1 00:00:00,000 --> 00:00:02,410 The following content is provided under a Creative 2 00:00:02,410 --> 00:00:03,820 Commons license. 3 00:00:03,820 --> 00:00:06,850 Your support will help MIT OpenCourseWare continue to 4 00:00:06,850 --> 00:00:10,520 offer high quality educational resources for free. 5 00:00:10,520 --> 00:00:13,390 To make a donation, or view additional materials from 6 00:00:13,390 --> 00:00:17,500 hundreds of MIT courses, visit MIT OpenCourseWare at 7 00:00:17,500 --> 00:00:20,500 ocw.mit.edu. 8 00:00:20,500 --> 00:00:23,290 PROFESSOR: So last time, there was a question that was asked 9 00:00:23,290 --> 00:00:26,610 about the order of magnitude rates. 10 00:00:26,610 --> 00:00:28,390 You asked the question about the 11 00:00:28,390 --> 00:00:31,630 concentration of the catalysts. 12 00:00:31,630 --> 00:00:34,452 So the rates that I gave you, which were the rates of 13 00:00:34,452 --> 00:00:36,050 reactions are basically order of magnitude. 14 00:00:36,050 --> 00:00:39,380 They're estimates and odd numbers that, they obviously 15 00:00:39,380 --> 00:00:41,800 depend, as you pointed out, depend on the concentration of 16 00:00:41,800 --> 00:00:42,400 the catalyst. 17 00:00:42,400 --> 00:00:46,370 Depends on the concentration of the reactants. 18 00:00:46,370 --> 00:00:50,500 Depends on where you are on those concentrations. 19 00:00:50,500 --> 00:00:54,660 As we saw when we did enzyme catalysis, is that depending 20 00:00:54,660 --> 00:00:59,000 on KM, the Michaelis constant, and the substrate 21 00:00:59,000 --> 00:01:02,350 concentration, if your substrate's concentration is 22 00:01:02,350 --> 00:01:08,400 much higher than KM, you are in the maximum velocity limit, 23 00:01:08,400 --> 00:01:12,730 where the rate depends on k2 times the initial 24 00:01:12,730 --> 00:01:14,370 concentration of enzyme. 25 00:01:14,370 --> 00:01:17,640 In that case the rate of the reaction depends just on the 26 00:01:17,640 --> 00:01:19,800 concentration of enzyme, and not on the 27 00:01:19,800 --> 00:01:21,870 concentration of substrate. 28 00:01:21,870 --> 00:01:27,410 But if you are at the small concentration of substrate 29 00:01:27,410 --> 00:01:31,630 relative to KM, then your rate depends linearly with the 30 00:01:31,630 --> 00:01:33,190 concentration of substrate. 31 00:01:33,190 --> 00:01:38,780 And the same thing happens with the non-enzymatic 32 00:01:38,780 --> 00:01:43,470 catalyst, that there will be some sort of mechanism that 33 00:01:43,470 --> 00:01:44,820 goes along with it. 34 00:01:44,820 --> 00:01:46,660 And most likely it's going to be second 35 00:01:46,660 --> 00:01:47,650 order in the catalyst. 36 00:01:47,650 --> 00:01:54,630 And the reactant if it's just the two body process. 37 00:01:54,630 --> 00:01:56,580 But it could be more complicated. 38 00:01:56,580 --> 00:01:57,660 So definitely those 39 00:01:57,660 --> 00:01:59,430 concentrations will go in there. 40 00:01:59,430 --> 00:02:04,430 But a priori, it's not clear how they will go in there. 41 00:02:04,430 --> 00:02:06,870 But the orders of magnitude are roughly right. 42 00:02:06,870 --> 00:02:10,450 And the basic idea is that you change your rate of reactions 43 00:02:10,450 --> 00:02:14,950 by 15, 18 orders of magnitude with the right catalyst, 44 00:02:14,950 --> 00:02:18,810 especially if it's a biological catalyst. 45 00:02:18,810 --> 00:02:22,280 One concept that I didn't go through last time, which is 46 00:02:22,280 --> 00:02:34,360 also quite useful, it's called the turnover number. 47 00:02:34,360 --> 00:02:36,630 Turnover number. 48 00:02:36,630 --> 00:02:40,920 And this is useful for any kind of catalyst. 49 00:02:40,920 --> 00:02:46,910 But for enzymatic, or enzyme catalysts, the 50 00:02:46,910 --> 00:02:51,380 turnover number is k cat. 51 00:02:51,380 --> 00:02:53,900 The number per second. 52 00:02:53,900 --> 00:02:55,300 It's a rate per second. 53 00:02:55,300 --> 00:03:01,760 How many times does a reactant get, the number of times that 54 00:03:01,760 --> 00:03:04,730 a reactant turns into a product 55 00:03:04,730 --> 00:03:08,300 per second, per catalyst. 56 00:03:08,300 --> 00:03:09,760 So let me write it out. 57 00:03:09,760 --> 00:03:34,330 So it's the number of product formed per second, per second, 58 00:03:34,330 --> 00:03:45,030 per enzyme, per molecule of enzyme. 59 00:03:45,030 --> 00:03:48,470 OK, so if one enzyme is going to be there, there's going to 60 00:03:48,470 --> 00:03:50,180 be some substrate going in. 61 00:03:50,180 --> 00:03:51,800 Coming in, the pocket coming out. 62 00:03:51,800 --> 00:03:54,990 Coming in, the pocket coming out, so the number of cycles 63 00:03:54,990 --> 00:04:00,250 of this process per second. is the turnover number. 64 00:04:00,250 --> 00:04:03,450 And so if you are where the concentration of substrate is 65 00:04:03,450 --> 00:04:06,190 very much larger than the Michaelis constant, where are 66 00:04:06,190 --> 00:04:10,000 you are at the saturation limit, then this turnover 67 00:04:10,000 --> 00:04:13,710 number doesn't depend on the substrate concentration. 68 00:04:13,710 --> 00:04:16,900 It's just k cat. 69 00:04:16,900 --> 00:04:20,860 So that's why k cat is important. 70 00:04:20,860 --> 00:04:33,850 As this turnover number in the maximum velocity limit. 71 00:04:33,850 --> 00:04:38,080 Because in the maximum velocity limit, then the 72 00:04:38,080 --> 00:04:43,080 process that's dominant is the second process, ES goes to 73 00:04:43,080 --> 00:04:44,190 enzyme plus product. 74 00:04:44,190 --> 00:04:47,170 The first process here is not the rate limiting step. 75 00:04:47,170 --> 00:04:48,170 This is the rate limiting step. 76 00:04:48,170 --> 00:04:50,840 Because the substrate concentration is very high. 77 00:04:50,840 --> 00:04:53,920 So this first part goes very high, goes very fast. 78 00:04:53,920 --> 00:04:59,550 And the number of product formed per second, per 79 00:04:59,550 --> 00:05:02,680 molecule of enzyme, the number of molecules formed for 80 00:05:02,680 --> 00:05:11,150 second, that's the rate of product formation. 81 00:05:11,150 --> 00:05:14,790 Rate of product formation per second. 82 00:05:14,790 --> 00:05:20,330 Divided by the concentration of the enzyme. 83 00:05:20,330 --> 00:05:22,700 If you normalize by the number of enzyme. 84 00:05:22,700 --> 00:05:25,240 So this becomes rate of product formation in moles per 85 00:05:25,240 --> 00:05:27,870 second, divided by moles of enzyme. 86 00:05:27,870 --> 00:05:31,410 It's the same thing as what we're saying here in words. 87 00:05:31,410 --> 00:05:36,090 And so the rate of product formation is dP/dt, and you 88 00:05:36,090 --> 00:05:39,260 divide that by the concentration of enzyme, which 89 00:05:39,260 --> 00:05:42,380 you can write as E0. dp/dt, and the maximum 90 00:05:42,380 --> 00:05:47,230 rate limit is k max. 91 00:05:47,230 --> 00:05:49,840 It's a constant divided by E0. 92 00:05:49,840 --> 00:05:56,130 And k max is k cat times E0 divided by E0. 93 00:05:56,130 --> 00:05:57,340 The E0's come out. 94 00:05:57,340 --> 00:06:00,320 And this is k cat. 95 00:06:00,320 --> 00:06:08,200 Which is basically k2. 96 00:06:08,200 --> 00:06:10,960 So that's the turnover number. 97 00:06:10,960 --> 00:06:12,290 Any questions on this turnover number? 98 00:06:12,290 --> 00:06:17,370 You see that a lot in catalyst literature and 99 00:06:17,370 --> 00:06:19,850 especially the enzymes. 100 00:06:19,850 --> 00:06:27,310 I have to get out these lecture notes. 101 00:06:27,310 --> 00:06:32,010 Send these back. 102 00:06:32,010 --> 00:06:35,650 OK, so the last topic I want to talk about 103 00:06:35,650 --> 00:06:38,240 is oscillating reactions. 104 00:06:38,240 --> 00:06:49,040 It's sort of like an interesting topic. 105 00:06:49,040 --> 00:06:51,940 And it's applicable not just to just chemistry, but it's 106 00:06:51,940 --> 00:06:54,450 basically playing with the differential equations. 107 00:06:54,450 --> 00:07:01,700 And you'll see this sort of stuff later if you keep doing 108 00:07:01,700 --> 00:07:05,100 science that involves coupled differential equations. 109 00:07:05,100 --> 00:07:09,330 It's just all over the place. 110 00:07:09,330 --> 00:07:14,230 So normally we have an equilibrium process. 111 00:07:14,230 --> 00:07:18,760 A goes to B, B goes back to A. And if you start out with 112 00:07:18,760 --> 00:07:24,130 something which is out of equilibrium, then you will 113 00:07:24,130 --> 00:07:26,700 eventually reach equilibrium at a rate which is dependent 114 00:07:26,700 --> 00:07:30,530 on those two rates. k1 plus k minus one. 115 00:07:30,530 --> 00:07:33,730 So B will to come up to some B equilibrium. 116 00:07:33,730 --> 00:07:38,910 And A will, if you start out with a lot of A, and a little 117 00:07:38,910 --> 00:07:43,760 bit of B, and concentration of A, will come down smoothly, 118 00:07:43,760 --> 00:07:48,280 monotonically, to the equilibrium state. 119 00:07:48,280 --> 00:07:53,790 But sometimes, if you're out of equilibrium, and this 120 00:07:53,790 --> 00:07:54,720 doesn't have to be chemistry. 121 00:07:54,720 --> 00:08:02,550 It could be anything that's out of equilibrium, your 122 00:08:02,550 --> 00:08:05,300 emotion's out of equilibrium. 123 00:08:05,300 --> 00:08:08,090 Stock market is out of equilibrium. 124 00:08:08,090 --> 00:08:10,840 Population dynamics are out of equilibrium. 125 00:08:10,840 --> 00:08:11,750 The weather. 126 00:08:11,750 --> 00:08:14,630 Whatever. 127 00:08:14,630 --> 00:08:17,350 So, you're out of equilibrium, let's say this is equilibrium 128 00:08:17,350 --> 00:08:24,590 for A, this is equilibrium for B, let me try to keep them 129 00:08:24,590 --> 00:08:28,140 somewhat the same as here, I messed up. 130 00:08:28,140 --> 00:08:31,980 You're out of equilibrium, you start down here for A. And 131 00:08:31,980 --> 00:08:36,240 you, instead of going linearly down or monotonically down, 132 00:08:36,240 --> 00:08:37,380 exponentially down to the 133 00:08:37,380 --> 00:08:40,590 equilibrium, instead you overshoot. 134 00:08:40,590 --> 00:08:42,190 And you keep going back and forth. 135 00:08:42,190 --> 00:08:44,870 Like a spring. 136 00:08:44,870 --> 00:08:46,990 And then B does the same thing. 137 00:08:46,990 --> 00:08:48,380 Overshoots, comes down. 138 00:08:48,380 --> 00:08:50,760 Overshoots, comes down. 139 00:08:50,760 --> 00:08:51,450 Like a pendulum. 140 00:08:51,450 --> 00:08:58,820 A lot of things work like that. 141 00:08:58,820 --> 00:09:00,490 The heart is something that works like that, right? 142 00:09:00,490 --> 00:09:00,790 It beats. 143 00:09:00,790 --> 00:09:04,160 It's not in equilibrium. 144 00:09:04,160 --> 00:09:08,050 Good thing it's not in equilibrium. 145 00:09:08,050 --> 00:09:11,640 So the heart is one example of something that oscillates, 146 00:09:11,640 --> 00:09:13,010 back and forth, back and forth. 147 00:09:13,010 --> 00:09:16,100 And so there are actually many examples of complicated 148 00:09:16,100 --> 00:09:19,690 processes that involve chemistry, that aren't at 149 00:09:19,690 --> 00:09:26,970 equilibrium, that instead go back and forth, in and out, of 150 00:09:26,970 --> 00:09:32,080 on one side or the other of equilibrium. 151 00:09:32,080 --> 00:09:35,460 Now, there's an important feature of, if you want to 152 00:09:35,460 --> 00:09:42,310 build a chemical process that looks like this. 153 00:09:42,310 --> 00:09:47,170 It's often very useful to have a particular step in the 154 00:09:47,170 --> 00:09:56,030 mechanism that's called an autocatalysis step. 155 00:09:56,030 --> 00:09:59,490 And that provides feedback. 156 00:09:59,490 --> 00:10:05,950 It's like, if my microphone here, the speaker which is 157 00:10:05,950 --> 00:10:11,450 here, I were to stand right in front of the speaker and my 158 00:10:11,450 --> 00:10:15,190 microphone would pick up the volume from the speaker, it 159 00:10:15,190 --> 00:10:17,450 would get amplified, and I get feedback. 160 00:10:17,450 --> 00:10:21,040 And that would be a bad thing. 161 00:10:21,040 --> 00:10:22,940 So this is a form of feedback here. 162 00:10:22,940 --> 00:10:26,320 You need some sort of feedback that the reaction knows that 163 00:10:26,320 --> 00:10:29,930 you're building up, that this is going too fast. 164 00:10:29,930 --> 00:10:32,220 And then there's a feedback process that 165 00:10:32,220 --> 00:10:33,080 brings it back up. 166 00:10:33,080 --> 00:10:36,070 And a feedback process that goes back and forth. 167 00:10:36,070 --> 00:10:40,130 And the way you get this feedback is by having a step 168 00:10:40,130 --> 00:10:53,780 in the mechanism that has both a molecule as part of the 169 00:10:53,780 --> 00:10:58,620 reactants and as part of the product. 170 00:10:58,620 --> 00:11:02,360 It's not the same as a chain reaction, where we talked 171 00:11:02,360 --> 00:11:06,490 about an intermediate being recycled and building up. 172 00:11:06,490 --> 00:11:10,140 This is an actual reactant and an actual product that's part 173 00:11:10,140 --> 00:11:16,550 of this step here. 174 00:11:16,550 --> 00:11:20,190 So this is an autocatalytic step. 175 00:11:20,190 --> 00:11:24,530 And let's just see what that step looks like. 176 00:11:24,530 --> 00:11:26,400 Suppose we just have that step. 177 00:11:26,400 --> 00:11:29,800 So we want to find out, as a function of time, if you start 178 00:11:29,800 --> 00:11:33,180 out with some A and B, what happens to the concentration 179 00:11:33,180 --> 00:11:34,660 of B as a function of time. 180 00:11:34,660 --> 00:11:36,530 The time dependence of that. 181 00:11:36,530 --> 00:11:37,920 Clearly it's going to build up. 182 00:11:37,920 --> 00:11:40,590 But how is it going to build up? 183 00:11:40,590 --> 00:11:46,000 So we need to solve for B as a function of time. 184 00:11:46,000 --> 00:11:48,270 So let's go ahead and write down our rates. 185 00:11:48,270 --> 00:11:53,300 The rate of the reaction is d[A]/dt, is, let's put a rate 186 00:11:53,300 --> 00:12:00,400 constant here. k, k times [A] times [B]. 187 00:12:00,400 --> 00:12:01,050 Now, [B] and [A] 188 00:12:01,050 --> 00:12:05,190 are related to each other through stoichiometry. 189 00:12:05,190 --> 00:12:11,300 The concentration of B is the concentration of B that he 190 00:12:11,300 --> 00:12:18,650 started out with, [B]0, then every time you destroy an A, 191 00:12:18,650 --> 00:12:21,800 you create 2B. 192 00:12:21,800 --> 00:12:25,710 So it's plus two times the amount of A that you've 193 00:12:25,710 --> 00:12:31,830 destroyed, [A]0 minus [A], [A]0 is what 194 00:12:31,830 --> 00:12:32,720 you started out with. 195 00:12:32,720 --> 00:12:33,670 This is what's left over. 196 00:12:33,670 --> 00:12:37,790 So the difference is what you've destroyed. 197 00:12:37,790 --> 00:12:40,410 But every time you destroy an A to form 2B, you 198 00:12:40,410 --> 00:12:43,170 also destroy a b. 199 00:12:43,170 --> 00:12:46,470 So you have to subtract away the B that you've destroyed. 200 00:12:46,470 --> 00:12:52,450 Which is [A]0 minus [A]. 201 00:12:52,450 --> 00:12:56,540 This is the 2B that you created by destroying A, and 202 00:12:56,540 --> 00:13:00,640 this is the B that you destroy by having to destroy A, for 203 00:13:00,640 --> 00:13:02,410 the reaction. 204 00:13:02,410 --> 00:13:05,770 You've got to bean-count correctly. 205 00:13:05,770 --> 00:13:07,030 Keep track of everything. 206 00:13:07,030 --> 00:13:11,270 So [B], then, is just, I'm going to drop those brackets. 207 00:13:11,270 --> 00:13:17,860 B0 plus A0 minus A. So you can plug that in here. 208 00:13:17,860 --> 00:13:28,890 And is equal to k times A, times B0 plus B0 minus A. And 209 00:13:28,890 --> 00:13:34,720 now you have a differential equation that only 210 00:13:34,720 --> 00:13:37,510 contains A and time. 211 00:13:37,510 --> 00:13:39,070 B0 is a constant. 212 00:13:39,070 --> 00:13:43,080 And the way you solve that is by partial fractions. 213 00:13:43,080 --> 00:13:49,050 Use partial fractions, solve for this. 214 00:13:49,050 --> 00:13:51,245 And I'm not going to go through the math 215 00:13:51,245 --> 00:13:52,140 of solving for it. 216 00:13:52,140 --> 00:13:59,440 It's not that interesting. 217 00:13:59,440 --> 00:14:03,520 Let me give you the answer. 218 00:14:03,520 --> 00:14:07,080 You end up with [B] as a function of time, is this 219 00:14:07,080 --> 00:14:16,040 function A0 plus B0 over one plus A0 over B0 times e to the 220 00:14:16,040 --> 00:14:24,440 minus k times A0 plus B0 plus time. 221 00:14:24,440 --> 00:14:26,010 And that's where the time component comes in. 222 00:14:26,010 --> 00:14:28,640 Whoops, I'm sorry. 223 00:14:28,640 --> 00:14:33,950 Usually I don't have this on, let's turn this off. 224 00:14:33,950 --> 00:14:41,660 That's where the time dependence comes in. 225 00:14:41,660 --> 00:14:45,460 And clearly, at t is equal to infinity, t equals infinity 226 00:14:45,460 --> 00:14:47,300 just goes to zero. 227 00:14:47,300 --> 00:14:50,570 So you have B as A0, plus B0, everything goes through to the 228 00:14:50,570 --> 00:14:54,050 product. t equals to zero, this is equal to one. 229 00:14:54,050 --> 00:15:00,590 And B is equal to just B0. 230 00:15:00,590 --> 00:15:05,490 Alright, and so the curve looks something like this. 231 00:15:05,490 --> 00:15:09,180 This is what it turns out to look like. 232 00:15:09,180 --> 00:15:11,920 We're starting with B0 here. 233 00:15:11,920 --> 00:15:14,630 There's an induction period where very little happens. 234 00:15:14,630 --> 00:15:17,560 It's like the dormant stage. 235 00:15:17,560 --> 00:15:20,340 So, locusts that are sitting in the ground for seventeen 236 00:15:20,340 --> 00:15:27,230 years waiting for something to happen. 237 00:15:27,230 --> 00:15:28,700 Dormant stage, the induction period. 238 00:15:28,700 --> 00:15:30,710 And suddenly something starts to happen. 239 00:15:30,710 --> 00:15:35,710 There's an inflection point. 240 00:15:35,710 --> 00:15:39,020 And you get to A0 plus B0. 241 00:15:39,020 --> 00:15:41,070 The locusts wake up. 242 00:15:41,070 --> 00:15:47,970 And go to maximum concentration. 243 00:15:47,970 --> 00:15:55,250 So there's an s-like shape that has an inflection point 244 00:15:55,250 --> 00:16:04,290 and an induction period. 245 00:16:04,290 --> 00:16:10,310 That's pretty typical of an auto-catalytic reaction. 246 00:16:10,310 --> 00:16:13,270 OK so let's apply it to a process now. 247 00:16:13,270 --> 00:16:15,340 And again, this is going to be coupled 248 00:16:15,340 --> 00:16:16,320 differential equations. 249 00:16:16,320 --> 00:16:18,890 Broadly, broadly applicable, the example that I'm 250 00:16:18,890 --> 00:16:20,560 going to give you. 251 00:16:20,560 --> 00:16:45,030 This mechanism has been used for all sorts of things. 252 00:16:45,030 --> 00:16:45,820 So what do we have here? 253 00:16:45,820 --> 00:16:51,560 We're going to have an island. 254 00:16:51,560 --> 00:16:56,310 In the ocean. 255 00:16:56,310 --> 00:16:58,780 Island here. 256 00:16:58,780 --> 00:17:03,920 And it's going to be sunny, so let's get some 257 00:17:03,920 --> 00:17:05,810 yellow chalk here. 258 00:17:05,810 --> 00:17:08,160 There's the sun. 259 00:17:08,160 --> 00:17:11,860 And every now and then it's going to rain. 260 00:17:11,860 --> 00:17:15,160 We need some white chalk for the rain. 261 00:17:15,160 --> 00:17:22,000 There's a cloud that comes in. 262 00:17:22,000 --> 00:17:26,600 That's going to be one of the reactants. 263 00:17:26,600 --> 00:17:31,820 And then some seeds blow in from the continent, and grass 264 00:17:31,820 --> 00:17:35,990 starts to grow on the island. 265 00:17:35,990 --> 00:17:38,410 Grass. 266 00:17:38,410 --> 00:17:41,590 And then there's a shipwreck and a couple of rabbits get on 267 00:17:41,590 --> 00:17:43,190 the island. 268 00:17:43,190 --> 00:17:55,040 And now we have rabbits. 269 00:17:55,040 --> 00:17:55,430 Rabbits. 270 00:17:55,430 --> 00:17:58,490 A little tail on the back. 271 00:17:58,490 --> 00:18:00,570 I hate rabbits. 272 00:18:00,570 --> 00:18:04,280 I have a little vegetable garden, a little tiny 273 00:18:04,280 --> 00:18:05,180 vegetable garden. 274 00:18:05,180 --> 00:18:06,990 And there are rabbits. 275 00:18:06,990 --> 00:18:10,710 And I have to fence in my vegetable garden and make it 276 00:18:10,710 --> 00:18:13,405 impermeable impenetrable to rabbits. 277 00:18:13,405 --> 00:18:15,990 And rabbits are very clever. 278 00:18:15,990 --> 00:18:18,740 And my vegetable garden looks like a fortress. 279 00:18:18,740 --> 00:18:21,910 It's got a green fence. 280 00:18:21,910 --> 00:18:23,380 Heavy duty things. 281 00:18:23,380 --> 00:18:24,010 It's terrible. 282 00:18:24,010 --> 00:18:25,070 Rabbits are awful. 283 00:18:25,070 --> 00:18:27,590 They're very cute but they're awful. 284 00:18:27,590 --> 00:18:29,400 Alright, so now what happens. 285 00:18:29,400 --> 00:18:34,060 Well, given the rain and the grass, it turns out that this 286 00:18:34,060 --> 00:18:38,140 island can only support 25 rabbits. 287 00:18:38,140 --> 00:18:39,200 If there are more rabbits than that, then 288 00:18:39,200 --> 00:18:40,540 they all eat too much. 289 00:18:40,540 --> 00:18:43,550 If there are less rabbits then that's fine. 290 00:18:43,550 --> 00:18:45,820 But 25 is about the maximum that can 291 00:18:45,820 --> 00:18:46,650 support on this island. 292 00:18:46,650 --> 00:18:51,020 The rabbits, unfortunately, are not that smart. 293 00:18:51,020 --> 00:18:52,850 They don't know that. 294 00:18:52,850 --> 00:18:56,380 They don't know that only 25 rabbits can live there. 295 00:18:56,380 --> 00:19:04,750 So, what happens is that the Year one, Year one there are 296 00:19:04,750 --> 00:19:06,490 ten rabbits. 297 00:19:06,490 --> 00:19:11,520 And the food condition is great. 298 00:19:11,520 --> 00:19:18,100 Lots of food, number of rabbits. 299 00:19:18,100 --> 00:19:20,060 Lots of food. 300 00:19:20,060 --> 00:19:22,810 So the bunnies, the rabbits make bunnies. 301 00:19:22,810 --> 00:19:23,950 Lots of rabbits. 302 00:19:23,950 --> 00:19:26,630 Year two, they kind of went overboard. 303 00:19:26,630 --> 00:19:29,240 Now there are 50 rabbits. 304 00:19:29,240 --> 00:19:30,950 The food is lousy. 305 00:19:30,950 --> 00:19:34,120 Terrible. 306 00:19:34,120 --> 00:19:36,700 Well, there's a feedback here. 307 00:19:36,700 --> 00:19:40,130 Feedback and rabbits start to die off. 308 00:19:40,130 --> 00:19:42,440 Don't reproduce as much. 309 00:19:42,440 --> 00:19:44,520 Year three, we're back to ten. 310 00:19:44,520 --> 00:19:47,040 Great food. 311 00:19:47,040 --> 00:19:48,290 Rabbits don't notice. 312 00:19:48,290 --> 00:19:50,250 They multiply. 313 00:19:50,250 --> 00:19:53,040 Food is terrible. 314 00:19:53,040 --> 00:19:55,390 And they just don't learn. 315 00:19:55,390 --> 00:19:56,750 Humans are a little bit like that. 316 00:19:56,750 --> 00:19:59,990 Except our cycle isn't one year, it's usually on the 317 00:19:59,990 --> 00:20:00,870 order of 30 years. 318 00:20:00,870 --> 00:20:03,870 It's like the memory of a generation disappearing. 319 00:20:03,870 --> 00:20:05,660 And we have to relearn the mistakes of the previous 320 00:20:05,660 --> 00:20:09,520 generation. 321 00:20:09,520 --> 00:20:10,440 Science is like that too. 322 00:20:10,440 --> 00:20:12,090 Science goes in cycles. 323 00:20:12,090 --> 00:20:13,800 20, 30-year cycles. 324 00:20:13,800 --> 00:20:17,310 Topics that were out of fashion 20 years ago, 25 years 325 00:20:17,310 --> 00:20:19,270 ago, come back. 326 00:20:19,270 --> 00:20:21,070 And suddenly everybody's excited about them. 327 00:20:21,070 --> 00:20:24,690 And all the ideas, of course we make progress. 328 00:20:24,690 --> 00:20:26,710 We go a little bit further than we did 20 years ago. 329 00:20:26,710 --> 00:20:28,630 But all the ideas that were out of fashion 20 330 00:20:28,630 --> 00:20:29,780 years ago come back. 331 00:20:29,780 --> 00:20:32,380 People get really excited, and they relearn all the mistakes 332 00:20:32,380 --> 00:20:34,250 that people made at the beginning. 333 00:20:34,250 --> 00:20:37,770 Technology has improved, we know more, and therefore we go 334 00:20:37,770 --> 00:20:39,370 a little bit further this cycle than we did 335 00:20:39,370 --> 00:20:40,060 the previous cycle. 336 00:20:40,060 --> 00:20:42,786 But it's really interesting to look at history of science and 337 00:20:42,786 --> 00:20:52,140 see this sort of cycle. 338 00:20:52,140 --> 00:20:55,240 So let's write a mechanism. 339 00:20:55,240 --> 00:20:59,540 Let's write a mechanism here. 340 00:20:59,540 --> 00:21:07,910 So we have rain plus grass makes more grass. 341 00:21:07,910 --> 00:21:09,240 At the rate k1. 342 00:21:09,240 --> 00:21:10,860 That's autocatalytic. 343 00:21:10,860 --> 00:21:12,430 It provides some feedback. 344 00:21:12,430 --> 00:21:21,520 Then we have grass plus rabbits make more rabbits. 345 00:21:21,520 --> 00:21:23,870 OK, let's do more because I don't know 346 00:21:23,870 --> 00:21:28,820 how much, more, more. 347 00:21:28,820 --> 00:21:30,900 Also autocatalytic. 348 00:21:30,900 --> 00:21:33,200 Then we have rabbits. 349 00:21:33,200 --> 00:21:39,700 Eventually, unfortunately, rabbits become dead rabbits. 350 00:21:39,700 --> 00:21:45,330 Rate k2 k3, and that's the termination of the process. 351 00:21:45,330 --> 00:21:47,580 OK, so now let's put some chemistry on this. 352 00:21:47,580 --> 00:21:51,220 Let's assume that instead of having objects, live objects 353 00:21:51,220 --> 00:21:54,440 here, we have chemical molecules. 354 00:21:54,440 --> 00:21:57,950 So A plus B goes to 2B. 355 00:21:57,950 --> 00:21:59,820 That's the first step. 356 00:21:59,820 --> 00:22:05,350 Then we have B plus C, C is the rabbits. 357 00:22:05,350 --> 00:22:10,960 And then C goes to some sort of products. 358 00:22:10,960 --> 00:22:15,450 Soil. 359 00:22:15,450 --> 00:22:18,020 This is a very famous mechanism. 360 00:22:18,020 --> 00:22:26,820 It's called the Lotka-Volterra mechanism. 361 00:22:26,820 --> 00:22:33,970 It's also called the predator-prey mechanism. 362 00:22:33,970 --> 00:22:36,920 In my case here, I made it sort of warm and fuzzy. 363 00:22:36,920 --> 00:22:38,480 I didn't have a predator in there. 364 00:22:38,480 --> 00:22:41,720 But you could change this. 365 00:22:41,720 --> 00:22:44,510 Instead of having rabbits and grass, we could 366 00:22:44,510 --> 00:22:47,170 have rabbits and foxes. 367 00:22:47,170 --> 00:22:54,580 Where A is the grass, if you take A to be the grass. 368 00:22:54,580 --> 00:22:57,880 B to be the rabbits. 369 00:22:57,880 --> 00:23:01,630 And C you be the foxes. 370 00:23:01,630 --> 00:23:02,660 Works the same. 371 00:23:02,660 --> 00:23:05,580 You have grass plus rabbits makes more rabbits. 372 00:23:05,580 --> 00:23:07,870 Rabbit plus foxes make more foxes. 373 00:23:07,870 --> 00:23:10,450 Eventually the foxes die off. 374 00:23:10,450 --> 00:23:12,720 Same idea. 375 00:23:12,720 --> 00:23:32,770 So now, let's solve this. 376 00:23:32,770 --> 00:23:38,820 Let's assume that A, the rain here, it's constant. 377 00:23:38,820 --> 00:23:40,090 There's a steady supply of rain. 378 00:23:40,090 --> 00:23:41,860 It doesn't go away. 379 00:23:41,860 --> 00:23:44,680 That makes sense in my example here. 380 00:23:44,680 --> 00:23:50,780 So the concentration of A is kept constant. 381 00:23:50,780 --> 00:23:53,590 In order to get oscillations to keep going, that turns out 382 00:23:53,590 --> 00:23:54,230 to be important. 383 00:23:54,230 --> 00:23:58,410 To have one of the reactants just keeping being 384 00:23:58,410 --> 00:23:59,500 reintroduced in the system. 385 00:23:59,500 --> 00:24:01,040 Because it gets used up. 386 00:24:01,040 --> 00:24:06,280 And when all the reactant gets used up, then you're done. 387 00:24:06,280 --> 00:24:10,060 So in the case of the heart, we have to keep feeding 388 00:24:10,060 --> 00:24:12,040 ourselves to produce energy. 389 00:24:12,040 --> 00:24:18,850 Otherwise the heart would stop. 390 00:24:18,850 --> 00:24:22,350 OK, now we want to know what is the concentration of B and 391 00:24:22,350 --> 00:24:26,350 C as a function of time. 392 00:24:26,350 --> 00:24:36,110 So we write down our kinetic equations. dB/dt gets produced 393 00:24:36,110 --> 00:24:39,800 through the first step, k1 A times B, gets destroyed in the 394 00:24:39,800 --> 00:24:47,850 second step. k2 B times C. dC/dt, C gets produced in the 395 00:24:47,850 --> 00:24:52,920 second step, k2 B times C. Gets destroyed in the third 396 00:24:52,920 --> 00:24:56,700 step. k3 times C, got to do your bean-counting correctly. 397 00:24:56,700 --> 00:24:58,870 And the strategy here that we're going to use to solve 398 00:24:58,870 --> 00:25:03,700 the problem is to assume that we're at steady state. 399 00:25:03,700 --> 00:25:06,200 We're going to find the solution at steady state. 400 00:25:06,200 --> 00:25:08,270 Then we're going to perturb away from steady state a 401 00:25:08,270 --> 00:25:09,800 little bit. 402 00:25:09,800 --> 00:25:13,560 And see whether this creates an oscillation. 403 00:25:13,560 --> 00:25:17,230 First we need to find out what the steady state solution is. 404 00:25:17,230 --> 00:25:21,950 So first, let's find steady state. 405 00:25:21,950 --> 00:25:30,740 Then perturb away from steady state. 406 00:25:30,740 --> 00:25:32,470 We're looking at a harmonic response, if 407 00:25:32,470 --> 00:25:33,840 you've heard that term. 408 00:25:33,840 --> 00:25:36,640 When it's going to look like a spring, close to its 409 00:25:36,640 --> 00:25:41,300 equilibrium state. 410 00:25:41,300 --> 00:25:44,290 That means that we're going to set it these two things equal 411 00:25:44,290 --> 00:25:47,630 to zero, for this steady state. 412 00:25:47,630 --> 00:25:48,970 Things are not changing. 413 00:25:48,970 --> 00:25:50,520 Concentration of B and C are not changing. 414 00:25:50,520 --> 00:25:53,970 A is not changing on purpose here. 415 00:25:53,970 --> 00:25:56,380 And so all these guys here are then steady state 416 00:25:56,380 --> 00:25:57,920 concentrations. 417 00:25:57,920 --> 00:26:02,450 If we set it equal to zero. 418 00:26:02,450 --> 00:26:10,110 We solve for A in this. 419 00:26:10,110 --> 00:26:11,930 For C in this process here. 420 00:26:11,930 --> 00:26:13,140 The B's cancel out here. 421 00:26:13,140 --> 00:26:20,160 We divide by B. And you solve for C as a function of A. So 422 00:26:20,160 --> 00:26:27,600 this one here gives you C steady state is equal to k1 423 00:26:27,600 --> 00:26:32,750 over k2 times the concentration of A. And this 424 00:26:32,750 --> 00:26:37,210 one here, now you put in the C and you solve for B, B steady 425 00:26:37,210 --> 00:26:58,740 state, is equal to k3 over k2. 426 00:26:58,740 --> 00:27:02,740 So the next step is to perturb away from equilibrium. 427 00:27:02,740 --> 00:27:03,820 Or from the steady state, rather. 428 00:27:03,820 --> 00:27:05,910 This is not equilibrium, it's not equilibrium because we 429 00:27:05,910 --> 00:27:18,060 keep adding A. We have to put some input into the system. 430 00:27:18,060 --> 00:27:19,890 So here we are. 431 00:27:19,890 --> 00:27:23,980 We've got some concentration of B here at the steady state. 432 00:27:23,980 --> 00:27:28,910 We have some concentration of C at the steady state. 433 00:27:28,910 --> 00:27:35,480 And now we're going to add some delta B to the system, or 434 00:27:35,480 --> 00:27:37,760 delta C. Or both. 435 00:27:37,760 --> 00:27:45,240 Delta C, we're going to ask the question, given our system 436 00:27:45,240 --> 00:27:47,630 here, how is it going to respond? 437 00:27:47,630 --> 00:27:51,280 It has a number of choices. 438 00:27:51,280 --> 00:27:58,300 It could respond by going back to the steady state in a 439 00:27:58,300 --> 00:28:03,250 monotonic fashion. 440 00:28:03,250 --> 00:28:07,420 Without overshooting. 441 00:28:07,420 --> 00:28:15,910 This is kind of like an overdamped system. 442 00:28:15,910 --> 00:28:18,280 It's like having a shock absorber on your car. 443 00:28:18,280 --> 00:28:21,100 You go over a bump, the car doesn't oscillate up and down. 444 00:28:21,100 --> 00:28:22,170 It's sort of damped. 445 00:28:22,170 --> 00:28:23,110 It's a damped oscillator. 446 00:28:23,110 --> 00:28:27,010 It goes back to the steady state. 447 00:28:27,010 --> 00:28:40,320 Or, you could, there we go, there's B, C. Or you could 448 00:28:40,320 --> 00:28:44,980 perturb, and it could just stay where it is. 449 00:28:44,980 --> 00:28:46,490 That could happen too. 450 00:28:46,490 --> 00:28:54,020 That's an inelastic response. 451 00:28:54,020 --> 00:28:58,320 Kind of like the supply and demand with oil. 452 00:28:58,320 --> 00:29:02,790 The price of oil goes from $10 a barrel to $120 a barrel. 453 00:29:02,790 --> 00:29:07,100 The use of oil doesn't seem to be affected very much by the 454 00:29:07,100 --> 00:29:07,585 increasing price. 455 00:29:07,585 --> 00:29:08,870 You're still using just as much oil. 456 00:29:08,870 --> 00:29:16,280 There's an inelastic response. 457 00:29:16,280 --> 00:29:20,690 What we're looking for is something that has an elastic 458 00:29:20,690 --> 00:29:22,940 response, or a harmonic response. 459 00:29:22,940 --> 00:29:34,010 We perturb it, it tries to get back to steady state. 460 00:29:34,010 --> 00:29:38,920 But because of feedback, and not being over-damped, if it's 461 00:29:38,920 --> 00:29:42,180 an oscillator, it goes up and down. 462 00:29:42,180 --> 00:29:51,540 Like a scale that is not well, well, like a car that has bad 463 00:29:51,540 --> 00:29:53,750 shock absorbers, all it has is the springs. 464 00:29:53,750 --> 00:29:57,720 Up and down. 465 00:29:57,720 --> 00:30:03,730 So what we want is, we want to solve for dB/dt as 466 00:30:03,730 --> 00:30:22,320 a function of time. 467 00:30:22,320 --> 00:30:24,580 We want, well, I'm going to do everything in green. 468 00:30:24,580 --> 00:30:27,040 Since I lost my white chalk. 469 00:30:27,040 --> 00:30:33,780 We want delta B, delta C as a function of time. 470 00:30:33,780 --> 00:30:39,120 So, that means that we start with B is B steady state plus 471 00:30:39,120 --> 00:30:45,280 delta B, C is C steady state plus delta C. And we put that 472 00:30:45,280 --> 00:30:46,340 in our equations. 473 00:30:46,340 --> 00:30:48,640 For dB/dt and dC/dt. 474 00:30:48,640 --> 00:30:57,190 So now dB/dt is the same thing as dB steady state plus delta 475 00:30:57,190 --> 00:31:01,730 B dt, which is d delta B / dt. 476 00:31:01,730 --> 00:31:04,930 Because this is a constant here. 477 00:31:04,930 --> 00:31:11,670 And that's equal to k1 times A, times B steady state plus 478 00:31:11,670 --> 00:31:19,110 delta B. Minus k2 times B steady state plus delta B 479 00:31:19,110 --> 00:31:23,940 times C steady state plus delta C. And then you have the 480 00:31:23,940 --> 00:31:30,090 same thing for dC/dt, it's going to look exactly 481 00:31:30,090 --> 00:31:34,330 identical. d delta C / dt. 482 00:31:34,330 --> 00:31:49,440 Thank you much, this is magic. k2 times B steady state plus 483 00:31:49,440 --> 00:31:57,340 delta B, times C steady state plus delta C, minus k3 times C 484 00:31:57,340 --> 00:32:00,830 steady state plus delta C. 485 00:32:00,830 --> 00:32:02,770 So these are two differential equations. 486 00:32:02,770 --> 00:32:04,260 And there are coupled differential equations. 487 00:32:04,260 --> 00:32:07,300 Because delta B is occurring here, here, and here. 488 00:32:07,300 --> 00:32:12,990 So the time derivative of delta C depends on delta B. 489 00:32:12,990 --> 00:32:17,530 And the time derivative of delta B depends on delta C. 490 00:32:17,530 --> 00:32:21,180 It's a coupled system. 491 00:32:21,180 --> 00:32:26,140 And if you actually expand this out and take away all the 492 00:32:26,140 --> 00:32:34,350 terms that cancel out, you end up with something that looks 493 00:32:34,350 --> 00:32:40,160 much simpler but is just as hard. d delta B / dt is equal 494 00:32:40,160 --> 00:32:52,900 to minus k3 delta C and d delta C / dt is equal to k1 A 495 00:32:52,900 --> 00:32:53,320 delta B. 496 00:32:53,320 --> 00:32:56,820 And then you really see that it's coupled, because the time 497 00:32:56,820 --> 00:33:00,190 derivative of delta B depends on delta C. And the time 498 00:33:00,190 --> 00:33:03,680 derivative of delta C depends on delta B. And there's the 499 00:33:03,680 --> 00:33:08,430 feedback sitting right here. 500 00:33:08,430 --> 00:33:09,380 And that's what you do. 501 00:33:09,380 --> 00:33:11,830 You learn how to solve in 18.03. 502 00:33:11,830 --> 00:33:13,460 So I'm not going to solve it here, I'm just going to give 503 00:33:13,460 --> 00:33:14,190 you the solutions. 504 00:33:14,190 --> 00:33:16,050 Because that's what we're interested in. 505 00:33:16,050 --> 00:33:18,300 After all, you're not expected to learn how 506 00:33:18,300 --> 00:33:19,740 to solve this equation. 507 00:33:19,740 --> 00:33:22,250 Although you should be able to put the solutions in and make 508 00:33:22,250 --> 00:33:33,350 sure that they do work out. 509 00:33:33,350 --> 00:33:37,190 And so the solutions are harmonic solutions. 510 00:33:37,190 --> 00:33:41,180 Delta B is a function of time is equal to delta B0. 511 00:33:41,180 --> 00:33:45,990 The amount of the perturbation times the cosine of omega t, 512 00:33:45,990 --> 00:33:49,450 where omega is the frequency of the oscillation. 513 00:33:49,450 --> 00:33:59,370 Minus some pre-factor here, k3 over k1 A to the 1/2 power. 514 00:33:59,370 --> 00:34:05,350 Times delta C0 sine of omega t. 515 00:34:05,350 --> 00:34:07,810 And then delta C looks the same. 516 00:34:07,810 --> 00:34:09,630 The same frequency. 517 00:34:09,630 --> 00:34:15,770 Delta C0, cosine of omega t. 518 00:34:15,770 --> 00:34:23,960 Plus k1 over A times k3 to the 1/2 power. 519 00:34:23,960 --> 00:34:29,820 Delta B0 sine of omega t, where omega, the frequency of 520 00:34:29,820 --> 00:34:34,630 the oscillation, is these rates. k1, k3 and A. So 521 00:34:34,630 --> 00:34:37,360 obviously, keeping the concentration of a constant is 522 00:34:37,360 --> 00:34:40,210 important because we don't want a time dependence here. 523 00:34:40,210 --> 00:34:42,680 The rain is constant. 524 00:34:42,680 --> 00:34:45,440 And then if you do that, then you have this oscillation. 525 00:34:45,440 --> 00:34:48,570 Up and down and up and down with this frequency here. 526 00:34:48,570 --> 00:34:50,770 You get exactly what we're trying to get. 527 00:34:50,770 --> 00:34:58,620 Which is this process right here. 528 00:34:58,620 --> 00:35:00,490 So let me give you an example of this. 529 00:35:00,490 --> 00:35:08,040 And the best way to do is to actually go and see a movie of 530 00:35:08,040 --> 00:35:20,730 a reaction that looks like this. 531 00:35:20,730 --> 00:35:27,540 This is the reaction that was optimized and created for 532 00:35:27,540 --> 00:35:30,220 demonstration purposes. 533 00:35:30,220 --> 00:35:35,240 So it's a pretty complicated reaction. 534 00:35:35,240 --> 00:35:40,910 There are ten steps to the mechanism. 535 00:35:40,910 --> 00:35:47,270 And if we gave you that mechanism to solve for the 536 00:35:47,270 --> 00:35:49,200 final exam, you would still be here probably 537 00:35:49,200 --> 00:35:51,360 two weeks form now. 538 00:35:51,360 --> 00:35:52,740 It's pretty complicated. 539 00:35:52,740 --> 00:35:54,780 You'd need a computer for sure. 540 00:35:54,780 --> 00:35:57,040 So, ten steps, well, that's what's known. 541 00:35:57,040 --> 00:35:58,520 There are probably more than that. 542 00:35:58,520 --> 00:36:01,180 It's probably more complicated than what we really know in 543 00:36:01,180 --> 00:36:06,210 terms of fishing out intermediates. 544 00:36:06,210 --> 00:36:10,460 And the basic reaction is, you start out with an oxide of 545 00:36:10,460 --> 00:36:15,290 iodine, which is clear. 546 00:36:15,290 --> 00:36:22,200 You create I2, which is going to look gold. 547 00:36:22,200 --> 00:36:25,140 In solution, and then it creates Ii minus, which is 548 00:36:25,140 --> 00:36:29,160 going to look deep blue. 549 00:36:29,160 --> 00:36:32,830 And then it'll feed back and start again. 550 00:36:32,830 --> 00:36:35,400 So we're going to have this cycle of products. 551 00:36:35,400 --> 00:36:38,250 And amongst those ten steps, or more than ten steps, there 552 00:36:38,250 --> 00:36:44,700 are a few autocatalytic steps, which provide the feedback. 553 00:36:44,700 --> 00:36:52,840 So now I have to start it. 554 00:36:52,840 --> 00:36:56,260 Let's go ahead and read. 555 00:37:09,960 --> 00:37:10,290 Alright. 556 00:37:10,290 --> 00:37:11,770 So here we go. 557 00:37:11,770 --> 00:37:18,660 So, there are two solutions, A and solution C. Actually, we 558 00:37:18,660 --> 00:37:21,420 mix three solutions together. 559 00:37:21,420 --> 00:37:24,040 The potassium iodide is what is going to give rise to the 560 00:37:24,040 --> 00:37:25,300 different colors. 561 00:37:25,300 --> 00:37:29,330 And now we're in the gold phase of the reaction. 562 00:37:29,330 --> 00:37:31,580 It's important to stir, otherwise it wouldn't, and 563 00:37:31,580 --> 00:37:33,740 then it turns deep blue. 564 00:37:33,740 --> 00:37:35,530 And then there's an induction period, where 565 00:37:35,530 --> 00:37:37,710 you wait for a while. 566 00:37:37,710 --> 00:37:43,520 And then eventually it should go clear. 567 00:37:43,520 --> 00:37:45,010 So it goes clear. 568 00:37:45,010 --> 00:37:46,340 And then it turns gold again. 569 00:37:46,340 --> 00:37:47,850 And then deep blue, and there's the oscillation. 570 00:37:47,850 --> 00:37:50,060 And then a long induction period, and then it goes 571 00:37:50,060 --> 00:37:54,150 through that cycle over and over again. 572 00:37:54,150 --> 00:37:57,260 And the recipe can be found pretty easily. 573 00:37:57,260 --> 00:38:01,570 And if you ever are in a lab, have access to doing this, 574 00:38:01,570 --> 00:38:04,040 it's a pretty easy reaction to put together. 575 00:38:04,040 --> 00:38:08,840 The only problem is that unless you use fresh hydrogen 576 00:38:08,840 --> 00:38:11,510 peroxide, it doesn't work. 577 00:38:11,510 --> 00:38:14,560 Because hydrogen peroxide tends to go bad over time. 578 00:38:14,560 --> 00:38:18,290 If you're going to do this, buy your hydrogen peroxide 579 00:38:18,290 --> 00:38:21,070 immediately before you try to do the experiment. 580 00:38:21,070 --> 00:38:22,420 And then it'll work. 581 00:38:22,420 --> 00:38:26,350 So what happens is that as long as there is a steady 582 00:38:26,350 --> 00:38:31,290 supply of hydrogen peroxide in here, it'll keep oscillating. 583 00:38:31,290 --> 00:38:34,340 But as you've used up the hydrogen peroxide, this is 584 00:38:34,340 --> 00:38:35,240 going to start to die. 585 00:38:35,240 --> 00:38:39,050 And eventually it'll have this pale blue solution. 586 00:38:39,050 --> 00:38:41,770 And it's like the heart stopping. 587 00:38:41,770 --> 00:38:43,640 If you want to start it up again, add more hydrogen 588 00:38:43,640 --> 00:38:50,360 peroxide and it'll start up again. 589 00:38:50,360 --> 00:38:52,450 OK, any questions? 590 00:38:52,450 --> 00:38:54,760 There's like an infinite number of these oscillating 591 00:38:54,760 --> 00:38:59,210 reactions that people have discovered. 592 00:38:59,210 --> 00:39:09,600 And optimized, that you can find in different places. 593 00:39:09,600 --> 00:39:13,000 Questions? 594 00:39:13,000 --> 00:39:18,330 Questions about the final. 595 00:39:18,330 --> 00:39:23,400 I didn't get to do a review, but you'll get that 596 00:39:23,400 --> 00:39:26,610 done with the TAs. 597 00:39:26,610 --> 00:39:26,870 Yes. 598 00:39:26,870 --> 00:39:33,590 STUDENT: [INAUDIBLE] 599 00:39:33,590 --> 00:39:35,780 PROFESSOR: Because there's enough of the equivalent of A, 600 00:39:35,780 --> 00:39:37,760 of the rain here. 601 00:39:37,760 --> 00:39:41,550 So it'll oscillate and there'll be some damping. 602 00:39:41,550 --> 00:39:45,060 So what will happen in this reaction is that the induction 603 00:39:45,060 --> 00:39:48,910 period will get longer, and longer, and longer. 604 00:39:48,910 --> 00:39:55,530 Because the w, this term right here, the frequency which 605 00:39:55,530 --> 00:39:58,590 depends on the concentration of A, that well 606 00:39:58,590 --> 00:39:59,600 get slower and slower. 607 00:39:59,600 --> 00:40:06,910 And eventually it'll just die. 608 00:40:06,910 --> 00:40:08,160 OK?