1 00:00:00,000 --> 00:00:02,400 The following content is provided under a Creative 2 00:00:02,400 --> 00:00:03,820 Commons license. 3 00:00:03,820 --> 00:00:06,840 Your support will help MIT OpenCourseWare continue to 4 00:00:06,840 --> 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 --> 00:00:17,590 hundreds of MIT courses, visit MIT OpenCourseWare at 7 00:00:17,590 --> 00:00:20,600 ocw.mit.edu. 8 00:00:20,600 --> 00:00:21,180 PROFESSOR: OK. 9 00:00:21,180 --> 00:00:24,410 So last time you started kinetics which is a completely 10 00:00:24,410 --> 00:00:26,640 different topic from thermodynamics. 11 00:00:26,640 --> 00:00:30,570 They're related and we'll see a relationship at some point. 12 00:00:30,570 --> 00:00:32,980 And you did first order kinetics. 13 00:00:32,980 --> 00:00:37,550 And today we're going to move on and go ahead and do more 14 00:00:37,550 --> 00:00:41,190 complicated kinetics and hopefully get to some 15 00:00:41,190 --> 00:00:44,890 interesting stuff in a couple lectures. 16 00:00:44,890 --> 00:00:48,080 Right now we just have to do the review of stuff that 17 00:00:48,080 --> 00:00:51,100 you've probably seen before. 18 00:00:51,100 --> 00:00:53,680 And so we're going to go reasonably fast. 19 00:00:53,680 --> 00:00:55,140 So you saw first order reactions. 20 00:00:55,140 --> 00:00:56,870 Today we're going to do second order reactions. 21 00:00:56,870 --> 00:01:00,440 At least begin with second order reactions. 22 00:01:00,440 --> 00:01:05,980 Second order kinetics. 23 00:01:05,980 --> 00:01:09,720 Of the form, and then there are two kinds. 24 00:01:09,720 --> 00:01:21,410 There's first order, rather, second order, in one reactant 25 00:01:21,410 --> 00:01:26,900 of the form A goes to products. 26 00:01:26,900 --> 00:01:32,700 And then you have first order in two reactants first order 27 00:01:32,700 --> 00:01:35,980 in two reactants. 28 00:01:35,980 --> 00:01:44,850 So I'll have the form A plus B goes of products. 29 00:01:44,850 --> 00:01:46,960 Where that's first order in A and first order in B, and this 30 00:01:46,960 --> 00:01:49,640 is second order in A. So you can think of this as A plus A 31 00:01:49,640 --> 00:01:53,210 goes to products if you want. 32 00:01:53,210 --> 00:01:56,070 And there's some rate constant k, 33 00:01:56,070 --> 00:01:57,860 associated with this reaction. 34 00:01:57,860 --> 00:01:59,380 And we do the rate analysis. 35 00:01:59,380 --> 00:02:02,330 We write the rate of this process. 36 00:02:02,330 --> 00:02:03,580 Minus dA/dt. 37 00:02:07,120 --> 00:02:10,440 And for the purpose of writing on the board, and you might 38 00:02:10,440 --> 00:02:12,030 want to do this also in your homework, when 39 00:02:12,030 --> 00:02:15,320 you're tired of writing. 40 00:02:15,320 --> 00:02:18,130 I'm going to skip the brackets. 41 00:02:18,130 --> 00:02:23,040 The little brackets that we usually put for concentration. 42 00:02:23,040 --> 00:02:24,780 I'm going to skip those, because it's just too much 43 00:02:24,780 --> 00:02:26,125 work to write them. 44 00:02:26,125 --> 00:02:30,510 And you can understand that A is a concentration of A. So 45 00:02:30,510 --> 00:02:34,250 this is equal to k A squared. 46 00:02:34,250 --> 00:02:39,490 Second order on A. And the units for k, it's important to 47 00:02:39,490 --> 00:02:43,280 keep track of your units, at the end of the calculations, 48 00:02:43,280 --> 00:02:46,030 often you want to make sure your units work out. 49 00:02:46,030 --> 00:02:48,730 So the units for k are going to be of this, A is 50 00:02:48,730 --> 00:02:50,400 in moles per liter. 51 00:02:50,400 --> 00:02:53,600 The units for k, you're going to have to be able to match 52 00:02:53,600 --> 00:02:55,200 the units on this side here. 53 00:02:55,200 --> 00:02:59,070 So the units for k are going to be liters squared per mole 54 00:02:59,070 --> 00:03:04,960 squared per second. 55 00:03:04,960 --> 00:03:06,830 To make the units match. 56 00:03:06,830 --> 00:03:11,140 Then you integrate this, on both sides, from zero to t, or 57 00:03:11,140 --> 00:03:18,630 from A0, the initial rate, to A. Or from zero to t and this 58 00:03:18,630 --> 00:03:22,180 doesn't go like this. 59 00:03:22,180 --> 00:03:32,470 So integrate with A0 to A. You've got dA over A squared, 60 00:03:32,470 --> 00:03:35,520 you put all the A's on one side, all the t's on the other 61 00:03:35,520 --> 00:03:43,160 side, from zero to t dt with a minus k on this side here. 62 00:03:43,160 --> 00:03:50,420 And then you get your rate equation, integrated rate 63 00:03:50,420 --> 00:03:52,630 equation for this. 64 00:03:52,630 --> 00:04:01,240 Which gives you one over A is equal to kt plus one over A0. 65 00:04:01,240 --> 00:04:03,600 So this gives you A as a function of time. 66 00:04:03,600 --> 00:04:06,610 And this is a convenient way to write it, because it's 67 00:04:06,610 --> 00:04:07,410 linear in time. 68 00:04:07,410 --> 00:04:09,650 So you always try to get things to be linear in time. 69 00:04:09,650 --> 00:04:15,360 Because then you can plot them as a straight line. 70 00:04:15,360 --> 00:04:20,460 Plot on this axis here you plot one over A. on this axis 71 00:04:20,460 --> 00:04:25,650 here you plot t as a function of time. 72 00:04:25,650 --> 00:04:34,420 And then you get a straight line where the slope is k. 73 00:04:34,420 --> 00:04:39,460 Gives you the rate, and the intercept is one over A0. 74 00:04:39,460 --> 00:04:39,720 Yes. 75 00:04:39,720 --> 00:04:45,450 STUDENT: [INAUDIBLE] 76 00:04:45,450 --> 00:04:51,030 PROFESSOR: k is moles per liter squared per second. 77 00:04:51,030 --> 00:04:53,760 You're right, liters per mole per second, yes. 78 00:04:53,760 --> 00:05:03,880 That is correct. because it has to work. 79 00:05:03,880 --> 00:05:05,000 Otherwise it doesn't work. 80 00:05:05,000 --> 00:05:08,860 So I need to turn on my brain. 81 00:05:08,860 --> 00:05:10,530 OK, think. 82 00:05:10,530 --> 00:05:15,250 Liters per mole per second. 83 00:05:15,250 --> 00:05:24,630 Thank you. 84 00:05:24,630 --> 00:05:28,210 And the other thing that you want to know is the half-life. 85 00:05:28,210 --> 00:05:30,050 What is the half-life. 86 00:05:30,050 --> 00:05:34,030 So you set 2 A0. 87 00:05:34,030 --> 00:05:39,600 So A0 over two, you look for the time where you get to A0 88 00:05:39,600 --> 00:05:43,310 over two, so you put A0 over two in here. 89 00:05:43,310 --> 00:05:47,110 And that's going to be equal to k times t 1/2 90 00:05:47,110 --> 00:05:48,270 plus one over A0. 91 00:05:48,270 --> 00:05:52,490 So you solve for the half-life and you get a half-life 92 00:05:52,490 --> 00:05:57,590 of one over k A0. 93 00:05:57,590 --> 00:06:00,520 So the half-life is inversely proportional to the amount of 94 00:06:00,520 --> 00:06:02,820 stuff you started out with. 95 00:06:02,820 --> 00:06:06,820 Unlike the first order reaction, where the half-life 96 00:06:06,820 --> 00:06:14,010 was independent of the amount that you started out with. 97 00:06:14,010 --> 00:06:15,680 So this was the easy one. 98 00:06:15,680 --> 00:06:18,760 The next one is a little bit more complicated. 99 00:06:18,760 --> 00:06:27,270 Which is when your first order in each of two reactants. 100 00:06:27,270 --> 00:06:30,500 Then your rate, your differential rate equation, 101 00:06:30,500 --> 00:06:32,320 looks like this. 102 00:06:32,320 --> 00:06:36,240 Because the k times A times B, and now you don't know what to 103 00:06:36,240 --> 00:06:38,960 do with B, a priori. 104 00:06:38,960 --> 00:06:43,740 So you want to rewrite this equation a little bit 105 00:06:43,740 --> 00:06:47,350 differently in terms of the amount of A that's used up. 106 00:06:47,350 --> 00:06:51,740 So you define A, x is equal to A0 minus A, this is the amount 107 00:06:51,740 --> 00:06:54,070 of A that's used up. 108 00:06:54,070 --> 00:06:56,120 This is what you started out with, this is 109 00:06:56,120 --> 00:06:56,990 what you're left with. 110 00:06:56,990 --> 00:07:01,440 And so the difference is what's being used up. 111 00:07:01,440 --> 00:07:09,050 And dx/dt is minus dA/dt. 112 00:07:09,050 --> 00:07:15,180 And by stoichiometry, what you've used up, of A, is also 113 00:07:15,180 --> 00:07:19,310 what you've used up of B. Because for every A that 114 00:07:19,310 --> 00:07:25,575 reacts, you have to use up one mole of B. For every mole of A 115 00:07:25,575 --> 00:07:28,510 that reacts, you use up one mole of B. And so you also 116 00:07:28,510 --> 00:07:34,780 have x to B0 minus B. So, you can plug this in here. 117 00:07:34,780 --> 00:07:37,420 And get a differential equation which is purely in 118 00:07:37,420 --> 00:07:39,450 terms of one variable, which is x. 119 00:07:39,450 --> 00:07:41,480 Right here, it looks like it's in terms of two variables, 120 00:07:41,480 --> 00:07:42,730 which makes it complicated. 121 00:07:42,730 --> 00:07:46,160 By doing this change of variables, you see that A and 122 00:07:46,160 --> 00:07:47,370 B actually related. 123 00:07:47,370 --> 00:07:51,150 Because of the reaction stoichiometry. 124 00:07:51,150 --> 00:07:59,010 So you can rewrite that as dx/dt is equal to k times A0 125 00:07:59,010 --> 00:08:02,230 minus x times B0 minus x. 126 00:08:02,230 --> 00:08:05,210 And now you have a differential equation in one 127 00:08:05,210 --> 00:08:09,170 variable, which with some tricks you can solve. 128 00:08:09,170 --> 00:08:13,350 So we want to have the integrated equation. 129 00:08:13,350 --> 00:08:15,540 So we take an integral of both sides. 130 00:08:15,540 --> 00:08:17,560 We put in all the x's on one side. 131 00:08:17,560 --> 00:08:19,500 All the times on the other side. 132 00:08:19,500 --> 00:08:28,670 We'll go from x equals zero to x, dx, A0 minus x times B0 133 00:08:28,670 --> 00:08:34,920 minus x is equal to k from zero to t dt. 134 00:08:34,920 --> 00:08:37,990 And now you have to dig back into the last time you took 135 00:08:37,990 --> 00:08:42,830 integration calculus, which for me was about 136 00:08:42,830 --> 00:08:44,600 two centuries ago. 137 00:08:44,600 --> 00:08:49,450 And figure out how to do this integral here. 138 00:08:49,450 --> 00:08:51,430 And the trick for doing this integral is 139 00:08:51,430 --> 00:09:01,640 to use partial fractions. 140 00:09:01,640 --> 00:09:04,510 So you use partial fractions to do this integral here. 141 00:09:04,510 --> 00:09:13,890 Which means that if you take this ratio, one over A0 minus 142 00:09:13,890 --> 00:09:20,570 x times B0 minus x and rewrite it as some number, n1 divided 143 00:09:20,570 --> 00:09:29,200 by A0 minus x, plus some number n2 divided by B0 minus 144 00:09:29,200 --> 00:09:35,210 x, you solve for n1 and n2, and you find that n1 here is 145 00:09:35,210 --> 00:09:43,600 equal to one over B0 minus A0 and n2 is minus one 146 00:09:43,600 --> 00:09:45,930 over B0 minus A0. 147 00:09:45,930 --> 00:09:48,840 And so you plug, now, this in here. 148 00:09:48,840 --> 00:09:52,250 And instead of having this complicated denominator, you 149 00:09:52,250 --> 00:09:55,560 have a sum of two integrals that you know how to do. 150 00:09:55,560 --> 00:10:00,690 Because they're basically of the form one over x. 151 00:10:00,690 --> 00:10:03,410 And in doing this, you also realize that you have to be 152 00:10:03,410 --> 00:10:06,870 careful because when A0 is equal to B0, when you have the 153 00:10:06,870 --> 00:10:09,930 same amount of A and B, then things blow up 154 00:10:09,930 --> 00:10:10,770 and you're in trouble. 155 00:10:10,770 --> 00:10:13,980 So that's going to be a special case. 156 00:10:13,980 --> 00:10:16,790 So always look out for special cases for these things. 157 00:10:16,790 --> 00:10:23,450 So you assume, then, that B0 is not the same as A0. 158 00:10:23,450 --> 00:10:25,800 And then you can go forward with solving it. 159 00:10:25,800 --> 00:10:29,510 So you integrate, and at the end of the process, I'm not 160 00:10:29,510 --> 00:10:33,620 going to go through it, it's really complicated, you get 161 00:10:33,620 --> 00:10:34,520 something that looks like this. 162 00:10:34,520 --> 00:10:46,420 A0 minus B0 log of A B0 over A0 B. And you have your 163 00:10:46,420 --> 00:10:46,980 equation here. 164 00:10:46,980 --> 00:10:49,660 Now, we don't really have a good way to plot it against 165 00:10:49,660 --> 00:10:51,350 one variable. 166 00:10:51,350 --> 00:10:54,970 And the usual thing is look at specific cases. 167 00:10:54,970 --> 00:10:56,340 And limiting cases. 168 00:10:56,340 --> 00:11:00,160 And there's one limiting case we already brought up. 169 00:11:00,160 --> 00:11:02,340 Which is when you started with the same amount of material. 170 00:11:02,340 --> 00:11:04,650 What does it look like if you have the same amount of 171 00:11:04,650 --> 00:11:06,830 material to begin with? 172 00:11:06,830 --> 00:11:11,200 So if you have A0 equal to B0, you start out with the same 173 00:11:11,200 --> 00:11:15,680 amount of stuff, but if you start out with the same amount 174 00:11:15,680 --> 00:11:18,310 of stuff, then this doesn't look so different from, at 175 00:11:18,310 --> 00:11:21,260 least mathematically, from this one right here. 176 00:11:21,260 --> 00:11:24,570 If you start out with A0 is equal to B0, and for every 177 00:11:24,570 --> 00:11:27,345 mole of A that you use up you use a mole of B, then 178 00:11:27,345 --> 00:11:30,560 throughout the whole reaction, the concentration of A and the 179 00:11:30,560 --> 00:11:33,480 concentration of B are going to be the same. 180 00:11:33,480 --> 00:11:38,010 So for the whole reaction, if you started with this, A is 181 00:11:38,010 --> 00:11:41,290 equal to B for all times. 182 00:11:41,290 --> 00:11:42,000 That makes it easy. 183 00:11:42,000 --> 00:11:45,030 Because now you can go back and instead of writing A times 184 00:11:45,030 --> 00:11:47,320 B, you can write A times A, which is A squared, which is 185 00:11:47,320 --> 00:11:47,890 what we had here. 186 00:11:47,890 --> 00:11:51,870 And then you have the whole thing solved. 187 00:11:51,870 --> 00:11:57,150 So in that case here you just have minus dA/dt is equal to k 188 00:11:57,150 --> 00:12:06,610 A squared, and one over A is equal to kt plus one over A0. 189 00:12:06,610 --> 00:12:09,430 You don't even have to do any math, you just look at it. 190 00:12:09,430 --> 00:12:10,790 So that's one case. 191 00:12:10,790 --> 00:12:14,340 Another case is if one of the reactants is in much higher 192 00:12:14,340 --> 00:12:16,380 concentration than the other reactant. 193 00:12:16,380 --> 00:12:18,180 And that's called flooding. 194 00:12:18,180 --> 00:12:21,960 You basically flood the system with one reactant. 195 00:12:21,960 --> 00:12:27,670 And that's something that we'll use again, hopefully by 196 00:12:27,670 --> 00:12:29,710 the end of class today. 197 00:12:29,710 --> 00:12:33,740 Let's say that we take, so this is another limiting case. 198 00:12:33,740 --> 00:12:35,860 Let's say we take A0 to be much bigger than B0. 199 00:12:35,860 --> 00:12:39,010 So we flood the system with A0. 200 00:12:39,010 --> 00:12:45,550 As a result, the concentration of A doesn't change 201 00:12:45,550 --> 00:12:47,470 very much in my pot. 202 00:12:47,470 --> 00:12:50,260 It's hugely concentrated in A, there's a 203 00:12:50,260 --> 00:12:53,190 little bit of B around. 204 00:12:53,190 --> 00:12:54,150 Again, with the process. 205 00:12:54,150 --> 00:12:57,600 If I use all of the B up, the difference in A is going to be 206 00:12:57,600 --> 00:12:58,850 very small. 207 00:12:58,850 --> 00:13:00,840 So at the end of the process, I'm basically still going to 208 00:13:00,840 --> 00:13:04,240 have A0 left in the pot. 209 00:13:04,240 --> 00:13:07,230 So during the whole process, during the whole time period, 210 00:13:07,230 --> 00:13:13,280 I might as well assume that A is equal to A0. 211 00:13:13,280 --> 00:13:14,650 And that makes my life much easier. 212 00:13:14,650 --> 00:13:17,850 Because now if I write my differential equation in terms 213 00:13:17,850 --> 00:13:24,950 of B instead of A, so the rate of destruction of B, k A times 214 00:13:24,950 --> 00:13:30,160 B, instead of writing A here, it's pretty much constant for 215 00:13:30,160 --> 00:13:30,730 the whole time. 216 00:13:30,730 --> 00:13:34,310 I'm just going to write A0. 217 00:13:34,310 --> 00:13:37,650 So now, if k times A0 is a constant, and this looks 218 00:13:37,650 --> 00:13:42,160 awfully like a first order reaction. 219 00:13:42,160 --> 00:13:44,430 So I can solve for it. 220 00:13:44,430 --> 00:13:48,800 And I get that, so I can just write the answer because I've 221 00:13:48,800 --> 00:13:49,590 done this already. 222 00:13:49,590 --> 00:13:50,940 I don't have to do it again. 223 00:13:50,940 --> 00:13:56,760 The concentration of B then goes like B0 e to the minus k 224 00:13:56,760 --> 00:13:59,080 prime t, that's the first order reaction. 225 00:13:59,080 --> 00:14:03,670 Where k prime here is this new rate constant, this new 226 00:14:03,670 --> 00:14:10,860 number, which is k, times A0. 227 00:14:10,860 --> 00:14:14,260 So that's easy to solve also. 228 00:14:14,260 --> 00:14:15,910 So always go to the limiting cases, because 229 00:14:15,910 --> 00:14:18,150 they tend to easy. 230 00:14:18,150 --> 00:14:25,120 And if you were to go to the full solution and put in this 231 00:14:25,120 --> 00:14:29,220 limiting case, then you'd find that you get the right answer 232 00:14:29,220 --> 00:14:30,840 this way as well. 233 00:14:30,840 --> 00:14:32,890 You can directly go to the easy way of doing it, or you 234 00:14:32,890 --> 00:14:35,170 can go through the whole process of solving it and 235 00:14:35,170 --> 00:14:39,140 putting the approximation in there. 236 00:14:39,140 --> 00:14:41,530 And do the cancellations and get this. 237 00:14:41,530 --> 00:14:43,710 But this is much easier. 238 00:14:43,710 --> 00:14:46,590 Just writing the answer down is always much easier. 239 00:14:46,590 --> 00:14:48,540 So it's a pseudo first order reaction. 240 00:14:48,540 --> 00:14:53,320 We call this a pseudo first order reaction. 241 00:14:53,320 --> 00:14:56,460 So we're done with the simple stuff now. 242 00:14:56,460 --> 00:15:04,960 Any questions on first order and second order reactions? 243 00:15:04,960 --> 00:15:07,990 So the next step is, you've got a reaction. 244 00:15:07,990 --> 00:15:11,130 It could be a gas phase reaction, it could be a 245 00:15:11,130 --> 00:15:13,080 solution phase reaction. 246 00:15:13,080 --> 00:15:17,000 There's some quantity, some property, of the solution 247 00:15:17,000 --> 00:15:17,760 that's going to change. 248 00:15:17,760 --> 00:15:20,000 That's going to allow you to follow it as 249 00:15:20,000 --> 00:15:21,420 a function of time. 250 00:15:21,420 --> 00:15:22,920 And that property could be many things. 251 00:15:22,920 --> 00:15:24,730 It could be spectroscopic. 252 00:15:24,730 --> 00:15:27,770 It could be that there's an absorption in the visible that 253 00:15:27,770 --> 00:15:30,980 changes as the concentration of one of 254 00:15:30,980 --> 00:15:33,760 your reactants changes. 255 00:15:33,760 --> 00:15:37,210 Or one of the products could have an absorption band that 256 00:15:37,210 --> 00:15:38,600 you could follow in time. 257 00:15:38,600 --> 00:15:39,640 Or you could using infrared spectroscopy to 258 00:15:39,640 --> 00:15:41,090 follow it in time. 259 00:15:41,090 --> 00:15:44,360 Or if you have a reaction in the gas phase, and you have 260 00:15:44,360 --> 00:15:46,730 more products or less products than the reactants, and the 261 00:15:46,730 --> 00:15:48,224 pressure is going to change in time if you 262 00:15:48,224 --> 00:15:49,730 have a finite volume. 263 00:15:49,730 --> 00:15:52,840 So there's usually some quantity that you can use to 264 00:15:52,840 --> 00:15:59,150 follow the reaction in time to extract out data. 265 00:15:59,150 --> 00:16:03,140 And then from that data, that you want to know what are the 266 00:16:03,140 --> 00:16:04,790 kinetics of this reaction. 267 00:16:04,790 --> 00:16:05,790 Because eventually you're going to 268 00:16:05,790 --> 00:16:07,770 try to find a mechanism. 269 00:16:07,770 --> 00:16:10,440 You're going to try to find a mechanism that's consistent 270 00:16:10,440 --> 00:16:12,520 with the data. 271 00:16:12,520 --> 00:16:13,780 So you get data. 272 00:16:13,780 --> 00:16:18,980 Then you want extract out of the rate constants and orders. 273 00:16:18,980 --> 00:16:21,790 So let's assume that you've found a way to get data. 274 00:16:21,790 --> 00:16:24,270 And now you've got to analyze your data. 275 00:16:24,270 --> 00:16:30,980 And suppose that you've found a way to measure the reactant 276 00:16:30,980 --> 00:16:33,150 concentration as a function of time. 277 00:16:33,150 --> 00:16:38,210 And let's say that in the first case, the simplest case 278 00:16:38,210 --> 00:16:43,100 is that you have one reactant. 279 00:16:43,100 --> 00:16:50,140 So you have one reactant, A. So A goes to products. 280 00:16:50,140 --> 00:16:53,540 And you've managed to extract A as a function of time. 281 00:16:53,540 --> 00:16:57,670 Well, the obvious thing to do is to plot A versus time and 282 00:16:57,670 --> 00:16:59,670 see what it fits like. 283 00:16:59,670 --> 00:17:03,520 So you take A versus time, and you plot it. 284 00:17:03,520 --> 00:17:07,620 And you know that if you plot log A versus time and it's a 285 00:17:07,620 --> 00:17:10,515 straight line, well, that's going to be 286 00:17:10,515 --> 00:17:12,730 a first order process. 287 00:17:12,730 --> 00:17:16,320 Plot log A versus time in a straight line, you know that's 288 00:17:16,320 --> 00:17:18,290 going to be first order. 289 00:17:18,290 --> 00:17:20,040 If it doesn't go to a straight line you know 290 00:17:20,040 --> 00:17:21,100 it's not first order. 291 00:17:21,100 --> 00:17:26,810 So then you go ahead and plot one over A versus time. 292 00:17:26,810 --> 00:17:32,630 And if it's a straight line, you know it's second order. 293 00:17:32,630 --> 00:17:35,230 And if it's not second order, it's not a straight line. 294 00:17:35,230 --> 00:17:38,140 It's not a straight line, you look for some other order. 295 00:17:38,140 --> 00:17:39,590 So that's one way to do it. 296 00:17:39,590 --> 00:17:44,060 And you've got to have enough points on your graph, because 297 00:17:44,060 --> 00:17:48,780 if I were to plot a, let's say this is my data point, if I 298 00:17:48,780 --> 00:17:52,700 have a second order process, at the beginning it's going to 299 00:17:52,700 --> 00:17:54,570 look an awful lot like a first order process. 300 00:17:54,570 --> 00:17:56,460 It's not until after a while that it's 301 00:17:56,460 --> 00:17:59,130 going to start to deviate. 302 00:17:59,130 --> 00:18:03,460 So you've got to have enough points down in time to make 303 00:18:03,460 --> 00:18:06,810 sure that you can differentiate between a 304 00:18:06,810 --> 00:18:10,300 straight line and a line that's not straight. 305 00:18:10,300 --> 00:18:15,170 And usually, that's often a mistake that 306 00:18:15,170 --> 00:18:16,330 experimentalists make. 307 00:18:16,330 --> 00:18:17,500 They look at the beginning. 308 00:18:17,500 --> 00:18:21,050 They say, oh, it's a straight line, work is done. 309 00:18:21,050 --> 00:18:22,730 Go home. 310 00:18:22,730 --> 00:18:30,930 But usually you need to have a good amount of the reactant 311 00:18:30,930 --> 00:18:33,180 consumed before you can tell the difference between first 312 00:18:33,180 --> 00:18:37,700 and second order. 313 00:18:37,700 --> 00:18:41,000 So you'll have an opportunity to do this on the homework. 314 00:18:41,000 --> 00:18:46,470 This kind of exercise of extracting the order of a 315 00:18:46,470 --> 00:18:48,340 simple reaction. 316 00:18:48,340 --> 00:18:51,520 Another way to do it if you have a simple reaction is to 317 00:18:51,520 --> 00:18:53,340 look at half-lives. 318 00:18:53,340 --> 00:19:00,230 That would be the half-life method. 319 00:19:00,230 --> 00:19:02,700 If you can measure A as a function of time, then you 320 00:19:02,700 --> 00:19:07,060 know when you've gotten A over two. 321 00:19:07,060 --> 00:19:13,500 So, you know that if I look at the half-life versus the 322 00:19:13,500 --> 00:19:16,870 concentration, the initial concentration, of my reactant, 323 00:19:16,870 --> 00:19:18,370 that tells me something about the order. 324 00:19:18,370 --> 00:19:22,720 Because we saw that for our first order, t 1/2 was 325 00:19:22,720 --> 00:19:30,530 independent of the initial concentration. 326 00:19:30,530 --> 00:19:34,640 And that for a second order, t 1/2 was proportional to one 327 00:19:34,640 --> 00:19:39,620 over the inverse of the initial concentration. 328 00:19:39,620 --> 00:19:44,580 So if you plot t 1/2 versus A0, or have a few A0 versus t 329 00:19:44,580 --> 00:19:47,640 1/2s, then you can tell the difference between first order 330 00:19:47,640 --> 00:19:51,770 and a second order reaction, and see which one fits. 331 00:19:51,770 --> 00:19:59,710 Sometimes to get even more solid numbers, because from 332 00:19:59,710 --> 00:20:01,450 here you can also extract k. 333 00:20:01,450 --> 00:20:04,670 If you have a bunch of points, of t 1/2 versus A0, you can 334 00:20:04,670 --> 00:20:06,470 extract k, the rate constant. 335 00:20:06,470 --> 00:20:08,040 Not just the order, but also the rate 336 00:20:08,040 --> 00:20:10,040 constant out of this data. 337 00:20:10,040 --> 00:20:14,650 You can use multiple lifetimes if you have enough data. 338 00:20:14,650 --> 00:20:15,910 Multiple lifetimes. 339 00:20:15,910 --> 00:20:21,540 So you can define, you can define a t 3/4. 340 00:20:21,540 --> 00:20:26,470 Which is the amount of time it takes for the concentration of 341 00:20:26,470 --> 00:20:30,800 A to be 1/4 of what you started out with. 342 00:20:30,800 --> 00:20:34,400 So 3/4 is gone. 343 00:20:34,400 --> 00:20:39,010 And then you can put that into your first order rate law. 344 00:20:39,010 --> 00:20:46,530 So when you have log of A over A0 is equal to minus kt, you 345 00:20:46,530 --> 00:20:51,765 get that t 3/4, we can solve for t 3/4, and you get that 346 00:20:51,765 --> 00:20:57,670 that's equal to two log two, over k. 347 00:20:57,670 --> 00:20:58,970 And then you can do the same thing for a 348 00:20:58,970 --> 00:21:01,440 second order process. 349 00:21:01,440 --> 00:21:05,390 So this is first order. 350 00:21:05,390 --> 00:21:10,190 You plug in t 3/4 and A is equal to 1/4 A0, and you solve 351 00:21:10,190 --> 00:21:16,240 for t 3/4 for a second order process. 352 00:21:16,240 --> 00:21:21,880 And you get that this is equal to three over A0 times k. 353 00:21:21,880 --> 00:21:27,030 So it's the same functional form as these two. 354 00:21:27,030 --> 00:21:28,960 But there's a pre-factor that's different here. 355 00:21:28,960 --> 00:21:31,420 So here there's a two that comes in there. 356 00:21:31,420 --> 00:21:34,500 And here there's there's a three that comes in there. 357 00:21:34,500 --> 00:21:37,910 And so there's an obvious way to tell, then, if you have the 358 00:21:37,910 --> 00:21:39,630 t 1/2 and the t 3/4 signs. 359 00:21:39,630 --> 00:21:43,640 Basically, you follow, instead of having many reactions, 360 00:21:43,640 --> 00:21:46,130 instead of having many reactions to do, with 361 00:21:46,130 --> 00:21:49,050 different A0's here you can do one reaction. 362 00:21:49,050 --> 00:21:52,450 If you do one reaction and you watch the reactant go away, 363 00:21:52,450 --> 00:21:53,670 and you time it. 364 00:21:53,670 --> 00:21:57,430 When 1/2 of it is gone, that's one time, then you keep going, 365 00:21:57,430 --> 00:22:00,060 like 3/4 is gone, that's another time. 366 00:22:00,060 --> 00:22:03,210 Then you can take the ratio of those two times. 367 00:22:03,210 --> 00:22:10,800 Of t 3/4 versus t 1/2. t 3/4 versus t 1/2. 368 00:22:10,800 --> 00:22:13,050 And if it's a first order process, the ratio 369 00:22:13,050 --> 00:22:19,090 here is just two. 370 00:22:19,090 --> 00:22:24,840 And if you take t 3/4 over t 1/2 and it's a second order 371 00:22:24,840 --> 00:22:32,990 process, the ratio is three. 372 00:22:32,990 --> 00:22:38,010 So with one experiment, then, you can extract out the order. 373 00:22:38,010 --> 00:22:39,990 You can't extract out, well, you can extract 374 00:22:39,990 --> 00:22:41,630 out the rate constant. 375 00:22:41,630 --> 00:22:44,150 If you know the order then you know which equation fits, and 376 00:22:44,150 --> 00:22:46,230 you can extract out the the rate constant, 377 00:22:46,230 --> 00:22:46,980 with a big error bar. 378 00:22:46,980 --> 00:22:50,590 You're always better off doing many multiple lifetimes of 379 00:22:50,590 --> 00:22:55,540 different A0's or many of these just to get more 380 00:22:55,540 --> 00:22:58,470 statistics in the result. 381 00:22:58,470 --> 00:23:00,780 So this is the simple process. 382 00:23:00,780 --> 00:23:02,640 And you always try, if you have something complicated, 383 00:23:02,640 --> 00:23:05,920 you always try to bring it back to a one component 384 00:23:05,920 --> 00:23:08,580 process by doing something like flooding. 385 00:23:08,580 --> 00:23:14,470 So if you have five different reactants, if you make four of 386 00:23:14,470 --> 00:23:19,290 them in very large quantities and keep one of them in very 387 00:23:19,290 --> 00:23:22,170 small quantities, then all the four that are in large 388 00:23:22,170 --> 00:23:25,140 quantities are basically constant over the process. 389 00:23:25,140 --> 00:23:28,410 And you basically are looking at only one 390 00:23:28,410 --> 00:23:31,090 reactant going away. 391 00:23:31,090 --> 00:23:34,270 And then you can use these methods to figure out what the 392 00:23:34,270 --> 00:23:39,070 order is for that one reactant. 393 00:23:39,070 --> 00:23:45,890 So, questions about simple one reactant sort of processes. 394 00:23:45,890 --> 00:23:48,250 It's pretty straightforward. 395 00:23:48,250 --> 00:24:02,480 So now, let's say we have more complicated reactions. 396 00:24:02,480 --> 00:24:05,210 There are two ways that we can deal with that. 397 00:24:05,210 --> 00:24:07,360 The first, I already mentioned, which is the 398 00:24:07,360 --> 00:24:09,850 flooding which we'll get back to. 399 00:24:09,850 --> 00:24:13,880 And another way is called, so we have 400 00:24:13,880 --> 00:24:18,110 some complicated reactions. 401 00:24:18,110 --> 00:24:20,400 Complex reactions, with multiple reactants, that's A 402 00:24:20,400 --> 00:24:24,100 plus B plus C goes to products. 403 00:24:24,100 --> 00:24:24,600 And there could be some 404 00:24:24,600 --> 00:24:28,270 stoichiometry in front of there. 405 00:24:28,270 --> 00:24:29,970 So one of the ways to deal with that is called the 406 00:24:29,970 --> 00:24:33,160 initial rate method. 407 00:24:33,160 --> 00:24:38,630 We want to find out orders and rate constants. 408 00:24:38,630 --> 00:24:46,630 Initial rate method. 409 00:24:46,630 --> 00:24:53,270 So if I look at minus dA/dt, one of the reactants or the 410 00:24:53,270 --> 00:24:57,300 rate of the reaction near time t equals zero. 411 00:24:57,300 --> 00:24:59,310 That's the initial rate. 412 00:24:59,310 --> 00:25:00,820 Right as the reaction starts. 413 00:25:00,820 --> 00:25:03,450 I mix everything together and I, just as after I mix it 414 00:25:03,450 --> 00:25:06,640 together, I watch the process of A disappearing or B 415 00:25:06,640 --> 00:25:08,530 disappearing or C disappearing. 416 00:25:08,530 --> 00:25:13,260 And so in reality what I'm doing is minus delta A / delta 417 00:25:13,260 --> 00:25:18,090 t near t equals zero, where delta t is a small interval. 418 00:25:18,090 --> 00:25:20,910 So there's not much change in delta A. Experimentally 419 00:25:20,910 --> 00:25:21,460 that's what I do. 420 00:25:21,460 --> 00:25:24,410 And that's pretty much, essentially getting this 421 00:25:24,410 --> 00:25:26,550 number out. 422 00:25:26,550 --> 00:25:28,590 And we're going to call that the initial rate. 423 00:25:28,590 --> 00:25:32,990 And the initial rate is k, and at the beginning I haven't 424 00:25:32,990 --> 00:25:33,790 used up anything. 425 00:25:33,790 --> 00:25:37,200 So all the initial concentrations are there A0 to 426 00:25:37,200 --> 00:25:41,280 the alpha, B0 to the beta, C0 to the gamma. 427 00:25:41,280 --> 00:25:46,480 Et cetera if you have more reactants. 428 00:25:46,480 --> 00:25:47,310 So you measure this. 429 00:25:47,310 --> 00:25:55,320 You measure this R0, and then you repeat the same process 430 00:25:55,320 --> 00:25:59,880 with a new concentration of one of those three reactants. 431 00:25:59,880 --> 00:26:03,690 So with A0 prime, let's say. 432 00:26:03,690 --> 00:26:09,260 And then you get a new R0, R0 prime. 433 00:26:09,260 --> 00:26:14,340 Then you take the ratios of these R0 and R0 primes, R0 434 00:26:14,340 --> 00:26:16,640 divided by R0 prime. 435 00:26:16,640 --> 00:26:24,640 So R0 is k A0 to the alpha, B0 to the beta, C0 to the gamma, 436 00:26:24,640 --> 00:26:29,990 then you have k A0 prime to the alpha. 437 00:26:29,990 --> 00:26:34,210 B0 to the beta, C0 to the gamma, 438 00:26:34,210 --> 00:26:40,240 and the k's you disappear. 439 00:26:40,240 --> 00:26:42,250 The B0's disappear. 440 00:26:42,250 --> 00:26:43,150 The C0's disappear. 441 00:26:43,150 --> 00:26:45,210 The only thing that we've changed is the concentration 442 00:26:45,210 --> 00:26:46,450 of A0 to begin with. 443 00:26:46,450 --> 00:26:50,230 So the ratios of these initial rates is the ratios of the 444 00:26:50,230 --> 00:26:56,310 A0's to the alpha power, to the order in terms of A0. 445 00:26:56,310 --> 00:27:00,330 So if you're clever about your choice of ratios, then you can 446 00:27:00,330 --> 00:27:02,690 get alpha pretty easily. 447 00:27:02,690 --> 00:27:16,280 So if you choose, so if you now choose A0 prime to be 448 00:27:16,280 --> 00:27:27,770 equal to 1/2 A0, then you measure R0 over R0 prime. 449 00:27:27,770 --> 00:27:35,560 And if you get one, then you know that that alpha's zero. 450 00:27:35,560 --> 00:27:38,250 Alpha has to be zero here. 451 00:27:38,250 --> 00:27:39,610 Then you know alpha is zero, that gives you the order. 452 00:27:39,610 --> 00:27:41,100 It's a zero order reaction. 453 00:27:41,100 --> 00:27:44,740 If you get that R0 over R0 prime is square root of two, 454 00:27:44,740 --> 00:27:51,460 or rather 1/2, then a square root of two, square root of 2, 455 00:27:51,460 --> 00:27:54,950 then you know that alpha is 1/2. 456 00:27:54,950 --> 00:27:56,490 And that's a half order. 457 00:27:56,490 --> 00:28:00,060 We haven't seen any half order reactions yet, but we will. 458 00:28:00,060 --> 00:28:04,320 Those are indicative of a complicated mechanisms. 459 00:28:04,320 --> 00:28:07,940 But that's what you would get out of this experiment. 460 00:28:07,940 --> 00:28:15,230 If you get that it's equal to two, if you get that this 461 00:28:15,230 --> 00:28:16,880 ratio is equal to two, then you know that alpha is equal 462 00:28:16,880 --> 00:28:19,500 to one, et cetera. 463 00:28:19,500 --> 00:28:24,880 So it's a pretty easy way to get the order. 464 00:28:24,880 --> 00:28:26,410 Then you repeat the experiment. 465 00:28:26,410 --> 00:28:30,535 Now instead of changing A0, you keep A0 constant 466 00:28:30,535 --> 00:28:31,600 and you change B0. 467 00:28:31,600 --> 00:28:38,710 Or, you can use flooding or isolation, which is the next 468 00:28:38,710 --> 00:28:40,760 process to get the other orders. 469 00:28:40,760 --> 00:28:43,400 And eventually you get the rate constant. 470 00:28:43,400 --> 00:28:46,720 Once you have all the orders, and you have R0, then you have 471 00:28:46,720 --> 00:28:50,070 the rate constant. 472 00:28:50,070 --> 00:28:52,540 OK, so that's one way of doing it. 473 00:28:52,540 --> 00:28:55,260 And a second way of doing it is the 474 00:28:55,260 --> 00:28:57,360 way we already mentioned. 475 00:28:57,360 --> 00:29:02,500 To solve the second order reaction in two components. 476 00:29:02,500 --> 00:29:06,300 Which is to flood the reaction with 477 00:29:06,300 --> 00:29:08,110 everything except for one. 478 00:29:08,110 --> 00:29:14,430 So this is called flooding or isolation. 479 00:29:14,430 --> 00:29:20,920 Basically, you isolate one reactant and watch it. 480 00:29:20,920 --> 00:29:22,460 You're trying to get back to a system, 481 00:29:22,460 --> 00:29:23,840 which is a simple system. 482 00:29:23,840 --> 00:29:25,990 Which is a system of one reaction. 483 00:29:25,990 --> 00:29:31,090 So let's say you take A0 to be much smaller than all the 484 00:29:31,090 --> 00:29:31,910 other species. 485 00:29:31,910 --> 00:29:35,150 You flood with B and C, and you isolate A0. 486 00:29:35,150 --> 00:29:41,510 And then your rate minus dA/dt, it's going 487 00:29:41,510 --> 00:29:43,250 to be A to the alpha. 488 00:29:43,250 --> 00:29:47,840 And instead of B, well, during that process B's going to say 489 00:29:47,840 --> 00:29:49,030 pretty much constant. 490 00:29:49,030 --> 00:29:51,110 Because it's hugely concentrated in B. You can 491 00:29:51,110 --> 00:29:51,930 replace B with B0. 492 00:29:51,930 --> 00:29:54,810 You can replace C with C0. 493 00:29:54,810 --> 00:29:58,720 So now you have an effective constant, an effective rate 494 00:29:58,720 --> 00:30:06,890 constant, and then you have a process which is effectively, 495 00:30:06,890 --> 00:30:09,820 or pseudo, one reactant. 496 00:30:09,820 --> 00:30:14,380 Then you can use these methods here, you can plot A versus 497 00:30:14,380 --> 00:30:17,410 time, you can find path lines, et cetera, to gather 498 00:30:17,410 --> 00:30:19,440 the order for it. 499 00:30:19,440 --> 00:30:28,240 Then you can get alpha and k prime. 500 00:30:28,240 --> 00:30:30,100 And if you can change B0 and C0, then you 501 00:30:30,100 --> 00:30:33,220 get k out of this. 502 00:30:33,220 --> 00:30:36,340 So this is basically all fairly straightforward, just 503 00:30:36,340 --> 00:30:42,440 tedious experimentation to get all these numbers out. 504 00:30:42,440 --> 00:30:44,240 OK, any questions on this? 505 00:30:44,240 --> 00:30:46,080 You'll get experience on the homework. 506 00:30:46,080 --> 00:30:49,470 There's likely to be a question on the final where 507 00:30:49,470 --> 00:30:52,680 you're given data and asked to extract out 508 00:30:52,680 --> 00:30:59,030 orders and rate constants. 509 00:30:59,030 --> 00:31:00,730 So let's move on now. 510 00:31:00,730 --> 00:31:07,260 So, so far we've looked at first and second order 511 00:31:07,260 --> 00:31:10,480 elementary processes, and we've looked at taking data 512 00:31:10,480 --> 00:31:13,390 and extracting out rate and rate constant. 513 00:31:13,390 --> 00:31:19,370 And the next step is to build mechanisms. 514 00:31:19,370 --> 00:31:31,300 So a mechanism is when you take a complicated reaction, 515 00:31:31,300 --> 00:31:37,820 like A plus B plus C goes to D plus E. And you break it up 516 00:31:37,820 --> 00:31:39,050 into elementary steps. 517 00:31:39,050 --> 00:31:40,060 What's an elementary step? 518 00:31:40,060 --> 00:31:43,400 An elementary step is a step which 519 00:31:43,400 --> 00:31:45,440 happens in a single reaction. 520 00:31:45,440 --> 00:31:51,690 So I could hypothesize that this complicated reaction 521 00:31:51,690 --> 00:31:55,590 happens in three steps, where I need to have a molecule of A 522 00:31:55,590 --> 00:32:00,220 and a molecule of B collide with each other to first form 523 00:32:00,220 --> 00:32:03,630 an intermediate F. Then I want a molecule of F plus a 524 00:32:03,630 --> 00:32:07,850 molecule of B to collide to form intermediate G plus a 525 00:32:07,850 --> 00:32:12,250 product D. Then have the intermediate G plus reactant C 526 00:32:12,250 --> 00:32:17,390 collide together to form the product E. So this set of 527 00:32:17,390 --> 00:32:20,020 elementary steps, where at each step you have a collision 528 00:32:20,020 --> 00:32:25,390 of two or three molecules together, three is not so 529 00:32:25,390 --> 00:32:29,520 common but two is very common, those elementary steps are 530 00:32:29,520 --> 00:32:33,660 called the steps of the mechanisms. 531 00:32:33,660 --> 00:32:44,440 And these elementary steps you can define something called 532 00:32:44,440 --> 00:32:53,130 molecularity, which is the number of species that you 533 00:32:53,130 --> 00:32:56,460 need to collide with each other in one of these 534 00:32:56,460 --> 00:32:57,510 elementary steps. 535 00:32:57,510 --> 00:33:00,210 So the molecularity here would be two, you need two 536 00:33:00,210 --> 00:33:01,160 molecules to react. 537 00:33:01,160 --> 00:33:03,240 Here it's two, here it's two. 538 00:33:03,240 --> 00:33:06,200 If I have an elementary step which is a zero order in one 539 00:33:06,200 --> 00:33:09,490 reactant, then the molecularity would be one. 540 00:33:09,490 --> 00:33:12,690 Or I could have A plus A, the same molecules have to collide 541 00:33:12,690 --> 00:33:13,400 with each other. 542 00:33:13,400 --> 00:33:15,330 Molecularity would be two. 543 00:33:15,330 --> 00:33:18,800 And the molecularity and the order of the 544 00:33:18,800 --> 00:33:20,600 reaction are connected. 545 00:33:20,600 --> 00:33:23,790 So if you have something which is a molecularity of one, then 546 00:33:23,790 --> 00:33:25,730 it's going to be a first order reaction. 547 00:33:25,730 --> 00:33:29,570 One reactant is just sitting by itself, falls apart, like 548 00:33:29,570 --> 00:33:31,350 in radioactive decay. 549 00:33:31,350 --> 00:33:33,270 Molecularity of one, that's a first order of process. 550 00:33:33,270 --> 00:33:36,510 If I need to have two molecules come together, then 551 00:33:36,510 --> 00:33:37,790 it's a second order process. 552 00:33:37,790 --> 00:33:40,900 If I have to have three molecules collide at the same 553 00:33:40,900 --> 00:33:44,520 time together, molecularity of three, then it's going to 554 00:33:44,520 --> 00:33:49,650 depend on the concentration of all three at the same time. 555 00:33:49,650 --> 00:33:52,600 That's called a ternary reaction, and those are really 556 00:33:52,600 --> 00:33:54,140 quite rare. 557 00:33:54,140 --> 00:33:57,300 Termolecular reactions, you need to have your 558 00:33:57,300 --> 00:34:01,820 concentrations very, very high to statistically get an event 559 00:34:01,820 --> 00:34:07,370 happening where all three molecules collide together. 560 00:34:07,370 --> 00:34:12,750 So three body reaction is hard and anything higher than three 561 00:34:12,750 --> 00:34:15,050 body is essentially impossible. 562 00:34:15,050 --> 00:34:19,880 So that limits your choices, which is nice. 563 00:34:19,880 --> 00:34:21,940 So that's the mechanism. 564 00:34:21,940 --> 00:34:23,700 And so what we're going to do next is go through some 565 00:34:23,700 --> 00:34:24,200 mechanisms. 566 00:34:24,200 --> 00:34:28,730 Some simple mechanisms and build up the complexity. 567 00:34:28,730 --> 00:34:33,910 Any questions about mechanisms here? 568 00:34:33,910 --> 00:34:37,430 What we're doing here is, we're formulating a framework. 569 00:34:37,430 --> 00:34:39,715 Where we can go back and look at things that are more 570 00:34:39,715 --> 00:34:42,530 complicated, like chain reactions or explosions or 571 00:34:42,530 --> 00:34:47,680 enzymatic reactions, and know when to apply approximations 572 00:34:47,680 --> 00:34:49,530 and et cetera. 573 00:34:49,530 --> 00:34:52,750 So we basically, here, are just laying 574 00:34:52,750 --> 00:34:58,210 down the ground rules. 575 00:34:58,210 --> 00:35:03,820 So let's go to our first example of a mechanism, a more 576 00:35:03,820 --> 00:35:05,320 complicated reaction. 577 00:35:05,320 --> 00:35:08,050 And what we're going to do is we're going to extract out 578 00:35:08,050 --> 00:35:12,340 integrated rate laws out of all these mechanisms. 579 00:35:12,340 --> 00:35:15,710 And see what it looks like, as a function of time. 580 00:35:15,710 --> 00:35:20,610 So the first one we're going to do is 581 00:35:20,610 --> 00:35:22,340 called parallel reactions. 582 00:35:22,340 --> 00:35:23,280 Simple mechanism. 583 00:35:23,280 --> 00:35:31,030 In this case here I have one reactant, and that reactant 584 00:35:31,030 --> 00:35:34,040 has a choice. 585 00:35:34,040 --> 00:35:37,030 You can think of it as a radioactive decay, an atom 586 00:35:37,030 --> 00:35:41,050 decaying in two different channels. 587 00:35:41,050 --> 00:35:44,670 So it can decay into B, or it can decay into C. There are 588 00:35:44,670 --> 00:35:50,470 two rate constants, k1 and k2. 589 00:35:50,470 --> 00:35:54,360 So you can write it like this, or you can write it as A goes 590 00:35:54,360 --> 00:35:58,740 to B plus C. This is how you would write a reaction. 591 00:35:58,740 --> 00:36:01,290 And this is how you would write your mechanism. 592 00:36:01,290 --> 00:36:10,090 A goes to B and A goes to C. Each elementary step, these 593 00:36:10,090 --> 00:36:12,550 are the elementary steps and this is the complex reaction, 594 00:36:12,550 --> 00:36:15,440 each elementary step is unimolecular. 595 00:36:15,440 --> 00:36:18,480 It's a first order process. 596 00:36:18,480 --> 00:36:20,890 So in all of these examples, the first thing you do is you 597 00:36:20,890 --> 00:36:24,330 write your rate law. 598 00:36:24,330 --> 00:36:28,190 The rate at which A gets created or destroyed. 599 00:36:28,190 --> 00:36:31,240 And there are two paths. 600 00:36:31,240 --> 00:36:34,310 It gets destroyed and into B, with a rate which is 601 00:36:34,310 --> 00:36:37,540 proportional to the concentration of A, and it 602 00:36:37,540 --> 00:36:43,040 gets destroyed into C, proportional to the 603 00:36:43,040 --> 00:36:49,670 concentration of A. So you write down all the ways that A 604 00:36:49,670 --> 00:36:51,100 can get destroyed. 605 00:36:51,100 --> 00:36:52,160 There are two ways here. 606 00:36:52,160 --> 00:36:54,660 Two channels. 607 00:36:54,660 --> 00:36:59,270 This one happens to be fairly easy to solve. k1 plus k2 608 00:36:59,270 --> 00:37:00,970 times A, and you've seen this before. 609 00:37:00,970 --> 00:37:03,590 It's minus dA/dt as a constant times A. That's 610 00:37:03,590 --> 00:37:04,720 a first order process. 611 00:37:04,720 --> 00:37:06,530 So you can just write down the answer. 612 00:37:06,530 --> 00:37:08,420 You don't need to do any math here. 613 00:37:08,420 --> 00:37:11,680 You recognize that we just call this one k prime. 614 00:37:11,680 --> 00:37:15,010 And that the rate is A as a function of time. 615 00:37:15,010 --> 00:37:22,185 This is A0 e to the minus k1 plus k2 times t. 616 00:37:22,185 --> 00:37:24,520 And everything you've learned about plotting first order 617 00:37:24,520 --> 00:37:28,370 processes et cetera, is applicable here, where the 618 00:37:28,370 --> 00:37:32,920 rate constant is the sum of these two. 619 00:37:32,920 --> 00:37:34,580 So that's for the reactant. 620 00:37:34,580 --> 00:37:36,730 The products are also interesting to plot as a 621 00:37:36,730 --> 00:37:40,260 function of time, to see how they are related to each other 622 00:37:40,260 --> 00:37:41,650 in terms of their concentrations. 623 00:37:41,650 --> 00:37:52,400 So let me go through this also. 624 00:37:52,400 --> 00:37:54,420 When things get more complicated we'll quickly go 625 00:37:54,420 --> 00:37:56,110 and make approximations. 626 00:37:56,110 --> 00:37:59,380 But for now, we can still do everything exactly. 627 00:37:59,380 --> 00:38:04,960 So you write down your rate law for the product. dB/dt is 628 00:38:04,960 --> 00:38:14,270 equal to k1 A. dC/dt is equal to k2 times A. The formation 629 00:38:14,270 --> 00:38:17,420 of B depends linearly on A. The formation of C depends 630 00:38:17,420 --> 00:38:21,050 linearly on A, because they're both first order processes. 631 00:38:21,050 --> 00:38:25,800 To make B and C. And you integrate. 632 00:38:25,800 --> 00:38:31,180 You integrate here from zero to B, dB. 633 00:38:31,180 --> 00:38:38,430 Is equal from zero to t, A dt, k1. 634 00:38:38,430 --> 00:38:39,650 A is a function of time. 635 00:38:39,650 --> 00:38:41,570 And we've already solved for that. 636 00:38:41,570 --> 00:38:43,700 It's this exponential up there. 637 00:38:43,700 --> 00:38:51,370 So you plug in here A of time. 638 00:38:51,370 --> 00:38:53,430 And you turn the crank and you integrate, and it's an 639 00:38:53,430 --> 00:38:53,990 exponential. 640 00:38:53,990 --> 00:38:56,020 So it's not so hard to integrate. 641 00:38:56,020 --> 00:39:00,280 And you get that B is a function of time is k1 times 642 00:39:00,280 --> 00:39:10,480 A0 over k1 plus k2 times one minus e to the minus k1 plus 643 00:39:10,480 --> 00:39:13,680 k2 times the time. 644 00:39:13,680 --> 00:39:15,730 Things are already starting to get a little bit more 645 00:39:15,730 --> 00:39:18,190 messy in the math. 646 00:39:18,190 --> 00:39:19,700 And then to get C, you actually 647 00:39:19,700 --> 00:39:20,470 don't need to do anything. 648 00:39:20,470 --> 00:39:23,280 Because you notice that the only difference between B and 649 00:39:23,280 --> 00:39:26,670 C here is replacing k2 with k1. 650 00:39:26,670 --> 00:39:28,360 So don't worry about doing any math. 651 00:39:28,360 --> 00:39:34,100 Just write down the answer. k2 A0, you interchange k1 and k2 652 00:39:34,100 --> 00:39:36,890 at every step. 653 00:39:36,890 --> 00:39:42,000 One minus e to the minus k1 plus k2 times the time. 654 00:39:42,000 --> 00:39:45,480 The only difference is up here in the k2 term. 655 00:39:45,480 --> 00:39:48,090 And those are your equations for k1 and k2. 656 00:39:48,090 --> 00:39:53,610 And what you find, is the ratio of B to C is a constant. 657 00:39:53,610 --> 00:39:56,980 If I divide B by C, everything cancels out except for the k1 658 00:39:56,980 --> 00:39:58,690 and the k2 here. 659 00:39:58,690 --> 00:40:02,960 Is equal to k1 over k2. 660 00:40:02,960 --> 00:40:11,280 And that is called the branching ratio. 661 00:40:11,280 --> 00:40:13,390 The branching ratio, because there are two branches out of 662 00:40:13,390 --> 00:40:15,210 the reactions. 663 00:40:15,210 --> 00:40:18,900 And this gives you the ratio of which one is more likely to 664 00:40:18,900 --> 00:40:20,030 happen than the other one. 665 00:40:20,030 --> 00:40:22,510 And so if k1 is much bigger than k2, the 666 00:40:22,510 --> 00:40:25,400 rate is per unit time. 667 00:40:25,400 --> 00:40:30,420 The units of k1 are per second or per minute or per hour. 668 00:40:30,420 --> 00:40:33,730 So if this is big, if k1 is big, then mostly you're going 669 00:40:33,730 --> 00:40:38,830 from A to B, and only a little bit of C is formed. 670 00:40:38,830 --> 00:40:40,830 And the ratio of B and C is always constant. 671 00:40:40,830 --> 00:40:47,500 And so you can plot, then, you can plot the result. 672 00:40:47,500 --> 00:40:57,900 You can sketch out the result. 673 00:40:57,900 --> 00:41:03,520 So you know that A is going to come down exponentially. 674 00:41:03,520 --> 00:41:06,250 Time on this axis here, concentrations 675 00:41:06,250 --> 00:41:10,010 on this axis here. 676 00:41:10,010 --> 00:41:13,080 So this is A as a function of time. 677 00:41:13,080 --> 00:41:15,510 That's that equation up here, in exponential decay. 678 00:41:15,510 --> 00:41:21,490 And the quantity of A. And B and C are going to come up in 679 00:41:21,490 --> 00:41:27,120 time, also, with this exponential format here. 680 00:41:27,120 --> 00:41:30,370 B is going to saturate at this ratio right here, k1 A0 681 00:41:30,370 --> 00:41:31,650 divided by k1 plus k2. 682 00:41:31,650 --> 00:41:35,560 C is going to saturate at this quantity here. 683 00:41:35,560 --> 00:41:38,070 So they're going to start both at zero. 684 00:41:38,070 --> 00:41:52,410 And B is eventually going to go to k1 A0 over k1, plus k2. 685 00:41:52,410 --> 00:42:08,550 And C, eventually, will go to k2 A0 over k1 plus k2. 686 00:42:08,550 --> 00:42:14,140 And this, and the ratio of these two lines at every point 687 00:42:14,140 --> 00:42:22,760 is k1 over k2. 688 00:42:22,760 --> 00:42:25,960 So, a simple question, for instance, that you might be of 689 00:42:25,960 --> 00:42:32,520 the type that you might be asked to look at is, suppose 690 00:42:32,520 --> 00:42:39,220 that k1 is 1/10 of k2. 691 00:42:39,220 --> 00:42:41,630 Which would you expect? 692 00:42:41,630 --> 00:42:48,340 Would you expect to have, let's see. 693 00:42:48,340 --> 00:42:51,610 So C is in green here. 694 00:42:51,610 --> 00:43:05,810 C, B. Or do you expect, so A comes down. 695 00:43:05,810 --> 00:43:10,030 B comes up. 696 00:43:10,030 --> 00:43:17,280 C comes up like this, or do you expect the last choice, 697 00:43:17,280 --> 00:43:39,380 I'm going to put the last choice here, so this is, let's 698 00:43:39,380 --> 00:43:41,710 call this choice number one. 699 00:43:41,710 --> 00:43:44,450 Choice number two. 700 00:43:44,450 --> 00:43:49,240 Choice number three. 701 00:43:49,240 --> 00:43:54,670 So k1, the rate k1 is 1/10 of the rate k2. 702 00:43:54,670 --> 00:43:57,560 That tells you something about the branching ratio. 703 00:43:57,560 --> 00:44:00,960 So do you expect this one here to be the right one? 704 00:44:00,960 --> 00:44:05,880 How many people think this is the right one? 705 00:44:05,880 --> 00:44:06,870 What about this one here? 706 00:44:06,870 --> 00:44:09,350 How many people think this is the right one? 707 00:44:09,350 --> 00:44:11,130 One person. 708 00:44:11,130 --> 00:44:13,500 What about this one here? 709 00:44:13,500 --> 00:44:15,760 So the branching ratio is the ratio of the two. 710 00:44:15,760 --> 00:44:16,750 It's 1/10. 711 00:44:16,750 --> 00:44:20,670 So this is approximately, in my sketch, poor sketch, 712 00:44:20,670 --> 00:44:22,350 granted, but this is approximately 10 713 00:44:22,350 --> 00:44:24,840 times bigger than this. 714 00:44:24,840 --> 00:44:26,550 So that's the ratio that you'd expect. 715 00:44:26,550 --> 00:44:31,490 And it's the right, here k1 is the rate into B. It's slower 716 00:44:31,490 --> 00:44:38,420 than the rate into C. So, you got it right. 717 00:44:38,420 --> 00:44:41,120 So for more complicated, we have a more complicated 718 00:44:41,120 --> 00:44:43,740 process, we're going to ask you the same sort of stuff. 719 00:44:43,740 --> 00:44:48,070 And it won't be as straightforward. 720 00:44:48,070 --> 00:44:52,140 Any questions on this beginning here? 721 00:44:52,140 --> 00:44:59,370 Next time we're going to finish with the parallel first 722 00:44:59,370 --> 00:45:01,120 and second order processes. 723 00:45:01,120 --> 00:45:05,280 And hopefully we'll get done with the complex reactions and 724 00:45:05,280 --> 00:45:10,740 mechanisms and move on to, I forgot 725 00:45:10,740 --> 00:45:11,740 what's next on the list. 726 00:45:11,740 --> 00:45:17,150 But some explosions or chain reactions.