363 00:22:12 --> 00:22:18 So if you were to plot it, here you're plotting 364 00:22:18 --> 00:22:20 the concentration of A. 365 00:22:20 --> 00:22:26 If you were to plot the log, of the concentration of A, at long 366 00:22:26 --> 00:22:29 times you'd expect it to be first order. 367 00:22:29 --> 00:22:31 So something linear. 368 00:22:31 --> 00:22:35 But at early times you'd expect it to be second order. 369 00:22:35 --> 00:22:37 So you'd expect to see something that's nonlinear, 370 00:22:37 --> 00:22:40 then switching to something linear. 371 00:22:40 --> 00:22:48 And if you were to plot it as one over A, then you'd expect 372 00:22:48 --> 00:22:50 to start off with something that's linear, at second 373 00:22:50 --> 00:22:53 order, early times. 374 00:22:53 --> 00:22:58 But instead of keeping, being second order, as you deplete 375 00:22:58 --> 00:23:05 the amount of A, the branching into B becomes important. 376 00:23:05 --> 00:23:09 And you become second order. 377 00:23:09 --> 00:23:14 So this is second order at the beginning. 378 00:23:14 --> 00:23:23 And this becomes first order. 379 00:23:23 --> 00:23:26 So, any questions here? 380 00:23:26 --> 00:23:29 The importance of doing this problem here was to lay out 381 00:23:29 --> 00:23:32 basically a systematic way of looking at the problem. 382 00:23:32 --> 00:23:39 You lay out your rate equations. 383 00:23:39 --> 00:23:40 Like this. 384 00:23:40 --> 00:23:45 You solve what you can, in this case you solve for A. 385 00:23:45 --> 00:23:46 And it's complicated. 386 00:23:46 --> 00:23:49 So you want to learn something about what the data 387 00:23:49 --> 00:23:50 might look like. 388 00:23:50 --> 00:23:51 And you look at limiting cases. 389 00:23:51 --> 00:23:53 Two obvious limiting cases. 390 00:23:53 --> 00:23:56 One rate is faster than the other. 391 00:23:56 --> 00:23:59 You pick the first one first, then go through it. 392 00:23:59 --> 00:24:00 And then use your intuition. 393 00:24:00 --> 00:24:04 Every point along the way, your intuition can tell you what 394 00:24:04 --> 00:24:06 you expect the result to be. 395 00:24:06 --> 00:24:12 So in this case here, our intuition told us that if we 396 00:24:12 --> 00:24:15 took the rate into B to be much faster than that into C, than 397 00:24:15 --> 00:24:19 we expected to see something that was largely first order. 398 00:24:19 --> 00:24:22 And when you we the approximation in our exact 399 00:24:22 --> 00:24:25 solution, in fact it does look like it's first order. 400 00:24:25 --> 00:24:26 So always use your intuition. 401 00:24:26 --> 00:24:29 Because the math is going to get pretty hairy. 402 00:24:29 --> 00:24:32 The algebra might make it complicated. 403 00:24:32 --> 00:24:34 And it's really easy to make a plus sign into a minus sign 404 00:24:34 --> 00:24:35 somewhere along the way. 405 00:24:35 --> 00:24:37 Just like I did here. 406 00:24:37 --> 00:24:43 So if I hadn't fixed my mistake here, and I had gone and put in 407 00:24:43 --> 00:24:46 my Taylor series here, I would have gotten in trouble. 408 00:24:46 --> 00:24:48 Because this minus sign would have been a plus. 409 00:24:48 --> 00:24:50 And these k2 A0's would not have canceled out. 410 00:24:50 --> 00:24:53 And I wouldn't have gotten the right result, which my 411 00:24:53 --> 00:24:59 intuition had told me should be a second order process. 412 00:24:59 --> 00:25:03 And so if you have a problem, let's say you're on an exam and 413 00:25:03 --> 00:25:05 you're doing algebra, you're trying to, and you end up with 414 00:25:05 --> 00:25:08 a result that doesn't match what your intuition tells you, 415 00:25:08 --> 00:25:10 and you don't have time to fix it. 416 00:25:10 --> 00:25:10 Tell us. 417 00:25:10 --> 00:25:13 You know, I know this is wrong because I know it's supposed 418 00:25:13 --> 00:25:14 to be second order. 419 00:25:14 --> 00:25:15 But I don't know where I went wrong. 420 00:25:15 --> 00:25:18 And that's really important. 421 00:25:18 --> 00:25:21 Just tell us. 422 00:25:21 --> 00:25:27 That means that you're thinking. 423 00:25:27 --> 00:25:28 Let's get a little bit more complicated. 424 00:25:28 --> 00:25:35 Any questions? 425 00:25:35 --> 00:25:38 Consecutive series reactions. 426 00:25:38 --> 00:25:41 These were parallel reactions and now. 427 00:25:41 --> 00:25:43 Basically what we're doing here is we're building up a 428 00:25:43 --> 00:25:47 toolkit of simple mechanisms. 429 00:25:47 --> 00:25:51 And then we'll be able to put these mechanisms together 430 00:25:51 --> 00:25:57 to make something more complicated. 431 00:25:57 --> 00:26:04 So the next kind of mechanism, series mechanism, 432 00:26:04 --> 00:26:09 series reactions, so we have our reaction. 433 00:26:09 --> 00:26:12 Which is A goes to C. 434 00:26:12 --> 00:26:15 And the mechanism that's been thought for this reaction is 435 00:26:15 --> 00:26:18 that there's an intermediate. 436 00:26:18 --> 00:26:21 That first you have A goes to B, some intermediate, which 437 00:26:21 --> 00:26:26 gets used up to form the final product, C, with 438 00:26:26 --> 00:26:28 a rate, k2, here. 439 00:26:28 --> 00:26:33 And we can either write the mechanism in two steps. 440 00:26:33 --> 00:26:41 Or we can write it as A goes to B goes to C, like this. 441 00:26:41 --> 00:26:45 And see both ways of writing it. 442 00:26:45 --> 00:26:46 And we want to solve this. 443 00:26:46 --> 00:26:48 And we want to look at special cases, and we just want to 444 00:26:48 --> 00:26:51 understand the implications of this mechanism. 445 00:26:51 --> 00:26:53 So the first thing to do, just like we did before, is to 446 00:26:53 --> 00:26:57 write all the rate laws. 447 00:26:57 --> 00:27:01 Before we got going solving anything. 448 00:27:01 --> 00:27:05 So we start with A, minus dA/dt. 449 00:27:05 --> 00:27:07 There's only one way that gets used up. 450 00:27:07 --> 00:27:11 Through the formation of B in a first order process. 451 00:27:11 --> 00:27:12 Easy. 452 00:27:12 --> 00:27:13 This is easy to solve. 453 00:27:13 --> 00:27:17 We know the answer is going to be exponential in it. 454 00:27:17 --> 00:27:24 For B, dB/dt is equal to, well it can be formed through my 455 00:27:24 --> 00:27:27 first order in A, so that's k1 times A. 456 00:27:27 --> 00:27:31 But it also could get destroyed by forming C. 457 00:27:31 --> 00:27:35 So we've got to get all the paths out of A, here on the 458 00:27:35 --> 00:27:37 right side of this equation. 459 00:27:37 --> 00:27:41 So it can be destroyed in a process, which 460 00:27:41 --> 00:27:42 is first order in B. 461 00:27:42 --> 00:27:46 So we're going to take both k1 and k2 to be first 462 00:27:46 --> 00:27:50 order initially. 463 00:27:50 --> 00:27:51 Then we'll make one of them second order and things will 464 00:27:51 --> 00:27:54 become very complicated and we'll throw up our hands. 465 00:27:54 --> 00:27:57 But for now, let's not throw up our hands quite yet. 466 00:27:57 --> 00:28:02 OK and for dC/dt, there's only one way it can be formed is 467 00:28:02 --> 00:28:05 first order by destroying B. 468 00:28:05 --> 00:28:08 And we have a couple of differential equations here. 469 00:28:08 --> 00:28:16 This differential equation for C, depends on B here. 470 00:28:16 --> 00:28:21 The differential for B depends on A here. 471 00:28:21 --> 00:28:22 So that means that things are going to get a little 472 00:28:22 --> 00:28:28 bit more complicated. 473 00:28:28 --> 00:28:30 The easy one to solve is always the first one here. 474 00:28:30 --> 00:28:31 That's first order. 475 00:28:31 --> 00:28:33 So we just write down the answer. 476 00:28:33 --> 00:28:38 A is equal to A0, e to the minus k1 t. 477 00:28:38 --> 00:28:42 We know the answer to that one. 478 00:28:42 --> 00:28:46 Then we have to work on B. 479 00:28:46 --> 00:28:48 So we want to find out integrated rate laws for every 480 00:28:48 --> 00:28:59 one of these chemical species. 481 00:28:59 --> 00:29:01 And always there are tricks involved. 482 00:29:01 --> 00:29:03 Partial fractions, whatever. 483 00:29:03 --> 00:29:09 So we write down B now. dB/dt plus, let me rearrange it a 484 00:29:09 --> 00:29:12 little bit, plus k2 times B, putting the minus k2 B 485 00:29:12 --> 00:29:14 on the other side here. 486 00:29:14 --> 00:29:16 Is equal to k1 times A. 487 00:29:16 --> 00:29:18 But I'm not going to keep this A there, because now I know 488 00:29:18 --> 00:29:20 what A is in terms of time. 489 00:29:20 --> 00:29:28 And a constant. k1 times A0 e to the minus k1 times time. 490 00:29:28 --> 00:29:31 OK, that's my differential equation for B. 491 00:29:31 --> 00:29:35 Got to solve this thing. 492 00:29:35 --> 00:29:39 Well, the last time I did differential equations in 493 00:29:39 --> 00:29:40 college was a century ago. 494 00:29:40 --> 00:29:42 And I wasn't kidding. 495 00:29:42 --> 00:29:43 It really was last century. 496 00:29:43 --> 00:29:46 In fact, it was last millennium. 497 00:29:46 --> 00:29:49 That was a long time ago. 498 00:29:49 --> 00:29:50 So how do you do this? 499 00:29:50 --> 00:29:52 Well, the trick here, there's a trick. 500 00:29:52 --> 00:29:56 And the trick is to multiply the whole thing, both 501 00:29:56 --> 00:30:01 sides, by e to the k2 t. 502 00:30:01 --> 00:30:04 And you multiply this side by e to the k2 t and you multiply 503 00:30:04 --> 00:30:09 this side by e to the k2 t, to solve this equation. 504 00:30:09 --> 00:30:16 So once you do that, then you have e to the k2 t times 505 00:30:16 --> 00:30:21 dB/dt plus k2 times B. 506 00:30:21 --> 00:30:22 This side here. 507 00:30:22 --> 00:30:30 And then you have is equal to k1 times A0 e to the minus, e 508 00:30:30 --> 00:30:39 to the k2 minus k1 times time. 509 00:30:39 --> 00:30:43 OK, so the reason why this trick works is because this 510 00:30:43 --> 00:30:51 guy right here can be nicely written as d/dt of 511 00:30:51 --> 00:30:56 B times e to the k2 t. 512 00:30:56 --> 00:30:58 It's in nice simple form, so when you integrate, it'll 513 00:30:58 --> 00:31:01 be very easy to integrate. 514 00:31:01 --> 00:31:09 And then the other side, you still have k1 A0 e to the k2 515 00:31:09 --> 00:31:12 minus k1 times the time. 516 00:31:12 --> 00:31:15 So now you integrate both sides with respect to time, The 517 00:31:15 --> 00:31:19 integral of something d/dt with respect to time, it's just 518 00:31:19 --> 00:31:20 going to be itself right here. 519 00:31:20 --> 00:31:22 So it's very simple. 520 00:31:22 --> 00:31:24 So you integrate both sides from zero to time, dt. 521 00:31:24 --> 00:31:27 522 00:31:27 --> 00:31:31 Zero to t, dt on both sides. 523 00:31:31 --> 00:31:37 And on this side here, you end up with B, e to the k2 times 524 00:31:37 --> 00:31:42 time minus B, e to the k2 times zero. 525 00:31:42 --> 00:31:47 Just taking the upper limit minus the lower limit. 526 00:31:47 --> 00:31:53 And that's just equal to B0, at time to equal zero. 527 00:31:53 --> 00:31:58 And so, and at time to equal zero, B0 is equal to zero 528 00:31:58 --> 00:32:01 because we haven't made any intermediates. 529 00:32:01 --> 00:32:06 So this goes away. 530 00:32:06 --> 00:32:08 And then we have an equals sign. 531 00:32:08 --> 00:32:15 And on the other side, we have k1 A0 over k2 minus k1, it's 532 00:32:15 --> 00:32:18 just the integral of this thing, here with a k2 minus k1 533 00:32:18 --> 00:32:24 goes in the denominator. e to the k2 minus k1 times 534 00:32:24 --> 00:32:28 time minus one. 535 00:32:28 --> 00:32:33 So now you have your integral form for the amount of B that 536 00:32:33 --> 00:32:36 gets created at any time. 537 00:32:36 --> 00:32:38 And I want to keep that on the board. 538 00:32:38 --> 00:32:45 Because I'm going to use it later. k1 A0 over k1 minus 539 00:32:45 --> 00:32:54 k2 times e to the minus k1 times time minus e to 540 00:32:54 --> 00:33:01 the minus k2 times time. 541 00:33:01 --> 00:33:03 OK, we're two thirds of the way through. 542 00:33:03 --> 00:33:07 Now we've got to find C. 543 00:33:07 --> 00:33:10 And for the last component, you don't want to be 544 00:33:10 --> 00:33:12 doing this kind of stuff. 545 00:33:12 --> 00:33:14 You don't want to waste your time solving differential 546 00:33:14 --> 00:33:20 equations, because stoichiometry can help you out. 547 00:33:20 --> 00:33:22 At the end of the day, you know that everything that started 548 00:33:22 --> 00:33:25 out as A, in terms of quantities, is going 549 00:33:25 --> 00:33:27 to end up at C. 550 00:33:27 --> 00:33:29 You know that if you end up with a concentration A0 at the 551 00:33:29 --> 00:33:33 beginning, at the end of the process the concentration of C, 552 00:33:33 --> 00:33:37 at infinite time, is also going to be A0. 553 00:33:37 --> 00:33:41 If you know that there's a correlation, there's a 554 00:33:41 --> 00:33:46 stoichiometry that relates A and C together. 555 00:33:46 --> 00:33:47 And B. 556 00:33:47 --> 00:33:53 You have conservation of mass here, basically. 557 00:33:53 --> 00:33:55 Conservation of atoms. 558 00:33:55 --> 00:33:58 That has to come into play 559 00:33:58 --> 00:34:07 So, for C, we're just going to do algebra instead of calculus. 560 00:34:07 --> 00:34:10 So let's write down the stoichiometry here. 561 00:34:10 --> 00:34:15 The amount of C at any time is what it's going 562 00:34:15 --> 00:34:17 to be at the end. 563 00:34:17 --> 00:34:19 Which is A0, very end of the process. 564 00:34:19 --> 00:34:21 It's going to be A0. 565 00:34:21 --> 00:34:25 But before we get to the end, there's still stuff 566 00:34:25 --> 00:34:28 that's left in A and B. 567 00:34:28 --> 00:34:30 And that's going to decrease the amount of C. 568 00:34:30 --> 00:34:35 So minus whatever's still in A and B. 569 00:34:35 --> 00:34:37 That's the stoichiometry. 570 00:34:37 --> 00:34:38 That's one way of writing it. 571 00:34:38 --> 00:34:40 You could also write it in a different way. 572 00:34:40 --> 00:34:47 Which is that the amount of C is equal to the amount of 573 00:34:47 --> 00:34:50 A that you have used up. 574 00:34:50 --> 00:34:53 A0 is what you started out with. 575 00:34:53 --> 00:34:54 A is what's left over. 576 00:34:54 --> 00:34:57 So this is the amount of A used up. 577 00:34:57 --> 00:34:59 It's not all quite in C yet. 578 00:34:59 --> 00:35:02 Some of it is stuck in B. 579 00:35:02 --> 00:35:05 So it's the amount of A used up, minus the amount 580 00:35:05 --> 00:35:08 that's stuck in B. 581 00:35:08 --> 00:35:09 Those two are the same. 582 00:35:09 --> 00:35:10 I hope. 583 00:35:10 --> 00:35:12 A0 minus A, yeah, that's right. 584 00:35:12 --> 00:35:13 Minus B, yeah, that's right. 585 00:35:13 --> 00:35:18 So this is used up. 586 00:35:18 --> 00:35:24 And then stuck in B. 587 00:35:24 --> 00:35:26 That's the hard part. 588 00:35:26 --> 00:35:29 Putting down the stoichiometry. 589 00:35:29 --> 00:35:31 Once you've done the hard part, then if you solve the 590 00:35:31 --> 00:35:32 other two then it's easy. 591 00:35:32 --> 00:35:32 You just plug in. 592 00:35:32 --> 00:35:35 It's just algebra. 593 00:35:35 --> 00:35:37 And you get something complicated. 594 00:35:37 --> 00:35:43 C is equal to complicated. 595 00:35:43 --> 00:35:49 It's in the notes, I'm not going to write it down. 596 00:35:49 --> 00:35:51 So we've solved the problem exactly. 597 00:35:51 --> 00:35:53 And now you're going to do an experiment. 598 00:35:53 --> 00:35:55 And the experiment you're going to do is not going to follow 599 00:35:55 --> 00:35:57 this whole thing, exactly. 600 00:35:57 --> 00:35:59 It's going to look at limited cases. 601 00:35:59 --> 00:36:02 That's what you can do. 602 00:36:02 --> 00:36:04 And so the first thing to look at is to look at 603 00:36:04 --> 00:36:06 the initial times. 604 00:36:06 --> 00:36:08 The very beginning of the process, how are 605 00:36:08 --> 00:36:11 things appearing. 606 00:36:11 --> 00:36:33 And then we'll look at the late times. 607 00:36:33 --> 00:36:35 So initial times. 608 00:36:35 --> 00:36:46 Near t equals zero. 609 00:36:46 --> 00:36:50 Well, every one of these things has an exponential in it. 610 00:36:50 --> 00:36:52 And whenever we see exponentials and we look at 611 00:36:52 --> 00:36:56 something at the early times, times that are faster than one 612 00:36:56 --> 00:36:59 over k1, or faster than one over k2, it tells us 613 00:36:59 --> 00:37:02 use a Taylor series. 614 00:37:02 --> 00:37:07 So use the approximation e to the minus kt is approximately 615 00:37:07 --> 00:37:15 equal to one minus kt plus kt squared over two 616 00:37:15 --> 00:37:16 plus et cetera. 617 00:37:16 --> 00:37:20 And keep as many terms as you need in the Taylor series 618 00:37:20 --> 00:37:21 to get something sensible. 619 00:37:21 --> 00:37:23 So if everything comes out of zero at the end, you know you 620 00:37:23 --> 00:37:26 haven't kept enough terms in t. 621 00:37:26 --> 00:37:28 Things have canceled out too much. 622 00:37:28 --> 00:37:31 And you want something that is time dependent. 623 00:37:31 --> 00:37:33 Because you've got a change in time, right? 624 00:37:33 --> 00:37:36 It's a very common mistake to say only keep one here and then 625 00:37:36 --> 00:37:39 everything cancels out you get zero, and you say, well, B is 626 00:37:39 --> 00:37:41 equal to zero at all times. 627 00:37:41 --> 00:37:46 That's nonsensical. 628 00:37:46 --> 00:37:49 And I don't know a priori how much time to keep. 629 00:37:49 --> 00:37:53 So let's go to second order. 630 00:37:53 --> 00:38:00 So if you plug this Taylor series into A, and get 631 00:38:00 --> 00:38:02 something that's time dependent, you only need 632 00:38:02 --> 00:38:04 to go to first order. 633 00:38:04 --> 00:38:08 One minus k1 times time plus et cetera. 634 00:38:08 --> 00:38:14 Plug it into b, plug your Taylor series into these two 635 00:38:14 --> 00:38:22 guys here, get k1 A0 over k2 minus k1, you only need 636 00:38:22 --> 00:38:24 it to go to first order. 637 00:38:24 --> 00:38:27 The one's cancel out but the times don't cancel out. 638 00:38:27 --> 00:38:33 Minus k1 plus k2, that is, unless k1 and k2 are the same. 639 00:38:33 --> 00:38:36 So we're making the approximation that k1 640 00:38:36 --> 00:38:38 is different than k2. 641 00:38:38 --> 00:38:41 Times the time. 642 00:38:41 --> 00:38:44 And when we look at C, this complicated thing that I didn't 643 00:38:44 --> 00:38:48 write down, if you were just to stop at first order in C, 644 00:38:48 --> 00:38:50 you get C equals zero. 645 00:38:50 --> 00:38:52 You know that can't be right. 646 00:38:52 --> 00:38:55 You know that C is going to increase in time. 647 00:38:55 --> 00:38:56 It can't be equal to zero at all times. 648 00:38:56 --> 00:39:00 So we have to go to second order for C0. 649 00:39:00 --> 00:39:08 And it's approximately equal to A0 times k1 k2 over two 650 00:39:08 --> 00:39:11 times the time squared. 651 00:39:11 --> 00:39:13 Let me rewrite this a little bit closer in, because 652 00:39:13 --> 00:39:15 I need some room here. 653 00:39:15 --> 00:39:23 So c is approximately equal to A0 k1 is k2 over two 654 00:39:23 --> 00:39:24 times time squared. 655 00:39:24 --> 00:39:26 OK, and then we can start plotting out what this is 656 00:39:26 --> 00:39:27 going to look like then. 657 00:39:27 --> 00:39:30 So we have the concentrations time on the axis here. 658 00:39:30 --> 00:39:34 Let's start with the concentrating of A. 659 00:39:34 --> 00:39:38 A is dropping down linearly in time. 660 00:39:38 --> 00:39:41 It's dropping down linearly in time here. 661 00:39:41 --> 00:39:46 B is increasing linearly in time, at the beginning. 662 00:39:46 --> 00:39:51 So B is going up linearly in time, like this. 663 00:39:51 --> 00:39:55 And C is increasing quadratically in time. 664 00:39:55 --> 00:39:58 So it's pretty flat, coming up first. 665 00:39:58 --> 00:40:03 And then it starts to curve up. 666 00:40:03 --> 00:40:05 It's our first approximation. 667 00:40:05 --> 00:40:10 Then, we go to late times. t equals infinity. 668 00:40:10 --> 00:40:11 So t equals infinity where am I going to write this? 669 00:40:11 --> 00:40:33 Have to move this up. 670 00:40:33 --> 00:40:35 Let's go, t goes to infinity. 671 00:40:35 --> 00:40:38 Well, here I know that I'm allowed to go to zero. 672 00:40:38 --> 00:40:42 Or allowed to go to constant, because infinity is forever. 673 00:40:42 --> 00:40:46 So at t infinity, I know that I'm going to use up all of A. 674 00:40:46 --> 00:40:48 So I know that A is going to be equal to 0. 675 00:40:48 --> 00:40:50 I know I'm going to use up all of B. 676 00:40:50 --> 00:40:53 So I know B is going to go to zero. 677 00:40:53 --> 00:40:54 Somehow. 678 00:40:54 --> 00:40:58 And I know that C is going to go to A0. 679 00:40:58 --> 00:41:07 So C is going to go to A0 at infinity. 680 00:41:07 --> 00:41:17 So now on my graph here, I know that at infinity, A is going to 681 00:41:17 --> 00:41:20 go down to zero like this, somehow. 682 00:41:20 --> 00:41:23 Probably exponentially, because it is exponential in time. 683 00:41:23 --> 00:41:25 So it's going to go to zero exponentially. 684 00:41:25 --> 00:41:27 B is going to go down to zero exponentially. 685 00:41:27 --> 00:41:30 There are the exponentials right here. 686 00:41:30 --> 00:41:33 So B is going to go down to zero exponentially. 687 00:41:33 --> 00:41:35 Something like this. 688 00:41:35 --> 00:41:39 And C is going to saturate to A0. 689 00:41:39 --> 00:41:40 Exponentially rise up to A0. 690 00:41:40 --> 00:41:42 So this was A0 right here. 691 00:41:42 --> 00:41:45 It's going to go like this. 692 00:41:45 --> 00:41:47 So then I connect the dots. 693 00:41:47 --> 00:41:50 And I know that C is going to start quadratically. 694 00:41:50 --> 00:41:53 And then it's going to rise up an exponent like this. 695 00:41:53 --> 00:41:57 I know that B is going to start off linearly, and 696 00:41:57 --> 00:41:59 then it has to go to zero. 697 00:41:59 --> 00:42:03 So it has to go through some sort of maximum. 698 00:42:03 --> 00:42:08 And A is going to go down exponentially, 699 00:42:08 --> 00:42:10 exponentially like this. 700 00:42:10 --> 00:42:12 So the interesting part of this diagram is that 701 00:42:12 --> 00:42:13 B goes to a maximum. 702 00:42:13 --> 00:42:21 There is some point, there's some time, that we can call t 703 00:42:21 --> 00:42:24 sub B max, where B is a maxima. 704 00:42:24 --> 00:42:27 And that's an interesting point. 705 00:42:27 --> 00:42:29 That's something that we can try to figure out 706 00:42:29 --> 00:42:31 experimentally what it is. 707 00:42:31 --> 00:42:34 And we could get some parameters, we could maybe 708 00:42:34 --> 00:42:36 extract some information. 709 00:42:36 --> 00:42:42 There's a maximum B, concentration of B, 710 00:42:42 --> 00:42:45 that gets formed. 711 00:42:45 --> 00:42:47 And how do you solve that? 712 00:42:47 --> 00:42:52 You set dB/dt equal to zero. 713 00:42:52 --> 00:42:55 So you have your equation for B right here. 714 00:42:55 --> 00:42:58 You set it equal to zero. and you solve. 715 00:42:58 --> 00:43:00 You get your maximum time. 716 00:43:00 --> 00:43:02 It's made up of rate constant. 717 00:43:02 --> 00:43:04 You get your maximum concentration, which is also 718 00:43:04 --> 00:43:07 made up of rate constants. 719 00:43:07 --> 00:43:09 And you can get rate constants out of this. 720 00:43:09 --> 00:43:10 Question. 721 00:43:10 --> 00:43:16 STUDENT: [INAUDIBLE] 722 00:43:16 --> 00:43:17 PROFESSOR: Yes. 723 00:43:17 --> 00:43:23 STUDENT: [INAUDIBLE] 724 00:43:23 --> 00:43:24 PROFESSOR: That is a coincidence. 725 00:43:24 --> 00:43:31 That is my poor sketching. 726 00:43:31 --> 00:43:37 It should probably be somewhere like, in the middle here. 727 00:43:37 --> 00:43:41 Actually, it completely depends on the rates. 728 00:43:41 --> 00:43:44 And we're going to see limiting cases where this maximum can 729 00:43:44 --> 00:43:51 occur here, or it can occur somewhere down here. 730 00:43:51 --> 00:43:57 So this is the complete case. 731 00:43:57 --> 00:44:04 So, limiting cases. the important limiting cases. 732 00:44:04 --> 00:44:06 We've made an approximation along the way that k1 733 00:44:06 --> 00:44:07 is different than k2. 734 00:44:07 --> 00:44:10 So that has to be a case that needs to be worked out. 735 00:44:10 --> 00:44:17 And it's likely to be easier than the full solution. 736 00:44:17 --> 00:44:22 So one of the limiting cases is k1 is equal to k2 and I'll 737 00:44:22 --> 00:44:27 leave that for homework. 738 00:44:27 --> 00:44:31 You can do that yourselves. 739 00:44:31 --> 00:44:34 It's good exercise. 740 00:44:34 --> 00:44:35 What about another limiting case? 741 00:44:35 --> 00:44:37 Well, you have two other limiting cases. 742 00:44:37 --> 00:44:39 One, you've got two rates. 743 00:44:39 --> 00:44:45 So, one is bigger than the other. 744 00:44:45 --> 00:44:50 No information on this board here. 745 00:44:50 --> 00:44:54 You start off with one case. 746 00:44:54 --> 00:45:03 So k1 greater than k2. k1 greater than k2. 747 00:45:03 --> 00:45:05 What does it mean? k1 greater than k2. 748 00:45:05 --> 00:45:08 It means that the rate determining step, or the 749 00:45:08 --> 00:45:12 limiting step, in this reaction, is the second step. 750 00:45:12 --> 00:45:16 k2 is very slow compared to k1. 751 00:45:16 --> 00:45:21 And sometimes I like to think of these things as pipes. 752 00:45:21 --> 00:45:25 And connecting vessels. 753 00:45:25 --> 00:45:31 So let's say we have the amount of liquid in the top vessel 754 00:45:31 --> 00:45:32 is the concentration of A. 755 00:45:32 --> 00:45:36 The amount of liquid in the middle vessel, let me get my 756 00:45:36 --> 00:45:46 color scheme to be consistent here. 757 00:45:46 --> 00:45:48 Then we have a middle vessel, which is the concentration of 758 00:45:48 --> 00:45:53 B, and a final vessel, which is the concentration of C. 759 00:45:53 --> 00:45:59 And we start out with some sort of amount of A in here. 760 00:45:59 --> 00:46:03 Started with a lot of A, and then these vessels are 761 00:46:03 --> 00:46:04 connecting with pipes. 762 00:46:04 --> 00:46:10 So there's a very thick, big pipe that connects 763 00:46:10 --> 00:46:11 A and B together. 764 00:46:11 --> 00:46:16 The rate is fast, going from A to B. 765 00:46:16 --> 00:46:19 And the rate going from B to C is slow, so there's a skinny 766 00:46:19 --> 00:46:26 pipe connecting B and C together. 767 00:46:26 --> 00:46:29 Then I turn on the system, I set time to equal zero, 768 00:46:29 --> 00:46:31 poof, what happens? 769 00:46:31 --> 00:46:33 There's the big pipe connecting A and B. 770 00:46:33 --> 00:46:37 The whole amount of a here gets transferred to B, suddenly. 771 00:46:37 --> 00:46:38 Then it gets stuck here. 772 00:46:38 --> 00:46:41 And then it dribbles out from B to C. 773 00:46:41 --> 00:46:47 So what do we expect, if we were to plot our quantities 774 00:46:47 --> 00:46:49 as a function of time. 775 00:46:49 --> 00:46:52 What we'd expect then, then we're going to make sure that 776 00:46:52 --> 00:46:57 the math works out, is that A is going to decrease 777 00:46:57 --> 00:46:58 really fast. 778 00:46:58 --> 00:47:02 It's going to go, poof. 779 00:47:02 --> 00:47:08 First order in time with a very fast rate. 780 00:47:08 --> 00:47:11 And all of this is going to go straight into B. 781 00:47:11 --> 00:47:13 So we expect B to go linearly. 782 00:47:13 --> 00:47:14 Remember, at the beginning it's linear. 783 00:47:14 --> 00:47:17 Linear goes up. 784 00:47:17 --> 00:47:20 Almost all the way up to A0. 785 00:47:20 --> 00:47:22 Because it can't hardly get out. 786 00:47:22 --> 00:47:26 Reaches its maximum, then it goes from B to C through 787 00:47:26 --> 00:47:28 this skinny, skinny pipe. 788 00:47:28 --> 00:47:32 First order rate, with a rate k2, so we're exponentially 789 00:47:32 --> 00:47:38 going down, slowly, slowly, slowly, with a rate k2. 790 00:47:38 --> 00:47:44 And then C starts quadratically. 791 00:47:44 --> 00:47:46 And then very quickly, once everything's into 792 00:47:46 --> 00:47:49 B, A is forgotten. 793 00:47:49 --> 00:47:51 A doesn't matter any more. 794 00:47:51 --> 00:47:54 Basically what you're looking at is a transfer of B into 795 00:47:54 --> 00:47:56 C through this little pipe. 796 00:47:56 --> 00:47:59 And you know that's going to be a first order of rate. 797 00:47:59 --> 00:48:01 First order of process, B to C's first order of 798 00:48:01 --> 00:48:02 process with rate k2. 799 00:48:02 --> 00:48:05 So you expect, then, C to go up in the first order 800 00:48:05 --> 00:48:13 process to A0 with rate k2. 801 00:48:13 --> 00:48:18 There's the rate k2 here. 802 00:48:18 --> 00:48:22 And the rate k1 here. 803 00:48:22 --> 00:48:25 So when you turn the crank on your approximation in the math, 804 00:48:25 --> 00:48:32 what you'd better see is that at the end, B as a function of 805 00:48:32 --> 00:48:36 time should look like a first order process with rate k2. 806 00:48:36 --> 00:48:39 C as a function of time should look like a first order process 807 00:48:39 --> 00:48:41 with rate k2, going to A0. 808 00:48:41 --> 00:48:45 And you can write the answer, you should be able to write the 809 00:48:45 --> 00:48:51 answer, for C, it's going to be C is approximately, it's going 810 00:48:51 --> 00:48:53 to go to A0, there's no other choice. 811 00:48:53 --> 00:48:56 Everything that was in A gets transferred to B0, so at time 812 00:48:56 --> 00:48:58 is infinity it's going to be A0, and it's going to be a 813 00:48:58 --> 00:49:01 first order process with rate k2. 814 00:49:01 --> 00:49:06 One minus e to the minus k2 times times. 815 00:49:06 --> 00:49:07 I don't even have to do the math. 816 00:49:07 --> 00:49:10 I know that's the answer. 817 00:49:10 --> 00:49:14 If you don't believe me, then do the math. 818 00:49:14 --> 00:49:16 You should do it. 819 00:49:16 --> 00:49:16 But that's the answer. 820 00:49:16 --> 00:49:18 It has to be the answer. 821 00:49:18 --> 00:49:26 And if you look at B, the answer has to be that it's 822 00:49:26 --> 00:49:32 approximately looking like B is disappearing with a rate 823 00:49:32 --> 00:49:37 constant that's k2, e to the minus k2, and the maximum 824 00:49:37 --> 00:49:38 here is very close to A0. 825 00:49:38 --> 00:49:41 So I'm going to be putting A0 here. 826 00:49:41 --> 00:49:44 Close enough. 827 00:49:44 --> 00:49:48 So as long as I ignore the very initial time where A suddenly 828 00:49:48 --> 00:49:51 dumps into B, then everything after that very early time 829 00:49:51 --> 00:49:55 here is just looking like B goes to C. 830 00:49:55 --> 00:49:58 This is not zero, this is t. 831 00:49:58 --> 00:50:00 And A just disappears quickly. 832 00:50:00 --> 00:50:06 And we can write it as A is equal to A0 e to the 833 00:50:06 --> 00:50:09 minus k1 times the time. 834 00:50:09 --> 00:50:12 That's an easy approximation. 835 00:50:12 --> 00:50:17 And you've just got to make sure that this fits. 836 00:50:17 --> 00:50:23 So the other limiting case, which I'm also going to leave 837 00:50:23 --> 00:50:27 as a homework, because it's so straightforward, is to have k2 838 00:50:27 --> 00:50:32 greater than k1, and you should make sure that you can predict 839 00:50:32 --> 00:50:38 it. and that the math works out. 840 00:50:38 --> 00:50:39 OK, any questions? 841 00:50:39 --> 00:50:42 Next time we'll do reversible reactions. 842 00:50:42 --> 00:50:46 And some more approximations. 843 00:50:46 --> 00:50:47