1 00:00:00,090 --> 00:00:02,430 The following content is provided under a Creative 2 00:00:02,430 --> 00:00:03,820 Commons license. 3 00:00:03,820 --> 00:00:06,060 Your support will help MIT OpenCourseWare 4 00:00:06,060 --> 00:00:10,150 continue to offer high quality educational resources for free. 5 00:00:10,150 --> 00:00:12,690 To make a donation or to view additional materials 6 00:00:12,690 --> 00:00:16,435 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:16,435 --> 00:00:17,060 at ocw.mit.edu. 8 00:00:34,582 --> 00:00:35,790 CATHERINE DRENNAN: All right. 9 00:00:35,790 --> 00:00:37,280 So 10 more seconds. 10 00:01:02,230 --> 00:01:02,730 OK. 11 00:01:06,060 --> 00:01:07,275 Let's quiet down. 12 00:01:10,080 --> 00:01:14,800 So that is the-- you got the 70% with the right answer. 13 00:01:14,800 --> 00:01:18,676 If people can just yell out what was wrong with number one. 14 00:01:18,676 --> 00:01:19,490 AUDIENCE: Sig figs. 15 00:01:19,490 --> 00:01:20,940 CATHERINE DRENNAN: Sig figs. 16 00:01:20,940 --> 00:01:23,400 And what about numbers three and four? 17 00:01:23,400 --> 00:01:25,502 What equation was that using? 18 00:01:25,502 --> 00:01:26,460 AUDIENCE: Second order. 19 00:01:26,460 --> 00:01:27,190 CATHERINE DRENNAN: Second order. 20 00:01:27,190 --> 00:01:27,930 That's right. 21 00:01:27,930 --> 00:01:30,390 So this is a good clicker question 22 00:01:30,390 --> 00:01:34,260 for a problem that's going to be coming up on exam four 23 00:01:34,260 --> 00:01:37,890 and also on the final exam to a larger degree. 24 00:01:37,890 --> 00:01:40,560 On the final exam we have equation sheets 25 00:01:40,560 --> 00:01:44,530 that have all the equations from the whole semester, 26 00:01:44,530 --> 00:01:47,940 and so you need to figure out and remember which equation 27 00:01:47,940 --> 00:01:50,280 goes with which problem. 28 00:01:50,280 --> 00:01:54,300 It doesn't say, oh, here is the expression for first order. 29 00:01:54,300 --> 00:01:56,730 Here's the equation for second order. 30 00:01:56,730 --> 00:01:58,800 You need to look and remember which 31 00:01:58,800 --> 00:02:01,290 equation goes with which thing. 32 00:02:01,290 --> 00:02:03,120 But in terms of first order, what's 33 00:02:03,120 --> 00:02:06,360 a way that you can remember what a first order 34 00:02:06,360 --> 00:02:07,260 equation would be? 35 00:02:07,260 --> 00:02:10,199 What's missing from first order? 36 00:02:10,199 --> 00:02:12,240 Yeah, the concentration in the material. 37 00:02:12,240 --> 00:02:14,700 So for first order it's independent 38 00:02:14,700 --> 00:02:18,120 of the original concentration in the material, which 39 00:02:18,120 --> 00:02:20,700 is why we can use first order equations 40 00:02:20,700 --> 00:02:22,470 for nuclear chemistry. 41 00:02:22,470 --> 00:02:26,070 Because the rate of decay of radioactive nuclei 42 00:02:26,070 --> 00:02:28,320 are independent of the nuclei around them, 43 00:02:28,320 --> 00:02:30,060 so it's a first order process. 44 00:02:30,060 --> 00:02:32,440 So when you think a little bit about these equations, 45 00:02:32,440 --> 00:02:34,950 you should be able to identify which equation 46 00:02:34,950 --> 00:02:36,670 goes with which problem. 47 00:02:36,670 --> 00:02:40,110 And so you can identify the type of first order problems also. 48 00:02:40,110 --> 00:02:42,810 And for the second order it almost always says, 49 00:02:42,810 --> 00:02:45,210 for this second order process, which 50 00:02:45,210 --> 00:02:47,880 gives you a nice hint that that's a second order process. 51 00:02:47,880 --> 00:02:50,440 And then you just have to identify the equation. 52 00:02:50,440 --> 00:02:50,940 OK. 53 00:02:50,940 --> 00:02:52,350 So today is more kinetics. 54 00:02:52,350 --> 00:02:53,840 So we're in the kinetics unit. 55 00:02:53,840 --> 00:02:56,430 We'll be in the kinetics unit for the rest of the semester. 56 00:02:56,430 --> 00:02:58,710 It's our last unit. 57 00:02:58,710 --> 00:03:01,470 So today I think is one of the most important lectures 58 00:03:01,470 --> 00:03:03,930 in terms of the kinetics unit because we're 59 00:03:03,930 --> 00:03:06,000 talking about reaction mechanisms, 60 00:03:06,000 --> 00:03:09,360 and that is really an important part of kinetics. 61 00:03:09,360 --> 00:03:12,270 So investigating reaction mechanisms. 62 00:03:12,270 --> 00:03:16,805 So if you're going to describe how a reaction takes place-- 63 00:03:16,805 --> 00:03:19,080 often reactions don't occur in one step. 64 00:03:19,080 --> 00:03:22,800 It's really uncommon for reactions to occur in one step. 65 00:03:22,800 --> 00:03:25,560 So you want to describe the different steps. 66 00:03:25,560 --> 00:03:27,570 And you describe those steps, which are also 67 00:03:27,570 --> 00:03:30,270 called elementary reactions. 68 00:03:30,270 --> 00:03:32,460 So you break down a complex reaction 69 00:03:32,460 --> 00:03:34,620 into a series of steps, and then you 70 00:03:34,620 --> 00:03:37,110 try to figure out if that mechanism, 71 00:03:37,110 --> 00:03:41,370 if those series of steps, are consistent with the data 72 00:03:41,370 --> 00:03:44,620 that you've collected on this particular reaction. 73 00:03:44,620 --> 00:03:47,850 And over here I just show you some steps 74 00:03:47,850 --> 00:03:52,950 in the natural biosynthesis of a vitamin biotin. 75 00:03:52,950 --> 00:03:56,460 Biotin is important vitamin for us. 76 00:03:56,460 --> 00:03:59,670 It's also used a lot in feedstock. 77 00:03:59,670 --> 00:04:04,560 And so people buy a lot of biotin to put in feed stock, 78 00:04:04,560 --> 00:04:07,150 and so there's a huge market of biotin. 79 00:04:07,150 --> 00:04:09,690 Now right now it's made by what is estimated 80 00:04:09,690 --> 00:04:13,510 to be a 13 step organic synthesis which produces-- 81 00:04:13,510 --> 00:04:14,640 that's a lot of steps. 82 00:04:14,640 --> 00:04:16,290 It's really expensive, and there's 83 00:04:16,290 --> 00:04:20,519 a huge amount of organic waste associated with making biotin. 84 00:04:20,519 --> 00:04:24,255 And we're talking about in the tons level of waste. 85 00:04:24,255 --> 00:04:26,130 So researchers have been trying to figure out 86 00:04:26,130 --> 00:04:29,520 the mechanism by which nature makes biotin 87 00:04:29,520 --> 00:04:32,520 because that would be a lot more environmentally friendly. 88 00:04:32,520 --> 00:04:34,460 So when you're thinking about this, 89 00:04:34,460 --> 00:04:37,800 what you want to say is, OK, if we write out a mechanism 90 00:04:37,800 --> 00:04:40,860 is it consistent with the experimental data 91 00:04:40,860 --> 00:04:43,080 and are there fast and slow steps? 92 00:04:43,080 --> 00:04:45,341 Because if there's a step that's really slow maybe 93 00:04:45,341 --> 00:04:46,590 you can do something about it. 94 00:04:46,590 --> 00:04:48,420 Maybe if you're talking about an enzyme 95 00:04:48,420 --> 00:04:52,263 you could re-engineer the enzyme so it would be a better enzyme. 96 00:04:52,263 --> 00:04:55,340 Maybe natural selection didn't particularly work. 97 00:04:55,340 --> 00:04:57,710 Maybe the cell doesn't really need as much biotin 98 00:04:57,710 --> 00:05:00,900 as we do commercially, so there was no need to make it fast. 99 00:05:00,900 --> 00:05:02,525 But now we have a need to make it fast, 100 00:05:02,525 --> 00:05:06,740 so maybe we can do some evolution of this enzyme 101 00:05:06,740 --> 00:05:08,401 and design something to be better. 102 00:05:08,401 --> 00:05:10,400 So when you're talking about reaction mechanisms 103 00:05:10,400 --> 00:05:12,620 you want to know what's fast, what's slow. 104 00:05:12,620 --> 00:05:14,960 And if you want to use that product for something, 105 00:05:14,960 --> 00:05:17,270 you want to figure out how you can change things 106 00:05:17,270 --> 00:05:19,040 to have a better mechanism-- maybe 107 00:05:19,040 --> 00:05:23,210 avoid some really slow steps so that you can do better. 108 00:05:23,210 --> 00:05:26,690 Today is also World AIDS Day. 109 00:05:26,690 --> 00:05:30,530 And understanding the mechanism of HIV protease 110 00:05:30,530 --> 00:05:33,140 was really essential in designing inhibitors 111 00:05:33,140 --> 00:05:34,320 against that enzyme. 112 00:05:34,320 --> 00:05:36,230 And if you inhibit the enzyme you 113 00:05:36,230 --> 00:05:38,250 stop the development of the disease. 114 00:05:38,250 --> 00:05:40,010 So this was a very important thing, 115 00:05:40,010 --> 00:05:42,650 and we have some pretty good molecules 116 00:05:42,650 --> 00:05:43,760 to treat AIDS right now. 117 00:05:43,760 --> 00:05:46,134 And some of the challenges have moved on to other things, 118 00:05:46,134 --> 00:05:48,920 like less good health care in parts of the world where 119 00:05:48,920 --> 00:05:50,050 this has affected. 120 00:05:50,050 --> 00:05:53,510 So I feel like AIDS has kind of taken a backseat to hearing 121 00:05:53,510 --> 00:05:55,940 about Ebola recently, but AIDS is still 122 00:05:55,940 --> 00:05:58,730 a very important problem and one that smart people 123 00:05:58,730 --> 00:06:00,230 like you could address. 124 00:06:00,230 --> 00:06:03,040 So again, understanding reaction mechanisms 125 00:06:03,040 --> 00:06:05,360 is very, very important. 126 00:06:05,360 --> 00:06:05,860 All right. 127 00:06:05,860 --> 00:06:10,190 So let's go to a simpler problem and a simpler reaction 128 00:06:10,190 --> 00:06:11,960 mechanism. 129 00:06:11,960 --> 00:06:14,210 We'll go to our friend over here where 130 00:06:14,210 --> 00:06:17,780 we have two molecules of NO reacting 131 00:06:17,780 --> 00:06:23,720 with one molecule of O2 going to two molecules of NO2. 132 00:06:23,720 --> 00:06:26,990 So someone measured some rates for this 133 00:06:26,990 --> 00:06:30,430 and came up with the following rate law 134 00:06:30,430 --> 00:06:32,690 where you have a rate constant, which is just 135 00:06:32,690 --> 00:06:35,390 called k obs for observed. 136 00:06:35,390 --> 00:06:38,450 And we'll talk about that more in a little bit. 137 00:06:38,450 --> 00:06:42,200 And they discovered it is second order with respect to NO 138 00:06:42,200 --> 00:06:45,620 and first order with respect to O2. 139 00:06:45,620 --> 00:06:47,330 So there's a couple of things right away 140 00:06:47,330 --> 00:06:49,040 that we can ask about this. 141 00:06:49,040 --> 00:06:54,200 One is, what is the overall order then of this reaction 142 00:06:54,200 --> 00:06:55,820 from this experimental data? 143 00:06:55,820 --> 00:06:56,616 What would that be? 144 00:06:56,616 --> 00:06:57,359 AUDIENCE: Three. 145 00:06:57,359 --> 00:06:58,400 CATHERINE DRENNAN: Three. 146 00:06:58,400 --> 00:06:59,990 Again, if some of you missed it. 147 00:06:59,990 --> 00:07:01,580 2 plus 1 is 3. 148 00:07:01,580 --> 00:07:04,250 This is not where you want to lose your points on the exam. 149 00:07:04,250 --> 00:07:06,680 There's going to be some tricky significant figures 150 00:07:06,680 --> 00:07:08,390 in the kinetics units. 151 00:07:08,390 --> 00:07:10,420 Save your points to lose there. 152 00:07:10,420 --> 00:07:13,810 Count 2 plus 1 and get 3. 153 00:07:13,810 --> 00:07:18,650 So is one step likely to have three different things come 154 00:07:18,650 --> 00:07:20,390 together at the same time? 155 00:07:20,390 --> 00:07:22,970 So how likely is it that all three things are 156 00:07:22,970 --> 00:07:26,720 going to merge in one step? 157 00:07:26,720 --> 00:07:29,540 No, it's not very likely, so it's no. 158 00:07:29,540 --> 00:07:32,680 But if it did work that way, what would it be called? 159 00:07:32,680 --> 00:07:34,815 What molecular reaction? 160 00:07:34,815 --> 00:07:35,656 Do you remember? 161 00:07:35,656 --> 00:07:37,104 AUDIENCE: [INAUDIBLE]. 162 00:07:37,104 --> 00:07:39,020 CATHERINE DRENNAN: So if you have three things 163 00:07:39,020 --> 00:07:39,890 it's a termolecular. 164 00:07:42,620 --> 00:07:43,930 So they're rare. 165 00:07:43,930 --> 00:07:46,480 So that's not how this works. 166 00:07:46,480 --> 00:07:48,020 All right. 167 00:07:48,020 --> 00:07:51,680 So let's look at some rate laws and try to write a rate 168 00:07:51,680 --> 00:07:54,332 law for a reaction. 169 00:07:54,332 --> 00:07:56,040 We're going to take our overall reaction, 170 00:07:56,040 --> 00:07:57,770 we're going to divide it to two steps 171 00:07:57,770 --> 00:08:00,500 and write a rate law for that and see if that's 172 00:08:00,500 --> 00:08:03,710 consistent with the experiment. 173 00:08:03,710 --> 00:08:05,210 So we finally got this up here. 174 00:08:05,210 --> 00:08:06,710 Sorry the handwriting isn't perfect. 175 00:08:06,710 --> 00:08:08,990 I got here early but then-- the first time I'm 176 00:08:08,990 --> 00:08:10,640 going to use the boards a lot today. 177 00:08:10,640 --> 00:08:11,480 Squeegee was gone. 178 00:08:11,480 --> 00:08:13,420 There's always something. 179 00:08:13,420 --> 00:08:17,270 So we're going to break this reaction down into two steps. 180 00:08:17,270 --> 00:08:20,187 So in the first step we have our two molecules of NO 181 00:08:20,187 --> 00:08:25,160 coming together to form an intermediate N2O2. 182 00:08:25,160 --> 00:08:28,220 And this is a reversible step. 183 00:08:28,220 --> 00:08:30,620 So in the second step of the reaction 184 00:08:30,620 --> 00:08:33,080 we have our O2 molecule coming in, 185 00:08:33,080 --> 00:08:35,570 reacting with our intermediate and forming 186 00:08:35,570 --> 00:08:38,049 two molecules of NO2. 187 00:08:38,049 --> 00:08:40,100 And so this is really pretty common 188 00:08:40,100 --> 00:08:42,380 that when you have a multistep reaction that you 189 00:08:42,380 --> 00:08:46,730 form an intermediate and your intermediate goes away. 190 00:08:46,730 --> 00:08:50,000 So now we can think about how we would write the rate law 191 00:08:50,000 --> 00:08:51,645 for this particular mechanism. 192 00:08:54,380 --> 00:08:56,440 So starting up here. 193 00:08:56,440 --> 00:08:58,130 This is being written as a series 194 00:08:58,130 --> 00:09:01,760 of steps, which were also called elementary reactions. 195 00:09:01,760 --> 00:09:04,850 And for an elementary reaction it occurs exactly as written. 196 00:09:04,850 --> 00:09:08,470 That's its definition of an elementary reaction or a step. 197 00:09:08,470 --> 00:09:12,560 So now we can write the rate law for the forward direction 198 00:09:12,560 --> 00:09:15,730 of this exactly as it is written. 199 00:09:15,730 --> 00:09:21,320 So that would be writing-- and these are my little K's 200 00:09:21,320 --> 00:09:22,880 for rate constants. 201 00:09:22,880 --> 00:09:29,950 So we have K1 times the concentration of NO to the 2. 202 00:09:29,950 --> 00:09:31,690 So we're writing it exactly as written. 203 00:09:31,690 --> 00:09:33,210 I said that for an overall reaction 204 00:09:33,210 --> 00:09:35,870 you can't just look at the stoichiometry. 205 00:09:35,870 --> 00:09:38,630 You have to think about experiment. 206 00:09:38,630 --> 00:09:41,450 But for an elementary reaction you can write it just 207 00:09:41,450 --> 00:09:43,420 from the stoichiometry, and so this 208 00:09:43,420 --> 00:09:46,130 is how we would write it just from the stoichiometry. 209 00:09:46,130 --> 00:09:48,920 So what would be the order of that reaction? 210 00:09:48,920 --> 00:09:51,090 Just the forward reaction? 211 00:09:51,090 --> 00:09:51,590 2. 212 00:09:54,300 --> 00:09:57,274 And that makes it a what kind of molecular reaction? 213 00:09:57,274 --> 00:09:58,190 AUDIENCE: Bimolecular. 214 00:09:58,190 --> 00:09:59,481 CATHERINE DRENNAN: Bimolecular. 215 00:10:01,930 --> 00:10:03,110 Bimolecular. 216 00:10:03,110 --> 00:10:04,050 OK. 217 00:10:04,050 --> 00:10:08,490 So let's write out the rate law for the reverse reaction now. 218 00:10:08,490 --> 00:10:10,740 And again, exactly as written. 219 00:10:10,740 --> 00:10:16,680 So we're going to have K minus 1 times the concentration 220 00:10:16,680 --> 00:10:18,660 of our intermediate N2O2. 221 00:10:21,240 --> 00:10:25,690 And that would be the rate law for the reverse reaction. 222 00:10:25,690 --> 00:10:30,320 So what would be the order for this reaction now? 223 00:10:30,320 --> 00:10:31,040 1. 224 00:10:31,040 --> 00:10:31,540 Right. 225 00:10:31,540 --> 00:10:33,390 We only have one thing in here. 226 00:10:33,390 --> 00:10:35,610 And what do you call an x molecular 227 00:10:35,610 --> 00:10:37,183 when there's one thing? 228 00:10:37,183 --> 00:10:38,170 AUDIENCE: Uni. 229 00:10:38,170 --> 00:10:39,128 CATHERINE DRENNAN: Uni. 230 00:10:41,460 --> 00:10:43,820 Unimolecular. 231 00:10:43,820 --> 00:10:45,540 For step two now you're good at this. 232 00:10:45,540 --> 00:10:46,720 Let's do a clicker question. 233 00:11:04,970 --> 00:11:05,470 All right. 234 00:11:05,470 --> 00:11:06,312 10 more seconds. 235 00:11:22,230 --> 00:11:23,670 Yup. 236 00:11:23,670 --> 00:11:26,470 So we're just going to write this exactly as written. 237 00:11:26,470 --> 00:11:28,960 It doesn't say it's a reversible reaction. 238 00:11:28,960 --> 00:11:32,780 So it's K2-- and I'll put this down here. 239 00:11:32,780 --> 00:11:38,520 K2-- our rate constant for the second step-- 240 00:11:38,520 --> 00:11:41,700 and then times the reactants. 241 00:11:41,700 --> 00:11:47,280 So we have O2 and our intermediate N2O2. 242 00:11:52,020 --> 00:11:56,660 So what would be the order of this reaction or this step? 243 00:11:56,660 --> 00:11:57,690 Two? 244 00:11:57,690 --> 00:11:59,550 We have two things in there. 245 00:11:59,550 --> 00:12:03,000 And again, we call that bimolecular. 246 00:12:03,000 --> 00:12:06,940 So we have uni-, bi-, and ter- molecular reactions for order 247 00:12:06,940 --> 00:12:09,160 of 1, 2, and 3. 248 00:12:09,160 --> 00:12:09,660 All right. 249 00:12:09,660 --> 00:12:11,550 So we've written this out. 250 00:12:15,030 --> 00:12:19,360 Now what we want to do is we want to write out 251 00:12:19,360 --> 00:12:21,950 an overall rate law for this. 252 00:12:21,950 --> 00:12:23,430 So we've written out the rate laws 253 00:12:23,430 --> 00:12:26,190 for all the individual steps. 254 00:12:26,190 --> 00:12:32,360 And we'll put that-- Yeah, I guess I can put it-- maybe 255 00:12:32,360 --> 00:12:33,787 I'll try to write it down here. 256 00:12:33,787 --> 00:12:35,370 I was going to have this all organized 257 00:12:35,370 --> 00:12:37,740 but then the boards weren't cooperating today. 258 00:12:37,740 --> 00:12:40,560 So I'm going to write the overall rate. 259 00:12:40,560 --> 00:12:41,500 Let me try it here. 260 00:12:46,650 --> 00:12:51,390 And this is for formation of N2O2. 261 00:12:51,390 --> 00:12:53,790 And we're going to put a 2 in there, 262 00:12:53,790 --> 00:12:55,740 and I'll explain that in a minute. 263 00:12:55,740 --> 00:12:59,040 Then our K2-- so we're just writing this now 264 00:12:59,040 --> 00:13:08,860 from the last step-- times O2 and our intermediate N2O2. 265 00:13:08,860 --> 00:13:15,210 So the overall rate of forming NO2 266 00:13:15,210 --> 00:13:18,240 is the 2 because 2 moles are formed. 267 00:13:18,240 --> 00:13:23,070 So as these guys disappear-- as O2 and N2O2, 268 00:13:23,070 --> 00:13:24,892 our intermediate, disappear-- and O's 269 00:13:24,892 --> 00:13:26,600 going to form twice as fast because there 270 00:13:26,600 --> 00:13:28,120 are two of them being formed. 271 00:13:28,120 --> 00:13:30,060 So we put a 2 in there. 272 00:13:30,060 --> 00:13:32,400 And then we just have a rate order or rate 273 00:13:32,400 --> 00:13:36,660 law for the last step, K2 O2 intermediate. 274 00:13:36,660 --> 00:13:42,220 So you can always write the rate for the overall reaction 275 00:13:42,220 --> 00:13:43,592 from that last step. 276 00:13:43,592 --> 00:13:46,050 Although sometimes we'll see if they're fast and slow steps 277 00:13:46,050 --> 00:13:49,570 you can also write it from the rate determining step. 278 00:13:49,570 --> 00:13:53,460 But we're not done here because we have an intermediate 279 00:13:53,460 --> 00:13:54,910 in this expression. 280 00:13:54,910 --> 00:13:59,160 And in a rate law you cannot have an intermediate in there. 281 00:13:59,160 --> 00:14:02,880 You need to solve for the rate in terms of your rate 282 00:14:02,880 --> 00:14:05,940 constants, your reactant concentration, and your product 283 00:14:05,940 --> 00:14:07,050 concentration. 284 00:14:07,050 --> 00:14:08,740 So we need to get rid of this. 285 00:14:08,740 --> 00:14:13,020 We need to solve for this. 286 00:14:13,020 --> 00:14:14,590 So how are we going to do that? 287 00:14:14,590 --> 00:14:17,350 How are we going to solve for this? 288 00:14:17,350 --> 00:14:20,350 So we want to think about now, what 289 00:14:20,350 --> 00:14:23,640 is the net rate for this formation? 290 00:14:23,640 --> 00:14:25,950 How is that intermediate being formed, 291 00:14:25,950 --> 00:14:30,130 and how is it being consumed given those steps up here? 292 00:14:30,130 --> 00:14:35,415 So it's only being formed in the forward direction 293 00:14:35,415 --> 00:14:38,160 of the first step. 294 00:14:38,160 --> 00:14:42,000 And let me just grab this. 295 00:14:42,000 --> 00:14:47,970 So the forward direction of the first step, which is K1 times 296 00:14:47,970 --> 00:14:52,330 NO to the 2. 297 00:14:52,330 --> 00:14:55,150 And so this is where our intermediate is formed. 298 00:14:58,780 --> 00:15:03,010 Our intermediate is going away in two different steps. 299 00:15:03,010 --> 00:15:07,570 So it's going away in the reverse direction of step one. 300 00:15:07,570 --> 00:15:11,410 So it's decomposing in that step. 301 00:15:11,410 --> 00:15:13,730 And so it's decomposing at the rate 302 00:15:13,730 --> 00:15:16,980 of K minus 1 times its concentration. 303 00:15:20,280 --> 00:15:24,090 The intermediate is also being consumed in the second step. 304 00:15:24,090 --> 00:15:27,210 It's reacting with oxygen and being consumed. 305 00:15:27,210 --> 00:15:36,510 So here it decays and here it is consumed. 306 00:15:36,510 --> 00:15:38,640 And it's being consumed by the reaction 307 00:15:38,640 --> 00:15:44,046 K2 times the concentration of O2 times its concentration. 308 00:15:47,670 --> 00:15:51,050 So that's the second step. 309 00:15:51,050 --> 00:15:54,230 So now we just need to take that equation 310 00:15:54,230 --> 00:15:58,850 and solve for the intermediate N2O2. 311 00:15:58,850 --> 00:16:02,690 But we don't know net rate over there. 312 00:16:02,690 --> 00:16:05,090 We have too many variables right now. 313 00:16:05,090 --> 00:16:08,600 So at this point we have to use what's known as a steady state 314 00:16:08,600 --> 00:16:10,490 approximation. 315 00:16:10,490 --> 00:16:14,885 So steady state approximation and pretty much everything-- 316 00:16:14,885 --> 00:16:16,880 I'm talking about reaction mechanisms-- 317 00:16:16,880 --> 00:16:19,560 we're going to use the steady state approximation. 318 00:16:19,560 --> 00:16:23,450 And that is that the rate of formation of intermediates 319 00:16:23,450 --> 00:16:26,810 equals the rate at which they go away. 320 00:16:26,810 --> 00:16:29,900 So the net rate is 0. 321 00:16:29,900 --> 00:16:36,230 So we can set this whole thing now equal to 0. 322 00:16:36,230 --> 00:16:38,570 So steady state approximation net rate 323 00:16:38,570 --> 00:16:44,060 is 0, or the rate at which an intermediate forms equals 324 00:16:44,060 --> 00:16:49,190 the rate at which that intermediate decays. 325 00:16:49,190 --> 00:16:52,140 So I can rearrange this equation now, 326 00:16:52,140 --> 00:16:54,410 and I'm going to bring the terms that have 327 00:16:54,410 --> 00:16:57,530 the intermediate to one side. 328 00:16:57,530 --> 00:16:59,970 So I'm going to put them over here. 329 00:16:59,970 --> 00:17:03,830 So we have the term at which it decays-- 330 00:17:03,830 --> 00:17:07,430 K minus 1 times our concentration 331 00:17:07,430 --> 00:17:09,599 of our intermediate. 332 00:17:09,599 --> 00:17:12,260 And it had a negative but I brought it over here 333 00:17:12,260 --> 00:17:15,260 to this side with a 0, so now it's positive. 334 00:17:15,260 --> 00:17:19,250 And I'm going to bring the same for the rate law at which it's 335 00:17:19,250 --> 00:17:20,869 consumed over. 336 00:17:20,869 --> 00:17:27,680 So that's K2 times our intermediate concentration 337 00:17:27,680 --> 00:17:30,770 and our oxygen concentration. 338 00:17:30,770 --> 00:17:32,570 And now on the other side we'll have 339 00:17:32,570 --> 00:17:34,730 the rate at which the intermediate is formed, 340 00:17:34,730 --> 00:17:39,080 which is K1 No to the 2. 341 00:17:39,080 --> 00:17:41,750 So this is another way to express the steady state 342 00:17:41,750 --> 00:17:43,010 approximation. 343 00:17:43,010 --> 00:17:46,070 The rate at which the intermediate goes away 344 00:17:46,070 --> 00:17:48,240 equals the rate at which it's formed. 345 00:17:48,240 --> 00:17:49,640 That's the steady state. 346 00:17:49,640 --> 00:17:52,730 There's no sort of flux in the intermediate. 347 00:17:52,730 --> 00:17:56,270 It's being formed and going away at equal rates. 348 00:17:56,270 --> 00:17:59,240 So now we can use this to solve for the intermediate. 349 00:17:59,240 --> 00:18:00,440 Now we're set. 350 00:18:00,440 --> 00:18:02,780 Now we can solve for the intermediate. 351 00:18:02,780 --> 00:18:06,060 And so let's do that over here. 352 00:18:06,060 --> 00:18:09,680 So I'm going to pull out-- I had a straight line here. 353 00:18:09,680 --> 00:18:12,150 This one's a little crooked. 354 00:18:12,150 --> 00:18:13,610 I used to write a lot on the board 355 00:18:13,610 --> 00:18:17,090 and used to evaluate professors by their good handwriting 356 00:18:17,090 --> 00:18:20,630 and my ratings were always-- my overall rating was limited 357 00:18:20,630 --> 00:18:22,527 by my handwriting to a large degree 358 00:18:22,527 --> 00:18:24,110 and so I stopped writing on the board. 359 00:18:24,110 --> 00:18:26,450 But now they've got rid of that as a criteria, 360 00:18:26,450 --> 00:18:28,790 so I can write on the board again. 361 00:18:28,790 --> 00:18:32,030 So I'm going to pull out the concentration 362 00:18:32,030 --> 00:18:37,820 of the intermediate-- our N2O2. 363 00:18:37,820 --> 00:18:41,560 and I'll pull it out of the expression leaving 364 00:18:41,560 --> 00:18:52,410 K minus 1 and K2 times O2. 365 00:18:52,410 --> 00:18:54,290 So I just pulled out the concentration 366 00:18:54,290 --> 00:18:56,690 of the intermediate over here. 367 00:18:56,690 --> 00:18:58,700 And then we leave the other side the same. 368 00:18:58,700 --> 00:19:03,950 Our K1 times our NO squared. 369 00:19:03,950 --> 00:19:07,010 So now I can solve for the concentration 370 00:19:07,010 --> 00:19:09,060 of the intermediate. 371 00:19:09,060 --> 00:19:20,780 N2O2 equals K1 times NO squared over K minus 1 372 00:19:20,780 --> 00:19:26,830 plus K2 times O2. 373 00:19:26,830 --> 00:19:27,890 Great. 374 00:19:27,890 --> 00:19:30,020 So we've solved for the concentration 375 00:19:30,020 --> 00:19:31,840 of the intermediate now. 376 00:19:31,840 --> 00:19:37,580 Now we can take this and bring it back over here 377 00:19:37,580 --> 00:19:40,430 and put that whole term into this. 378 00:19:40,430 --> 00:19:43,850 And then we'll have a rate law that is expressed only 379 00:19:43,850 --> 00:19:48,010 in terms of our rate constants and our reactants 380 00:19:48,010 --> 00:19:48,676 and/or products. 381 00:19:51,290 --> 00:19:58,520 So let's do more on the PowerPoint here. 382 00:19:58,520 --> 00:20:02,310 So this is what we just came up with. 383 00:20:02,310 --> 00:20:06,290 So the concentration of our intermediate K1 times 384 00:20:06,290 --> 00:20:11,060 NO-- one of our reactants second order-- over K minus 1 385 00:20:11,060 --> 00:20:15,500 plus K2 times the concentration of oxygen. 386 00:20:15,500 --> 00:20:21,620 Now I'm going to plug that into this expression, which we just 387 00:20:21,620 --> 00:20:26,720 got from writing the rate law for the last step using a 2 388 00:20:26,720 --> 00:20:29,850 because we have two molecules of product being formed. 389 00:20:29,850 --> 00:20:32,880 So we're going to plug it in there, 390 00:20:32,880 --> 00:20:34,830 and that's going to give us this. 391 00:20:34,830 --> 00:20:39,840 So we have our 2, we have a K1 from here, a K2 from here, 392 00:20:39,840 --> 00:20:42,600 we have our oxygen concentration, 393 00:20:42,600 --> 00:20:46,980 we have our NO overconcentration second order over K minus 1 394 00:20:46,980 --> 00:20:52,220 plus K2 times concentration of oxygen. 395 00:20:52,220 --> 00:20:55,920 So that might be the complete answer to some problems 396 00:20:55,920 --> 00:20:59,100 but here we were given an experimental rate 397 00:20:59,100 --> 00:21:03,900 law, which was second order in NO and first order in O2. 398 00:21:03,900 --> 00:21:06,720 That does not match this. 399 00:21:06,720 --> 00:21:08,850 Term has an O2 at the bottom. 400 00:21:08,850 --> 00:21:11,250 This one doesn't have an O2 at the bottom. 401 00:21:11,250 --> 00:21:13,050 These are not the same. 402 00:21:13,050 --> 00:21:15,960 So that must mean that the mechanism has 403 00:21:15,960 --> 00:21:19,010 fast and slow steps and that we need 404 00:21:19,010 --> 00:21:25,560 to reconsider this expression in terms of fast and slow steps. 405 00:21:25,560 --> 00:21:30,220 So let's go back now and think about our mechanism. 406 00:21:30,220 --> 00:21:33,810 And so if we come over here-- let's 407 00:21:33,810 --> 00:21:42,570 say that the first step-- let's bring this down again. 408 00:21:42,570 --> 00:21:46,250 Let's say the first step is fast. 409 00:21:46,250 --> 00:21:48,470 And we already said that it was reversible, 410 00:21:48,470 --> 00:21:51,250 but we'll put that down too. 411 00:21:51,250 --> 00:21:53,460 Is fast and reversible. 412 00:21:53,460 --> 00:21:56,820 And step two is slow. 413 00:21:56,820 --> 00:22:00,120 So I'm just going to propose that these are true, 414 00:22:00,120 --> 00:22:04,680 and then we will recalculate what the rate law would 415 00:22:04,680 --> 00:22:09,020 be if you have a fast reversible step followed by a slow step 416 00:22:09,020 --> 00:22:10,715 and see if that agrees with experiment. 417 00:22:16,440 --> 00:22:21,720 So to do that we have to consider 418 00:22:21,720 --> 00:22:27,010 what it means if we have a fast step followed by a slow step. 419 00:22:27,010 --> 00:22:31,740 So let's introduce a term-- very important term-- which 420 00:22:31,740 --> 00:22:33,840 is the rate determining step. 421 00:22:33,840 --> 00:22:37,650 Also known as the rate limiting step. 422 00:22:37,650 --> 00:22:42,660 So the slow step of a reaction, if it's truly a very slow step, 423 00:22:42,660 --> 00:22:45,912 is going to govern the overall rate of the reaction. 424 00:22:45,912 --> 00:22:47,370 So let's think about this a minute. 425 00:22:47,370 --> 00:22:51,660 So I told you that the extra problems for exam four 426 00:22:51,660 --> 00:22:52,770 are long. 427 00:22:52,770 --> 00:22:54,020 They're very long. 428 00:22:54,020 --> 00:22:56,770 Sorry about that, but there were a lot of problems 429 00:22:56,770 --> 00:22:58,770 and I wanted to get you ready for exam four. 430 00:22:58,770 --> 00:23:02,530 You also have problem set nine due tomorrow. 431 00:23:02,530 --> 00:23:04,710 So you've got a lot of problems to do. 432 00:23:04,710 --> 00:23:08,840 So after class today I feel that many of you 433 00:23:08,840 --> 00:23:11,970 are going to be really inspired to get 434 00:23:11,970 --> 00:23:14,490 started on those problems. 435 00:23:14,490 --> 00:23:16,010 And you may run out of here. 436 00:23:16,010 --> 00:23:18,839 You might leap over the chair in front of you and race 437 00:23:18,839 --> 00:23:19,380 out the door. 438 00:23:19,380 --> 00:23:21,370 You may clear the table on the way out 439 00:23:21,370 --> 00:23:25,520 because you're in a real hurry to start those extra problems. 440 00:23:25,520 --> 00:23:28,022 You will run to the library to look for a table, 441 00:23:28,022 --> 00:23:29,730 but all the tables will already be taken. 442 00:23:29,730 --> 00:23:32,160 How did your classmates get out of class so fast 443 00:23:32,160 --> 00:23:34,340 to get all of the tables in the library? 444 00:23:34,340 --> 00:23:36,450 And they're already finishing problem set nine 445 00:23:36,450 --> 00:23:39,450 and starting on the extra problems. 446 00:23:39,450 --> 00:23:41,910 So you race back to your dorm, but all the tables 447 00:23:41,910 --> 00:23:43,890 in the downstairs of the dorm are also 448 00:23:43,890 --> 00:23:46,230 filled with 5.111 students. 449 00:23:46,230 --> 00:23:48,672 So finally, on the fifth floor of one of the dorms 450 00:23:48,672 --> 00:23:50,880 you find an empty table, and then you're really fast. 451 00:23:50,880 --> 00:23:52,440 You got the problems out, you got your pencils out, 452 00:23:52,440 --> 00:23:54,898 you got your calculator out, you've got old equation sheets 453 00:23:54,898 --> 00:23:56,110 out, and ready to go. 454 00:23:56,110 --> 00:23:58,170 It's like two seconds. 455 00:23:58,170 --> 00:24:00,900 So it took you 40 minutes to find a free table 456 00:24:00,900 --> 00:24:04,590 at MIT that didn't already have a 5.111 student sitting 457 00:24:04,590 --> 00:24:07,860 and doing those extra problems and problem set nine. 458 00:24:07,860 --> 00:24:10,230 So it was 40 minutes plus 2 seconds 459 00:24:10,230 --> 00:24:12,850 to actually start doing the problems. 460 00:24:12,850 --> 00:24:16,310 So the rate determining or rate limiting step 461 00:24:16,310 --> 00:24:17,760 was finding your table. 462 00:24:17,760 --> 00:24:21,240 40 minutes plus 2 seconds is pretty much 40 minutes, 463 00:24:21,240 --> 00:24:23,700 and that's what happens in these reaction mechanisms. 464 00:24:23,700 --> 00:24:25,980 If you have a really slow step that 465 00:24:25,980 --> 00:24:29,650 governs the overall rate of the reaction. 466 00:24:29,650 --> 00:24:32,190 Now a lot of you know also about rate determining steps 467 00:24:32,190 --> 00:24:36,399 because some of you may be the rate determining step 468 00:24:36,399 --> 00:24:37,440 in your group of friends. 469 00:24:40,650 --> 00:24:44,320 The rate at which you get to dinner and eat 470 00:24:44,320 --> 00:24:47,020 is determined by you being ready to go. 471 00:24:49,714 --> 00:24:51,630 Some are in the room like me, yeah, that's me. 472 00:24:51,630 --> 00:24:54,730 You know who you are. 473 00:24:54,730 --> 00:24:56,190 So I'm saying that what you gotta 474 00:24:56,190 --> 00:25:02,200 do to not let yourself be that person-- that rate determining 475 00:25:02,200 --> 00:25:08,850 person, the RDS in your group of friends-- you need to get sleep 476 00:25:08,850 --> 00:25:11,740 and you need to eat well and you need to make sure 477 00:25:11,740 --> 00:25:15,150 you got your ATP. 478 00:25:15,150 --> 00:25:18,280 And that means, of course, to get 479 00:25:18,280 --> 00:25:23,170 enough sleep you gotta start problems sets early, especially 480 00:25:23,170 --> 00:25:25,640 the extra problems because they're long. 481 00:25:28,355 --> 00:25:30,820 Rate determining steps. 482 00:25:30,820 --> 00:25:31,752 Very important. 483 00:25:36,870 --> 00:25:40,510 So let's get back to our example. 484 00:25:40,510 --> 00:25:43,410 And we made a proposal. 485 00:25:43,410 --> 00:25:46,480 We made a proposal that step two was 486 00:25:46,480 --> 00:25:47,830 going to be rate determining. 487 00:25:47,830 --> 00:25:49,380 That was slow. 488 00:25:49,380 --> 00:25:53,320 We made the proposal that step one was fast and reversible. 489 00:25:53,320 --> 00:25:55,920 Step two was slow. 490 00:25:55,920 --> 00:25:58,620 So what that's going to mean then 491 00:25:58,620 --> 00:26:04,240 is that our rate law for the first step-- that's fast. 492 00:26:04,240 --> 00:26:06,250 That's a big number. 493 00:26:06,250 --> 00:26:09,760 The rate for the second step is slow. 494 00:26:09,760 --> 00:26:11,590 Rate determining. 495 00:26:11,590 --> 00:26:18,210 So that will mean then that K1 is going to be greater than K2 496 00:26:18,210 --> 00:26:21,767 times O2, so we can drop out our concentrations here 497 00:26:21,767 --> 00:26:22,600 to think about them. 498 00:26:22,600 --> 00:26:23,830 They're going to be the same. 499 00:26:23,830 --> 00:26:24,590 So what's left? 500 00:26:24,590 --> 00:26:27,660 That means the rate constant for that reverse step is very fast. 501 00:26:27,660 --> 00:26:28,940 It's a big number. 502 00:26:28,940 --> 00:26:33,790 That's going to be big compared to K2 times O2. 503 00:26:33,790 --> 00:26:36,210 So now we can go back and look at our expression 504 00:26:36,210 --> 00:26:37,840 for the intermediate. 505 00:26:37,840 --> 00:26:41,430 And we note that K minus 1 is in the bottom 506 00:26:41,430 --> 00:26:46,700 as well as K2 times O2 is in the bottom of the expression. 507 00:26:46,700 --> 00:26:51,590 So if K minus 1 now is really big compared to K2 times 508 00:26:51,590 --> 00:26:54,200 O2-- again, that's the fast step; that's 509 00:26:54,200 --> 00:26:57,640 the slow step-- then this term pretty much doesn't matter 510 00:26:57,640 --> 00:26:59,640 and it can drop out. 511 00:26:59,640 --> 00:27:04,750 Because this is really small compared to that. 512 00:27:04,750 --> 00:27:09,640 And if we drop out this term we can rewrite the expression 513 00:27:09,640 --> 00:27:10,810 like this. 514 00:27:10,810 --> 00:27:15,010 The concentration of our intermediate rate constant K1 515 00:27:15,010 --> 00:27:18,860 NO squared over K minus 1. 516 00:27:18,860 --> 00:27:21,280 And now we can rearrange this equation, 517 00:27:21,280 --> 00:27:25,390 bringing the concentrations to one side and our rate constants 518 00:27:25,390 --> 00:27:26,780 to the other side. 519 00:27:26,780 --> 00:27:31,410 So we have our intermediate N2O2 over NO squared 520 00:27:31,410 --> 00:27:35,080 equals rate constant K1 over K minus 1. 521 00:27:35,080 --> 00:27:42,700 What does rate constant K1 over rate constant K minus 1 equal? 522 00:27:42,700 --> 00:27:44,700 AUDIENCE: [INAUDIBLE]. 523 00:27:44,700 --> 00:27:48,210 CATHERINE DRENNAN: It equals the equilibrium constant K1 524 00:27:48,210 --> 00:27:50,420 for the first step. 525 00:27:50,420 --> 00:27:55,450 So if you have a fast reversible step followed by a slow step, 526 00:27:55,450 --> 00:27:58,480 the first step is basically in equilibrium. 527 00:27:58,480 --> 00:28:03,900 We can make that approximation that it is in equilibrium 528 00:28:03,900 --> 00:28:07,540 and solve it by thinking about equilibrium expressions. 529 00:28:07,540 --> 00:28:09,040 Fantastic. 530 00:28:09,040 --> 00:28:12,590 We can be back to equilibrium expressions. 531 00:28:12,590 --> 00:28:15,600 So let's think about this a little bit more. 532 00:28:15,600 --> 00:28:17,880 Here I have a pretty picture for you. 533 00:28:17,880 --> 00:28:20,590 So here we have our reactants forming our intermediates. 534 00:28:20,590 --> 00:28:23,500 The intermediates are also going back and forming our reactants. 535 00:28:23,500 --> 00:28:25,840 Reactants are forming the intermediates and then 536 00:28:25,840 --> 00:28:26,770 back again. 537 00:28:26,770 --> 00:28:28,540 Fast reversible. 538 00:28:28,540 --> 00:28:32,130 Every once in a while an intermediate gets siphoned off 539 00:28:32,130 --> 00:28:36,040 to products, but if this is a really, really slow step 540 00:28:36,040 --> 00:28:37,270 it doesn't happen very often. 541 00:28:37,270 --> 00:28:40,570 It doesn't really affect this very much. 542 00:28:40,570 --> 00:28:42,890 And so basically, this is in equilibrium. 543 00:28:42,890 --> 00:28:44,900 It's like this part doesn't even really matter. 544 00:28:44,900 --> 00:28:46,440 It doesn't play in. 545 00:28:46,440 --> 00:28:49,960 So when we have a fast reversible step followed 546 00:28:49,960 --> 00:28:52,840 by a slow step, we can assume the first step is 547 00:28:52,840 --> 00:28:56,610 in equilibrium and we can solve for our intermediate 548 00:28:56,610 --> 00:29:00,860 using equilibrium expressions. 549 00:29:00,860 --> 00:29:02,150 So let's do that. 550 00:29:02,150 --> 00:29:05,260 Let's take our equilibrium expression for the first step 551 00:29:05,260 --> 00:29:09,100 and now plug it in to our rate law. 552 00:29:09,100 --> 00:29:12,040 So we can substitute this step now 553 00:29:12,040 --> 00:29:14,740 or we can have rate constants or we can 554 00:29:14,740 --> 00:29:16,056 have an equilibrium constant. 555 00:29:16,056 --> 00:29:16,930 You can write either. 556 00:29:16,930 --> 00:29:18,550 These are equivalent. 557 00:29:18,550 --> 00:29:21,900 And we can put those back into this, 558 00:29:21,900 --> 00:29:25,980 which was our original overall rate that we wrote. 559 00:29:25,980 --> 00:29:28,440 We weren't done though because we had an intermediate. 560 00:29:28,440 --> 00:29:33,070 So we can plug this now in for our concentration 561 00:29:33,070 --> 00:29:35,130 of the intermediate. 562 00:29:35,130 --> 00:29:43,600 And so now we get 2 times K1 K2 O2 NO squared over K minus 1. 563 00:29:43,600 --> 00:29:46,610 Or we could just put that with the big equilibrium constant 564 00:29:46,610 --> 00:29:50,560 and get rid of our little K1 over K minus 1. 565 00:29:50,560 --> 00:29:52,920 Both of those are equivalent. 566 00:29:52,920 --> 00:29:59,340 And now we can take all of our K terms and call them K obs. 567 00:29:59,340 --> 00:30:02,430 So K obs is just the experimental rate constant. 568 00:30:02,430 --> 00:30:06,070 It's the collection of rate constants that are measured. 569 00:30:06,070 --> 00:30:09,040 And we often-- when we measure things, 570 00:30:09,040 --> 00:30:12,520 we can't distinguish K1 from K2. 571 00:30:12,520 --> 00:30:14,740 We sometimes try to do that, and that's 572 00:30:14,740 --> 00:30:15,840 a little more complicated. 573 00:30:15,840 --> 00:30:17,980 But in this case, all that was given 574 00:30:17,980 --> 00:30:20,850 was an overall K observed. 575 00:30:20,850 --> 00:30:27,310 And this was our experimental rate law K observed times O2 576 00:30:27,310 --> 00:30:29,560 first order NO second order. 577 00:30:29,560 --> 00:30:34,210 And now we see this expression agrees with this rate. 578 00:30:34,210 --> 00:30:37,170 So the fact that we have a good agreement 579 00:30:37,170 --> 00:30:42,100 means that a mechanism with this fast reversible step followed 580 00:30:42,100 --> 00:30:46,110 by a slow step gives rise to a rate law that's 581 00:30:46,110 --> 00:30:47,600 consistent with experiment. 582 00:30:47,600 --> 00:30:49,510 It doesn't prove that's the right mechanism. 583 00:30:49,510 --> 00:30:52,680 It's very hard to prove mechanisms are right, 584 00:30:52,680 --> 00:30:54,630 but at least it's consistent. 585 00:30:54,630 --> 00:30:59,290 So we can say this is a good guess, a good proposal, 586 00:30:59,290 --> 00:31:00,110 for our mechanism. 587 00:31:02,660 --> 00:31:04,578 So let's look at another example. 588 00:31:08,430 --> 00:31:11,570 So in this example we have NO again. 589 00:31:11,570 --> 00:31:13,940 We have two molecules of NO, and now we 590 00:31:13,940 --> 00:31:18,500 have Br2 going to two molecules of NOBr. 591 00:31:18,500 --> 00:31:21,660 And we're told that the experimental rate is 592 00:31:21,660 --> 00:31:27,610 K obs times NO first order, Br2 first order 593 00:31:27,610 --> 00:31:31,310 and asked, for this proposed mechanism, 594 00:31:31,310 --> 00:31:35,440 which would be the slow step to give rise 595 00:31:35,440 --> 00:31:39,050 to that experimental data? 596 00:31:39,050 --> 00:31:42,290 So the first thing that we would want to do with all of these 597 00:31:42,290 --> 00:31:48,400 is to write the rate laws for each individual step. 598 00:31:48,400 --> 00:31:51,290 So for the forward step we have one molecule 599 00:31:51,290 --> 00:31:58,420 of NO reacting with Br2 with rate constant K2. 600 00:31:58,420 --> 00:32:03,560 So we get rate constant K1 times the concentration of NO times 601 00:32:03,560 --> 00:32:05,650 the concentration of Br2. 602 00:32:05,650 --> 00:32:08,580 Again, this is a step or an elementary reaction, 603 00:32:08,580 --> 00:32:13,960 so we write the rate law just based on the stoichiometry 604 00:32:13,960 --> 00:32:15,610 here. 605 00:32:15,610 --> 00:32:19,210 Now we can do the same thing for the reverse rate. 606 00:32:19,210 --> 00:32:23,570 So we have K minus 1 times the concentration 607 00:32:23,570 --> 00:32:26,330 of our intermediate. 608 00:32:26,330 --> 00:32:30,320 So in step two our intermediate, which is formed in step one, 609 00:32:30,320 --> 00:32:33,460 is reacting with the second molecule NO, 610 00:32:33,460 --> 00:32:35,680 forming our product. 611 00:32:35,680 --> 00:32:38,630 And we can write the rate for this as well. 612 00:32:38,630 --> 00:32:43,520 K2 times the concentration of our intermediate, NOBr2, 613 00:32:43,520 --> 00:32:45,700 times the concentration of NO. 614 00:32:45,700 --> 00:32:47,440 So again, these are steps. 615 00:32:47,440 --> 00:32:49,460 They're elementary reactions so we 616 00:32:49,460 --> 00:32:52,600 can write the rate law based on the stoichiometry 617 00:32:52,600 --> 00:32:55,820 in that proposed step. 618 00:32:55,820 --> 00:32:59,620 So now we can write the overall rate law 619 00:32:59,620 --> 00:33:02,380 for the formation of NOBr, and we can just 620 00:33:02,380 --> 00:33:04,860 write it from the second step like we did before. 621 00:33:04,860 --> 00:33:06,020 Again, this is an example. 622 00:33:06,020 --> 00:33:07,820 We're forming two molecules of product 623 00:33:07,820 --> 00:33:09,260 so there's a two in there. 624 00:33:09,260 --> 00:33:10,900 We have K2. 625 00:33:10,900 --> 00:33:12,130 It's basically just this. 626 00:33:12,130 --> 00:33:15,800 K2 times our intermediate, NOBr2, 627 00:33:15,800 --> 00:33:18,370 times the concentration of NO. 628 00:33:18,370 --> 00:33:21,450 But once again, we're not done because there 629 00:33:21,450 --> 00:33:23,780 is an intermediate in the expression 630 00:33:23,780 --> 00:33:26,710 and you can't have an intermediate in your rate law. 631 00:33:26,710 --> 00:33:31,180 You need to solve for the rate law in terms of rate constants, 632 00:33:31,180 --> 00:33:33,530 reactants, and products. 633 00:33:33,530 --> 00:33:37,360 So we need to now solve for our intermediate 634 00:33:37,360 --> 00:33:40,550 in terms of things that are allowed in the overall rate 635 00:33:40,550 --> 00:33:42,080 law. 636 00:33:42,080 --> 00:33:44,420 And so we want to think again about what 637 00:33:44,420 --> 00:33:49,480 is the change in concentration of our intermediate 638 00:33:49,480 --> 00:33:52,550 So we can do the same thing that we did before? 639 00:33:52,550 --> 00:33:56,960 So the intermediate is being formed in the first step. 640 00:33:56,960 --> 00:34:01,440 So we have the rate law for the first step. 641 00:34:01,440 --> 00:34:07,480 K1 times NO times the concentration of Br2. 642 00:34:07,480 --> 00:34:13,120 The intermediate is also decaying in the reverse part 643 00:34:13,120 --> 00:34:15,429 of the first step. 644 00:34:15,429 --> 00:34:20,920 So that's minus the reverse rate minus K minus 1 times 645 00:34:20,920 --> 00:34:23,540 the intermediate here. 646 00:34:23,540 --> 00:34:27,020 And then it's being consumed in the second step. 647 00:34:27,020 --> 00:34:32,870 So it's going away by the rate K2 times the concentration 648 00:34:32,870 --> 00:34:36,610 of the intermediate times the concentration of NO. 649 00:34:36,610 --> 00:34:40,040 So again, this is exactly what we did with the first example. 650 00:34:40,040 --> 00:34:43,480 We think about the change, how it's being formed, 651 00:34:43,480 --> 00:34:47,679 and the two different ways that it's being consumed. 652 00:34:47,679 --> 00:34:51,159 We can again use a steady state approximation 653 00:34:51,159 --> 00:34:55,730 and set all of that equal to 0. 654 00:34:55,730 --> 00:35:00,920 So let's do that and we'll solve again for the intermediate. 655 00:35:00,920 --> 00:35:05,150 So on this next slide now-- I just put those things up there. 656 00:35:05,150 --> 00:35:06,980 This is what you were just copying down. 657 00:35:06,980 --> 00:35:09,810 If you didn't finish it's still here. 658 00:35:09,810 --> 00:35:12,290 And here is the steady state approximation. 659 00:35:12,290 --> 00:35:14,170 So again, the steady straight approximation 660 00:35:14,170 --> 00:35:16,480 is the net rate of formation of your intermediate 661 00:35:16,480 --> 00:35:18,940 equals the net rate of it's going away. 662 00:35:18,940 --> 00:35:21,190 Net rate is 0. 663 00:35:21,190 --> 00:35:25,150 So rearranging then we can bring the two terms that 664 00:35:25,150 --> 00:35:29,380 involve the decay or the consumption 665 00:35:29,380 --> 00:35:32,720 of our intermediate on one side and then set them equal 666 00:35:32,720 --> 00:35:35,910 to the rate at which that intermediate is formed. 667 00:35:35,910 --> 00:35:39,370 And then, we can pull out our terms for our intermediate. 668 00:35:39,370 --> 00:35:45,882 So we pull out NoBr2, leaving K minus 1, leaving K2 and NO, 669 00:35:45,882 --> 00:35:51,020 and set it equal to the rate law for the first step 670 00:35:51,020 --> 00:35:52,970 in the forward direction. 671 00:35:52,970 --> 00:35:54,670 We can solve for the intermediate. 672 00:35:54,670 --> 00:35:57,180 Take this, divide it by this term. 673 00:35:57,180 --> 00:36:00,500 So we have K1 times NO times Br2 over K 674 00:36:00,500 --> 00:36:04,540 minus 1 plus K2 times NO. 675 00:36:04,540 --> 00:36:05,900 Now we can take this. 676 00:36:05,900 --> 00:36:08,210 We are done solving for our intermediate. 677 00:36:08,210 --> 00:36:10,730 We have no more intermediates in there, 678 00:36:10,730 --> 00:36:13,790 so we can now plug this back in to the expression we 679 00:36:13,790 --> 00:36:15,080 had before. 680 00:36:15,080 --> 00:36:19,360 So we can take this, plug it in here, 681 00:36:19,360 --> 00:36:21,460 and that gives us this formation. 682 00:36:21,460 --> 00:36:28,540 So we have 2 times K1, NO times NO-- NO squared-- Br2 on top, 683 00:36:28,540 --> 00:36:32,398 K minus 1 on the bottom plus K2 times NO. 684 00:36:35,480 --> 00:36:38,030 Now we were asked, what are the fast 685 00:36:38,030 --> 00:36:39,840 and what are the slow steps? 686 00:36:39,840 --> 00:36:42,410 So now we want to take this and think about, 687 00:36:42,410 --> 00:36:45,470 if there's different fast and slow steps, 688 00:36:45,470 --> 00:36:49,520 is it consistent then with the experiment? 689 00:36:49,520 --> 00:36:54,110 So first, let's consider if the first step was slow 690 00:36:54,110 --> 00:36:58,250 and the second step was fast-- or i.e., 691 00:36:58,250 --> 00:37:03,860 if we have K2 NO greater than K minus 1. 692 00:37:03,860 --> 00:37:07,280 And this is a clicker question, so why don't you tell me how 693 00:37:07,280 --> 00:37:10,820 this then, using this, changes? 694 00:37:27,730 --> 00:37:28,230 OK. 695 00:37:28,230 --> 00:37:28,780 10 seconds. 696 00:37:44,300 --> 00:37:44,870 OK. 697 00:37:44,870 --> 00:37:49,690 Let's now think about why that's true. 698 00:37:49,690 --> 00:37:54,430 This involved doing a couple of steps in your head. 699 00:37:54,430 --> 00:37:57,700 So if we have a first step that's slow 700 00:37:57,700 --> 00:38:00,370 and a second step that's fast, the second step 701 00:38:00,370 --> 00:38:03,940 involves the K2 times the concentration of NO. 702 00:38:03,940 --> 00:38:04,920 That's the second step. 703 00:38:04,920 --> 00:38:05,860 That's fast. 704 00:38:05,860 --> 00:38:09,020 That's going to be a big number compared to K minus 1. 705 00:38:09,020 --> 00:38:11,710 So if we look, both are on the bottom here. 706 00:38:11,710 --> 00:38:14,810 And if this term, K2 NO, is much, much bigger 707 00:38:14,810 --> 00:38:20,770 than K minus 1, then we can say K minus 1 goes way. 708 00:38:20,770 --> 00:38:25,120 If we get rid of K minus 1, we can simplify 709 00:38:25,120 --> 00:38:27,070 the expression even more. 710 00:38:27,070 --> 00:38:30,820 We can get rid of our K2s and we can get rid 711 00:38:30,820 --> 00:38:34,745 of one of our NOs, which gives us this. 712 00:38:37,420 --> 00:38:40,300 So saying that the first step is slow 713 00:38:40,300 --> 00:38:42,130 and the second step is fast gives us 714 00:38:42,130 --> 00:38:44,620 a very different equation. 715 00:38:44,620 --> 00:38:47,420 A lot of things cancel out. 716 00:38:47,420 --> 00:38:51,240 So we can also write that expression as K 717 00:38:51,240 --> 00:38:56,050 obs times NO times Br2. 718 00:38:56,050 --> 00:39:02,000 And the overall order of that reaction would be what? 719 00:39:02,000 --> 00:39:04,226 Yell it out. 720 00:39:04,226 --> 00:39:05,170 AUDIENCE: Two. 721 00:39:05,170 --> 00:39:07,090 CATHERINE DRENNAN: Yes. 722 00:39:07,090 --> 00:39:10,030 So this is what we would get for a first step that's slow; 723 00:39:10,030 --> 00:39:11,860 second step that's fast. 724 00:39:11,860 --> 00:39:16,450 Now let's consider if the first step is fast 725 00:39:16,450 --> 00:39:19,300 and the second step is slow. 726 00:39:19,300 --> 00:39:22,510 So if the first step is fast that means K minus 1, 727 00:39:22,510 --> 00:39:25,000 the rate constant for the reverse step, 728 00:39:25,000 --> 00:39:29,910 is going to be a lot bigger than K2 times NO. 729 00:39:29,910 --> 00:39:32,530 And so now we can look up at this expression 730 00:39:32,530 --> 00:39:35,890 and say, OK, if this is much bigger than this 731 00:39:35,890 --> 00:39:38,710 then that cancels out. 732 00:39:38,710 --> 00:39:41,890 And then we're left with this expression, 733 00:39:41,890 --> 00:39:44,410 which I can put down here. 734 00:39:44,410 --> 00:39:48,190 So that leaves us-- we can't cancel any more at this point. 735 00:39:48,190 --> 00:39:51,190 So that leaves us with 2 times K1 times 736 00:39:51,190 --> 00:39:57,950 K2-- these-- NO squared Br2 over K minus 1. 737 00:39:57,950 --> 00:40:01,390 So assuming different things about how fast and slow 738 00:40:01,390 --> 00:40:06,070 the steps are gives you very different rate laws. 739 00:40:06,070 --> 00:40:08,890 We can also write that to make it look a little simpler 740 00:40:08,890 --> 00:40:14,530 as K obs, but you'll note that the overall order is 741 00:40:14,530 --> 00:40:15,560 very different. 742 00:40:15,560 --> 00:40:18,330 So what's the overall order here? 743 00:40:18,330 --> 00:40:19,000 Three. 744 00:40:19,000 --> 00:40:20,080 Right. 745 00:40:20,080 --> 00:40:25,060 So let's remind ourselves what the experimental rate law was. 746 00:40:25,060 --> 00:40:29,380 And it was NO first order Br2 first order. 747 00:40:29,380 --> 00:40:35,170 So that means that this one would be consistent. 748 00:40:35,170 --> 00:40:40,210 So the mechanism is likely to involve a slow first step 749 00:40:40,210 --> 00:40:42,670 and a fast second step. 750 00:40:42,670 --> 00:40:45,400 And so that's how you do a lot of these problems. 751 00:40:45,400 --> 00:40:47,830 You think about what is going to change when you have 752 00:40:47,830 --> 00:40:50,350 different fast and slow steps. 753 00:40:50,350 --> 00:40:52,150 One of them will be more consistent 754 00:40:52,150 --> 00:40:54,310 with the experimental data and one of them 755 00:40:54,310 --> 00:40:56,210 will not be consistent. 756 00:40:56,210 --> 00:41:01,690 OK Let's do one more fast example. 757 00:41:01,690 --> 00:41:07,720 Here we have rate law for two molecules of ozone O3 going 758 00:41:07,720 --> 00:41:11,010 to three molecules of O2. 759 00:41:11,010 --> 00:41:14,290 And ozone has been in the news a lot recently. 760 00:41:14,290 --> 00:41:17,020 So we want to keep our ozone layer. 761 00:41:17,020 --> 00:41:21,070 We don't want it to go away. 762 00:41:21,070 --> 00:41:25,690 So we have O3 going to O2 plus O, 763 00:41:25,690 --> 00:41:28,450 and you're forming an intermediate O. 764 00:41:28,450 --> 00:41:33,010 That intermediate is reacting with O3, forming 765 00:41:33,010 --> 00:41:35,510 our two molecules of O2. 766 00:41:35,510 --> 00:41:39,130 So let's just write out what our rate 767 00:41:39,130 --> 00:41:41,270 is for the forward reaction. 768 00:41:41,270 --> 00:41:47,430 So we have K1 times the concentration of O3. 769 00:41:47,430 --> 00:41:53,770 For the reverse we have K minus 1 times concentration 770 00:41:53,770 --> 00:41:59,120 of O2 times the concentration of our intermediate O. 771 00:41:59,120 --> 00:42:02,860 For the next one we have K2 times our intermediate O 772 00:42:02,860 --> 00:42:06,640 times the concentration of O3. 773 00:42:06,640 --> 00:42:09,400 So now we're told that there's a fast reversible step 774 00:42:09,400 --> 00:42:11,300 and a slow step. 775 00:42:11,300 --> 00:42:15,650 So the rate will be determined by the slow step. 776 00:42:15,650 --> 00:42:19,420 So we can write out the rate of formation of O2 777 00:42:19,420 --> 00:42:21,160 based on the slow step, which happens 778 00:42:21,160 --> 00:42:23,770 to be the second step, which is what we've done all along. 779 00:42:23,770 --> 00:42:26,560 So there's not really a huge change right now. 780 00:42:26,560 --> 00:42:29,230 So the formation-- again, two molecules of O2 781 00:42:29,230 --> 00:42:31,090 were formed so we have a 2. 782 00:42:31,090 --> 00:42:33,100 We have K2 times the concentration 783 00:42:33,100 --> 00:42:37,120 of our intermediate O times the concentration of O3. 784 00:42:37,120 --> 00:42:39,910 But again, O is an intermediate so we 785 00:42:39,910 --> 00:42:42,640 need to solve for it in terms of our products 786 00:42:42,640 --> 00:42:45,310 reactants of rate constants. 787 00:42:45,310 --> 00:42:48,730 But now we're told something about fast and slow steps 788 00:42:48,730 --> 00:42:51,250 right up front. 789 00:42:51,250 --> 00:42:53,560 And so if we have a fast reversible step 790 00:42:53,560 --> 00:42:56,260 followed by a slow step, how can we 791 00:42:56,260 --> 00:42:58,990 solve for our concentration of our intermediate 792 00:42:58,990 --> 00:43:01,660 in a simpler way than we've been doing? 793 00:43:01,660 --> 00:43:04,000 What do we use? 794 00:43:04,000 --> 00:43:05,874 Or what can we use? 795 00:43:05,874 --> 00:43:07,310 AUDIENCE: [INAUDIBLE]. 796 00:43:07,310 --> 00:43:10,900 CATHERINE DRENNAN: We can use the equilibrium expression. 797 00:43:10,900 --> 00:43:12,130 So we can put that in. 798 00:43:12,130 --> 00:43:15,070 We can say our equilibrium expression products 799 00:43:15,070 --> 00:43:20,020 over reactants equals little K1 over K minus 1, 800 00:43:20,020 --> 00:43:23,260 or equilibrium constant K1. 801 00:43:23,260 --> 00:43:29,380 Solve for O and get either big equilibrium constant K1 802 00:43:29,380 --> 00:43:33,070 or our little rate constant K1 over K minus 1, 803 00:43:33,070 --> 00:43:36,130 and we have O3 over O2 here. 804 00:43:36,130 --> 00:43:39,820 So this was a lot simpler than doing all of that. 805 00:43:39,820 --> 00:43:42,760 So again, if you have a fast reversible step followed 806 00:43:42,760 --> 00:43:45,460 by a slow step you can solve for the concentration 807 00:43:45,460 --> 00:43:48,160 of your intermediate using an equilibrium expression which 808 00:43:48,160 --> 00:43:49,660 you all know how to write. 809 00:43:49,660 --> 00:43:51,610 So that makes your life easier. 810 00:43:51,610 --> 00:43:54,610 Then we can substitute that back in 811 00:43:54,610 --> 00:43:58,810 and we are able to put this back. 812 00:43:58,810 --> 00:44:02,200 And we'll solve it for O and we'll plug in our K1 813 00:44:02,200 --> 00:44:03,390 over K minus 1. 814 00:44:03,390 --> 00:44:04,330 O3. 815 00:44:04,330 --> 00:44:05,300 We had an O3. 816 00:44:05,300 --> 00:44:06,680 So that's squared over O2. 817 00:44:09,520 --> 00:44:14,650 Or we could write that in terms of K obs O3 concentration 818 00:44:14,650 --> 00:44:17,710 to the 2 over O2. 819 00:44:17,710 --> 00:44:21,550 So let's end with some fun, thinking about what 820 00:44:21,550 --> 00:44:23,440 we would observe here. 821 00:44:23,440 --> 00:44:24,940 First, the order and then what would 822 00:44:24,940 --> 00:44:26,980 happen if we double things. 823 00:44:26,980 --> 00:44:29,800 So what is the order with respect to O3? 824 00:44:29,800 --> 00:44:31,196 You can yell that out. 825 00:44:31,196 --> 00:44:32,340 AUDIENCE: [INAUDIBLE]. 826 00:44:32,340 --> 00:44:33,298 CATHERINE DRENNAN: Yup. 827 00:44:33,298 --> 00:44:36,256 What is the order for O2? 828 00:44:36,256 --> 00:44:38,852 AUDIENCE: [INAUDIBLE]. 829 00:44:38,852 --> 00:44:40,060 CATHERINE DRENNAN: Oh, sorry. 830 00:44:40,060 --> 00:44:40,990 I had something here. 831 00:44:40,990 --> 00:44:42,700 Double this what happens? 832 00:44:42,700 --> 00:44:44,024 AUDIENCE: [INAUDIBLE]. 833 00:44:44,024 --> 00:44:45,190 CATHERINE DRENNAN: The rate. 834 00:44:45,190 --> 00:44:46,930 Well, four times. 835 00:44:46,930 --> 00:44:47,720 Order here. 836 00:44:47,720 --> 00:44:49,750 Some people yelled it out. 837 00:44:49,750 --> 00:44:50,260 Oh, no. 838 00:44:50,260 --> 00:44:50,760 Oh, man. 839 00:44:50,760 --> 00:44:51,760 It's a clicker question. 840 00:44:51,760 --> 00:44:53,009 I forgot about that. 841 00:44:53,009 --> 00:44:53,550 Don't listen. 842 00:44:53,550 --> 00:44:55,675 But luckily, everyone yelled out different answers. 843 00:45:06,071 --> 00:45:06,570 All right. 844 00:45:06,570 --> 00:45:07,433 10 more seconds. 845 00:45:22,210 --> 00:45:23,440 OK. 846 00:45:23,440 --> 00:45:24,730 Yup. 847 00:45:24,730 --> 00:45:27,760 So it's minus 1. 848 00:45:27,760 --> 00:45:31,090 And so if you double it, it will half. 849 00:45:31,090 --> 00:45:36,910 And then finally, the overall order would be 1 because again, 850 00:45:36,910 --> 00:45:39,010 the overall order is the sum. 851 00:45:39,010 --> 00:45:41,950 So 2 minus 1 is 1. 852 00:45:41,950 --> 00:45:45,415 And last clicker question. 853 00:45:56,070 --> 00:45:56,570 All right. 854 00:45:56,570 --> 00:45:58,760 10 more seconds. 855 00:45:58,760 --> 00:46:01,370 I know you're really in a hurry to do those extra problems 856 00:46:01,370 --> 00:46:03,480 for exam four. 857 00:46:03,480 --> 00:46:05,300 I'm the rate determining step. 858 00:46:05,300 --> 00:46:07,430 I am with my clicker question. 859 00:46:07,430 --> 00:46:09,495 I admit it. 860 00:46:09,495 --> 00:46:11,037 AUDIENCE: Yay. 861 00:46:11,037 --> 00:46:12,245 CATHERINE DRENNAN: All right. 862 00:46:14,790 --> 00:46:15,290 All right. 863 00:46:15,290 --> 00:46:16,100 See you Wednesday. 864 00:46:16,100 --> 00:46:18,890 Remember, final clicker competition 865 00:46:18,890 --> 00:46:22,430 of the year before the finals.