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 --> 00:00:06 Your support will help MIT OpenCourseWare continue to 5 00:00:06 --> 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 --> 00:00:24 PROFESSOR: So last time we were starting more complicated 10 00:00:24 --> 00:00:37 mechanisms and we looked at series reactions. 11 00:00:37 --> 00:00:40 A goes B, which is an intermediate. 12 00:00:40 --> 00:00:43 Goes to C with rates k1 and k2. 13 00:00:43 --> 00:00:49 And we solved the problem exactly. 14 00:00:49 --> 00:00:52 Which turned out to be a little bit complicated. 15 00:00:52 --> 00:00:58 And then we were looking at special cases. 16 00:00:58 --> 00:01:05 And I left the special case of k1 equals k2 to do 17 00:01:05 --> 00:01:08 as a homework problem. 18 00:01:08 --> 00:01:14 And then on the board we did k1 much greater than k2. 19 00:01:14 --> 00:01:22 Which looks like, if k1 is much greater than k2, the rate into 20 00:01:22 --> 00:01:28 k1 is much greater than k2, so using the analogy of buckets 21 00:01:28 --> 00:01:37 and pipes, it's a big pipe between k1 and between A and B. 22 00:01:37 --> 00:01:39 And a little pipe between B and C and you started with 23 00:01:39 --> 00:01:41 a bunch of liquid on top. 24 00:01:41 --> 00:01:45 And you basically dump it into here. 25 00:01:45 --> 00:01:49 And then you slowly drip into C here. 26 00:01:49 --> 00:01:55 So the first thing that happens is that A gets dumped, it 27 00:01:55 --> 00:01:56 becomes B very quickly. 28 00:01:56 --> 00:01:58 With a rate constant k1. 29 00:01:58 --> 00:02:02 In a first order fashion. 30 00:02:02 --> 00:02:07 So A is A0 e to the minus k1 times time. 31 00:02:07 --> 00:02:10 Then B comes up sharply. 32 00:02:10 --> 00:02:18 And then drops down with rate constant k2 slowly. 33 00:02:18 --> 00:02:24 So B is approximately A0 e to the minus k2 times time. 34 00:02:24 --> 00:02:30 And then C has a very small quadratic rise, followed by 35 00:02:30 --> 00:02:33 the exponential rise to saturation that you'd expect. 36 00:02:33 --> 00:02:37 So C is approximately a0 times one minus e to the 37 00:02:37 --> 00:02:40 minus k2 times time. 38 00:02:40 --> 00:02:43 The rate constant k2 dominates the long times. 39 00:02:43 --> 00:02:45 And everything it's looking like it's pseudo first 40 00:02:45 --> 00:02:47 order at long times. 41 00:02:47 --> 00:02:51 So you can do this either by thinking about it, solve the 42 00:02:51 --> 00:02:55 problem by thinking about it, which is perfectly fine. 43 00:02:55 --> 00:02:58 Or you can do it by solving it exactly, as 44 00:02:58 --> 00:03:00 we have done last time. 45 00:03:00 --> 00:03:02 And then putting in the right approximations. 46 00:03:02 --> 00:03:05 Which in this case was Taylor's approximation, 47 00:03:05 --> 00:03:08 if I remember right. 48 00:03:08 --> 00:03:12 And know just by looking at the equation and putting the right 49 00:03:12 --> 00:03:14 approximation, you get it out. 50 00:03:14 --> 00:03:22 And a third approximation, which is the opposite one, k2 51 00:03:22 --> 00:03:28 much less than k1, so k1 much less than k2, the way that 52 00:03:28 --> 00:03:39 that's going to look, we can do it by inspection. 53 00:03:39 --> 00:03:43 So basically if we think of buckets and pipes, that's a 54 00:03:43 --> 00:03:48 very thin pipe connecting the bucket for A and B, so we drip 55 00:03:48 --> 00:03:51 from A to B, then we have this very fat pipe 56 00:03:51 --> 00:03:53 connecting B and C. 57 00:03:53 --> 00:03:56 So as soon as a little B gets in there, it 58 00:03:56 --> 00:03:58 immediately becomes C. 59 00:03:58 --> 00:04:03 So, the rate determining step is the first one, A goes to B. 60 00:04:03 --> 00:04:06 And so k1, we expect that to be the rate that 61 00:04:06 --> 00:04:09 dominates the long time. 62 00:04:09 --> 00:04:17 And we never expect to get any significant amount of B to pile 63 00:04:17 --> 00:04:19 up in the middle bucket, because this pipe is much 64 00:04:19 --> 00:04:22 bigger than that one here. 65 00:04:22 --> 00:04:28 Therefore, what we expect to see is a slow decay in A, with 66 00:04:28 --> 00:04:33 rate constant k1 dominating the long times, A is approximately 67 00:04:33 --> 00:04:40 A0 e to the minus k1 times the time. 68 00:04:40 --> 00:04:46 We don't expect B to amount to very much at all. 69 00:04:46 --> 00:04:49 It'll rise up a little bit at the very early times, and 70 00:04:49 --> 00:04:52 it'll pretty much stay very, very small. 71 00:04:52 --> 00:04:53 Not quite constant. 72 00:04:53 --> 00:05:01 And in fact if you look at the full result, and you put in 73 00:05:01 --> 00:05:05 your approximations, and I'll let you do that, you would find 74 00:05:05 --> 00:05:11 that B is essentially related to A through a constant term 75 00:05:11 --> 00:05:15 where k1 is very small compared to k2. 76 00:05:15 --> 00:05:17 So this is a very small number here. 77 00:05:17 --> 00:05:19 B is equal to a very small number times A. 78 00:05:19 --> 00:05:24 So it basically follows A, but at a very very, low level. 79 00:05:24 --> 00:05:33 And then you would find that C rises up in approximately e to 80 00:05:33 --> 00:05:37 the minus k1 times t, so with rate constant k1. 81 00:05:37 --> 00:05:45 One minus e to the minus k1 times the time. 82 00:05:45 --> 00:05:49 So the dominant rate is k1, and in this case here this is an 83 00:05:49 --> 00:05:52 approximation that we're going to see again. 84 00:05:52 --> 00:05:56 We're going to revisit this approximation here when we get 85 00:05:56 --> 00:05:58 to more complicated mechanisms, it's going to become this 86 00:05:58 --> 00:06:01 steady state approximation. 87 00:06:01 --> 00:06:03 Where the amount of the intermediate, or the 88 00:06:03 --> 00:06:07 concentration of intermediate, is low at all times. 89 00:06:07 --> 00:06:09 And doesn't change very fast. 90 00:06:09 --> 00:06:11 The slope here is very slow. 91 00:06:11 --> 00:06:14 Because k1 is so slow coming down, and B is related to A, 92 00:06:14 --> 00:06:19 which means it's related to this slow decrease in A, the 93 00:06:19 --> 00:06:21 amount of B is small and the rate of change in B 94 00:06:21 --> 00:06:24 is also very small. 95 00:06:24 --> 00:06:28 And hopefully we'll get to that very soon in this lecture. 96 00:06:28 --> 00:06:31 OK, so that's a summary of what we did last time. 97 00:06:31 --> 00:06:36 Any questions? 98 00:06:36 --> 00:06:50 So the next thing now is to go to the next mechanism. 99 00:06:50 --> 00:06:51 Which is, so far we've looked at reactions that 100 00:06:51 --> 00:06:53 proceed all the way. 101 00:06:53 --> 00:06:55 We haven't looked at equilibrium yet. 102 00:06:55 --> 00:06:58 But we know thermodynamics is all about equilibrium. 103 00:06:58 --> 00:07:01 So now we're going to connect kinetics and thermodynamics. 104 00:07:01 --> 00:07:05 We're going to look at a mechanism which is an 105 00:07:05 --> 00:07:09 equilibrium mechanism. 106 00:07:09 --> 00:07:16 Equilibrium or reversible reactions. 107 00:07:16 --> 00:07:21 Where you have A goes to B, with summary constant k1, but 108 00:07:21 --> 00:07:24 you can have the reverse process B goes back to A with 109 00:07:24 --> 00:07:27 the rate constant k minus one. 110 00:07:27 --> 00:07:35 Or you can write is as A k1, k minus one to B, where k1 could 111 00:07:35 --> 00:07:39 be k forward and k minus one could be k backwards. 112 00:07:39 --> 00:07:42 Also written as k sub f, and this is also 113 00:07:42 --> 00:07:46 written as k sub b. 114 00:07:46 --> 00:07:49 So now we have more information than we had before. 115 00:07:49 --> 00:07:51 Because now we know this is an equilibrium problem. 116 00:07:51 --> 00:07:54 And we already know about equilibrium 117 00:07:54 --> 00:07:55 from thermodynamics. 118 00:07:55 --> 00:08:01 And we know that at equilibrium, in addition to all 119 00:08:01 --> 00:08:03 the rate laws that we're going to write down to solve the 120 00:08:03 --> 00:08:07 problem, we also know that at equilibrium, the equilibrium 121 00:08:07 --> 00:08:11 constant is related to the equilibrium concentrations. 122 00:08:11 --> 00:08:14 To the ratio of the equilibrium concentrations. 123 00:08:14 --> 00:08:19 This is something additional that we have. 124 00:08:19 --> 00:08:20 So let's write down everything we know. 125 00:08:20 --> 00:08:24 And our goal is to find out the time, the dynamics 126 00:08:24 --> 00:08:25 of this problem. 127 00:08:25 --> 00:08:28 Suppose you start at some time t equals zero, at a certain 128 00:08:28 --> 00:08:31 amount of A0 in your pot, a certain amount of B0, how 129 00:08:31 --> 00:08:33 do you get to equilibrium. 130 00:08:33 --> 00:08:37 What's the rate, what does it look like? 131 00:08:37 --> 00:08:41 So the first thing is to write down the rates. 132 00:08:41 --> 00:08:47 So the forward rate minus dA/dt, which is the forward 133 00:08:47 --> 00:08:52 rate, R sub forward is equal to k1 times A. 134 00:08:52 --> 00:08:57 And the backwards rate, minus dB/dt, which we can write as R 135 00:08:57 --> 00:09:11 sub backwards, that's k minus one times B. 136 00:09:11 --> 00:09:16 At equilibrium, we have no dynamics. 137 00:09:16 --> 00:09:19 Or no ensemble dynamics. 138 00:09:19 --> 00:09:21 The pot looks constant. 139 00:09:21 --> 00:09:24 It doesn't look like anything's going on in there. 140 00:09:24 --> 00:09:26 The concentration of A doesn't change. 141 00:09:26 --> 00:09:27 The concentration of B doesn't change. 142 00:09:27 --> 00:09:32 They're constant at the equilibrium concentration. 143 00:09:32 --> 00:09:36 So that implies that the rate of formation of B must be 144 00:09:36 --> 00:09:39 equal to the rate of destruction of B. 145 00:09:39 --> 00:09:43 Or, in other words, that the rate forward at equilibrium has 146 00:09:43 --> 00:09:46 to be equal the rate backwards. 147 00:09:46 --> 00:09:50 Otherwise, there'd be changes in concentrations. 148 00:09:50 --> 00:09:54 So if that's true, then that means that at equilibrium, k1 149 00:09:54 --> 00:10:02 A equilibrium has to be equal to k minus one B equilibrium. 150 00:10:02 --> 00:10:10 Which means that we can get this ratio of B to A. 151 00:10:10 --> 00:10:16 And see that this is equal to k1 over k minus one. 152 00:10:16 --> 00:10:19 So we've just, this is pretty major here. 153 00:10:19 --> 00:10:23 Looks pretty simple, but we've taken an equilibrium 154 00:10:23 --> 00:10:24 property here. 155 00:10:24 --> 00:10:27 And related to kinetics. 156 00:10:27 --> 00:10:34 To a rate property. 157 00:10:34 --> 00:10:37 So here we have kinetics. 158 00:10:37 --> 00:10:46 Here we have thermo. 159 00:10:46 --> 00:10:49 And this is going to turn out to be, this kind of ratio 160 00:10:49 --> 00:10:52 between forward rates and backwards rates, related to 161 00:10:52 --> 00:10:55 equilibrium constants is going to be valid, not just for this 162 00:10:55 --> 00:10:58 very simple mechanism we have here with just two species. 163 00:10:58 --> 00:11:01 But it's going to be valid for more complicated. 164 00:11:01 --> 00:11:03 And we have two species reacting to form another 165 00:11:03 --> 00:11:09 species, or bimolecular stuff, et cetera. 166 00:11:09 --> 00:11:11 OK, so now we've laid the groundwork, let's 167 00:11:11 --> 00:11:12 look at the dynamics. 168 00:11:12 --> 00:11:14 Let's look at the time evolution. 169 00:11:14 --> 00:11:19 So that means that we need to take, well this 170 00:11:19 --> 00:11:19 is not quite right. 171 00:11:19 --> 00:11:22 I shouldn't write minus dA/dt. 172 00:11:22 --> 00:11:26 This is minus dA/dt rate forward in the absence 173 00:11:26 --> 00:11:38 of reverse process. 174 00:11:38 --> 00:11:42 It says just the destruction of A and not the creation of A. 175 00:11:42 --> 00:11:48 So if you look at the full kinetics, the full mechanism, 176 00:11:48 --> 00:11:51 this also, the rate backwards is minus dB/dt, in the absence 177 00:11:51 --> 00:11:55 of forming B back through the reverse process. 178 00:11:55 --> 00:11:58 So both of these are in the absence of their 179 00:11:58 --> 00:11:59 reverse process. 180 00:11:59 --> 00:12:06 The full kinetics, for the destruction of A. 181 00:12:06 --> 00:12:10 A is the rate backwards, or the rate forwards, which is k1 182 00:12:10 --> 00:12:12 times A, and then the rate backwards, which is 183 00:12:12 --> 00:12:14 the creation of A. 184 00:12:14 --> 00:12:15 That's the two rates together. 185 00:12:15 --> 00:12:18 Minus k minus one times B. 186 00:12:18 --> 00:12:20 There's the formation of A here, this is 187 00:12:20 --> 00:12:21 the destruction of A. 188 00:12:21 --> 00:12:24 This is the full differential equation for the 189 00:12:24 --> 00:12:24 full mechanism. 190 00:12:24 --> 00:12:26 Not just one part of it. 191 00:12:26 --> 00:12:28 This one here is just for the first step, and that one here 192 00:12:28 --> 00:12:31 is for the second step. 193 00:12:31 --> 00:12:34 This is what we need to solve. 194 00:12:34 --> 00:12:40 But now we have a lot of help by what we've already done. 195 00:12:40 --> 00:12:44 So, what we can do is well, the problem here is that's it looks 196 00:12:44 --> 00:12:47 hard because there's B included in here. 197 00:12:47 --> 00:12:48 This is not just A, there's B in here. 198 00:12:48 --> 00:12:50 But there's only two species. 199 00:12:50 --> 00:12:52 And stoichiometry is going to help us out. 200 00:12:52 --> 00:12:55 Remember last time, or a few times ago, we said the last 201 00:12:55 --> 00:12:57 species is the easy one. 202 00:12:57 --> 00:13:00 Because it's always related to the concentration of all the 203 00:13:00 --> 00:13:02 other species by stoichiometry. 204 00:13:02 --> 00:13:04 And the last one here is B. 205 00:13:04 --> 00:13:07 So we know that B is related to A. 206 00:13:07 --> 00:13:11 B is whatever you started out with. 207 00:13:11 --> 00:13:14 Plus whatever you used up for A. 208 00:13:14 --> 00:13:17 That's one way of writing it. 209 00:13:17 --> 00:13:23 So you start with this amount, and you create this amount. 210 00:13:23 --> 00:13:24 By having A react. 211 00:13:24 --> 00:13:27 This is what you started with, A, and this is what's left. 212 00:13:27 --> 00:13:30 The difference has to go into B. 213 00:13:30 --> 00:13:34 So stroke stoichiometry give you a relationship, which you 214 00:13:34 --> 00:13:41 can use to plug into your differential equation here. 215 00:13:41 --> 00:13:44 And now, after rearranging your differential equation, putting 216 00:13:44 --> 00:13:50 B in there, you can rewrite this as k1 plus k minus one 217 00:13:50 --> 00:14:01 times A minus k minus one times B0 plus A0. 218 00:14:01 --> 00:14:03 Something which depends on A, something which 219 00:14:03 --> 00:14:07 is constant here. 220 00:14:07 --> 00:14:08 OK, and what else do we know? 221 00:14:08 --> 00:14:10 We know something about equilibrium. 222 00:14:10 --> 00:14:14 We know something about when we reach a state where the rate 223 00:14:14 --> 00:14:16 forward is equal to the rate backward. 224 00:14:16 --> 00:14:21 When the rate forward is equal to the rate backwards, the 225 00:14:21 --> 00:14:24 rate of change of A is zero. 226 00:14:24 --> 00:14:25 A is not changing. 227 00:14:25 --> 00:14:27 The concentration of A is not changing. 228 00:14:27 --> 00:14:38 So at equilibrium, dA/dt is equal to zero. 229 00:14:38 --> 00:14:43 And just like in this example here, this is going to turn out 230 00:14:43 --> 00:14:45 to be related to an approximation called the steady 231 00:14:45 --> 00:14:47 state approximation, which we're going to see 232 00:14:47 --> 00:14:51 later this morning. 233 00:14:51 --> 00:14:55 Writing at equilibrium that the change in the concentration of 234 00:14:55 --> 00:14:58 one of the species is equal to zero, we're going to use that 235 00:14:58 --> 00:15:00 as an approximation also later, when we get to more 236 00:15:00 --> 00:15:02 complicated mechanisms. 237 00:15:02 --> 00:15:04 This is going to be the equilibrium approximation. 238 00:15:04 --> 00:15:08 And this is going to be the steady state approximation. 239 00:15:08 --> 00:15:12 Let's keep going with our, we're solving this guy here. 240 00:15:12 --> 00:15:14 So at equilibrium, dA/dt is equal to zero. 241 00:15:14 --> 00:15:27 And then we can replace our concentrations here with, and 242 00:15:27 --> 00:15:30 so if you write equal to zero here, then we can write it 243 00:15:30 --> 00:15:32 as equilibrium right here. 244 00:15:32 --> 00:15:34 And we'll be able to solve for A equilibrium in terms of 245 00:15:34 --> 00:15:40 k1, k minus one, B0, and A0. 246 00:15:40 --> 00:15:44 And then if you do that, you get A equilibrium is equal to 247 00:15:44 --> 00:15:53 k minus one over k1 plus k minus one times B0 plus A0. 248 00:15:53 --> 00:16:00 And we're going to need that. 249 00:16:00 --> 00:16:05 Because now we're going to be able to plug this, so there's 250 00:16:05 --> 00:16:07 A0 plus B0 sitting here. 251 00:16:07 --> 00:16:09 And we're going to replace A0 plus B0 in terms 252 00:16:09 --> 00:16:12 of A equilibrium. 253 00:16:12 --> 00:16:15 There's a minus sign here. 254 00:16:15 --> 00:16:21 And between this A and this A equilibrium here, so by 255 00:16:21 --> 00:16:27 rewriting, by plugging in A B0 A equilibrium instead of B0 256 00:16:27 --> 00:16:36 plus A0, we can rewrite that equation as dA/dt is equal to 257 00:16:36 --> 00:16:45 k1 plus k minus one times A minus A equilibrium. 258 00:16:45 --> 00:16:48 This is nice because, this quantity, A minus A 259 00:16:48 --> 00:16:52 equilibrium, describes the difference between where 260 00:16:52 --> 00:16:55 are you are and where you want to be. 261 00:16:55 --> 00:16:56 It's a nice variable to have. 262 00:16:56 --> 00:16:59 It's going to change in time and at infinite time, this 263 00:16:59 --> 00:17:01 is going to go to zero. 264 00:17:01 --> 00:17:03 A is going to go to equilibrium. 265 00:17:03 --> 00:17:05 A equilibrium is a constant. 266 00:17:05 --> 00:17:07 Now we have a differential equation that relates 267 00:17:07 --> 00:17:10 A to this difference. 268 00:17:10 --> 00:17:12 But A equilibrium is a constant. 269 00:17:12 --> 00:17:16 So there's nothing that forbids us from just writing minus A 270 00:17:16 --> 00:17:20 equilibrium here. d of a constant dt is zero. 271 00:17:20 --> 00:17:24 So I'm just basically subtracting zero here. 272 00:17:24 --> 00:17:26 So and I have something of the form dx/dt is equal 273 00:17:26 --> 00:17:28 to constant times x. 274 00:17:28 --> 00:17:30 And I know how to write that. 275 00:17:30 --> 00:17:32 That looks just like a first order process. 276 00:17:32 --> 00:17:39 I know the solution is A minus A equilibrium is equal to my 277 00:17:39 --> 00:17:42 initial A, my initial state. 278 00:17:42 --> 00:17:52 A0 times e to the minus k1 plus k minus one times time. 279 00:17:52 --> 00:17:55 And I've solved the problem. 280 00:17:55 --> 00:18:00 I've described how, as a function of time, how A, 281 00:18:00 --> 00:18:04 the concentration of A, gets equilibrium. 282 00:18:04 --> 00:18:06 And if you want to know B, it's just the same thing. 283 00:18:06 --> 00:18:09 You replace B here and B0 here. 284 00:18:09 --> 00:18:11 You put k minus one here, k1 here, but it's the same thing. 285 00:18:11 --> 00:18:17 286 00:18:17 --> 00:18:34 So if you were to sketch it out as a function of time, there's, 287 00:18:34 --> 00:18:37 eventually you'll get to the concentration A equilibrium. 288 00:18:37 --> 00:18:40 If you're above it, you're going to decay in a first 289 00:18:40 --> 00:18:44 order fashion, with a rate constant k1. 290 00:18:44 --> 00:18:51 A minus A equilibrium is A0 minus A equilibrium e to the 291 00:18:51 --> 00:18:58 minus k prime times t, where k prime is k1 plus k minus one. 292 00:18:58 --> 00:19:02 And this is when A is greater than A0, A equilibrium. 293 00:19:02 --> 00:19:03 A0 is greater than equilibrium. 294 00:19:03 --> 00:19:08 And if you start below, you're going to come up like this. 295 00:19:08 --> 00:19:11 A0 is less than A equilibrium. 296 00:19:11 --> 00:19:14 And both rates come into here, in a very simple way. 297 00:19:14 --> 00:19:17 Just the sum of the two. 298 00:19:17 --> 00:19:22 So experimental, if you want to find out these rates, you can 299 00:19:22 --> 00:19:27 measure their equilibrium concentrations. 300 00:19:27 --> 00:19:34 Measure k equilibrium by obtaining the equilibrium 301 00:19:34 --> 00:19:36 concentrations of B and A. 302 00:19:36 --> 00:19:40 And that gives you the ratio of k1 plus k minus one. 303 00:19:40 --> 00:19:43 And then you can just start the process with some 304 00:19:43 --> 00:19:44 concentration of A. 305 00:19:44 --> 00:19:46 Which is different in equilibrium. 306 00:19:46 --> 00:19:55 Then watching it in time, extract out, observe and 307 00:19:55 --> 00:19:58 measure k1 plus k minus one. 308 00:19:58 --> 00:19:59 In this kinetic equation. 309 00:19:59 --> 00:20:01 At kinetic relation. 310 00:20:01 --> 00:20:06 And then you have two results, two pieces of data. 311 00:20:06 --> 00:20:09 You've measured k prime, which gives you the sum of the two. 312 00:20:09 --> 00:20:11 And you've measured K equilibrium, which gives 313 00:20:11 --> 00:20:12 you the ratio of the two. 314 00:20:12 --> 00:20:16 And you've got your two rate constants out. 315 00:20:16 --> 00:20:21 OK, any questions for the equilibrium problem? 316 00:20:21 --> 00:20:22 We're slowly -- yes. 317 00:20:22 --> 00:20:27 STUDENT: [INAUDIBLE] 318 00:20:27 --> 00:20:29 PROFESSOR: There is not yet a relationship between 319 00:20:29 --> 00:20:31 k1 and k minus one. 320 00:20:31 --> 00:20:38 It depends on the problem. 321 00:20:38 --> 00:20:45 And we're going to study the, when we get to potential 322 00:20:45 --> 00:20:52 barriers, we look at, we'll look at this issue. 323 00:20:52 --> 00:20:54 OK, it's a good question. 324 00:20:54 --> 00:20:59 We're going to do that probably next time. 325 00:20:59 --> 00:21:01 So let's get going, then, with reversible reactions. 326 00:21:01 --> 00:21:03 So now we have, let's make it a little bit more complicated. 327 00:21:03 --> 00:21:07 Let's, instead of two species, let's make it three species. 328 00:21:07 --> 00:21:14 So we have A plus B goes to C, where the rate constant k1, and 329 00:21:14 --> 00:21:23 then C goes to A plus to B with a rate constant k minus one 330 00:21:23 --> 00:21:34 that you can rewrite as A plus B goes to C, k1, k minus one, 331 00:21:34 --> 00:21:36 and then you write down your, you want to solve this. 332 00:21:36 --> 00:21:38 So the first thing you do is your write down your 333 00:21:38 --> 00:21:43 differential equations. dA/dt is equal to, put all the rates 334 00:21:43 --> 00:21:49 in there. k1 A B minus k minus one times C. 335 00:21:49 --> 00:21:57 And at equilibrium, we set that equal to zero. 336 00:21:57 --> 00:22:01 So you can write equilibrium here, equilibrium here, 337 00:22:01 --> 00:22:04 equilibrium here. 338 00:22:04 --> 00:22:11 And when you bring this term over to the other side, you 339 00:22:11 --> 00:22:17 find, then that at equilibrium the ratio of the forward rate 340 00:22:17 --> 00:22:21 to the backward rates is equal to C equilibrium divided 341 00:22:21 --> 00:22:23 by A equilibrium. 342 00:22:23 --> 00:22:24 The equilibrium, which, you know is the 343 00:22:24 --> 00:22:26 equilibrium constant. 344 00:22:26 --> 00:22:30 So again, for this slightly more complicated problem, the 345 00:22:30 --> 00:22:32 ratio of the forward and backward rates are related to 346 00:22:32 --> 00:22:39 the equilibrium constant. 347 00:22:39 --> 00:22:41 So now you want to solve this. 348 00:22:41 --> 00:22:42 Well, we're not even going to try. 349 00:22:42 --> 00:22:44 Because it's going to be too complicated. 350 00:22:44 --> 00:22:48 So as soon as you get away from first order kinetics and go to 351 00:22:48 --> 00:22:55 second order kinetics with multiple steps, it's a mess. 352 00:22:55 --> 00:22:59 So instead you try to find approximations. 353 00:22:59 --> 00:23:00 In this case here there's one we can use. 354 00:23:00 --> 00:23:02 Or a limiting case, at least. 355 00:23:02 --> 00:23:09 One we can use, which is an obvious one, which is flooding. 356 00:23:09 --> 00:23:11 As a limiting case. 357 00:23:11 --> 00:23:16 If we want to isolate a small part of the problem, 358 00:23:16 --> 00:23:17 we can use flooding. 359 00:23:17 --> 00:23:22 Because we can overwhelm the system with either A or B. 360 00:23:22 --> 00:23:27 Take B0 much greater than A0 and C0. 361 00:23:27 --> 00:23:29 So over the course of the reaction of the process, 362 00:23:29 --> 00:23:33 the concentration of B hardly changes. 363 00:23:33 --> 00:23:39 And so when we write our kinetic equation, our rate law, 364 00:23:39 --> 00:23:46 k1 A B minus k minus one C, we could put a little naught here, 365 00:23:46 --> 00:23:47 because we know the concentration of B 366 00:23:47 --> 00:23:49 is not changing. 367 00:23:49 --> 00:23:53 And so now we're left with the problem we had before. 368 00:23:53 --> 00:23:56 Which was a reversible first order process which 369 00:23:56 --> 00:24:00 we have just solved. 370 00:24:00 --> 00:24:04 And you go through the experiment and you extract out 371 00:24:04 --> 00:24:08 k1 times B0 and k minus one. 372 00:24:08 --> 00:24:10 You do the experiment again with a different concentration 373 00:24:10 --> 00:24:12 of B0 to start out with. 374 00:24:12 --> 00:24:14 And you can extract out k1. 375 00:24:14 --> 00:24:18 So this is a process of getting the rates by using a 376 00:24:18 --> 00:24:21 simple approximation here. 377 00:24:21 --> 00:24:26 Simple limiting case. 378 00:24:26 --> 00:24:30 Any questions? 379 00:24:30 --> 00:24:32 So we've just finished putting all the building 380 00:24:32 --> 00:24:35 blocks together now. 381 00:24:35 --> 00:24:43 Let me remind you what these building blocks are. 382 00:24:43 --> 00:24:45 We have a bunch of approximations under our belt. 383 00:24:45 --> 00:24:55 Flooding, we've looked at rates being much faster than others. 384 00:24:55 --> 00:25:01 And we have three simple mechanisms, we've looked 385 00:25:01 --> 00:25:03 at parallel reactions. 386 00:25:03 --> 00:25:06 We looked at series reactions. 387 00:25:06 --> 00:25:13 And we've looked at reversible reactions. 388 00:25:13 --> 00:25:15 So a complicated mechanism will basically put these 389 00:25:15 --> 00:25:19 three building blocks together in a series. 390 00:25:19 --> 00:25:22 And they'll all be happening at the same time somehow. 391 00:25:22 --> 00:25:27 And obviously, since it was, since we threw up our hands, 392 00:25:27 --> 00:25:31 just with a simple reversible process like here, it's clear 393 00:25:31 --> 00:25:35 that we're going to throw up our hands a lot. 394 00:25:35 --> 00:25:37 When we write down these mechanisms. 395 00:25:37 --> 00:25:38 So we're going to need approximations. 396 00:25:38 --> 00:25:40 We're going to need something that we'll automatically 397 00:25:40 --> 00:25:43 go to, and say this is going to be too hard. 398 00:25:43 --> 00:25:45 I'm not even going to try. 399 00:25:45 --> 00:25:50 Let's do an approximation. 400 00:25:50 --> 00:25:52 And two approximations that we're going to talk about are 401 00:25:52 --> 00:25:55 ones I already mentioned at the beginning. 402 00:25:55 --> 00:26:01 Which is the steady state approximation, where one rate, 403 00:26:01 --> 00:26:07 the rate to go into the intermediate is very slow, or 404 00:26:07 --> 00:26:10 the rate to get out of the intermediate is very fast. 405 00:26:10 --> 00:26:14 And the equilibrium approximation where the system 406 00:26:14 --> 00:26:19 sets up a very fast equilibrium and you're allowed to use 407 00:26:19 --> 00:26:22 thermodynamics to help you out in solving the problem. 408 00:26:22 --> 00:26:23 These are the two approximations that we'll 409 00:26:23 --> 00:26:27 use when we put these things together. 410 00:26:27 --> 00:26:30 OK, so let's look at the first, simple, more 411 00:26:30 --> 00:26:32 complicated mechanism. 412 00:26:32 --> 00:26:34 To see where these approximations come in. 413 00:26:34 --> 00:26:36 So the first one is going to be a series. 414 00:26:36 --> 00:26:38 We're going to start putting these things together. 415 00:26:38 --> 00:26:40 First one is a series reversible. 416 00:26:40 --> 00:26:41 Those two together. 417 00:26:41 --> 00:26:46 Series reversible. 418 00:26:46 --> 00:26:50 So we have first, a reversible process. 419 00:26:50 --> 00:26:54 Everything is first order here. k minus one goes to B. 420 00:26:54 --> 00:26:57 And then B goes to C with some rate constant 421 00:26:57 --> 00:27:01 k2, all first order. 422 00:27:01 --> 00:27:03 If we were to turn the crank, we'd say, oh, I've got to write 423 00:27:03 --> 00:27:04 down all my rate laws here. 424 00:27:04 --> 00:27:09 Minus dA/dt is all the ways that I destroy and create B, so 425 00:27:09 --> 00:27:18 there's a k1 times A minus k minus one times B. dB/dt, all 426 00:27:18 --> 00:27:22 the ways that create and destroy B is that you create it 427 00:27:22 --> 00:27:25 through destruction of A. 428 00:27:25 --> 00:27:29 I destroy it through the backwards rate to make A. 429 00:27:29 --> 00:27:32 And I destroy it by making the product, C. 430 00:27:32 --> 00:27:33 There's the intermediate. 431 00:27:33 --> 00:27:35 And then for C, that's pretty simple. 432 00:27:35 --> 00:27:38 There's only one channel into C, and that's the destruction 433 00:27:38 --> 00:27:41 of B to form C, k2 B. 434 00:27:41 --> 00:27:42 All the rate laws. 435 00:27:42 --> 00:27:45 I write everything I know here. 436 00:27:45 --> 00:27:48 Couple of differential equations. 437 00:27:48 --> 00:27:52 We know it's going to be hard to solve. 438 00:27:52 --> 00:27:58 So, let's remind ourselves of what I just erased. 439 00:27:58 --> 00:28:08 Which was the case where the intermediate concentration 440 00:28:08 --> 00:28:14 was always very small and didn't change very fast. 441 00:28:14 --> 00:28:20 And the process was dominated by k1 here. 442 00:28:20 --> 00:28:22 That was the rate limiting step. 443 00:28:22 --> 00:28:29 Remember that our first example this morning. 444 00:28:29 --> 00:28:34 So this is where the concentration of B is 445 00:28:34 --> 00:28:36 roughly constant over time. 446 00:28:36 --> 00:28:41 It's small. 447 00:28:41 --> 00:28:45 And over any small time period, it's roughly constant. 448 00:28:45 --> 00:28:52 Which means that dB/dt is roughly equal to zero. 449 00:28:52 --> 00:28:53 If you look at a long term, obviously it's going to 450 00:28:53 --> 00:28:56 change as you go from this point to that point. 451 00:28:56 --> 00:28:58 In fact, there's a relationship between A which is 452 00:28:58 --> 00:28:59 changing in time and B. 453 00:28:59 --> 00:29:02 But A is also changing in time very slowly. 454 00:29:02 --> 00:29:05 Because the rate k1 is very small. 455 00:29:05 --> 00:29:08 So dB/dt, which is related to this rate k1, is going to be 456 00:29:08 --> 00:29:10 very slowly changing in time. 457 00:29:10 --> 00:29:16 We can approximate it at zero. 458 00:29:16 --> 00:29:19 What are the ways in terms of matching these rate constants 459 00:29:19 --> 00:29:22 that we can get to the approximation. 460 00:29:22 --> 00:29:28 To a diagram that looks like that? 461 00:29:28 --> 00:29:34 But we don't want B to pile up. 462 00:29:34 --> 00:29:38 That means we have to have the rates out of the intermediate 463 00:29:38 --> 00:29:41 to be much faster, at least one of the rates out of the 464 00:29:41 --> 00:29:43 intermediate to be much faster than the rate that creates 465 00:29:43 --> 00:29:45 the intermediate. 466 00:29:45 --> 00:29:50 So the different ways of creating that approximation are 467 00:29:50 --> 00:29:53 if, and the length of the arrows now is going to be 468 00:29:53 --> 00:29:55 proportional to the rate. 469 00:29:55 --> 00:29:58 We want it to be very hard to make B. 470 00:29:58 --> 00:30:03 And as soon as we make B, we want it to go away. 471 00:30:03 --> 00:30:05 So one of the ways to do that is to have the reverse 472 00:30:05 --> 00:30:09 rate much faster than the initial rate. 473 00:30:09 --> 00:30:12 And we could have a slow rate into C, that's fine. 474 00:30:12 --> 00:30:16 We could have, again, a very slow rate into B. 475 00:30:16 --> 00:30:20 We could have a slow reverse rate and a fast rate into C. 476 00:30:20 --> 00:30:21 Perfectly fine. 477 00:30:21 --> 00:30:22 We could have both. 478 00:30:22 --> 00:30:28 Fast rate out of B through A, or out of B through C. 479 00:30:28 --> 00:30:31 As long as this first rate into the B is small compared to one 480 00:30:31 --> 00:30:33 of those two rates, we're never going to pile up B. 481 00:30:33 --> 00:30:37 It's going to go away as soon as we make it. 482 00:30:37 --> 00:30:42 So in all these cases, k2 plus k minus one is 483 00:30:42 --> 00:30:46 much bigger than k1. 484 00:30:46 --> 00:30:49 When we have that situation, then we can make the 485 00:30:49 --> 00:30:51 approximation up here. 486 00:30:51 --> 00:30:54 That the rate of change in B is very small. 487 00:30:54 --> 00:30:56 Almost zero, which means it's basically a constant. 488 00:30:56 --> 00:31:01 And instead of writing B in this case here, we're going to 489 00:31:01 --> 00:31:04 write it as B steady state. 490 00:31:04 --> 00:31:08 Which is basically a constant in terms of solving the 491 00:31:08 --> 00:31:10 differential equations. 492 00:31:10 --> 00:31:14 And that's going to make our life much easier. 493 00:31:14 --> 00:31:19 Because instead of having to solve these coupled 494 00:31:19 --> 00:31:22 differential equations, we're just going to have to solve 495 00:31:22 --> 00:31:24 coupled algebraic equations. 496 00:31:24 --> 00:31:29 Which is really messy, but less hard. 497 00:31:29 --> 00:31:34 If you're a bean-counter than this is heaven. 498 00:31:34 --> 00:31:37 So now we're going to put steady state, this is 499 00:31:37 --> 00:31:38 going to be a constant. 500 00:31:38 --> 00:31:40 And this is going to be equal to zero. 501 00:31:40 --> 00:31:42 And then here this is going to be steady state here. 502 00:31:42 --> 00:31:44 This is going to be steady state here. 503 00:31:44 --> 00:31:47 And the first thing to do now is that we've eliminated 504 00:31:47 --> 00:31:48 this differential equation. 505 00:31:48 --> 00:31:54 We now have an algebraic equation where we can 506 00:31:54 --> 00:32:01 solve for B steady state. 507 00:32:01 --> 00:32:07 So if you recognize that you're in this situation here, 508 00:32:07 --> 00:32:08 forget about trying to solve. 509 00:32:08 --> 00:32:13 Immediately go to the process of putting in a constant for B. 510 00:32:13 --> 00:32:16 Setting dB/dt for the intermediate equal to zero. 511 00:32:16 --> 00:32:22 And then starting to turn the crank on the algebra. 512 00:32:22 --> 00:32:26 So let me go ahead and turn the crank here. 513 00:32:26 --> 00:32:30 Go through the steps, which is basically typical 514 00:32:30 --> 00:32:37 of these problems. 515 00:32:37 --> 00:32:38 So you turn the crank. 516 00:32:38 --> 00:32:42 You solve for the steady state concentration. 517 00:32:42 --> 00:32:50 You get, in terms of A, over k minus one, plus k2. 518 00:32:50 --> 00:32:55 Then you plug this back in here. 519 00:32:55 --> 00:33:01 And you get a new differential equation, minus dA/dt is equal 520 00:33:01 --> 00:33:06 to k1 times A minus k minus one times B steady state, so 521 00:33:06 --> 00:33:09 we plug this in here. 522 00:33:09 --> 00:33:12 And you get, so there's A sitting here. 523 00:33:12 --> 00:33:13 A sitting here. 524 00:33:13 --> 00:33:16 It's going to be of the form, effective rate times A. 525 00:33:16 --> 00:33:18 It's going to be basically a first order form. 526 00:33:18 --> 00:33:21 Which is what we'd expect from sketching the 527 00:33:21 --> 00:33:24 diagram just like that. 528 00:33:24 --> 00:33:30 So you get minus dA/dt is an effective rate. k1 times 529 00:33:30 --> 00:33:34 k2 over k1 plus k2. 530 00:33:34 --> 00:33:36 Times A. 531 00:33:36 --> 00:33:39 This is k prime. 532 00:33:39 --> 00:33:41 So if you were to do an experiment under these 533 00:33:41 --> 00:33:47 conditions, you'd find that A behaves, it's basically a 534 00:33:47 --> 00:33:48 pseudo first order problem. 535 00:33:48 --> 00:33:53 With a funny rate that contains all the elementary rate 536 00:33:53 --> 00:33:55 constants as part of it. 537 00:33:55 --> 00:33:59 And when you do the same thing for C, dC/dt, you 538 00:33:59 --> 00:34:01 find that depends on A. 539 00:34:01 --> 00:34:07 It's k1, you plug these steady states in here. k1 k2 divided 540 00:34:07 --> 00:34:11 by k1 plus k2 times A. 541 00:34:11 --> 00:34:13 This is k prime again. 542 00:34:13 --> 00:34:16 So the problem, and it's first order. 543 00:34:16 --> 00:34:18 So the problem looks like effectively you're 544 00:34:18 --> 00:34:20 going from A to C. 545 00:34:20 --> 00:34:21 Forget about the intermediate. 546 00:34:21 --> 00:34:24 With an effective rate k prime. 547 00:34:24 --> 00:34:30 Where k prime contains these rates. 548 00:34:30 --> 00:34:32 So this is steady state approximation. 549 00:34:32 --> 00:34:36 It's the prototypical problem. 550 00:34:36 --> 00:34:38 Questions on steady state, before we go to the 551 00:34:38 --> 00:34:44 next approximation. 552 00:34:44 --> 00:34:44 Alright. 553 00:34:44 --> 00:34:53 So now, as promised, the next approximation is that where are 554 00:34:53 --> 00:34:54 set up the fast equilibrium. 555 00:34:54 --> 00:34:59 And you can use thermo to help you out. 556 00:34:59 --> 00:35:11 Equilibrium approximation A goes to B. 557 00:35:11 --> 00:35:13 This is a fast process. 558 00:35:13 --> 00:35:21 And then k2 out of B is a slow process. 559 00:35:21 --> 00:35:24 Little arrow here, and two big arrows here. 560 00:35:24 --> 00:35:27 You put your A in your flask. 561 00:35:27 --> 00:35:30 Immediately you set up the equilibrium. 562 00:35:30 --> 00:35:32 And you slowly dribble out of that. 563 00:35:32 --> 00:35:40 If you want to do it as a function of buckets and, so you 564 00:35:40 --> 00:35:42 have a big pipe connecting two buckets that are just 565 00:35:42 --> 00:35:43 offset from each other. 566 00:35:43 --> 00:35:48 So there's, in this case here B is favored over A. 567 00:35:48 --> 00:35:51 Because it's a little bit lower than this guy here. 568 00:35:51 --> 00:35:53 And then there's a little bucket, there's a little 569 00:35:53 --> 00:35:58 tube, comes out of here. 570 00:35:58 --> 00:36:02 So there's a little dribble into C over time. 571 00:36:02 --> 00:36:05 First thing that happens is you set up your equilibrium. 572 00:36:05 --> 00:36:08 And then you slowly extract stuff through 573 00:36:08 --> 00:36:10 a little tube into C. 574 00:36:10 --> 00:36:14 So what do you expect this to look like? 575 00:36:14 --> 00:36:22 Well, you expect A to really slowly come out. 576 00:36:22 --> 00:36:26 Because the rate limiting step is the rate from B to C. 577 00:36:26 --> 00:36:30 So k2 is going to be the way that A is going to come out. 578 00:36:30 --> 00:36:32 You expect it to slowly go away. 579 00:36:32 --> 00:36:35 You expect B to get created very fast, to some 580 00:36:35 --> 00:36:37 equilibrium amount. 581 00:36:37 --> 00:36:42 And then also to follow the same rate k2, slow 582 00:36:42 --> 00:36:44 k2, to disappear. 583 00:36:44 --> 00:36:48 And you expect C to come up. 584 00:36:48 --> 00:36:50 You're basically in a first order process 585 00:36:50 --> 00:36:53 to saturation of A0. 586 00:36:53 --> 00:36:56 So when you expect the dominant rate to be k2, and the fast 587 00:36:56 --> 00:37:00 dynamics to happen at very, very early times, and then 588 00:37:00 --> 00:37:05 everything to follow first order kinetics. 589 00:37:05 --> 00:37:07 So now let's do the math and make sure that it agrees with 590 00:37:07 --> 00:37:11 what we've just assumed that it's going to look like. 591 00:37:11 --> 00:37:14 So equilibrium is happening. 592 00:37:14 --> 00:37:20 You can assume that at any time after the initial very fast 593 00:37:20 --> 00:37:21 process of getting equilibrium. 594 00:37:21 --> 00:37:25 So after some initial time, the ratio of B over A is always 595 00:37:25 --> 00:37:27 going to be a constant. 596 00:37:27 --> 00:37:30 The dribble out of B here is not going to be fast enough 597 00:37:30 --> 00:37:33 to change that ratio. 598 00:37:33 --> 00:37:39 As soon as you get a little bit of B out here, immediately the 599 00:37:39 --> 00:37:44 ratios, the amounts rearrange to keep the ratio constant. 600 00:37:44 --> 00:37:48 So now when I look at the rate of the reaction, the rate of 601 00:37:48 --> 00:37:53 formation of C, k2 times B, well, in terms of A, it's 602 00:37:53 --> 00:37:57 k2 times K equilibrium. 603 00:37:57 --> 00:38:00 Times A. 604 00:38:00 --> 00:38:06 Which is k2 times k1 over k minus one times A. 605 00:38:06 --> 00:38:11 And you can immediately see that C is behaving, or the 606 00:38:11 --> 00:38:15 reaction is behaving, like a first order process. 607 00:38:15 --> 00:38:18 With an effective rate which contains all three rates 608 00:38:18 --> 00:38:23 in this ratio here. 609 00:38:23 --> 00:38:25 So it looks like A goes to C, with some k prime 610 00:38:25 --> 00:38:28 where this is k prime. 611 00:38:28 --> 00:38:33 It looks like A goes to C with an effective rate 612 00:38:33 --> 00:38:36 constant k prime. 613 00:38:36 --> 00:38:36 OK. 614 00:38:36 --> 00:38:39 Questions? 615 00:38:39 --> 00:38:40 We're going to do some examples. 616 00:38:40 --> 00:38:43 And then we're going to do chain reactions next time. 617 00:38:43 --> 00:38:50 We're one lecture behind. 618 00:38:50 --> 00:38:50 Alright. 619 00:38:50 --> 00:38:53 Let's do some reactions. 620 00:38:53 --> 00:39:01 Some examples. 621 00:39:01 --> 00:39:04 I'm going to skip the first example, which is 622 00:39:04 --> 00:39:08 the example with the chaperone, in the notes. 623 00:39:08 --> 00:39:12 And I'm going to go directly to the gas decomposition. 624 00:39:12 --> 00:39:15 The two examples are basically very similar in their use of 625 00:39:15 --> 00:39:18 the steady state approximation. 626 00:39:18 --> 00:39:23 So I'll let you read over the first example. 627 00:39:23 --> 00:39:26 So this example here is going to give us the Lindemann 628 00:39:26 --> 00:39:29 mechanism, for which Mr. Lindemann got a Nobel Prize 629 00:39:29 --> 00:39:33 many many, years ago. 630 00:39:33 --> 00:39:35 Basically, it's looking at decomposition of a 631 00:39:35 --> 00:39:44 gas phase molecule. 632 00:39:44 --> 00:39:52 Where you have a molecule, A, that breaks down into products. 633 00:39:52 --> 00:39:58 And the observation before Mr. Lindemann got around, was 634 00:39:58 --> 00:40:06 that it looked like a first order process. 635 00:40:06 --> 00:40:10 It looked like A is just falling apart on its own 636 00:40:10 --> 00:40:13 into these products. 637 00:40:13 --> 00:40:16 Mr. Lindemann got around and said, well, why would A 638 00:40:16 --> 00:40:18 just want to fall apart. 639 00:40:18 --> 00:40:21 There's something a little bit odd about that. 640 00:40:21 --> 00:40:24 Stable molecule, why would it want to fall apart. 641 00:40:24 --> 00:40:29 And so he hypothesized a mechanism. 642 00:40:29 --> 00:40:33 And it went like this. 643 00:40:33 --> 00:40:36 That A actually collides with another molecule, which 644 00:40:36 --> 00:40:39 could be a bystander. 645 00:40:39 --> 00:40:42 Could be a chaperone molecule that just happens to be there. 646 00:40:42 --> 00:40:43 Or it could be another molecule of A. 647 00:40:43 --> 00:40:46 There's a collision. 648 00:40:46 --> 00:40:50 There's a collision that creates a vibrationally 649 00:40:50 --> 00:40:52 excited version of A. 650 00:40:52 --> 00:40:53 Of A sitting there. 651 00:40:53 --> 00:40:53 It collides. 652 00:40:53 --> 00:40:56 It suddenly starts to be really vibrationally excited. 653 00:40:56 --> 00:40:58 Bonds vibrate all over the place. 654 00:40:58 --> 00:41:01 Atoms wiggle. 655 00:41:01 --> 00:41:04 There's the collision partner that goes away. 656 00:41:04 --> 00:41:08 So kinetic energy is transferred from M and A to 657 00:41:08 --> 00:41:11 the vibrational loads of A. 658 00:41:11 --> 00:41:15 In a process which is reversible. k1, k minus one, 659 00:41:15 --> 00:41:18 because this excited A, this vibrationally excited A, 660 00:41:18 --> 00:41:21 could also collide with a molecule and cool down. 661 00:41:21 --> 00:41:25 The energy in the vibrations could be transferred back to 662 00:41:25 --> 00:41:29 kinetic energy for a reverse process. 663 00:41:29 --> 00:41:34 But if you wait long enough, or if this vibrationally excited 664 00:41:34 --> 00:41:39 A, with these atoms widely moving around, that has more of 665 00:41:39 --> 00:41:42 a chance of falling apart than this guy here. 666 00:41:42 --> 00:41:49 So A star could fall apart into products. 667 00:41:49 --> 00:41:51 With a rate k2. 668 00:41:51 --> 00:41:52 So he said, this is actually an important step. 669 00:41:52 --> 00:41:54 It's not that A just suddenly falls apart. 670 00:41:54 --> 00:41:58 But this has to happen. 671 00:41:58 --> 00:42:01 So all the observations that supported a first order 672 00:42:01 --> 00:42:05 process, which basically only supported one elementary 673 00:42:05 --> 00:42:13 reactions, might not be right. 674 00:42:13 --> 00:42:16 So let's assume his mechanism, and let's look at what we need 675 00:42:16 --> 00:42:18 to do as an approximation. 676 00:42:18 --> 00:42:21 And where that leads us in terms of trying to 677 00:42:21 --> 00:42:28 experimentally confirm that this mechanism is plausible. 678 00:42:28 --> 00:42:31 Not prove it, but just to make sure that it's consistent with, 679 00:42:31 --> 00:42:43 the data's consistent with the hypothesis. 680 00:42:43 --> 00:42:50 So let's compare our rates here. 681 00:42:50 --> 00:42:51 So we have a collision. 682 00:42:51 --> 00:42:55 A and M collide, and there's a certain probability on the 683 00:42:55 --> 00:42:59 collision that kinetic energy will be transferred to 684 00:42:59 --> 00:43:00 vibrational energy. 685 00:43:00 --> 00:43:02 That turns out to be a moderate probability. 686 00:43:02 --> 00:43:09 So k1 is reasonably fast. 687 00:43:09 --> 00:43:11 The reverse process, you have something that's vibrationally 688 00:43:11 --> 00:43:14 excited colliding with a molecule. 689 00:43:14 --> 00:43:18 There's a probability that that excitation is going to get 690 00:43:18 --> 00:43:20 turned back into kinetic energy. 691 00:43:20 --> 00:43:22 And then these two molecules will fly apart with 692 00:43:22 --> 00:43:23 high velocity. 693 00:43:23 --> 00:43:25 And that tends to be much more highly probable 694 00:43:25 --> 00:43:28 then the first process. 695 00:43:28 --> 00:43:30 So this is faster. 696 00:43:30 --> 00:43:32 Then the excited molecule's waiting around. 697 00:43:32 --> 00:43:34 And there's a probability that there's it's going 698 00:43:34 --> 00:43:35 to just fall apart. 699 00:43:35 --> 00:43:38 And that turns out to be really slow. 700 00:43:38 --> 00:43:49 So we're going to assume that the last step is a slow step. 701 00:43:49 --> 00:43:50 So what does it look like? 702 00:43:50 --> 00:43:55 Well, we have the creation is fast. 703 00:43:55 --> 00:43:57 So the intermediate is fast. 704 00:43:57 --> 00:44:02 But the destruction through the reverse process is much faster. 705 00:44:02 --> 00:44:05 We have at least one step, out of the intermediate, which 706 00:44:05 --> 00:44:07 is faster than the step into the intermediate. 707 00:44:07 --> 00:44:08 That's all we care about. 708 00:44:08 --> 00:44:11 It doesn't really matter that this is slow in terms of using 709 00:44:11 --> 00:44:12 the steady state approximation. 710 00:44:12 --> 00:44:17 We just want the step getting out of the intermediate, a 711 00:44:17 --> 00:44:20 backwards step in this case here, to be faster than the 712 00:44:20 --> 00:44:22 state into the intermediate. 713 00:44:22 --> 00:44:26 So this allows this, this fact that this is faster, that 714 00:44:26 --> 00:44:30 allows us to use a steady state approximation. 715 00:44:30 --> 00:44:33 So now we can go ahead and solve the problem. 716 00:44:33 --> 00:44:38 So we write down all the rates minus dA/dt. 717 00:44:38 --> 00:44:44 It's k1 A. 718 00:44:44 --> 00:44:46 Let's start with the rate of the action. 719 00:44:46 --> 00:44:47 Let's take a product. 720 00:44:47 --> 00:44:48 Let's take C. 721 00:44:48 --> 00:44:50 Let's say this goes to product C here. 722 00:44:50 --> 00:44:55 So dC/dt is equal to k2 times A, that's the 723 00:44:55 --> 00:45:00 rate of the reaction. 724 00:45:00 --> 00:45:03 And we're going to look at this rate of reaction. 725 00:45:03 --> 00:45:04 And see how it changes. 726 00:45:04 --> 00:45:08 What it looks like under limiting conditions. 727 00:45:08 --> 00:45:13 The rate of reaction. 728 00:45:13 --> 00:45:15 Let's look at the intermediate. 729 00:45:15 --> 00:45:16 Because this is what we're going to solve first, because 730 00:45:16 --> 00:45:19 this is going to turn out to be an algebraic equation. 731 00:45:19 --> 00:45:20 Because we're going to apply the steady 732 00:45:20 --> 00:45:23 state approximations. 733 00:45:23 --> 00:45:28 Intermediate d A star / dt is, we can create it through 734 00:45:28 --> 00:45:30 the forward process. 735 00:45:30 --> 00:45:33 We destroy it through the backwards process. 736 00:45:33 --> 00:45:35 A star times M. 737 00:45:35 --> 00:45:37 And we destroy it through the final process. 738 00:45:37 --> 00:45:40 A star. 739 00:45:40 --> 00:45:43 We apply the steady state approximation, steady 740 00:45:43 --> 00:45:46 state, steady state. 741 00:45:46 --> 00:45:54 We solve for A steady state algebraically. k1 A M over 742 00:45:54 --> 00:46:00 M times k minus 1 plus k2. 743 00:46:00 --> 00:46:04 And then we plug that back into the rate of the reaction. 744 00:46:04 --> 00:46:05 The rate of appearance of C. 745 00:46:05 --> 00:46:07 Which is what we're measuring experimentally. 746 00:46:07 --> 00:46:08 We're looking at the products. 747 00:46:08 --> 00:46:10 Measuring the appearance of products. 748 00:46:10 --> 00:46:11 So they're destruction. 749 00:46:11 --> 00:46:21 This is also equal to the destruction of A. 750 00:46:21 --> 00:46:31 And so we plug this into the, we write our rate of reaction. 751 00:46:31 --> 00:46:34 Which we measure, experimentally. 752 00:46:34 --> 00:46:37 We find this is k1 times k2. 753 00:46:37 --> 00:46:44 Times, we plug in for, and here, what we do here, we solve 754 00:46:44 --> 00:46:48 A in terms of A steady state. 755 00:46:48 --> 00:46:53 A is equal to something A steady state. 756 00:46:53 --> 00:47:04 Well, let's see, A is equal to M k minus one plus k2 times, 757 00:47:04 --> 00:47:06 well this is actually, I have this wrong up here, 758 00:47:06 --> 00:47:11 this is d A star. 759 00:47:11 --> 00:47:15 So we don't need to do this here. 760 00:47:15 --> 00:47:18 The appearance of C is, A star here. 761 00:47:18 --> 00:47:24 So we plug in A steady state here. 762 00:47:24 --> 00:47:29 A star steady state, and this is A star steady state. 763 00:47:29 --> 00:47:36 The intermediate is what we're solving a steady state for. 764 00:47:36 --> 00:47:43 So this is k1 k2 A M over M k minus one plus k2. 765 00:47:43 --> 00:47:48 We just have the extra k2 that appears when we multiply 766 00:47:48 --> 00:47:52 A steady state star with the k2 there. 767 00:47:52 --> 00:47:55 And then we can look at the limiting cases. 768 00:47:55 --> 00:47:58 There are two limiting cases, which we can tell by 769 00:47:58 --> 00:47:59 looking at the denominator. 770 00:47:59 --> 00:48:01 If we have one term bigger than the other. 771 00:48:01 --> 00:48:03 Only two choices here. 772 00:48:03 --> 00:48:07 So the first case is where M k minus one is 773 00:48:07 --> 00:48:10 much bigger than k2. 774 00:48:10 --> 00:48:11 Experimentally, what does that mean? 775 00:48:11 --> 00:48:14 It means that this in the gas phase now. 776 00:48:14 --> 00:48:17 It means that regardless of what these rates are here, 777 00:48:17 --> 00:48:21 we've managed to make the concentration of M high enough 778 00:48:21 --> 00:48:23 so that this becomes true. 779 00:48:23 --> 00:48:26 There's always an M that's going to be high enough so that 780 00:48:26 --> 00:48:28 the multiplication of these two is going to be 781 00:48:28 --> 00:48:29 bigger than that. 782 00:48:29 --> 00:48:29 What does that mean? 783 00:48:29 --> 00:48:33 It means that the pressure is high. 784 00:48:33 --> 00:48:34 Concentration of M is high. 785 00:48:34 --> 00:48:40 This is a high pressure case. 786 00:48:40 --> 00:48:45 And then the other one is M k minus one is much less than k2. 787 00:48:45 --> 00:48:47 M is very small. 788 00:48:47 --> 00:48:51 Low concentration, low pressure. 789 00:48:51 --> 00:48:54 The partial pressure of M is very low. 790 00:48:54 --> 00:48:55 This could be A. 791 00:48:55 --> 00:48:57 M could be A, or it could be some other molecule 792 00:48:57 --> 00:49:00 that's in the mix. 793 00:49:00 --> 00:49:02 And by high pressure we mean something on the order 794 00:49:02 --> 00:49:04 of one bar, let's say. 795 00:49:04 --> 00:49:07 That by low pressure we mean something around 796 00:49:07 --> 00:49:12 10 to the minus 4 bar. 797 00:49:12 --> 00:49:16 So now we can go and put these limiting cases in our equation 798 00:49:16 --> 00:49:19 here. k minus one is much bigger than k2, so we 799 00:49:19 --> 00:49:21 can ignore k2 here. 800 00:49:21 --> 00:49:25 And then we have something that looks like the M's disappear. 801 00:49:25 --> 00:49:30 We have something that looks like k1 k2 over k minus 802 00:49:30 --> 00:49:37 one times A is the rate of reaction. 803 00:49:37 --> 00:49:38 First order. 804 00:49:38 --> 00:49:39 Looks like it's first order in A. 805 00:49:39 --> 00:49:41 This is what people had observed. 806 00:49:41 --> 00:49:42 Great. 807 00:49:42 --> 00:49:45 The problem is that they didn't know to go to low pressure. 808 00:49:45 --> 00:49:48 They were doing all their experiments at 809 00:49:48 --> 00:49:49 atmospheric pressure. 810 00:49:49 --> 00:49:50 They were getting a first order rate. 811 00:49:50 --> 00:49:52 They thought they had solved the problem. 812 00:49:52 --> 00:49:54 It's just A falling apart by itself. 813 00:49:54 --> 00:49:57 Until you go to low pressure, where now this term 814 00:49:57 --> 00:49:59 dominates the denominator. 815 00:49:59 --> 00:50:03 Or rather, this one dominates the denominator. 816 00:50:03 --> 00:50:12 And now you have something of the form k1 A times M. 817 00:50:12 --> 00:50:14 This dominates, this term goes away. 818 00:50:14 --> 00:50:19 The k2 goes away and you have k1 times A times M left over 819 00:50:19 --> 00:50:25 as the rate of the reaction. 820 00:50:25 --> 00:50:25 Second order. 821 00:50:25 --> 00:50:27 There are two species here. 822 00:50:27 --> 00:50:27 A times M. 823 00:50:27 --> 00:50:30 M could be A or it could be some chaperone. 824 00:50:30 --> 00:50:31 A squared. 825 00:50:31 --> 00:50:33 Second order process. 826 00:50:33 --> 00:50:34 So if you're at low pressure, you see 827 00:50:34 --> 00:50:35 something at second order. 828 00:50:35 --> 00:50:37 You've discovered something about the mechanisms 829 00:50:37 --> 00:50:39 you didn't expect. 830 00:50:39 --> 00:50:42 And then you get a Nobel Prize. 831 00:50:42 --> 00:50:44 Alright, next time we'll do chain reactions.