1 00:00:00,130 --> 00:00:01,690 The following content is provided 2 00:00:01,690 --> 00:00:03,820 under a Creative Commons license. 3 00:00:03,820 --> 00:00:06,850 Your support will help MIT OpenCourseWare continue 4 00:00:06,850 --> 00:00:10,520 to offer high quality educational resources for free. 5 00:00:10,520 --> 00:00:13,200 To make a donation, or view additional materials 6 00:00:13,200 --> 00:00:17,160 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:17,160 --> 00:00:20,027 at ocw.mit.edu. 8 00:00:20,027 --> 00:00:21,860 PROFESSOR: So, any questions from last time? 9 00:00:21,860 --> 00:00:27,090 We started doing reaction mechanisms. 10 00:00:27,090 --> 00:00:33,430 We did first order parallel reactions. 11 00:00:33,430 --> 00:00:35,280 So one of the things that we'll be stressing 12 00:00:35,280 --> 00:00:39,630 is that the mechanisms that we write down 13 00:00:39,630 --> 00:00:46,560 are ways to try to understand a complicated chemical process. 14 00:00:46,560 --> 00:00:53,510 And you take data, you can find intermediates sometimes. 15 00:00:53,510 --> 00:00:56,740 And you get rate laws from the data. 16 00:00:56,740 --> 00:00:58,220 You infer rate laws. 17 00:00:58,220 --> 00:01:01,470 And from these rate laws, you think up 18 00:01:01,470 --> 00:01:08,890 of a reaction mechanism and you make sure that what you measure 19 00:01:08,890 --> 00:01:10,460 is consistent with your mechanism 20 00:01:10,460 --> 00:01:15,051 that you make up to explain the complicated chemical process. 21 00:01:15,051 --> 00:01:16,800 And one really important thing to remember 22 00:01:16,800 --> 00:01:20,860 is that when you come up with a mechanism 23 00:01:20,860 --> 00:01:25,800 and you've got a bunch of data that supports it, it's great. 24 00:01:25,800 --> 00:01:29,270 But it doesn't mean that you've solve the problem. 25 00:01:29,270 --> 00:01:31,400 It doesn't mean that you've proven 26 00:01:31,400 --> 00:01:35,280 that the way that the chemistry proceeds 27 00:01:35,280 --> 00:01:36,780 is really the way it proceeds. 28 00:01:36,780 --> 00:01:40,822 Because there's always the possibility 29 00:01:40,822 --> 00:01:42,780 that there's an intermediate somewhere in there 30 00:01:42,780 --> 00:01:45,930 that you haven't been able to measure. 31 00:01:45,930 --> 00:01:48,280 And a lot of people got tripped up 32 00:01:48,280 --> 00:01:51,500 over the course of the last century, 33 00:01:51,500 --> 00:01:55,700 of making up mechanisms, getting a lot of data that supports 34 00:01:55,700 --> 00:02:01,760 the mechanism and missing something really important. 35 00:02:01,760 --> 00:02:04,350 So it's a common mistake, that people 36 00:02:04,350 --> 00:02:07,660 think they've proven a particular chemical 37 00:02:07,660 --> 00:02:12,880 mechanism just because they have data that supports it. 38 00:02:12,880 --> 00:02:15,450 It's up to a certain point that you know what you have. 39 00:02:15,450 --> 00:02:20,330 And we'll see examples of that today, probably. 40 00:02:20,330 --> 00:02:22,660 And then certainly next time when 41 00:02:22,660 --> 00:02:25,000 we start making approximations in our kinetics. 42 00:02:25,000 --> 00:02:27,990 And see where things can really go wrong 43 00:02:27,990 --> 00:02:30,600 if you don't have enough data. 44 00:02:30,600 --> 00:02:32,330 If you don't try to fish out really, 45 00:02:32,330 --> 00:02:37,680 really, tiny intermediates somewhere along the way. 46 00:02:37,680 --> 00:02:40,510 So last time we did then parallel first order reactions. 47 00:02:40,510 --> 00:02:45,980 And today we're going to march through and do 48 00:02:45,980 --> 00:02:47,960 parallel first and second order. 49 00:02:47,960 --> 00:02:50,180 So one is first order, the other one is second order. 50 00:02:50,180 --> 00:02:52,260 Things are going to get more complicated. 51 00:02:52,260 --> 00:02:54,010 And what you're going to get is a flavor 52 00:02:54,010 --> 00:02:55,950 of how to solve the problems. 53 00:02:55,950 --> 00:02:57,319 Setting up the problems. 54 00:02:57,319 --> 00:02:59,860 We're not going to do a lot of the algebra here on the board, 55 00:02:59,860 --> 00:03:02,870 because otherwise we would spend the next three 56 00:03:02,870 --> 00:03:04,080 weeks doing algebra. 57 00:03:04,080 --> 00:03:06,095 And that would be no fun at all. 58 00:03:06,095 --> 00:03:07,470 That doesn't mean that you're not 59 00:03:07,470 --> 00:03:10,360 going to be doing any algebra on the homework. 60 00:03:10,360 --> 00:03:15,520 Because part of it is figuring how to do it. 61 00:03:15,520 --> 00:03:22,090 So, we have a reaction where A goes to B plus C. 62 00:03:22,090 --> 00:03:24,480 And the first thing we did was, the first mechanism 63 00:03:24,480 --> 00:03:26,640 we wrote was a parallel mechanism where 64 00:03:26,640 --> 00:03:34,240 we had A goes B and A goes to C. That's the mechanism. 65 00:03:34,240 --> 00:03:41,810 And one other way to write it is A goes to B or C, 66 00:03:41,810 --> 00:03:47,810 in a way that is more, sort of describes 67 00:03:47,810 --> 00:03:49,830 the branching process, that when we 68 00:03:49,830 --> 00:03:54,810 talk about a branching ratio, the ratio of amount of B to C, 69 00:03:54,810 --> 00:03:59,020 this looks like the branching out of two different paths. 70 00:03:59,020 --> 00:04:03,520 And this time we're going to do where this is first order 71 00:04:03,520 --> 00:04:07,620 and this is second order. 72 00:04:07,620 --> 00:04:13,220 With the rate k1 and a rate k2. 73 00:04:13,220 --> 00:04:15,150 And the way that you do all these problems, 74 00:04:15,150 --> 00:04:18,284 you have to be very systematic about it. 75 00:04:18,284 --> 00:04:19,700 The first thing you do is you have 76 00:04:19,700 --> 00:04:22,240 to write down all your rate laws. 77 00:04:22,240 --> 00:04:25,470 And make sure that you include everything. 78 00:04:25,470 --> 00:04:28,830 So you start by writing the rate law for the destruction of A. 79 00:04:28,830 --> 00:04:32,260 And then the rate law for the appearance of B and for C. 80 00:04:32,260 --> 00:04:33,940 So for a, we have dA/dt. 81 00:04:33,940 --> 00:04:36,090 And again, I'm dropping the brackets around the A, 82 00:04:36,090 --> 00:04:41,030 because I don't want to carry around all these extra symbols. 83 00:04:41,030 --> 00:04:47,990 So we destroy A by making B. So that's going to be first order. 84 00:04:47,990 --> 00:04:49,721 And then we also destroy A by making C, 85 00:04:49,721 --> 00:04:50,720 and that's second order. 86 00:04:50,720 --> 00:04:57,420 So we have two paths out of A. To get rid of A. 87 00:04:57,420 --> 00:05:02,660 Then we write down the appearance of B, dB/dt. 88 00:05:02,660 --> 00:05:08,710 It's only through one path, first order in A, 89 00:05:08,710 --> 00:05:13,430 and then the appearance of C, only one path 90 00:05:13,430 --> 00:05:16,305 which is second order. 91 00:05:16,305 --> 00:05:18,180 And these are all our differential equations. 92 00:05:18,180 --> 00:05:24,900 And now the trick is, how do you solve these three differential 93 00:05:24,900 --> 00:05:32,300 equations in a way that you get meaningful information out. 94 00:05:32,300 --> 00:05:36,490 Well, the first thing to do is to, 95 00:05:36,490 --> 00:05:41,532 so these two here depend on A. So the appearance of B 96 00:05:41,532 --> 00:05:43,490 depends on A, the appearance of C depends on A. 97 00:05:43,490 --> 00:05:45,490 But this one here only depends on A by itself. 98 00:05:45,490 --> 00:05:47,904 So this is the first one you're going to start out with. 99 00:05:47,904 --> 00:05:49,320 Because you can put all of the A's 100 00:05:49,320 --> 00:05:53,530 on one side and all the time on the other side and integrate. 101 00:05:53,530 --> 00:05:59,340 So then you solve, and you can rewrite this 102 00:05:59,340 --> 00:06:10,740 as minus dA/dt is equal to k1 times A times one plus k2 103 00:06:10,740 --> 00:06:16,630 over k1 times A. And then you put all the A's on one side, 104 00:06:16,630 --> 00:06:18,950 all the t's on the other side. 105 00:06:18,950 --> 00:06:30,580 And integrate from A0 to A minus dA over A times one plus k2 106 00:06:30,580 --> 00:06:39,430 over k1 times A. That is equal to k1 from zero to t dt. 107 00:06:39,430 --> 00:06:41,485 And then you have to solve this integral here. 108 00:06:41,485 --> 00:06:44,410 And there's a reason why I factored out the A here. 109 00:06:44,410 --> 00:06:45,910 It's because this is of the form you 110 00:06:45,910 --> 00:06:48,810 can use to use the trick of partial fractions, which 111 00:06:48,810 --> 00:06:51,900 we mentioned this before. 112 00:06:51,900 --> 00:07:01,650 So you use partial fractions to solve this. 113 00:07:01,650 --> 00:07:04,380 And I'm not going to do it on the board. 114 00:07:04,380 --> 00:07:06,270 You turn the crank, you basically 115 00:07:06,270 --> 00:07:09,560 write one over A times one plus k2 over k1 A 116 00:07:09,560 --> 00:07:13,184 is equal to n1 of A plus n2 over this part here, 117 00:07:13,184 --> 00:07:15,350 and you solve for n1 and n2 and you plug it back in, 118 00:07:15,350 --> 00:07:18,730 and you redo your integral, et cetera. 119 00:07:18,730 --> 00:07:19,802 And you get your answer. 120 00:07:19,802 --> 00:07:21,510 Which I'm going to write down because I'm 121 00:07:21,510 --> 00:07:29,170 going to use it later. k1 A0 over e 122 00:07:29,170 --> 00:07:36,195 to the k1 times t, k1 plus k2 A0. 123 00:07:36,195 --> 00:07:37,820 So one of the things we immediately see 124 00:07:37,820 --> 00:07:41,930 is even though this mechanism isn't very complicated, 125 00:07:41,930 --> 00:07:45,470 just two paths, one first order, one second order, the solutions 126 00:07:45,470 --> 00:07:53,080 start to get not so simple pretty quickly. 127 00:07:53,080 --> 00:07:55,590 So it depends, there's an exponential on the bottom here. 128 00:07:55,590 --> 00:07:58,700 Depends on the initial concentrations and both rate 129 00:07:58,700 --> 00:08:02,205 constants are in there. 130 00:08:02,205 --> 00:08:03,580 The first thing you want to do is 131 00:08:03,580 --> 00:08:09,900 you want to look at limiting cases. 132 00:08:09,900 --> 00:08:12,080 Just like before. 133 00:08:12,080 --> 00:08:15,520 We did, and the first limiting case 134 00:08:15,520 --> 00:08:18,300 we're going to look at, so interesting to look at 135 00:08:18,300 --> 00:08:22,110 is this guy right here. k1 and k2 A here. 136 00:08:22,110 --> 00:08:27,870 If one is bigger than the other, then things will cancel out. 137 00:08:27,870 --> 00:08:30,620 And we can get some intuition. 138 00:08:30,620 --> 00:08:33,270 So the first limiting case we're going to look at 139 00:08:33,270 --> 00:08:43,961 is where k2 A0, this term right here, is much smaller than k1. 140 00:08:43,961 --> 00:08:46,210 Now, we want to make sure that in these limiting cases 141 00:08:46,210 --> 00:08:49,160 that when I write something like this, that the units actually 142 00:08:49,160 --> 00:08:50,080 make sense. 143 00:08:50,080 --> 00:08:52,200 That I'm not saying, three apples 144 00:08:52,200 --> 00:08:55,000 are less than four oranges. 145 00:08:55,000 --> 00:08:57,870 So k1, it's first order, so the units 146 00:08:57,870 --> 00:09:01,950 are one over second. k2, it's a second order rate constant, 147 00:09:01,950 --> 00:09:07,534 so the units are one over second molar, times moles per liter. 148 00:09:07,534 --> 00:09:08,950 So the moles per liters cancel out 149 00:09:08,950 --> 00:09:10,920 and I have one over our second is less than one over second, 150 00:09:10,920 --> 00:09:12,160 so things are great. 151 00:09:12,160 --> 00:09:15,320 It I'm making the right kind of approximation here. 152 00:09:15,320 --> 00:09:17,110 What else is this approximation saying? 153 00:09:17,110 --> 00:09:23,650 This is saying that the rate into B, so this one is 154 00:09:23,650 --> 00:09:26,670 the faster one. k1, the rate into B, is the rate into, 155 00:09:26,670 --> 00:09:32,580 is k1, is faster than k2 times A, which is basically the rate 156 00:09:32,580 --> 00:09:41,055 into C. So this is saying that the rate into B, to form B, 157 00:09:41,055 --> 00:09:53,250 is faster than into C. So we can write down a sketch. 158 00:09:53,250 --> 00:09:57,970 We can sketch what we expect this approximation to be, then. 159 00:09:57,970 --> 00:10:04,370 And we know it's going to look something like this. 160 00:10:04,370 --> 00:10:10,290 We know that A is going to come down, exponentially. 161 00:10:10,290 --> 00:10:11,938 Without any structure to it. 162 00:10:11,938 --> 00:10:13,646 It's going to go either into B or into C, 163 00:10:13,646 --> 00:10:15,840 in some branching fractions. 164 00:10:15,840 --> 00:10:19,430 And we know that the rate into B is faster than into C, 165 00:10:19,430 --> 00:10:22,440 so B is going to come up like this. 166 00:10:22,440 --> 00:10:27,510 And C is going to come up, but slower. 167 00:10:27,510 --> 00:10:34,660 And there's going to be a ratio between these two guys. 168 00:10:34,660 --> 00:10:37,400 So we know the slope here is going to be slower 169 00:10:37,400 --> 00:10:40,140 then the slope for B. And we can check that out, 170 00:10:40,140 --> 00:10:41,980 within our approximation. 171 00:10:41,980 --> 00:10:47,511 We can write, at t equals zero, find out 172 00:10:47,511 --> 00:10:48,760 what these initial slopes are. 173 00:10:48,760 --> 00:10:52,780 For the creation of B and C. So dB/dt 174 00:10:52,780 --> 00:10:58,150 is the slope of the creation of B at the beginning. 175 00:10:58,150 --> 00:11:02,930 At t equals zero. 176 00:11:02,930 --> 00:11:05,740 And that's just the rate law. dB/dt at t 177 00:11:05,740 --> 00:11:11,830 equals zero is k1 A0. 178 00:11:11,830 --> 00:11:21,320 And dC/dt, the initial slope, t equals zero, is k1 A0 squared. 179 00:11:21,320 --> 00:11:28,430 Or k2 A0 squared, rather. 180 00:11:28,430 --> 00:11:36,890 So I'm going to write it as k2 A0 squared, like this. 181 00:11:36,890 --> 00:11:40,130 We said k1 is much bigger than k2 times A0. 182 00:11:40,130 --> 00:11:41,870 There's the same A0 here. 183 00:11:41,870 --> 00:11:43,730 So we see, just by writing this down, 184 00:11:43,730 --> 00:11:48,904 that our intuition that because the rate into B 185 00:11:48,904 --> 00:11:51,320 is much faster than into A, that it should look like this. 186 00:11:51,320 --> 00:11:54,700 Well it's borne out just like this very simple sort 187 00:11:54,700 --> 00:11:58,640 of looking at the initial rate where the slope here 188 00:11:58,640 --> 00:12:04,220 is much smaller than the slope here. 189 00:12:04,220 --> 00:12:10,240 And then we can take the approximation in here. 190 00:12:10,240 --> 00:12:13,330 And see what it implies. 191 00:12:13,330 --> 00:12:18,100 So let me do that here, let me just use my green chalk here. 192 00:12:18,100 --> 00:12:21,520 So k2 A0 is much less than k1. 193 00:12:21,520 --> 00:12:27,760 So this is going to go away. k2 A0 is much less than k1. 194 00:12:27,760 --> 00:12:29,260 Well, this is building up from zero. 195 00:12:29,260 --> 00:12:31,530 So this is always bigger than one here. 196 00:12:31,530 --> 00:12:33,171 So I can get rid of this as well. 197 00:12:33,171 --> 00:12:34,670 Because there's the k1 sitting here. 198 00:12:34,670 --> 00:12:39,100 And k1 is much bigger than k2 A0. 199 00:12:39,100 --> 00:12:44,280 Once I got rid of all these two guys, then the k1's cancel out. 200 00:12:44,280 --> 00:12:45,980 And this is looking very nice. 201 00:12:45,980 --> 00:12:50,110 Because now I can take that e to the k1 t and put it upstairs. 202 00:12:50,110 --> 00:12:56,380 A is equal to approximately A0 e to the minus k1 times 203 00:12:56,380 --> 00:13:00,320 time in this approximation. 204 00:13:00,320 --> 00:13:02,870 It looks like it's first order coming down, 205 00:13:02,870 --> 00:13:05,670 with a rate constant k1. 206 00:13:05,670 --> 00:13:10,250 It looks like basically the branching into C 207 00:13:10,250 --> 00:13:13,874 is nonexistent as far as, at least in this approximation, 208 00:13:13,874 --> 00:13:14,790 as far A is concerned. 209 00:13:14,790 --> 00:13:19,970 It's going to look like this thing here. 210 00:13:19,970 --> 00:13:23,650 It's going to look like first order 211 00:13:23,650 --> 00:13:26,120 A goes to B with rate constant k1. 212 00:13:26,120 --> 00:13:28,330 So if you didn't know, if you didn't 213 00:13:28,330 --> 00:13:30,520 know that there was another substance, C, 214 00:13:30,520 --> 00:13:33,290 that was being formed along the way, if you didn't measure it, 215 00:13:33,290 --> 00:13:35,990 if you didn't do the analytical chemistry and sort of fish it 216 00:13:35,990 --> 00:13:41,870 out, and you just looked at the two major components, A and B, 217 00:13:41,870 --> 00:13:45,130 in this case here, you'd measure a rate out of A, 218 00:13:45,130 --> 00:13:48,110 it would look like it's first order. 219 00:13:48,110 --> 00:13:50,380 It's great. 220 00:13:50,380 --> 00:13:51,220 Got my mechanism. 221 00:13:51,220 --> 00:13:54,150 A goes to B, period. 222 00:13:54,150 --> 00:13:57,195 And you would know there was a minor component that 223 00:13:57,195 --> 00:14:02,050 was being formed at the same time. 224 00:14:02,050 --> 00:14:15,560 OK, let's do another approximation. 225 00:14:15,560 --> 00:14:24,917 Let's do the other case, where k2 A0 is much bigger than k1. 226 00:14:24,917 --> 00:14:26,500 And, in this case here, you have to be 227 00:14:26,500 --> 00:14:27,583 a little bit more careful. 228 00:14:27,583 --> 00:14:30,780 You have to add that you're going 229 00:14:30,780 --> 00:14:32,300 to look at this at early times. 230 00:14:32,300 --> 00:14:36,240 And we're going to see why early times is important here. 231 00:14:36,240 --> 00:14:38,610 In just five minutes. 232 00:14:38,610 --> 00:14:41,180 And early times, and how you define early times. 233 00:14:41,180 --> 00:14:44,000 What does it mean for things to be close to time equals zero? 234 00:14:44,000 --> 00:14:48,250 Well, we have to have a reference time scale. 235 00:14:48,250 --> 00:14:51,690 The reactions, I'm going at a certain rate. 236 00:14:51,690 --> 00:14:54,150 And one second may be very slow. 237 00:14:54,150 --> 00:14:55,120 Or it may be very fast. 238 00:14:55,120 --> 00:14:57,640 Depending on the rate of the reaction. 239 00:14:57,640 --> 00:15:03,220 So k1 here defines the time scale of the problem. 240 00:15:03,220 --> 00:15:06,640 It's one over second, unit of one over second. 241 00:15:06,640 --> 00:15:10,250 And early times means that the times that I'm looking at 242 00:15:10,250 --> 00:15:15,620 compare to this rate, it's very small. 243 00:15:15,620 --> 00:15:19,960 So k1 t is less than one. 244 00:15:19,960 --> 00:15:23,520 That means that the time, during the time period that I'm 245 00:15:23,520 --> 00:15:28,680 looking at, hardly anything has happened to the branching of A 246 00:15:28,680 --> 00:15:32,360 to B. That that's what I mean by early time. 247 00:15:32,360 --> 00:15:36,910 So B is hardly being built up yet. k1 is our reference 248 00:15:36,910 --> 00:15:40,590 time. k1 is units of one over second. 249 00:15:40,590 --> 00:15:42,480 This has units of seconds. 250 00:15:42,480 --> 00:15:45,025 So it's all working out fine. 251 00:15:45,025 --> 00:15:47,650 In other words, I would have no idea how to define early times. 252 00:15:47,650 --> 00:15:49,590 I could say, it's in a year, it could be, 253 00:15:49,590 --> 00:15:51,320 if you're looking at plutonium decay, 254 00:15:51,320 --> 00:15:53,850 it's a 100,000 year half-life. 255 00:15:53,850 --> 00:15:55,460 A year would be an early time. 256 00:15:55,460 --> 00:15:58,760 So it would be really fast, compared to the process. 257 00:15:58,760 --> 00:16:01,075 But if you're looking at a reaction that 258 00:16:01,075 --> 00:16:04,450 takes a few nanoseconds than a year would be very, very long. 259 00:16:04,450 --> 00:16:06,430 So you really need your reference time 260 00:16:06,430 --> 00:16:11,030 somewhere in there. 261 00:16:11,030 --> 00:16:13,910 So this approximation means that essentially, no B 262 00:16:13,910 --> 00:16:16,150 is being created. 263 00:16:16,150 --> 00:16:19,400 While we're looking at the process. 264 00:16:19,400 --> 00:16:23,800 And so we might then very well expect 265 00:16:23,800 --> 00:16:26,165 that if we look at A and C, that the answer is going 266 00:16:26,165 --> 00:16:29,110 to be that we're going to see something that's second order. 267 00:16:29,110 --> 00:16:35,255 So we expect that the form of A, the way to write it 268 00:16:35,255 --> 00:16:39,140 is going to be one over A equals k1 t plus A0. 269 00:16:39,140 --> 00:16:42,950 So let's go ahead and take our complete solution 270 00:16:42,950 --> 00:16:46,620 and write it in the way that we might expect the answer to be. 271 00:16:46,620 --> 00:16:49,500 In terms of one over A rather than A here. 272 00:16:49,500 --> 00:16:56,760 So let's just invert it. e to the k1 times time, k1 plus k2 273 00:16:56,760 --> 00:17:07,900 A0, minus k2 A0 over k1 A0. 274 00:17:07,900 --> 00:17:11,190 This should be a minus sign here. 275 00:17:11,190 --> 00:17:14,157 There should be a minus sign here. 276 00:17:14,157 --> 00:17:15,740 It didn't matter for our approximation 277 00:17:15,740 --> 00:17:17,750 because we got rid of it anyway, but it 278 00:17:17,750 --> 00:17:24,120 would be a proper sign here is minus. 279 00:17:24,120 --> 00:17:27,390 And so now we need to make our approximation 280 00:17:27,390 --> 00:17:29,530 and figure out what this here. 281 00:17:29,530 --> 00:17:33,400 So if we take the approximation that k1 t is small, less 282 00:17:33,400 --> 00:17:36,530 than one, then what that should trigger in your mind 283 00:17:36,530 --> 00:17:38,770 if that if you have e to the something that small 284 00:17:38,770 --> 00:17:40,061 you should use a Taylor series. 285 00:17:40,061 --> 00:17:42,990 And expand this into a Taylor series. 286 00:17:42,990 --> 00:17:45,880 So we're going to see this multiple times. 287 00:17:45,880 --> 00:17:51,650 So we take e to the k1 t and expand it out as one 288 00:17:51,650 --> 00:17:55,750 plus k1 t plus et cetera. 289 00:17:55,750 --> 00:17:58,850 We're going to keep the first order terms in there. 290 00:17:58,850 --> 00:18:05,430 So we're going to plug this approximation into here. 291 00:18:05,430 --> 00:18:07,600 That's this approximation there. 292 00:18:07,600 --> 00:18:11,680 And then we're going to expand it out. 293 00:18:11,680 --> 00:18:21,510 So plug it in here, and, keeping it to first order in time, 294 00:18:21,510 --> 00:18:23,340 and multiplying it out, and I'm not 295 00:18:23,340 --> 00:18:24,940 going to do the intermediate steps, 296 00:18:24,940 --> 00:18:28,990 I'm just going to give you the stuff after doing 297 00:18:28,990 --> 00:18:32,870 the multiplications, we get k1 t squared. 298 00:18:32,870 --> 00:18:37,130 So that's basically k1 times one plus k1 t, 299 00:18:37,130 --> 00:18:45,600 and then you have plus k1 k2 A0 times t. 300 00:18:45,600 --> 00:18:50,350 That's this term here. k1 k2 A0 times t times that. 301 00:18:50,350 --> 00:18:54,190 And the k2 A0 times one gets subtracted out 302 00:18:54,190 --> 00:18:56,232 from this minus k2 A0 here. 303 00:18:56,232 --> 00:18:57,440 So that's all we have on top. 304 00:18:57,440 --> 00:19:00,304 And then there are terms of higher order in time. 305 00:19:00,304 --> 00:19:01,970 But we're not going to worry about this. 306 00:19:01,970 --> 00:19:05,670 We're going to stick to first order in time. 307 00:19:05,670 --> 00:19:08,420 Divided by k1 A0. 308 00:19:08,420 --> 00:19:11,990 And now you do your cancellations. 309 00:19:11,990 --> 00:19:14,530 We have our approximation at k1 times 310 00:19:14,530 --> 00:19:16,940 time is much less than one. 311 00:19:16,940 --> 00:19:20,600 So here we have k1 times k1 times time. 312 00:19:20,600 --> 00:19:22,980 So this term here means it's much less than this term 313 00:19:22,980 --> 00:19:24,540 here, because of our approximations. 314 00:19:24,540 --> 00:19:28,960 So we can get rid of this term here. 315 00:19:28,960 --> 00:19:29,920 Get rid of that. 316 00:19:29,920 --> 00:19:35,690 There's no reason for us to get rid of this one here. 317 00:19:35,690 --> 00:19:40,490 Because we've got k2 A0 there, which is much larger than k1. 318 00:19:40,490 --> 00:19:45,520 And so when you expand this out, all the k1's cancel out here. 319 00:19:45,520 --> 00:19:49,660 So so k1 cancels out that k1, we have one plus one. 320 00:19:49,660 --> 00:19:51,340 So this is basically one over A0. 321 00:19:51,340 --> 00:19:54,480 And then we have plus k2 A0 time over A0, 322 00:19:54,480 --> 00:20:07,060 so this ends up being equal to one over A0 plus k2 times time. 323 00:20:07,060 --> 00:20:09,310 Exactly what we expected. 324 00:20:09,310 --> 00:20:12,990 That if you put in the right approximation in the math, 325 00:20:12,990 --> 00:20:16,270 it behaves as your intuition would tell you. 326 00:20:16,270 --> 00:20:20,140 Which is that the branching into B nonexistent. 327 00:20:20,140 --> 00:20:23,500 And that you're looking at everything going to C. 328 00:20:23,500 --> 00:20:28,280 It looks second order in C. But in this case 329 00:20:28,280 --> 00:20:30,040 here, it's only early times. 330 00:20:30,040 --> 00:20:31,740 So if you keep going, what happens? 331 00:20:31,740 --> 00:20:34,990 If you keep going in time, beyond the time 332 00:20:34,990 --> 00:20:40,360 scale of this first order rate constant, 333 00:20:40,360 --> 00:20:45,470 and you wait long enough, then the amount of A 334 00:20:45,470 --> 00:20:46,800 keeps decreasing. 335 00:20:46,800 --> 00:20:50,800 Keeps getting smaller and smaller. 336 00:20:50,800 --> 00:20:53,580 So if you start the clock again, a little bit later, 337 00:20:53,580 --> 00:20:58,020 where A is much smaller, then k2 times 338 00:20:58,020 --> 00:21:01,030 the amount of A you have there, is no longer 339 00:21:01,030 --> 00:21:04,230 going to be much greater than k1. 340 00:21:04,230 --> 00:21:09,980 At some point along the way, k2 times A 341 00:21:09,980 --> 00:21:13,360 is going to be much less than k1. 342 00:21:13,360 --> 00:21:15,377 So at some point along the way, it's 343 00:21:15,377 --> 00:21:17,210 not going to be this approximation any more. 344 00:21:17,210 --> 00:21:19,400 It's going to be the previous one that we did. 345 00:21:19,400 --> 00:21:22,010 The one up there. 346 00:21:22,010 --> 00:21:24,510 So at early times we start with something 347 00:21:24,510 --> 00:21:29,420 that may look second order for A. Time goes on, time goes on, 348 00:21:29,420 --> 00:21:31,290 time goes on, A gets depleted. 349 00:21:31,290 --> 00:21:34,170 A gets depleted, and the rate k2 times 350 00:21:34,170 --> 00:21:38,510 A, which is sitting right here, this part here 351 00:21:38,510 --> 00:21:40,850 becomes smaller and smaller. 352 00:21:40,850 --> 00:21:43,210 And pretty soon this part wins. 353 00:21:43,210 --> 00:21:46,240 And it becomes first order. 354 00:21:46,240 --> 00:21:49,370 So what you'd expect to see, then, 355 00:21:49,370 --> 00:21:59,950 is if you were to plot as a function of A, 356 00:21:59,950 --> 00:22:02,860 you expect to see something that starts out as first order 357 00:22:02,860 --> 00:22:07,130 and then as A gets depleted, switches over to second order. 358 00:22:07,130 --> 00:22:09,220 Something that's second order here. 359 00:22:09,220 --> 00:22:11,550 And first order. 360 00:22:11,550 --> 00:22:17,720 This is the concentration of A. So if you were to plot it, 361 00:22:17,720 --> 00:22:20,260 here you're plotting the concentration of A. 362 00:22:20,260 --> 00:22:26,230 If you were to plot the log, of the concentration of A, 363 00:22:26,230 --> 00:22:29,500 at long times you'd expect it to be first order. 364 00:22:29,500 --> 00:22:31,960 So something linear. 365 00:22:31,960 --> 00:22:35,030 But at early times you'd expect it to be second order. 366 00:22:35,030 --> 00:22:37,320 So you'd expect to see something that's nonlinear, 367 00:22:37,320 --> 00:22:40,060 then switching to something linear. 368 00:22:40,060 --> 00:22:47,110 And if you were to plot it as one over A, 369 00:22:47,110 --> 00:22:49,610 then you'd expect to start off with something that's linear, 370 00:22:49,610 --> 00:22:53,650 at second order, early times. 371 00:22:53,650 --> 00:22:57,130 But instead of keeping, being second order, 372 00:22:57,130 --> 00:23:04,220 as you deplete the amount of A, the branching into B 373 00:23:04,220 --> 00:23:05,680 becomes important. 374 00:23:05,680 --> 00:23:09,710 And you become second order. 375 00:23:09,710 --> 00:23:14,380 So this is second order at the beginning. 376 00:23:14,380 --> 00:23:23,150 And this becomes first order. 377 00:23:23,150 --> 00:23:26,040 So, any questions here? 378 00:23:26,040 --> 00:23:27,600 The importance of doing this problem 379 00:23:27,600 --> 00:23:30,390 here was to lay out basically a systematic way 380 00:23:30,390 --> 00:23:32,800 of looking at the problem. 381 00:23:32,800 --> 00:23:39,080 You lay out your rate equations. 382 00:23:39,080 --> 00:23:40,030 Like this. 383 00:23:40,030 --> 00:23:45,230 You solve what you can, in this case you solve for A. 384 00:23:45,230 --> 00:23:46,360 And it's complicated. 385 00:23:46,360 --> 00:23:48,220 So you want to learn something about what 386 00:23:48,220 --> 00:23:50,630 the data might look like. 387 00:23:50,630 --> 00:23:51,980 And you look at limiting cases. 388 00:23:51,980 --> 00:23:53,530 Two obvious limiting cases. 389 00:23:53,530 --> 00:23:56,370 One rate is faster than the other. 390 00:23:56,370 --> 00:23:59,110 You pick the first one first, then go through it. 391 00:23:59,110 --> 00:24:00,860 And then use your intuition. 392 00:24:00,860 --> 00:24:03,830 Every point along the way, your intuition 393 00:24:03,830 --> 00:24:06,990 can tell you what you expect the result to be. 394 00:24:06,990 --> 00:24:09,690 So in this case here, our intuition 395 00:24:09,690 --> 00:24:13,160 told us that if we took the rate into B 396 00:24:13,160 --> 00:24:15,326 to be much faster than that into C, 397 00:24:15,326 --> 00:24:16,700 than we expected to see something 398 00:24:16,700 --> 00:24:19,430 that was largely first order. 399 00:24:19,430 --> 00:24:20,870 And when you we the approximation 400 00:24:20,870 --> 00:24:23,290 in our exact solution, in fact it does 401 00:24:23,290 --> 00:24:25,330 look like it's first order. 402 00:24:25,330 --> 00:24:26,600 So always use your intuition. 403 00:24:26,600 --> 00:24:29,760 Because the math is going to get pretty hairy. 404 00:24:29,760 --> 00:24:32,110 The algebra might make it complicated. 405 00:24:32,110 --> 00:24:34,090 And it's really easy to make a plus sign 406 00:24:34,090 --> 00:24:35,980 into a minus sign somewhere along the way. 407 00:24:35,980 --> 00:24:37,810 Just like I did here. 408 00:24:37,810 --> 00:24:41,820 So if I hadn't fixed my mistake here, 409 00:24:41,820 --> 00:24:45,219 and I had gone and put in my Taylor series here, 410 00:24:45,219 --> 00:24:46,510 I would have gotten in trouble. 411 00:24:46,510 --> 00:24:48,468 Because this minus sign would have been a plus. 412 00:24:48,468 --> 00:24:50,760 And these k2 A0's would not have canceled out. 413 00:24:50,760 --> 00:24:53,330 And I wouldn't have gotten the right result, which 414 00:24:53,330 --> 00:24:59,230 my intuition had told me should be a second order process. 415 00:24:59,230 --> 00:25:02,920 And so if you have a problem, let's say you're on an exam 416 00:25:02,920 --> 00:25:04,780 and you're doing algebra, you're trying to, 417 00:25:04,780 --> 00:25:06,160 and you end up with a result that 418 00:25:06,160 --> 00:25:08,624 doesn't match what your intuition tells you, 419 00:25:08,624 --> 00:25:10,040 and you don't have time to fix it. 420 00:25:10,040 --> 00:25:10,700 Tell us. 421 00:25:10,700 --> 00:25:12,517 You know, I know this is wrong because I 422 00:25:12,517 --> 00:25:14,100 know it's supposed to be second order. 423 00:25:14,100 --> 00:25:15,880 But I don't know where I went wrong. 424 00:25:15,880 --> 00:25:18,100 And that's really important. 425 00:25:18,100 --> 00:25:21,320 Just tell us. 426 00:25:21,320 --> 00:25:27,324 That means that you're thinking. 427 00:25:27,324 --> 00:25:28,990 Let's get a little bit more complicated. 428 00:25:28,990 --> 00:25:35,670 Any questions? 429 00:25:35,670 --> 00:25:38,790 Consecutive series reactions. 430 00:25:38,790 --> 00:25:41,140 These were parallel reactions and now. 431 00:25:41,140 --> 00:25:42,750 Basically what we're doing here is 432 00:25:42,750 --> 00:25:47,740 we're building up a toolkit of simple mechanisms. 433 00:25:47,740 --> 00:25:51,110 And then we'll be able to put these mechanisms together 434 00:25:51,110 --> 00:25:57,980 to make something more complicated. 435 00:25:57,980 --> 00:26:04,840 So the next kind of mechanism, series mechanism, series 436 00:26:04,840 --> 00:26:09,280 reactions, so we have our reaction. 437 00:26:09,280 --> 00:26:14,149 Which is A goes to C. And the mechanism 438 00:26:14,149 --> 00:26:15,690 that's been thought for this reaction 439 00:26:15,690 --> 00:26:18,170 is that there's an intermediate. 440 00:26:18,170 --> 00:26:22,090 That first you have A goes to B, some intermediate, which gets 441 00:26:22,090 --> 00:26:27,480 used up to form the final product, C, with a rate, k2, 442 00:26:27,480 --> 00:26:28,520 here. 443 00:26:28,520 --> 00:26:33,790 And we can either write the mechanism in two steps. 444 00:26:33,790 --> 00:26:41,180 Or we can write it as A goes to B goes to C, like this. 445 00:26:41,180 --> 00:26:45,067 And see both ways of writing it. 446 00:26:45,067 --> 00:26:46,150 And we want to solve this. 447 00:26:46,150 --> 00:26:47,691 And we want to look at special cases, 448 00:26:47,691 --> 00:26:50,390 and we just want to understand the implications 449 00:26:50,390 --> 00:26:51,230 of this mechanism. 450 00:26:51,230 --> 00:26:53,313 So the first thing to do, just like we did before, 451 00:26:53,313 --> 00:26:57,970 is to write all the rate laws. 452 00:26:57,970 --> 00:27:01,410 Before we got going solving anything. 453 00:27:01,410 --> 00:27:05,080 So we start with A, minus dA/dt. 454 00:27:05,080 --> 00:27:07,510 There's only one way that gets used up. 455 00:27:07,510 --> 00:27:11,270 Through the formation of B in a first order process. 456 00:27:11,270 --> 00:27:12,130 Easy. 457 00:27:12,130 --> 00:27:13,060 This is easy to solve. 458 00:27:13,060 --> 00:27:17,970 We know the answer is going to be exponential in it. 459 00:27:17,970 --> 00:27:23,240 For B, dB/dt is equal to, well it 460 00:27:23,240 --> 00:27:25,680 can be formed through my first order in A, 461 00:27:25,680 --> 00:27:28,480 so that's k1 times A. But it also 462 00:27:28,480 --> 00:27:31,000 could get destroyed by forming C. 463 00:27:31,000 --> 00:27:34,360 So we've got to get all the paths out 464 00:27:34,360 --> 00:27:37,120 of A, here on the right side of this equation. 465 00:27:37,120 --> 00:27:41,290 So it can be destroyed in a process, which 466 00:27:41,290 --> 00:27:42,795 is first order in B. So we're going 467 00:27:42,795 --> 00:27:49,537 to take both k1 and k2 to be first order initially. 468 00:27:49,537 --> 00:27:51,870 Then we'll make one of them second order and things will 469 00:27:51,870 --> 00:27:54,330 become very complicated and we'll throw up our hands. 470 00:27:54,330 --> 00:27:57,350 But for now, let's not throw up our hands quite yet. 471 00:27:57,350 --> 00:28:00,480 OK and for dC/dt, there's only one way 472 00:28:00,480 --> 00:28:05,810 it can be formed is first order by destroying B. 473 00:28:05,810 --> 00:28:08,070 And we have a couple of differential equations here. 474 00:28:08,070 --> 00:28:16,350 This differential equation for C, depends on B here. 475 00:28:16,350 --> 00:28:21,150 The differential for B depends on A here. 476 00:28:21,150 --> 00:28:24,650 So that means that things are going to get a little bit more 477 00:28:24,650 --> 00:28:27,935 complicated. 478 00:28:27,935 --> 00:28:30,060 The easy one to solve is always the first one here. 479 00:28:30,060 --> 00:28:31,380 That's first order. 480 00:28:31,380 --> 00:28:33,140 So we just write down the answer. 481 00:28:33,140 --> 00:28:38,800 A is equal to A0, e to the minus k1 t. 482 00:28:38,800 --> 00:28:42,030 We know the answer to that one. 483 00:28:42,030 --> 00:28:46,790 Then we have to work on B. So we want 484 00:28:46,790 --> 00:28:51,010 to find out integrated rate laws for every one of these chemical 485 00:28:51,010 --> 00:28:59,950 species. 486 00:28:59,950 --> 00:29:01,990 And always there are tricks involved. 487 00:29:01,990 --> 00:29:03,840 Partial fractions, whatever. 488 00:29:03,840 --> 00:29:08,480 So we write down B now. dB/dt plus, 489 00:29:08,480 --> 00:29:12,170 let me rearrange it a little bit, plus k2 times B, putting 490 00:29:12,170 --> 00:29:14,320 the minus k2 B on the other side here. 491 00:29:14,320 --> 00:29:16,502 Is equal to k1 times A. But I'm not 492 00:29:16,502 --> 00:29:18,210 going to keep this A there, because now I 493 00:29:18,210 --> 00:29:20,410 know what A is in terms of time. 494 00:29:20,410 --> 00:29:28,790 And a constant. k1 times A0 e to the minus k1 times time. 495 00:29:28,790 --> 00:29:30,340 OK, that's my differential equation 496 00:29:30,340 --> 00:29:35,370 for B. Got to solve this thing. 497 00:29:35,370 --> 00:29:39,270 Well, the last time I did differential equations 498 00:29:39,270 --> 00:29:40,610 in college was a century ago. 499 00:29:40,610 --> 00:29:42,250 And I wasn't kidding. 500 00:29:42,250 --> 00:29:43,590 It really was last century. 501 00:29:43,590 --> 00:29:46,920 In fact, it was last millennium. 502 00:29:46,920 --> 00:29:49,234 That was a long time ago. 503 00:29:49,234 --> 00:29:50,150 So how do you do this? 504 00:29:50,150 --> 00:29:52,940 Well, the trick here, there's a trick. 505 00:29:52,940 --> 00:29:58,310 And the trick is to multiply the whole thing, both sides, by e 506 00:29:58,310 --> 00:30:01,470 to the k2 t. 507 00:30:01,470 --> 00:30:03,510 And you multiply this side by e to the k2 t 508 00:30:03,510 --> 00:30:06,520 and you multiply this side by e to the k2 t, 509 00:30:06,520 --> 00:30:09,620 to solve this equation. 510 00:30:09,620 --> 00:30:11,270 So once you do that, then you have 511 00:30:11,270 --> 00:30:22,326 e to the k2 t times dB/dt plus k2 times B. This side here. 512 00:30:22,326 --> 00:30:30,200 And then you have is equal to k1 times A0 e to the minus, e 513 00:30:30,200 --> 00:30:39,380 to the k2 minus k1 times time. 514 00:30:39,380 --> 00:30:42,515 OK, so the reason why this trick works 515 00:30:42,515 --> 00:30:47,270 is because this guy right here can be nicely written 516 00:30:47,270 --> 00:30:56,430 as d/dt of B times e to the k2 t. 517 00:30:56,430 --> 00:30:58,500 It's in nice simple form, so when you integrate, 518 00:30:58,500 --> 00:31:01,254 it'll be very easy to integrate. 519 00:31:01,254 --> 00:31:02,670 And then the other side, you still 520 00:31:02,670 --> 00:31:12,350 have k1 A0 e to the k2 minus k1 times the time. 521 00:31:12,350 --> 00:31:14,530 So now you integrate both sides with respect 522 00:31:14,530 --> 00:31:18,180 to time, The integral of something d/dt with respect 523 00:31:18,180 --> 00:31:20,690 to time, it's just going to be itself right here. 524 00:31:20,690 --> 00:31:22,190 So it's very simple. 525 00:31:22,190 --> 00:31:24,540 So you integrate both sides from zero to time, dt. 526 00:31:27,170 --> 00:31:31,430 Zero to t, dt on both sides. 527 00:31:31,430 --> 00:31:34,480 And on this side here, you end up with B, 528 00:31:34,480 --> 00:31:42,340 e to the k2 times time minus B, e to the k2 times zero. 529 00:31:42,340 --> 00:31:47,440 Just taking the upper limit minus the lower limit. 530 00:31:47,440 --> 00:31:53,840 And that's just equal to B0, at time to equal zero. 531 00:31:53,840 --> 00:31:57,740 And so, and at time to equal zero, 532 00:31:57,740 --> 00:32:01,870 B0 is equal to zero because we haven't made any intermediates. 533 00:32:01,870 --> 00:32:06,480 So this goes away. 534 00:32:06,480 --> 00:32:08,410 And then we have an equals sign. 535 00:32:08,410 --> 00:32:15,079 And on the other side, we have k1 A0 over k2 minus k1, 536 00:32:15,079 --> 00:32:16,620 it's just the integral of this thing, 537 00:32:16,620 --> 00:32:21,580 here with a k2 minus k1 goes in the denominator. e 538 00:32:21,580 --> 00:32:28,290 to the k2 minus k1 times time minus one. 539 00:32:28,290 --> 00:32:33,150 So now you have your integral form for the amount of B 540 00:32:33,150 --> 00:32:36,820 that gets created at any time. 541 00:32:36,820 --> 00:32:38,980 And I want to keep that on the board. 542 00:32:38,980 --> 00:32:40,540 Because I'm going to use it later. 543 00:32:40,540 --> 00:32:52,250 k1 A0 over k1 minus k2 times e to the minus k1 times 544 00:32:52,250 --> 00:33:01,810 time minus e to the minus k2 times time. 545 00:33:01,810 --> 00:33:03,770 OK, we're two thirds of the way through. 546 00:33:03,770 --> 00:33:10,080 Now we've got to find C. And for the last component, 547 00:33:10,080 --> 00:33:12,290 you don't want to be doing this kind of stuff. 548 00:33:12,290 --> 00:33:14,830 You don't want to waste your time solving differential 549 00:33:14,830 --> 00:33:20,082 equations, because stoichiometry can help you out. 550 00:33:20,082 --> 00:33:22,040 At the end of the day, you know that everything 551 00:33:22,040 --> 00:33:25,340 that started out as A, in terms of quantities, 552 00:33:25,340 --> 00:33:28,240 is going to end up at C. You know that if you end up 553 00:33:28,240 --> 00:33:31,220 with a concentration A0 at the beginning, 554 00:33:31,220 --> 00:33:33,440 at the end of the process the concentration of C, 555 00:33:33,440 --> 00:33:37,210 at infinite time, is also going to be A0. 556 00:33:37,210 --> 00:33:41,470 If you know that there's a correlation, 557 00:33:41,470 --> 00:33:46,130 there's a stoichiometry that relates A and C together. 558 00:33:46,130 --> 00:33:53,200 And B. You have conservation of mass here, basically. 559 00:33:53,200 --> 00:33:55,430 Conservation of atoms. 560 00:33:55,430 --> 00:33:58,610 That has to come into play 561 00:33:58,610 --> 00:34:07,870 So, for C, we're just going to do algebra instead of calculus. 562 00:34:07,870 --> 00:34:10,620 So let's write down the stoichiometry here. 563 00:34:10,620 --> 00:34:15,370 The amount of C at any time is what 564 00:34:15,370 --> 00:34:17,200 it's going to be at the end. 565 00:34:17,200 --> 00:34:19,600 Which is A0, very end of the process. 566 00:34:19,600 --> 00:34:21,240 It's going to be A0. 567 00:34:21,240 --> 00:34:24,530 But before we get to the end, there's 568 00:34:24,530 --> 00:34:28,346 still stuff that's left in A and B. 569 00:34:28,346 --> 00:34:30,220 And that's going to decrease the amount of C. 570 00:34:30,220 --> 00:34:35,660 So minus whatever's still in A and B. 571 00:34:35,660 --> 00:34:37,712 That's the stoichiometry. 572 00:34:37,712 --> 00:34:38,920 That's one way of writing it. 573 00:34:38,920 --> 00:34:40,711 You could also write it in a different way. 574 00:34:40,711 --> 00:34:47,450 Which is that the amount of C is equal to the amount of A 575 00:34:47,450 --> 00:34:50,240 that you have used up. 576 00:34:50,240 --> 00:34:53,270 A0 is what you started out with. 577 00:34:53,270 --> 00:34:54,350 A is what's left over. 578 00:34:54,350 --> 00:34:57,210 So this is the amount of A used up. 579 00:34:57,210 --> 00:34:59,280 It's not all quite in C yet. 580 00:34:59,280 --> 00:35:02,510 Some of it is stuck in B. 581 00:35:02,510 --> 00:35:06,060 So it's the amount of A used up, minus the amount that's 582 00:35:06,060 --> 00:35:09,330 stuck in B. Those two are the same. 583 00:35:09,330 --> 00:35:10,580 I hope. 584 00:35:10,580 --> 00:35:12,220 A0 minus A, yeah, that's right. 585 00:35:12,220 --> 00:35:13,940 Minus B, yeah, that's right. 586 00:35:13,940 --> 00:35:18,180 So this is used up. 587 00:35:18,180 --> 00:35:26,770 And then stuck in B. That's the hard part. 588 00:35:26,770 --> 00:35:29,080 Putting down the stoichiometry. 589 00:35:29,080 --> 00:35:30,710 Once you've done the hard part, then 590 00:35:30,710 --> 00:35:32,180 if you solve the other two then it's easy. 591 00:35:32,180 --> 00:35:32,888 You just plug in. 592 00:35:32,888 --> 00:35:35,210 It's just algebra. 593 00:35:35,210 --> 00:35:37,910 And you get something complicated. 594 00:35:37,910 --> 00:35:43,920 C is equal to complicated. 595 00:35:43,920 --> 00:35:49,110 It's in the notes, I'm not going to write it down. 596 00:35:49,110 --> 00:35:51,680 So we've solved the problem exactly. 597 00:35:51,680 --> 00:35:53,389 And now you're going to do an experiment. 598 00:35:53,389 --> 00:35:54,930 And the experiment you're going to do 599 00:35:54,930 --> 00:35:57,100 is not going to follow this whole thing, exactly. 600 00:35:57,100 --> 00:35:59,840 It's going to look at limited cases. 601 00:35:59,840 --> 00:36:02,560 That's what you can do. 602 00:36:02,560 --> 00:36:04,020 And so the first thing to look at 603 00:36:04,020 --> 00:36:06,140 is to look at the initial times. 604 00:36:06,140 --> 00:36:07,840 The very beginning of the process, 605 00:36:07,840 --> 00:36:11,180 how are things appearing. 606 00:36:11,180 --> 00:36:33,240 And then we'll look at the late times. 607 00:36:33,240 --> 00:36:35,940 So initial times. 608 00:36:35,940 --> 00:36:46,040 Near t equals zero. 609 00:36:46,040 --> 00:36:50,780 Well, every one of these things has an exponential in it. 610 00:36:50,780 --> 00:36:53,470 And whenever we see exponentials and we look at something 611 00:36:53,470 --> 00:36:57,040 at the early times, times that are faster than one over k1, 612 00:36:57,040 --> 00:37:02,940 or faster than one over k2, it tells us use a Taylor series. 613 00:37:02,940 --> 00:37:06,210 So use the approximation e to the minus kt 614 00:37:06,210 --> 00:37:14,510 is approximately equal to one minus kt plus kt squared 615 00:37:14,510 --> 00:37:16,110 over two plus et cetera. 616 00:37:16,110 --> 00:37:20,040 And keep as many terms as you need in the Taylor series 617 00:37:20,040 --> 00:37:21,640 to get something sensible. 618 00:37:21,640 --> 00:37:23,610 So if everything comes out of zero at the end, 619 00:37:23,610 --> 00:37:26,210 you know you haven't kept enough terms in t. 620 00:37:26,210 --> 00:37:28,110 Things have canceled out too much. 621 00:37:28,110 --> 00:37:31,010 And you want something that is time dependent. 622 00:37:31,010 --> 00:37:33,210 Because you've got a change in time, right? 623 00:37:33,210 --> 00:37:36,330 It's a very common mistake to say only keep one here 624 00:37:36,330 --> 00:37:38,680 and then everything cancels out you get zero, 625 00:37:38,680 --> 00:37:41,940 and you say, well, B is equal to zero at all times. 626 00:37:41,940 --> 00:37:46,390 That's nonsensical. 627 00:37:46,390 --> 00:37:49,330 And I don't know a priori how much time to keep. 628 00:37:49,330 --> 00:37:53,340 So let's go to second order. 629 00:37:53,340 --> 00:38:00,150 So if you plug this Taylor series into A, 630 00:38:00,150 --> 00:38:02,230 and get something that's time dependent, 631 00:38:02,230 --> 00:38:04,070 you only need to go to first order. 632 00:38:04,070 --> 00:38:08,850 One minus k1 times time plus et cetera. 633 00:38:08,850 --> 00:38:14,000 Plug it into b, plug your Taylor series into these two 634 00:38:14,000 --> 00:38:22,450 guys here, get k1 A0 over k2 minus k1, 635 00:38:22,450 --> 00:38:24,640 you only need it to go to first order. 636 00:38:24,640 --> 00:38:27,720 The one's cancel out but the times don't cancel out. 637 00:38:27,720 --> 00:38:33,160 Minus k1 plus k2, that is, unless k1 and k2 are the same. 638 00:38:33,160 --> 00:38:35,130 So we're making the approximation 639 00:38:35,130 --> 00:38:38,930 that k1 is different than k2. 640 00:38:38,930 --> 00:38:41,430 Times the time. 641 00:38:41,430 --> 00:38:43,780 And when we look at C, this complicated thing 642 00:38:43,780 --> 00:38:46,680 that I didn't write down, if you were just 643 00:38:46,680 --> 00:38:50,832 to stop at first order in C, you get C equals zero. 644 00:38:50,832 --> 00:38:52,040 You know that can't be right. 645 00:38:52,040 --> 00:38:55,036 You know that C is going to increase in time. 646 00:38:55,036 --> 00:38:56,660 It can't be equal to zero at all times. 647 00:38:56,660 --> 00:39:00,950 So we have to go to second order for C0. 648 00:39:00,950 --> 00:39:08,560 And it's approximately equal to A0 times k1 k2 over two times 649 00:39:08,560 --> 00:39:11,459 the time squared. 650 00:39:11,459 --> 00:39:13,250 Let me rewrite this a little bit closer in, 651 00:39:13,250 --> 00:39:15,450 because I need some room here. 652 00:39:15,450 --> 00:39:20,540 So c is approximately equal to A0 k1 653 00:39:20,540 --> 00:39:24,807 is k2 over two times time squared. 654 00:39:24,807 --> 00:39:26,390 OK, and then we can start plotting out 655 00:39:26,390 --> 00:39:27,931 what this is going to look like then. 656 00:39:27,931 --> 00:39:30,960 So we have the concentrations time on the axis here. 657 00:39:30,960 --> 00:39:34,305 Let's start with the concentrating of A. A 658 00:39:34,305 --> 00:39:38,200 is dropping down linearly in time. 659 00:39:38,200 --> 00:39:41,850 It's dropping down linearly in time here. 660 00:39:41,850 --> 00:39:46,590 B is increasing linearly in time, at the beginning. 661 00:39:46,590 --> 00:39:51,190 So B is going up linearly in time, like this. 662 00:39:51,190 --> 00:39:55,210 And C is increasing quadratically in time. 663 00:39:55,210 --> 00:39:58,120 So it's pretty flat, coming up first. 664 00:39:58,120 --> 00:40:03,150 And then it starts to curve up. 665 00:40:03,150 --> 00:40:05,070 It's our first approximation. 666 00:40:05,070 --> 00:40:09,494 Then, we go to late times. t equals infinity. 667 00:40:09,494 --> 00:40:11,660 So t equals infinity where am I going to write this? 668 00:40:11,660 --> 00:40:33,010 Have to move this up. 669 00:40:33,010 --> 00:40:35,620 Let's go, t goes to infinity. 670 00:40:35,620 --> 00:40:38,160 Well, here I know that I'm allowed to go to zero. 671 00:40:38,160 --> 00:40:42,560 Or allowed to go to constant, because infinity is forever. 672 00:40:42,560 --> 00:40:46,180 So at t infinity, I know that I'm going to use up all of A. 673 00:40:46,180 --> 00:40:48,620 So I know that A is going to be equal to 0. 674 00:40:48,620 --> 00:40:50,820 I know I'm going to use up all of B. 675 00:40:50,820 --> 00:40:53,670 So I know B is going to go to zero. 676 00:40:53,670 --> 00:40:54,920 Somehow. 677 00:40:54,920 --> 00:40:58,950 And I know that C is going to go to A0. 678 00:40:58,950 --> 00:41:07,840 So C is going to go to A0 at infinity. 679 00:41:07,840 --> 00:41:17,390 So now on my graph here, I know that at infinity, A 680 00:41:17,390 --> 00:41:20,850 is going to go down to zero like this, somehow. 681 00:41:20,850 --> 00:41:23,960 Probably exponentially, because it is exponential in time. 682 00:41:23,960 --> 00:41:25,710 So it's going to go to zero exponentially. 683 00:41:25,710 --> 00:41:27,630 B is going to go down to zero exponentially. 684 00:41:27,630 --> 00:41:30,070 There are the exponentials right here. 685 00:41:30,070 --> 00:41:33,580 So B is going to go down to zero exponentially. 686 00:41:33,580 --> 00:41:35,500 Something like this. 687 00:41:35,500 --> 00:41:39,160 And C is going to saturate to A0. 688 00:41:39,160 --> 00:41:40,770 Exponentially rise up to A0. 689 00:41:40,770 --> 00:41:42,510 So this was A0 right here. 690 00:41:42,510 --> 00:41:45,730 It's going to go like this. 691 00:41:45,730 --> 00:41:47,160 So then I connect the dots. 692 00:41:47,160 --> 00:41:50,810 And I know that C is going to start quadratically. 693 00:41:50,810 --> 00:41:53,550 And then it's going to rise up an exponent like this. 694 00:41:53,550 --> 00:41:57,220 I know that B is going to start off linearly, 695 00:41:57,220 --> 00:41:59,470 and then it has to go to zero. 696 00:41:59,470 --> 00:42:03,030 So it has to go through some sort of maximum. 697 00:42:03,030 --> 00:42:08,120 And A is going to go down exponentially, 698 00:42:08,120 --> 00:42:10,640 exponentially like this. 699 00:42:10,640 --> 00:42:12,550 So the interesting part of this diagram 700 00:42:12,550 --> 00:42:13,870 is that B goes to a maximum. 701 00:42:13,870 --> 00:42:20,680 There is some point, there's some time, 702 00:42:20,680 --> 00:42:24,870 that we can call t sub B max, where B is a maxima. 703 00:42:24,870 --> 00:42:27,570 And that's an interesting point. 704 00:42:27,570 --> 00:42:30,250 That's something that we can try to figure out experimentally 705 00:42:30,250 --> 00:42:31,710 what it is. 706 00:42:31,710 --> 00:42:33,990 And we could get some parameters, 707 00:42:33,990 --> 00:42:36,810 we could maybe extract some information. 708 00:42:36,810 --> 00:42:42,470 There's a maximum B, concentration of B, 709 00:42:42,470 --> 00:42:45,190 that gets formed. 710 00:42:45,190 --> 00:42:47,860 And how do you solve that? 711 00:42:47,860 --> 00:42:52,590 You set dB/dt equal to zero. 712 00:42:52,590 --> 00:42:55,510 So you have your equation for B right here. 713 00:42:55,510 --> 00:42:58,820 You set it equal to zero. and you solve. 714 00:42:58,820 --> 00:43:00,290 You get your maximum time. 715 00:43:00,290 --> 00:43:02,390 It's made up of rate constant. 716 00:43:02,390 --> 00:43:04,190 You get your maximum concentration, 717 00:43:04,190 --> 00:43:07,120 which is also made up of rate constants. 718 00:43:07,120 --> 00:43:09,791 And you can get rate constants out of this. 719 00:43:09,791 --> 00:43:10,290 Question. 720 00:43:10,290 --> 00:43:16,555 STUDENT: [INAUDIBLE] 721 00:43:16,555 --> 00:43:17,180 PROFESSOR: Yes. 722 00:43:17,180 --> 00:43:22,705 STUDENT: [INAUDIBLE] 723 00:43:22,705 --> 00:43:24,080 PROFESSOR: That is a coincidence. 724 00:43:24,080 --> 00:43:31,380 That is my poor sketching. 725 00:43:31,380 --> 00:43:37,870 It should probably be somewhere like, in the middle here. 726 00:43:37,870 --> 00:43:41,359 Actually, it completely depends on the rates. 727 00:43:41,359 --> 00:43:42,900 And we're going to see limiting cases 728 00:43:42,900 --> 00:43:45,970 where this maximum can occur here, 729 00:43:45,970 --> 00:43:51,860 or it can occur somewhere down here. 730 00:43:51,860 --> 00:43:57,610 So this is the complete case. 731 00:43:57,610 --> 00:44:04,772 So, limiting cases. the important limiting cases. 732 00:44:04,772 --> 00:44:06,480 We've made an approximation along the way 733 00:44:06,480 --> 00:44:07,688 that k1 is different than k2. 734 00:44:07,688 --> 00:44:10,950 So that has to be a case that needs to be worked out. 735 00:44:10,950 --> 00:44:17,690 And it's likely to be easier than the full solution. 736 00:44:17,690 --> 00:44:22,300 So one of the limiting cases is k1 is equal to k2 737 00:44:22,300 --> 00:44:27,180 and I'll leave that for homework. 738 00:44:27,180 --> 00:44:31,010 You can do that yourselves. 739 00:44:31,010 --> 00:44:34,255 It's good exercise. 740 00:44:34,255 --> 00:44:35,630 What about another limiting case? 741 00:44:35,630 --> 00:44:37,740 Well, you have two other limiting cases. 742 00:44:37,740 --> 00:44:39,140 One, you've got two rates. 743 00:44:39,140 --> 00:44:45,540 So, one is bigger than the other. 744 00:44:45,540 --> 00:44:50,980 No information on this board here. 745 00:44:50,980 --> 00:44:54,550 You start off with one case. 746 00:44:54,550 --> 00:45:03,050 So k1 greater than k2. k1 greater than k2. 747 00:45:03,050 --> 00:45:05,840 What does it mean? k1 greater than k2. 748 00:45:05,840 --> 00:45:08,360 It means that the rate determining 749 00:45:08,360 --> 00:45:11,320 step, or the limiting step, in this reaction, 750 00:45:11,320 --> 00:45:16,330 is the second step. k2 is very slow compared to k1. 751 00:45:16,330 --> 00:45:21,850 And sometimes I like to think of these things as pipes. 752 00:45:21,850 --> 00:45:25,460 And connecting vessels. 753 00:45:25,460 --> 00:45:30,490 So let's say we have the amount of liquid 754 00:45:30,490 --> 00:45:32,530 in the top vessel is the concentration of A. 755 00:45:32,530 --> 00:45:34,870 The amount of liquid in the middle vessel, 756 00:45:34,870 --> 00:45:45,972 let me get my color scheme to be consistent here. 757 00:45:45,972 --> 00:45:47,430 Then we have a middle vessel, which 758 00:45:47,430 --> 00:45:50,380 is the concentration of B, and a final vessel, which 759 00:45:50,380 --> 00:45:53,650 is the concentration of C. And we start out 760 00:45:53,650 --> 00:45:59,250 with some sort of amount of A in here. 761 00:45:59,250 --> 00:46:02,940 Started with a lot of A, and then these vessels 762 00:46:02,940 --> 00:46:04,910 are connecting with pipes. 763 00:46:04,910 --> 00:46:07,150 So there's a very thick, big pipe 764 00:46:07,150 --> 00:46:11,050 that connects A and B together. 765 00:46:11,050 --> 00:46:16,430 The rate is fast, going from A to B. 766 00:46:16,430 --> 00:46:18,780 And the rate going from B to C is slow, 767 00:46:18,780 --> 00:46:26,930 so there's a skinny pipe connecting B and C together. 768 00:46:26,930 --> 00:46:29,080 Then I turn on the system, I set time 769 00:46:29,080 --> 00:46:31,680 to equal zero, poof, what happens? 770 00:46:31,680 --> 00:46:33,360 There's the big pipe connecting A and B. 771 00:46:33,360 --> 00:46:37,050 The whole amount of a here gets transferred to B, suddenly. 772 00:46:37,050 --> 00:46:38,090 Then it gets stuck here. 773 00:46:38,090 --> 00:46:41,830 And then it dribbles out from B to C. 774 00:46:41,830 --> 00:46:44,050 So what do we expect, if we were to plot 775 00:46:44,050 --> 00:46:49,100 our quantities as a function of time. 776 00:46:49,100 --> 00:46:51,820 What we'd expect then, then we're 777 00:46:51,820 --> 00:46:54,470 going to make sure that the math works out, 778 00:46:54,470 --> 00:46:58,350 is that A is going to decrease really fast. 779 00:46:58,350 --> 00:47:02,050 It's going to go, poof. 780 00:47:02,050 --> 00:47:08,770 First order in time with a very fast rate. 781 00:47:08,770 --> 00:47:11,680 And all of this is going to go straight into B. 782 00:47:11,680 --> 00:47:13,305 So we expect B to go linearly. 783 00:47:13,305 --> 00:47:14,930 Remember, at the beginning it's linear. 784 00:47:14,930 --> 00:47:17,040 Linear goes up. 785 00:47:17,040 --> 00:47:20,420 Almost all the way up to A0. 786 00:47:20,420 --> 00:47:22,400 Because it can't hardly get out. 787 00:47:22,400 --> 00:47:26,770 Reaches its maximum, then it goes from B to C 788 00:47:26,770 --> 00:47:28,590 through this skinny, skinny pipe. 789 00:47:28,590 --> 00:47:31,540 First order rate, with a rate k2, 790 00:47:31,540 --> 00:47:35,970 so we're exponentially going down, slowly, slowly, slowly, 791 00:47:35,970 --> 00:47:38,800 with a rate k2. 792 00:47:38,800 --> 00:47:44,280 And then C starts quadratically. 793 00:47:44,280 --> 00:47:48,370 And then very quickly, once everything's into B, 794 00:47:48,370 --> 00:47:49,880 A is forgotten. 795 00:47:49,880 --> 00:47:51,540 A doesn't matter any more. 796 00:47:51,540 --> 00:47:54,305 Basically what you're looking at is a transfer of B 797 00:47:54,305 --> 00:47:56,810 into C through this little pipe. 798 00:47:56,810 --> 00:47:59,100 And you know that's going to be a first order of rate. 799 00:47:59,100 --> 00:48:01,040 First order of process, B to C's first order 800 00:48:01,040 --> 00:48:02,690 of process with rate k2. 801 00:48:02,690 --> 00:48:05,830 So you expect, then, C to go up in the first order 802 00:48:05,830 --> 00:48:13,490 process to A0 with rate k2. 803 00:48:13,490 --> 00:48:18,450 There's the rate k2 here. 804 00:48:18,450 --> 00:48:22,650 And the rate k1 here. 805 00:48:22,650 --> 00:48:25,230 So when you turn the crank on your approximation in the math, 806 00:48:25,230 --> 00:48:31,050 what you'd better see is that at the end, 807 00:48:31,050 --> 00:48:32,890 B as a function of time should look 808 00:48:32,890 --> 00:48:36,400 like a first order process with rate k2. 809 00:48:36,400 --> 00:48:38,730 C as a function of time should look like a first order 810 00:48:38,730 --> 00:48:41,220 process with rate k2, going to A0. 811 00:48:41,220 --> 00:48:44,180 And you can write the answer, you 812 00:48:44,180 --> 00:48:47,320 should be able to write the answer, 813 00:48:47,320 --> 00:48:50,930 for C, it's going to be C is approximately, 814 00:48:50,930 --> 00:48:53,890 it's going to go to A0, there's no other choice. 815 00:48:53,890 --> 00:48:56,852 Everything that was in A gets transferred to B0, so at time 816 00:48:56,852 --> 00:48:58,560 is infinity it's going to be A0, and it's 817 00:48:58,560 --> 00:49:01,120 going to be a first order process with rate k2. 818 00:49:01,120 --> 00:49:06,100 One minus e to the minus k2 times times. 819 00:49:06,100 --> 00:49:07,830 I don't even have to do the math. 820 00:49:07,830 --> 00:49:10,580 I know that's the answer. 821 00:49:10,580 --> 00:49:14,420 If you don't believe me, then do the math. 822 00:49:14,420 --> 00:49:16,034 You should do it. 823 00:49:16,034 --> 00:49:16,950 But that's the answer. 824 00:49:16,950 --> 00:49:18,990 It has to be the answer. 825 00:49:18,990 --> 00:49:24,950 And if you look at B, the answer has 826 00:49:24,950 --> 00:49:31,640 to be that it's approximately looking like B is disappearing 827 00:49:31,640 --> 00:49:35,760 with a rate constant that's k2, e to the minus k2, 828 00:49:35,760 --> 00:49:38,550 and the maximum here is very close to A0. 829 00:49:38,550 --> 00:49:41,400 So I'm going to be putting A0 here. 830 00:49:41,400 --> 00:49:44,040 Close enough. 831 00:49:44,040 --> 00:49:47,270 So as long as I ignore the very initial time where 832 00:49:47,270 --> 00:49:50,580 A suddenly dumps into B, then everything after that very 833 00:49:50,580 --> 00:49:55,220 early time here is just looking like B goes to C. 834 00:49:55,220 --> 00:49:58,340 This is not zero, this is t. 835 00:49:58,340 --> 00:50:00,960 And A just disappears quickly. 836 00:50:00,960 --> 00:50:06,340 And we can write it as A is equal to A0 e to the minus 837 00:50:06,340 --> 00:50:09,530 k1 times the time. 838 00:50:09,530 --> 00:50:12,960 That's an easy approximation. 839 00:50:12,960 --> 00:50:17,360 And you've just got to make sure that this fits. 840 00:50:17,360 --> 00:50:22,320 So the other limiting case, which I'm also 841 00:50:22,320 --> 00:50:24,070 going to leave as a homework, because it's 842 00:50:24,070 --> 00:50:30,610 so straightforward, is to have k2 greater than k1, 843 00:50:30,610 --> 00:50:34,240 and you should make sure that you can predict it. 844 00:50:34,240 --> 00:50:38,170 and that the math works out. 845 00:50:38,170 --> 00:50:39,660 OK, any questions? 846 00:50:39,660 --> 00:50:42,420 Next time we'll do reversible reactions. 847 00:50:42,420 --> 00:50:46,010 And some more approximations.