WEBVTT 00:00:00.090 --> 00:00:02.430 The following content is provided under a Creative 00:00:02.430 --> 00:00:03.820 Commons license. 00:00:03.820 --> 00:00:06.060 Your support will help MIT OpenCourseWare 00:00:06.060 --> 00:00:10.150 continue to offer high quality educational resources for free. 00:00:10.150 --> 00:00:12.690 To make a donation or to view additional materials 00:00:12.690 --> 00:00:16.435 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:16.435 --> 00:00:17.060 at ocw.mit.edu. 00:00:34.582 --> 00:00:35.790 CATHERINE DRENNAN: All right. 00:00:35.790 --> 00:00:37.280 So 10 more seconds. 00:01:02.230 --> 00:01:02.730 OK. 00:01:06.060 --> 00:01:07.275 Let's quiet down. 00:01:10.080 --> 00:01:14.800 So that is the-- you got the 70% with the right answer. 00:01:14.800 --> 00:01:18.676 If people can just yell out what was wrong with number one. 00:01:18.676 --> 00:01:19.490 AUDIENCE: Sig figs. 00:01:19.490 --> 00:01:20.940 CATHERINE DRENNAN: Sig figs. 00:01:20.940 --> 00:01:23.400 And what about numbers three and four? 00:01:23.400 --> 00:01:25.502 What equation was that using? 00:01:25.502 --> 00:01:26.460 AUDIENCE: Second order. 00:01:26.460 --> 00:01:27.190 CATHERINE DRENNAN: Second order. 00:01:27.190 --> 00:01:27.930 That's right. 00:01:27.930 --> 00:01:30.390 So this is a good clicker question 00:01:30.390 --> 00:01:34.260 for a problem that's going to be coming up on exam four 00:01:34.260 --> 00:01:37.890 and also on the final exam to a larger degree. 00:01:37.890 --> 00:01:40.560 On the final exam we have equation sheets 00:01:40.560 --> 00:01:44.530 that have all the equations from the whole semester, 00:01:44.530 --> 00:01:47.940 and so you need to figure out and remember which equation 00:01:47.940 --> 00:01:50.280 goes with which problem. 00:01:50.280 --> 00:01:54.300 It doesn't say, oh, here is the expression for first order. 00:01:54.300 --> 00:01:56.730 Here's the equation for second order. 00:01:56.730 --> 00:01:58.800 You need to look and remember which 00:01:58.800 --> 00:02:01.290 equation goes with which thing. 00:02:01.290 --> 00:02:03.120 But in terms of first order, what's 00:02:03.120 --> 00:02:06.360 a way that you can remember what a first order 00:02:06.360 --> 00:02:07.260 equation would be? 00:02:07.260 --> 00:02:10.199 What's missing from first order? 00:02:10.199 --> 00:02:12.240 Yeah, the concentration in the material. 00:02:12.240 --> 00:02:14.700 So for first order it's independent 00:02:14.700 --> 00:02:18.120 of the original concentration in the material, which 00:02:18.120 --> 00:02:20.700 is why we can use first order equations 00:02:20.700 --> 00:02:22.470 for nuclear chemistry. 00:02:22.470 --> 00:02:26.070 Because the rate of decay of radioactive nuclei 00:02:26.070 --> 00:02:28.320 are independent of the nuclei around them, 00:02:28.320 --> 00:02:30.060 so it's a first order process. 00:02:30.060 --> 00:02:32.440 So when you think a little bit about these equations, 00:02:32.440 --> 00:02:34.950 you should be able to identify which equation 00:02:34.950 --> 00:02:36.670 goes with which problem. 00:02:36.670 --> 00:02:40.110 And so you can identify the type of first order problems also. 00:02:40.110 --> 00:02:42.810 And for the second order it almost always says, 00:02:42.810 --> 00:02:45.210 for this second order process, which 00:02:45.210 --> 00:02:47.880 gives you a nice hint that that's a second order process. 00:02:47.880 --> 00:02:50.440 And then you just have to identify the equation. 00:02:50.440 --> 00:02:50.940 OK. 00:02:50.940 --> 00:02:52.350 So today is more kinetics. 00:02:52.350 --> 00:02:53.840 So we're in the kinetics unit. 00:02:53.840 --> 00:02:56.430 We'll be in the kinetics unit for the rest of the semester. 00:02:56.430 --> 00:02:58.710 It's our last unit. 00:02:58.710 --> 00:03:01.470 So today I think is one of the most important lectures 00:03:01.470 --> 00:03:03.930 in terms of the kinetics unit because we're 00:03:03.930 --> 00:03:06.000 talking about reaction mechanisms, 00:03:06.000 --> 00:03:09.360 and that is really an important part of kinetics. 00:03:09.360 --> 00:03:12.270 So investigating reaction mechanisms. 00:03:12.270 --> 00:03:16.805 So if you're going to describe how a reaction takes place-- 00:03:16.805 --> 00:03:19.080 often reactions don't occur in one step. 00:03:19.080 --> 00:03:22.800 It's really uncommon for reactions to occur in one step. 00:03:22.800 --> 00:03:25.560 So you want to describe the different steps. 00:03:25.560 --> 00:03:27.570 And you describe those steps, which are also 00:03:27.570 --> 00:03:30.270 called elementary reactions. 00:03:30.270 --> 00:03:32.460 So you break down a complex reaction 00:03:32.460 --> 00:03:34.620 into a series of steps, and then you 00:03:34.620 --> 00:03:37.110 try to figure out if that mechanism, 00:03:37.110 --> 00:03:41.370 if those series of steps, are consistent with the data 00:03:41.370 --> 00:03:44.620 that you've collected on this particular reaction. 00:03:44.620 --> 00:03:47.850 And over here I just show you some steps 00:03:47.850 --> 00:03:52.950 in the natural biosynthesis of a vitamin biotin. 00:03:52.950 --> 00:03:56.460 Biotin is important vitamin for us. 00:03:56.460 --> 00:03:59.670 It's also used a lot in feedstock. 00:03:59.670 --> 00:04:04.560 And so people buy a lot of biotin to put in feed stock, 00:04:04.560 --> 00:04:07.150 and so there's a huge market of biotin. 00:04:07.150 --> 00:04:09.690 Now right now it's made by what is estimated 00:04:09.690 --> 00:04:13.510 to be a 13 step organic synthesis which produces-- 00:04:13.510 --> 00:04:14.640 that's a lot of steps. 00:04:14.640 --> 00:04:16.290 It's really expensive, and there's 00:04:16.290 --> 00:04:20.519 a huge amount of organic waste associated with making biotin. 00:04:20.519 --> 00:04:24.255 And we're talking about in the tons level of waste. 00:04:24.255 --> 00:04:26.130 So researchers have been trying to figure out 00:04:26.130 --> 00:04:29.520 the mechanism by which nature makes biotin 00:04:29.520 --> 00:04:32.520 because that would be a lot more environmentally friendly. 00:04:32.520 --> 00:04:34.460 So when you're thinking about this, 00:04:34.460 --> 00:04:37.800 what you want to say is, OK, if we write out a mechanism 00:04:37.800 --> 00:04:40.860 is it consistent with the experimental data 00:04:40.860 --> 00:04:43.080 and are there fast and slow steps? 00:04:43.080 --> 00:04:45.341 Because if there's a step that's really slow maybe 00:04:45.341 --> 00:04:46.590 you can do something about it. 00:04:46.590 --> 00:04:48.420 Maybe if you're talking about an enzyme 00:04:48.420 --> 00:04:52.263 you could re-engineer the enzyme so it would be a better enzyme. 00:04:52.263 --> 00:04:55.340 Maybe natural selection didn't particularly work. 00:04:55.340 --> 00:04:57.710 Maybe the cell doesn't really need as much biotin 00:04:57.710 --> 00:05:00.900 as we do commercially, so there was no need to make it fast. 00:05:00.900 --> 00:05:02.525 But now we have a need to make it fast, 00:05:02.525 --> 00:05:06.740 so maybe we can do some evolution of this enzyme 00:05:06.740 --> 00:05:08.401 and design something to be better. 00:05:08.401 --> 00:05:10.400 So when you're talking about reaction mechanisms 00:05:10.400 --> 00:05:12.620 you want to know what's fast, what's slow. 00:05:12.620 --> 00:05:14.960 And if you want to use that product for something, 00:05:14.960 --> 00:05:17.270 you want to figure out how you can change things 00:05:17.270 --> 00:05:19.040 to have a better mechanism-- maybe 00:05:19.040 --> 00:05:23.210 avoid some really slow steps so that you can do better. 00:05:23.210 --> 00:05:26.690 Today is also World AIDS Day. 00:05:26.690 --> 00:05:30.530 And understanding the mechanism of HIV protease 00:05:30.530 --> 00:05:33.140 was really essential in designing inhibitors 00:05:33.140 --> 00:05:34.320 against that enzyme. 00:05:34.320 --> 00:05:36.230 And if you inhibit the enzyme you 00:05:36.230 --> 00:05:38.250 stop the development of the disease. 00:05:38.250 --> 00:05:40.010 So this was a very important thing, 00:05:40.010 --> 00:05:42.650 and we have some pretty good molecules 00:05:42.650 --> 00:05:43.760 to treat AIDS right now. 00:05:43.760 --> 00:05:46.134 And some of the challenges have moved on to other things, 00:05:46.134 --> 00:05:48.920 like less good health care in parts of the world where 00:05:48.920 --> 00:05:50.050 this has affected. 00:05:50.050 --> 00:05:53.510 So I feel like AIDS has kind of taken a backseat to hearing 00:05:53.510 --> 00:05:55.940 about Ebola recently, but AIDS is still 00:05:55.940 --> 00:05:58.730 a very important problem and one that smart people 00:05:58.730 --> 00:06:00.230 like you could address. 00:06:00.230 --> 00:06:03.040 So again, understanding reaction mechanisms 00:06:03.040 --> 00:06:05.360 is very, very important. 00:06:05.360 --> 00:06:05.860 All right. 00:06:05.860 --> 00:06:10.190 So let's go to a simpler problem and a simpler reaction 00:06:10.190 --> 00:06:11.960 mechanism. 00:06:11.960 --> 00:06:14.210 We'll go to our friend over here where 00:06:14.210 --> 00:06:17.780 we have two molecules of NO reacting 00:06:17.780 --> 00:06:23.720 with one molecule of O2 going to two molecules of NO2. 00:06:23.720 --> 00:06:26.990 So someone measured some rates for this 00:06:26.990 --> 00:06:30.430 and came up with the following rate law 00:06:30.430 --> 00:06:32.690 where you have a rate constant, which is just 00:06:32.690 --> 00:06:35.390 called k obs for observed. 00:06:35.390 --> 00:06:38.450 And we'll talk about that more in a little bit. 00:06:38.450 --> 00:06:42.200 And they discovered it is second order with respect to NO 00:06:42.200 --> 00:06:45.620 and first order with respect to O2. 00:06:45.620 --> 00:06:47.330 So there's a couple of things right away 00:06:47.330 --> 00:06:49.040 that we can ask about this. 00:06:49.040 --> 00:06:54.200 One is, what is the overall order then of this reaction 00:06:54.200 --> 00:06:55.820 from this experimental data? 00:06:55.820 --> 00:06:56.616 What would that be? 00:06:56.616 --> 00:06:57.359 AUDIENCE: Three. 00:06:57.359 --> 00:06:58.400 CATHERINE DRENNAN: Three. 00:06:58.400 --> 00:06:59.990 Again, if some of you missed it. 00:06:59.990 --> 00:07:01.580 2 plus 1 is 3. 00:07:01.580 --> 00:07:04.250 This is not where you want to lose your points on the exam. 00:07:04.250 --> 00:07:06.680 There's going to be some tricky significant figures 00:07:06.680 --> 00:07:08.390 in the kinetics units. 00:07:08.390 --> 00:07:10.420 Save your points to lose there. 00:07:10.420 --> 00:07:13.810 Count 2 plus 1 and get 3. 00:07:13.810 --> 00:07:18.650 So is one step likely to have three different things come 00:07:18.650 --> 00:07:20.390 together at the same time? 00:07:20.390 --> 00:07:22.970 So how likely is it that all three things are 00:07:22.970 --> 00:07:26.720 going to merge in one step? 00:07:26.720 --> 00:07:29.540 No, it's not very likely, so it's no. 00:07:29.540 --> 00:07:32.680 But if it did work that way, what would it be called? 00:07:32.680 --> 00:07:34.815 What molecular reaction? 00:07:34.815 --> 00:07:35.656 Do you remember? 00:07:35.656 --> 00:07:37.104 AUDIENCE: [INAUDIBLE]. 00:07:37.104 --> 00:07:39.020 CATHERINE DRENNAN: So if you have three things 00:07:39.020 --> 00:07:39.890 it's a termolecular. 00:07:42.620 --> 00:07:43.930 So they're rare. 00:07:43.930 --> 00:07:46.480 So that's not how this works. 00:07:46.480 --> 00:07:48.020 All right. 00:07:48.020 --> 00:07:51.680 So let's look at some rate laws and try to write a rate 00:07:51.680 --> 00:07:54.332 law for a reaction. 00:07:54.332 --> 00:07:56.040 We're going to take our overall reaction, 00:07:56.040 --> 00:07:57.770 we're going to divide it to two steps 00:07:57.770 --> 00:08:00.500 and write a rate law for that and see if that's 00:08:00.500 --> 00:08:03.710 consistent with the experiment. 00:08:03.710 --> 00:08:05.210 So we finally got this up here. 00:08:05.210 --> 00:08:06.710 Sorry the handwriting isn't perfect. 00:08:06.710 --> 00:08:08.990 I got here early but then-- the first time I'm 00:08:08.990 --> 00:08:10.640 going to use the boards a lot today. 00:08:10.640 --> 00:08:11.480 Squeegee was gone. 00:08:11.480 --> 00:08:13.420 There's always something. 00:08:13.420 --> 00:08:17.270 So we're going to break this reaction down into two steps. 00:08:17.270 --> 00:08:20.187 So in the first step we have our two molecules of NO 00:08:20.187 --> 00:08:25.160 coming together to form an intermediate N2O2. 00:08:25.160 --> 00:08:28.220 And this is a reversible step. 00:08:28.220 --> 00:08:30.620 So in the second step of the reaction 00:08:30.620 --> 00:08:33.080 we have our O2 molecule coming in, 00:08:33.080 --> 00:08:35.570 reacting with our intermediate and forming 00:08:35.570 --> 00:08:38.049 two molecules of NO2. 00:08:38.049 --> 00:08:40.100 And so this is really pretty common 00:08:40.100 --> 00:08:42.380 that when you have a multistep reaction that you 00:08:42.380 --> 00:08:46.730 form an intermediate and your intermediate goes away. 00:08:46.730 --> 00:08:50.000 So now we can think about how we would write the rate law 00:08:50.000 --> 00:08:51.645 for this particular mechanism. 00:08:54.380 --> 00:08:56.440 So starting up here. 00:08:56.440 --> 00:08:58.130 This is being written as a series 00:08:58.130 --> 00:09:01.760 of steps, which were also called elementary reactions. 00:09:01.760 --> 00:09:04.850 And for an elementary reaction it occurs exactly as written. 00:09:04.850 --> 00:09:08.470 That's its definition of an elementary reaction or a step. 00:09:08.470 --> 00:09:12.560 So now we can write the rate law for the forward direction 00:09:12.560 --> 00:09:15.730 of this exactly as it is written. 00:09:15.730 --> 00:09:21.320 So that would be writing-- and these are my little K's 00:09:21.320 --> 00:09:22.880 for rate constants. 00:09:22.880 --> 00:09:29.950 So we have K1 times the concentration of NO to the 2. 00:09:29.950 --> 00:09:31.690 So we're writing it exactly as written. 00:09:31.690 --> 00:09:33.210 I said that for an overall reaction 00:09:33.210 --> 00:09:35.870 you can't just look at the stoichiometry. 00:09:35.870 --> 00:09:38.630 You have to think about experiment. 00:09:38.630 --> 00:09:41.450 But for an elementary reaction you can write it just 00:09:41.450 --> 00:09:43.420 from the stoichiometry, and so this 00:09:43.420 --> 00:09:46.130 is how we would write it just from the stoichiometry. 00:09:46.130 --> 00:09:48.920 So what would be the order of that reaction? 00:09:48.920 --> 00:09:51.090 Just the forward reaction? 00:09:51.090 --> 00:09:51.590 2. 00:09:54.300 --> 00:09:57.274 And that makes it a what kind of molecular reaction? 00:09:57.274 --> 00:09:58.190 AUDIENCE: Bimolecular. 00:09:58.190 --> 00:09:59.481 CATHERINE DRENNAN: Bimolecular. 00:10:01.930 --> 00:10:03.110 Bimolecular. 00:10:03.110 --> 00:10:04.050 OK. 00:10:04.050 --> 00:10:08.490 So let's write out the rate law for the reverse reaction now. 00:10:08.490 --> 00:10:10.740 And again, exactly as written. 00:10:10.740 --> 00:10:16.680 So we're going to have K minus 1 times the concentration 00:10:16.680 --> 00:10:18.660 of our intermediate N2O2. 00:10:21.240 --> 00:10:25.690 And that would be the rate law for the reverse reaction. 00:10:25.690 --> 00:10:30.320 So what would be the order for this reaction now? 00:10:30.320 --> 00:10:31.040 1. 00:10:31.040 --> 00:10:31.540 Right. 00:10:31.540 --> 00:10:33.390 We only have one thing in here. 00:10:33.390 --> 00:10:35.610 And what do you call an x molecular 00:10:35.610 --> 00:10:37.183 when there's one thing? 00:10:37.183 --> 00:10:38.170 AUDIENCE: Uni. 00:10:38.170 --> 00:10:39.128 CATHERINE DRENNAN: Uni. 00:10:41.460 --> 00:10:43.820 Unimolecular. 00:10:43.820 --> 00:10:45.540 For step two now you're good at this. 00:10:45.540 --> 00:10:46.720 Let's do a clicker question. 00:11:04.970 --> 00:11:05.470 All right. 00:11:05.470 --> 00:11:06.312 10 more seconds. 00:11:22.230 --> 00:11:23.670 Yup. 00:11:23.670 --> 00:11:26.470 So we're just going to write this exactly as written. 00:11:26.470 --> 00:11:28.960 It doesn't say it's a reversible reaction. 00:11:28.960 --> 00:11:32.780 So it's K2-- and I'll put this down here. 00:11:32.780 --> 00:11:38.520 K2-- our rate constant for the second step-- 00:11:38.520 --> 00:11:41.700 and then times the reactants. 00:11:41.700 --> 00:11:47.280 So we have O2 and our intermediate N2O2. 00:11:52.020 --> 00:11:56.660 So what would be the order of this reaction or this step? 00:11:56.660 --> 00:11:57.690 Two? 00:11:57.690 --> 00:11:59.550 We have two things in there. 00:11:59.550 --> 00:12:03.000 And again, we call that bimolecular. 00:12:03.000 --> 00:12:06.940 So we have uni-, bi-, and ter- molecular reactions for order 00:12:06.940 --> 00:12:09.160 of 1, 2, and 3. 00:12:09.160 --> 00:12:09.660 All right. 00:12:09.660 --> 00:12:11.550 So we've written this out. 00:12:15.030 --> 00:12:19.360 Now what we want to do is we want to write out 00:12:19.360 --> 00:12:21.950 an overall rate law for this. 00:12:21.950 --> 00:12:23.430 So we've written out the rate laws 00:12:23.430 --> 00:12:26.190 for all the individual steps. 00:12:26.190 --> 00:12:32.360 And we'll put that-- Yeah, I guess I can put it-- maybe 00:12:32.360 --> 00:12:33.787 I'll try to write it down here. 00:12:33.787 --> 00:12:35.370 I was going to have this all organized 00:12:35.370 --> 00:12:37.740 but then the boards weren't cooperating today. 00:12:37.740 --> 00:12:40.560 So I'm going to write the overall rate. 00:12:40.560 --> 00:12:41.500 Let me try it here. 00:12:46.650 --> 00:12:51.390 And this is for formation of N2O2. 00:12:51.390 --> 00:12:53.790 And we're going to put a 2 in there, 00:12:53.790 --> 00:12:55.740 and I'll explain that in a minute. 00:12:55.740 --> 00:12:59.040 Then our K2-- so we're just writing this now 00:12:59.040 --> 00:13:08.860 from the last step-- times O2 and our intermediate N2O2. 00:13:08.860 --> 00:13:15.210 So the overall rate of forming NO2 00:13:15.210 --> 00:13:18.240 is the 2 because 2 moles are formed. 00:13:18.240 --> 00:13:23.070 So as these guys disappear-- as O2 and N2O2, 00:13:23.070 --> 00:13:24.892 our intermediate, disappear-- and O's 00:13:24.892 --> 00:13:26.600 going to form twice as fast because there 00:13:26.600 --> 00:13:28.120 are two of them being formed. 00:13:28.120 --> 00:13:30.060 So we put a 2 in there. 00:13:30.060 --> 00:13:32.400 And then we just have a rate order or rate 00:13:32.400 --> 00:13:36.660 law for the last step, K2 O2 intermediate. 00:13:36.660 --> 00:13:42.220 So you can always write the rate for the overall reaction 00:13:42.220 --> 00:13:43.592 from that last step. 00:13:43.592 --> 00:13:46.050 Although sometimes we'll see if they're fast and slow steps 00:13:46.050 --> 00:13:49.570 you can also write it from the rate determining step. 00:13:49.570 --> 00:13:53.460 But we're not done here because we have an intermediate 00:13:53.460 --> 00:13:54.910 in this expression. 00:13:54.910 --> 00:13:59.160 And in a rate law you cannot have an intermediate in there. 00:13:59.160 --> 00:14:02.880 You need to solve for the rate in terms of your rate 00:14:02.880 --> 00:14:05.940 constants, your reactant concentration, and your product 00:14:05.940 --> 00:14:07.050 concentration. 00:14:07.050 --> 00:14:08.740 So we need to get rid of this. 00:14:08.740 --> 00:14:13.020 We need to solve for this. 00:14:13.020 --> 00:14:14.590 So how are we going to do that? 00:14:14.590 --> 00:14:17.350 How are we going to solve for this? 00:14:17.350 --> 00:14:20.350 So we want to think about now, what 00:14:20.350 --> 00:14:23.640 is the net rate for this formation? 00:14:23.640 --> 00:14:25.950 How is that intermediate being formed, 00:14:25.950 --> 00:14:30.130 and how is it being consumed given those steps up here? 00:14:30.130 --> 00:14:35.415 So it's only being formed in the forward direction 00:14:35.415 --> 00:14:38.160 of the first step. 00:14:38.160 --> 00:14:42.000 And let me just grab this. 00:14:42.000 --> 00:14:47.970 So the forward direction of the first step, which is K1 times 00:14:47.970 --> 00:14:52.330 NO to the 2. 00:14:52.330 --> 00:14:55.150 And so this is where our intermediate is formed. 00:14:58.780 --> 00:15:03.010 Our intermediate is going away in two different steps. 00:15:03.010 --> 00:15:07.570 So it's going away in the reverse direction of step one. 00:15:07.570 --> 00:15:11.410 So it's decomposing in that step. 00:15:11.410 --> 00:15:13.730 And so it's decomposing at the rate 00:15:13.730 --> 00:15:16.980 of K minus 1 times its concentration. 00:15:20.280 --> 00:15:24.090 The intermediate is also being consumed in the second step. 00:15:24.090 --> 00:15:27.210 It's reacting with oxygen and being consumed. 00:15:27.210 --> 00:15:36.510 So here it decays and here it is consumed. 00:15:36.510 --> 00:15:38.640 And it's being consumed by the reaction 00:15:38.640 --> 00:15:44.046 K2 times the concentration of O2 times its concentration. 00:15:47.670 --> 00:15:51.050 So that's the second step. 00:15:51.050 --> 00:15:54.230 So now we just need to take that equation 00:15:54.230 --> 00:15:58.850 and solve for the intermediate N2O2. 00:15:58.850 --> 00:16:02.690 But we don't know net rate over there. 00:16:02.690 --> 00:16:05.090 We have too many variables right now. 00:16:05.090 --> 00:16:08.600 So at this point we have to use what's known as a steady state 00:16:08.600 --> 00:16:10.490 approximation. 00:16:10.490 --> 00:16:14.885 So steady state approximation and pretty much everything-- 00:16:14.885 --> 00:16:16.880 I'm talking about reaction mechanisms-- 00:16:16.880 --> 00:16:19.560 we're going to use the steady state approximation. 00:16:19.560 --> 00:16:23.450 And that is that the rate of formation of intermediates 00:16:23.450 --> 00:16:26.810 equals the rate at which they go away. 00:16:26.810 --> 00:16:29.900 So the net rate is 0. 00:16:29.900 --> 00:16:36.230 So we can set this whole thing now equal to 0. 00:16:36.230 --> 00:16:38.570 So steady state approximation net rate 00:16:38.570 --> 00:16:44.060 is 0, or the rate at which an intermediate forms equals 00:16:44.060 --> 00:16:49.190 the rate at which that intermediate decays. 00:16:49.190 --> 00:16:52.140 So I can rearrange this equation now, 00:16:52.140 --> 00:16:54.410 and I'm going to bring the terms that have 00:16:54.410 --> 00:16:57.530 the intermediate to one side. 00:16:57.530 --> 00:16:59.970 So I'm going to put them over here. 00:16:59.970 --> 00:17:03.830 So we have the term at which it decays-- 00:17:03.830 --> 00:17:07.430 K minus 1 times our concentration 00:17:07.430 --> 00:17:09.599 of our intermediate. 00:17:09.599 --> 00:17:12.260 And it had a negative but I brought it over here 00:17:12.260 --> 00:17:15.260 to this side with a 0, so now it's positive. 00:17:15.260 --> 00:17:19.250 And I'm going to bring the same for the rate law at which it's 00:17:19.250 --> 00:17:20.869 consumed over. 00:17:20.869 --> 00:17:27.680 So that's K2 times our intermediate concentration 00:17:27.680 --> 00:17:30.770 and our oxygen concentration. 00:17:30.770 --> 00:17:32.570 And now on the other side we'll have 00:17:32.570 --> 00:17:34.730 the rate at which the intermediate is formed, 00:17:34.730 --> 00:17:39.080 which is K1 No to the 2. 00:17:39.080 --> 00:17:41.750 So this is another way to express the steady state 00:17:41.750 --> 00:17:43.010 approximation. 00:17:43.010 --> 00:17:46.070 The rate at which the intermediate goes away 00:17:46.070 --> 00:17:48.240 equals the rate at which it's formed. 00:17:48.240 --> 00:17:49.640 That's the steady state. 00:17:49.640 --> 00:17:52.730 There's no sort of flux in the intermediate. 00:17:52.730 --> 00:17:56.270 It's being formed and going away at equal rates. 00:17:56.270 --> 00:17:59.240 So now we can use this to solve for the intermediate. 00:17:59.240 --> 00:18:00.440 Now we're set. 00:18:00.440 --> 00:18:02.780 Now we can solve for the intermediate. 00:18:02.780 --> 00:18:06.060 And so let's do that over here. 00:18:06.060 --> 00:18:09.680 So I'm going to pull out-- I had a straight line here. 00:18:09.680 --> 00:18:12.150 This one's a little crooked. 00:18:12.150 --> 00:18:13.610 I used to write a lot on the board 00:18:13.610 --> 00:18:17.090 and used to evaluate professors by their good handwriting 00:18:17.090 --> 00:18:20.630 and my ratings were always-- my overall rating was limited 00:18:20.630 --> 00:18:22.527 by my handwriting to a large degree 00:18:22.527 --> 00:18:24.110 and so I stopped writing on the board. 00:18:24.110 --> 00:18:26.450 But now they've got rid of that as a criteria, 00:18:26.450 --> 00:18:28.790 so I can write on the board again. 00:18:28.790 --> 00:18:32.030 So I'm going to pull out the concentration 00:18:32.030 --> 00:18:37.820 of the intermediate-- our N2O2. 00:18:37.820 --> 00:18:41.560 and I'll pull it out of the expression leaving 00:18:41.560 --> 00:18:52.410 K minus 1 and K2 times O2. 00:18:52.410 --> 00:18:54.290 So I just pulled out the concentration 00:18:54.290 --> 00:18:56.690 of the intermediate over here. 00:18:56.690 --> 00:18:58.700 And then we leave the other side the same. 00:18:58.700 --> 00:19:03.950 Our K1 times our NO squared. 00:19:03.950 --> 00:19:07.010 So now I can solve for the concentration 00:19:07.010 --> 00:19:09.060 of the intermediate. 00:19:09.060 --> 00:19:20.780 N2O2 equals K1 times NO squared over K minus 1 00:19:20.780 --> 00:19:26.830 plus K2 times O2. 00:19:26.830 --> 00:19:27.890 Great. 00:19:27.890 --> 00:19:30.020 So we've solved for the concentration 00:19:30.020 --> 00:19:31.840 of the intermediate now. 00:19:31.840 --> 00:19:37.580 Now we can take this and bring it back over here 00:19:37.580 --> 00:19:40.430 and put that whole term into this. 00:19:40.430 --> 00:19:43.850 And then we'll have a rate law that is expressed only 00:19:43.850 --> 00:19:48.010 in terms of our rate constants and our reactants 00:19:48.010 --> 00:19:48.676 and/or products. 00:19:51.290 --> 00:19:58.520 So let's do more on the PowerPoint here. 00:19:58.520 --> 00:20:02.310 So this is what we just came up with. 00:20:02.310 --> 00:20:06.290 So the concentration of our intermediate K1 times 00:20:06.290 --> 00:20:11.060 NO-- one of our reactants second order-- over K minus 1 00:20:11.060 --> 00:20:15.500 plus K2 times the concentration of oxygen. 00:20:15.500 --> 00:20:21.620 Now I'm going to plug that into this expression, which we just 00:20:21.620 --> 00:20:26.720 got from writing the rate law for the last step using a 2 00:20:26.720 --> 00:20:29.850 because we have two molecules of product being formed. 00:20:29.850 --> 00:20:32.880 So we're going to plug it in there, 00:20:32.880 --> 00:20:34.830 and that's going to give us this. 00:20:34.830 --> 00:20:39.840 So we have our 2, we have a K1 from here, a K2 from here, 00:20:39.840 --> 00:20:42.600 we have our oxygen concentration, 00:20:42.600 --> 00:20:46.980 we have our NO overconcentration second order over K minus 1 00:20:46.980 --> 00:20:52.220 plus K2 times concentration of oxygen. 00:20:52.220 --> 00:20:55.920 So that might be the complete answer to some problems 00:20:55.920 --> 00:20:59.100 but here we were given an experimental rate 00:20:59.100 --> 00:21:03.900 law, which was second order in NO and first order in O2. 00:21:03.900 --> 00:21:06.720 That does not match this. 00:21:06.720 --> 00:21:08.850 Term has an O2 at the bottom. 00:21:08.850 --> 00:21:11.250 This one doesn't have an O2 at the bottom. 00:21:11.250 --> 00:21:13.050 These are not the same. 00:21:13.050 --> 00:21:15.960 So that must mean that the mechanism has 00:21:15.960 --> 00:21:19.010 fast and slow steps and that we need 00:21:19.010 --> 00:21:25.560 to reconsider this expression in terms of fast and slow steps. 00:21:25.560 --> 00:21:30.220 So let's go back now and think about our mechanism. 00:21:30.220 --> 00:21:33.810 And so if we come over here-- let's 00:21:33.810 --> 00:21:42.570 say that the first step-- let's bring this down again. 00:21:42.570 --> 00:21:46.250 Let's say the first step is fast. 00:21:46.250 --> 00:21:48.470 And we already said that it was reversible, 00:21:48.470 --> 00:21:51.250 but we'll put that down too. 00:21:51.250 --> 00:21:53.460 Is fast and reversible. 00:21:53.460 --> 00:21:56.820 And step two is slow. 00:21:56.820 --> 00:22:00.120 So I'm just going to propose that these are true, 00:22:00.120 --> 00:22:04.680 and then we will recalculate what the rate law would 00:22:04.680 --> 00:22:09.020 be if you have a fast reversible step followed by a slow step 00:22:09.020 --> 00:22:10.715 and see if that agrees with experiment. 00:22:16.440 --> 00:22:21.720 So to do that we have to consider 00:22:21.720 --> 00:22:27.010 what it means if we have a fast step followed by a slow step. 00:22:27.010 --> 00:22:31.740 So let's introduce a term-- very important term-- which 00:22:31.740 --> 00:22:33.840 is the rate determining step. 00:22:33.840 --> 00:22:37.650 Also known as the rate limiting step. 00:22:37.650 --> 00:22:42.660 So the slow step of a reaction, if it's truly a very slow step, 00:22:42.660 --> 00:22:45.912 is going to govern the overall rate of the reaction. 00:22:45.912 --> 00:22:47.370 So let's think about this a minute. 00:22:47.370 --> 00:22:51.660 So I told you that the extra problems for exam four 00:22:51.660 --> 00:22:52.770 are long. 00:22:52.770 --> 00:22:54.020 They're very long. 00:22:54.020 --> 00:22:56.770 Sorry about that, but there were a lot of problems 00:22:56.770 --> 00:22:58.770 and I wanted to get you ready for exam four. 00:22:58.770 --> 00:23:02.530 You also have problem set nine due tomorrow. 00:23:02.530 --> 00:23:04.710 So you've got a lot of problems to do. 00:23:04.710 --> 00:23:08.840 So after class today I feel that many of you 00:23:08.840 --> 00:23:11.970 are going to be really inspired to get 00:23:11.970 --> 00:23:14.490 started on those problems. 00:23:14.490 --> 00:23:16.010 And you may run out of here. 00:23:16.010 --> 00:23:18.839 You might leap over the chair in front of you and race 00:23:18.839 --> 00:23:19.380 out the door. 00:23:19.380 --> 00:23:21.370 You may clear the table on the way out 00:23:21.370 --> 00:23:25.520 because you're in a real hurry to start those extra problems. 00:23:25.520 --> 00:23:28.022 You will run to the library to look for a table, 00:23:28.022 --> 00:23:29.730 but all the tables will already be taken. 00:23:29.730 --> 00:23:32.160 How did your classmates get out of class so fast 00:23:32.160 --> 00:23:34.340 to get all of the tables in the library? 00:23:34.340 --> 00:23:36.450 And they're already finishing problem set nine 00:23:36.450 --> 00:23:39.450 and starting on the extra problems. 00:23:39.450 --> 00:23:41.910 So you race back to your dorm, but all the tables 00:23:41.910 --> 00:23:43.890 in the downstairs of the dorm are also 00:23:43.890 --> 00:23:46.230 filled with 5.111 students. 00:23:46.230 --> 00:23:48.672 So finally, on the fifth floor of one of the dorms 00:23:48.672 --> 00:23:50.880 you find an empty table, and then you're really fast. 00:23:50.880 --> 00:23:52.440 You got the problems out, you got your pencils out, 00:23:52.440 --> 00:23:54.898 you got your calculator out, you've got old equation sheets 00:23:54.898 --> 00:23:56.110 out, and ready to go. 00:23:56.110 --> 00:23:58.170 It's like two seconds. 00:23:58.170 --> 00:24:00.900 So it took you 40 minutes to find a free table 00:24:00.900 --> 00:24:04.590 at MIT that didn't already have a 5.111 student sitting 00:24:04.590 --> 00:24:07.860 and doing those extra problems and problem set nine. 00:24:07.860 --> 00:24:10.230 So it was 40 minutes plus 2 seconds 00:24:10.230 --> 00:24:12.850 to actually start doing the problems. 00:24:12.850 --> 00:24:16.310 So the rate determining or rate limiting step 00:24:16.310 --> 00:24:17.760 was finding your table. 00:24:17.760 --> 00:24:21.240 40 minutes plus 2 seconds is pretty much 40 minutes, 00:24:21.240 --> 00:24:23.700 and that's what happens in these reaction mechanisms. 00:24:23.700 --> 00:24:25.980 If you have a really slow step that 00:24:25.980 --> 00:24:29.650 governs the overall rate of the reaction. 00:24:29.650 --> 00:24:32.190 Now a lot of you know also about rate determining steps 00:24:32.190 --> 00:24:36.399 because some of you may be the rate determining step 00:24:36.399 --> 00:24:37.440 in your group of friends. 00:24:40.650 --> 00:24:44.320 The rate at which you get to dinner and eat 00:24:44.320 --> 00:24:47.020 is determined by you being ready to go. 00:24:49.714 --> 00:24:51.630 Some are in the room like me, yeah, that's me. 00:24:51.630 --> 00:24:54.730 You know who you are. 00:24:54.730 --> 00:24:56.190 So I'm saying that what you gotta 00:24:56.190 --> 00:25:02.200 do to not let yourself be that person-- that rate determining 00:25:02.200 --> 00:25:08.850 person, the RDS in your group of friends-- you need to get sleep 00:25:08.850 --> 00:25:11.740 and you need to eat well and you need to make sure 00:25:11.740 --> 00:25:15.150 you got your ATP. 00:25:15.150 --> 00:25:18.280 And that means, of course, to get 00:25:18.280 --> 00:25:23.170 enough sleep you gotta start problems sets early, especially 00:25:23.170 --> 00:25:25.640 the extra problems because they're long. 00:25:28.355 --> 00:25:30.820 Rate determining steps. 00:25:30.820 --> 00:25:31.752 Very important. 00:25:36.870 --> 00:25:40.510 So let's get back to our example. 00:25:40.510 --> 00:25:43.410 And we made a proposal. 00:25:43.410 --> 00:25:46.480 We made a proposal that step two was 00:25:46.480 --> 00:25:47.830 going to be rate determining. 00:25:47.830 --> 00:25:49.380 That was slow. 00:25:49.380 --> 00:25:53.320 We made the proposal that step one was fast and reversible. 00:25:53.320 --> 00:25:55.920 Step two was slow. 00:25:55.920 --> 00:25:58.620 So what that's going to mean then 00:25:58.620 --> 00:26:04.240 is that our rate law for the first step-- that's fast. 00:26:04.240 --> 00:26:06.250 That's a big number. 00:26:06.250 --> 00:26:09.760 The rate for the second step is slow. 00:26:09.760 --> 00:26:11.590 Rate determining. 00:26:11.590 --> 00:26:18.210 So that will mean then that K1 is going to be greater than K2 00:26:18.210 --> 00:26:21.767 times O2, so we can drop out our concentrations here 00:26:21.767 --> 00:26:22.600 to think about them. 00:26:22.600 --> 00:26:23.830 They're going to be the same. 00:26:23.830 --> 00:26:24.590 So what's left? 00:26:24.590 --> 00:26:27.660 That means the rate constant for that reverse step is very fast. 00:26:27.660 --> 00:26:28.940 It's a big number. 00:26:28.940 --> 00:26:33.790 That's going to be big compared to K2 times O2. 00:26:33.790 --> 00:26:36.210 So now we can go back and look at our expression 00:26:36.210 --> 00:26:37.840 for the intermediate. 00:26:37.840 --> 00:26:41.430 And we note that K minus 1 is in the bottom 00:26:41.430 --> 00:26:46.700 as well as K2 times O2 is in the bottom of the expression. 00:26:46.700 --> 00:26:51.590 So if K minus 1 now is really big compared to K2 times 00:26:51.590 --> 00:26:54.200 O2-- again, that's the fast step; that's 00:26:54.200 --> 00:26:57.640 the slow step-- then this term pretty much doesn't matter 00:26:57.640 --> 00:26:59.640 and it can drop out. 00:26:59.640 --> 00:27:04.750 Because this is really small compared to that. 00:27:04.750 --> 00:27:09.640 And if we drop out this term we can rewrite the expression 00:27:09.640 --> 00:27:10.810 like this. 00:27:10.810 --> 00:27:15.010 The concentration of our intermediate rate constant K1 00:27:15.010 --> 00:27:18.860 NO squared over K minus 1. 00:27:18.860 --> 00:27:21.280 And now we can rearrange this equation, 00:27:21.280 --> 00:27:25.390 bringing the concentrations to one side and our rate constants 00:27:25.390 --> 00:27:26.780 to the other side. 00:27:26.780 --> 00:27:31.410 So we have our intermediate N2O2 over NO squared 00:27:31.410 --> 00:27:35.080 equals rate constant K1 over K minus 1. 00:27:35.080 --> 00:27:42.700 What does rate constant K1 over rate constant K minus 1 equal? 00:27:42.700 --> 00:27:44.700 AUDIENCE: [INAUDIBLE]. 00:27:44.700 --> 00:27:48.210 CATHERINE DRENNAN: It equals the equilibrium constant K1 00:27:48.210 --> 00:27:50.420 for the first step. 00:27:50.420 --> 00:27:55.450 So if you have a fast reversible step followed by a slow step, 00:27:55.450 --> 00:27:58.480 the first step is basically in equilibrium. 00:27:58.480 --> 00:28:03.900 We can make that approximation that it is in equilibrium 00:28:03.900 --> 00:28:07.540 and solve it by thinking about equilibrium expressions. 00:28:07.540 --> 00:28:09.040 Fantastic. 00:28:09.040 --> 00:28:12.590 We can be back to equilibrium expressions. 00:28:12.590 --> 00:28:15.600 So let's think about this a little bit more. 00:28:15.600 --> 00:28:17.880 Here I have a pretty picture for you. 00:28:17.880 --> 00:28:20.590 So here we have our reactants forming our intermediates. 00:28:20.590 --> 00:28:23.500 The intermediates are also going back and forming our reactants. 00:28:23.500 --> 00:28:25.840 Reactants are forming the intermediates and then 00:28:25.840 --> 00:28:26.770 back again. 00:28:26.770 --> 00:28:28.540 Fast reversible. 00:28:28.540 --> 00:28:32.130 Every once in a while an intermediate gets siphoned off 00:28:32.130 --> 00:28:36.040 to products, but if this is a really, really slow step 00:28:36.040 --> 00:28:37.270 it doesn't happen very often. 00:28:37.270 --> 00:28:40.570 It doesn't really affect this very much. 00:28:40.570 --> 00:28:42.890 And so basically, this is in equilibrium. 00:28:42.890 --> 00:28:44.900 It's like this part doesn't even really matter. 00:28:44.900 --> 00:28:46.440 It doesn't play in. 00:28:46.440 --> 00:28:49.960 So when we have a fast reversible step followed 00:28:49.960 --> 00:28:52.840 by a slow step, we can assume the first step is 00:28:52.840 --> 00:28:56.610 in equilibrium and we can solve for our intermediate 00:28:56.610 --> 00:29:00.860 using equilibrium expressions. 00:29:00.860 --> 00:29:02.150 So let's do that. 00:29:02.150 --> 00:29:05.260 Let's take our equilibrium expression for the first step 00:29:05.260 --> 00:29:09.100 and now plug it in to our rate law. 00:29:09.100 --> 00:29:12.040 So we can substitute this step now 00:29:12.040 --> 00:29:14.740 or we can have rate constants or we can 00:29:14.740 --> 00:29:16.056 have an equilibrium constant. 00:29:16.056 --> 00:29:16.930 You can write either. 00:29:16.930 --> 00:29:18.550 These are equivalent. 00:29:18.550 --> 00:29:21.900 And we can put those back into this, 00:29:21.900 --> 00:29:25.980 which was our original overall rate that we wrote. 00:29:25.980 --> 00:29:28.440 We weren't done though because we had an intermediate. 00:29:28.440 --> 00:29:33.070 So we can plug this now in for our concentration 00:29:33.070 --> 00:29:35.130 of the intermediate. 00:29:35.130 --> 00:29:43.600 And so now we get 2 times K1 K2 O2 NO squared over K minus 1. 00:29:43.600 --> 00:29:46.610 Or we could just put that with the big equilibrium constant 00:29:46.610 --> 00:29:50.560 and get rid of our little K1 over K minus 1. 00:29:50.560 --> 00:29:52.920 Both of those are equivalent. 00:29:52.920 --> 00:29:59.340 And now we can take all of our K terms and call them K obs. 00:29:59.340 --> 00:30:02.430 So K obs is just the experimental rate constant. 00:30:02.430 --> 00:30:06.070 It's the collection of rate constants that are measured. 00:30:06.070 --> 00:30:09.040 And we often-- when we measure things, 00:30:09.040 --> 00:30:12.520 we can't distinguish K1 from K2. 00:30:12.520 --> 00:30:14.740 We sometimes try to do that, and that's 00:30:14.740 --> 00:30:15.840 a little more complicated. 00:30:15.840 --> 00:30:17.980 But in this case, all that was given 00:30:17.980 --> 00:30:20.850 was an overall K observed. 00:30:20.850 --> 00:30:27.310 And this was our experimental rate law K observed times O2 00:30:27.310 --> 00:30:29.560 first order NO second order. 00:30:29.560 --> 00:30:34.210 And now we see this expression agrees with this rate. 00:30:34.210 --> 00:30:37.170 So the fact that we have a good agreement 00:30:37.170 --> 00:30:42.100 means that a mechanism with this fast reversible step followed 00:30:42.100 --> 00:30:46.110 by a slow step gives rise to a rate law that's 00:30:46.110 --> 00:30:47.600 consistent with experiment. 00:30:47.600 --> 00:30:49.510 It doesn't prove that's the right mechanism. 00:30:49.510 --> 00:30:52.680 It's very hard to prove mechanisms are right, 00:30:52.680 --> 00:30:54.630 but at least it's consistent. 00:30:54.630 --> 00:30:59.290 So we can say this is a good guess, a good proposal, 00:30:59.290 --> 00:31:00.110 for our mechanism. 00:31:02.660 --> 00:31:04.578 So let's look at another example. 00:31:08.430 --> 00:31:11.570 So in this example we have NO again. 00:31:11.570 --> 00:31:13.940 We have two molecules of NO, and now we 00:31:13.940 --> 00:31:18.500 have Br2 going to two molecules of NOBr. 00:31:18.500 --> 00:31:21.660 And we're told that the experimental rate is 00:31:21.660 --> 00:31:27.610 K obs times NO first order, Br2 first order 00:31:27.610 --> 00:31:31.310 and asked, for this proposed mechanism, 00:31:31.310 --> 00:31:35.440 which would be the slow step to give rise 00:31:35.440 --> 00:31:39.050 to that experimental data? 00:31:39.050 --> 00:31:42.290 So the first thing that we would want to do with all of these 00:31:42.290 --> 00:31:48.400 is to write the rate laws for each individual step. 00:31:48.400 --> 00:31:51.290 So for the forward step we have one molecule 00:31:51.290 --> 00:31:58.420 of NO reacting with Br2 with rate constant K2. 00:31:58.420 --> 00:32:03.560 So we get rate constant K1 times the concentration of NO times 00:32:03.560 --> 00:32:05.650 the concentration of Br2. 00:32:05.650 --> 00:32:08.580 Again, this is a step or an elementary reaction, 00:32:08.580 --> 00:32:13.960 so we write the rate law just based on the stoichiometry 00:32:13.960 --> 00:32:15.610 here. 00:32:15.610 --> 00:32:19.210 Now we can do the same thing for the reverse rate. 00:32:19.210 --> 00:32:23.570 So we have K minus 1 times the concentration 00:32:23.570 --> 00:32:26.330 of our intermediate. 00:32:26.330 --> 00:32:30.320 So in step two our intermediate, which is formed in step one, 00:32:30.320 --> 00:32:33.460 is reacting with the second molecule NO, 00:32:33.460 --> 00:32:35.680 forming our product. 00:32:35.680 --> 00:32:38.630 And we can write the rate for this as well. 00:32:38.630 --> 00:32:43.520 K2 times the concentration of our intermediate, NOBr2, 00:32:43.520 --> 00:32:45.700 times the concentration of NO. 00:32:45.700 --> 00:32:47.440 So again, these are steps. 00:32:47.440 --> 00:32:49.460 They're elementary reactions so we 00:32:49.460 --> 00:32:52.600 can write the rate law based on the stoichiometry 00:32:52.600 --> 00:32:55.820 in that proposed step. 00:32:55.820 --> 00:32:59.620 So now we can write the overall rate law 00:32:59.620 --> 00:33:02.380 for the formation of NOBr, and we can just 00:33:02.380 --> 00:33:04.860 write it from the second step like we did before. 00:33:04.860 --> 00:33:06.020 Again, this is an example. 00:33:06.020 --> 00:33:07.820 We're forming two molecules of product 00:33:07.820 --> 00:33:09.260 so there's a two in there. 00:33:09.260 --> 00:33:10.900 We have K2. 00:33:10.900 --> 00:33:12.130 It's basically just this. 00:33:12.130 --> 00:33:15.800 K2 times our intermediate, NOBr2, 00:33:15.800 --> 00:33:18.370 times the concentration of NO. 00:33:18.370 --> 00:33:21.450 But once again, we're not done because there 00:33:21.450 --> 00:33:23.780 is an intermediate in the expression 00:33:23.780 --> 00:33:26.710 and you can't have an intermediate in your rate law. 00:33:26.710 --> 00:33:31.180 You need to solve for the rate law in terms of rate constants, 00:33:31.180 --> 00:33:33.530 reactants, and products. 00:33:33.530 --> 00:33:37.360 So we need to now solve for our intermediate 00:33:37.360 --> 00:33:40.550 in terms of things that are allowed in the overall rate 00:33:40.550 --> 00:33:42.080 law. 00:33:42.080 --> 00:33:44.420 And so we want to think again about what 00:33:44.420 --> 00:33:49.480 is the change in concentration of our intermediate 00:33:49.480 --> 00:33:52.550 So we can do the same thing that we did before? 00:33:52.550 --> 00:33:56.960 So the intermediate is being formed in the first step. 00:33:56.960 --> 00:34:01.440 So we have the rate law for the first step. 00:34:01.440 --> 00:34:07.480 K1 times NO times the concentration of Br2. 00:34:07.480 --> 00:34:13.120 The intermediate is also decaying in the reverse part 00:34:13.120 --> 00:34:15.429 of the first step. 00:34:15.429 --> 00:34:20.920 So that's minus the reverse rate minus K minus 1 times 00:34:20.920 --> 00:34:23.540 the intermediate here. 00:34:23.540 --> 00:34:27.020 And then it's being consumed in the second step. 00:34:27.020 --> 00:34:32.870 So it's going away by the rate K2 times the concentration 00:34:32.870 --> 00:34:36.610 of the intermediate times the concentration of NO. 00:34:36.610 --> 00:34:40.040 So again, this is exactly what we did with the first example. 00:34:40.040 --> 00:34:43.480 We think about the change, how it's being formed, 00:34:43.480 --> 00:34:47.679 and the two different ways that it's being consumed. 00:34:47.679 --> 00:34:51.159 We can again use a steady state approximation 00:34:51.159 --> 00:34:55.730 and set all of that equal to 0. 00:34:55.730 --> 00:35:00.920 So let's do that and we'll solve again for the intermediate. 00:35:00.920 --> 00:35:05.150 So on this next slide now-- I just put those things up there. 00:35:05.150 --> 00:35:06.980 This is what you were just copying down. 00:35:06.980 --> 00:35:09.810 If you didn't finish it's still here. 00:35:09.810 --> 00:35:12.290 And here is the steady state approximation. 00:35:12.290 --> 00:35:14.170 So again, the steady straight approximation 00:35:14.170 --> 00:35:16.480 is the net rate of formation of your intermediate 00:35:16.480 --> 00:35:18.940 equals the net rate of it's going away. 00:35:18.940 --> 00:35:21.190 Net rate is 0. 00:35:21.190 --> 00:35:25.150 So rearranging then we can bring the two terms that 00:35:25.150 --> 00:35:29.380 involve the decay or the consumption 00:35:29.380 --> 00:35:32.720 of our intermediate on one side and then set them equal 00:35:32.720 --> 00:35:35.910 to the rate at which that intermediate is formed. 00:35:35.910 --> 00:35:39.370 And then, we can pull out our terms for our intermediate. 00:35:39.370 --> 00:35:45.882 So we pull out NoBr2, leaving K minus 1, leaving K2 and NO, 00:35:45.882 --> 00:35:51.020 and set it equal to the rate law for the first step 00:35:51.020 --> 00:35:52.970 in the forward direction. 00:35:52.970 --> 00:35:54.670 We can solve for the intermediate. 00:35:54.670 --> 00:35:57.180 Take this, divide it by this term. 00:35:57.180 --> 00:36:00.500 So we have K1 times NO times Br2 over K 00:36:00.500 --> 00:36:04.540 minus 1 plus K2 times NO. 00:36:04.540 --> 00:36:05.900 Now we can take this. 00:36:05.900 --> 00:36:08.210 We are done solving for our intermediate. 00:36:08.210 --> 00:36:10.730 We have no more intermediates in there, 00:36:10.730 --> 00:36:13.790 so we can now plug this back in to the expression we 00:36:13.790 --> 00:36:15.080 had before. 00:36:15.080 --> 00:36:19.360 So we can take this, plug it in here, 00:36:19.360 --> 00:36:21.460 and that gives us this formation. 00:36:21.460 --> 00:36:28.540 So we have 2 times K1, NO times NO-- NO squared-- Br2 on top, 00:36:28.540 --> 00:36:32.398 K minus 1 on the bottom plus K2 times NO. 00:36:35.480 --> 00:36:38.030 Now we were asked, what are the fast 00:36:38.030 --> 00:36:39.840 and what are the slow steps? 00:36:39.840 --> 00:36:42.410 So now we want to take this and think about, 00:36:42.410 --> 00:36:45.470 if there's different fast and slow steps, 00:36:45.470 --> 00:36:49.520 is it consistent then with the experiment? 00:36:49.520 --> 00:36:54.110 So first, let's consider if the first step was slow 00:36:54.110 --> 00:36:58.250 and the second step was fast-- or i.e., 00:36:58.250 --> 00:37:03.860 if we have K2 NO greater than K minus 1. 00:37:03.860 --> 00:37:07.280 And this is a clicker question, so why don't you tell me how 00:37:07.280 --> 00:37:10.820 this then, using this, changes? 00:37:27.730 --> 00:37:28.230 OK. 00:37:28.230 --> 00:37:28.780 10 seconds. 00:37:44.300 --> 00:37:44.870 OK. 00:37:44.870 --> 00:37:49.690 Let's now think about why that's true. 00:37:49.690 --> 00:37:54.430 This involved doing a couple of steps in your head. 00:37:54.430 --> 00:37:57.700 So if we have a first step that's slow 00:37:57.700 --> 00:38:00.370 and a second step that's fast, the second step 00:38:00.370 --> 00:38:03.940 involves the K2 times the concentration of NO. 00:38:03.940 --> 00:38:04.920 That's the second step. 00:38:04.920 --> 00:38:05.860 That's fast. 00:38:05.860 --> 00:38:09.020 That's going to be a big number compared to K minus 1. 00:38:09.020 --> 00:38:11.710 So if we look, both are on the bottom here. 00:38:11.710 --> 00:38:14.810 And if this term, K2 NO, is much, much bigger 00:38:14.810 --> 00:38:20.770 than K minus 1, then we can say K minus 1 goes way. 00:38:20.770 --> 00:38:25.120 If we get rid of K minus 1, we can simplify 00:38:25.120 --> 00:38:27.070 the expression even more. 00:38:27.070 --> 00:38:30.820 We can get rid of our K2s and we can get rid 00:38:30.820 --> 00:38:34.745 of one of our NOs, which gives us this. 00:38:37.420 --> 00:38:40.300 So saying that the first step is slow 00:38:40.300 --> 00:38:42.130 and the second step is fast gives us 00:38:42.130 --> 00:38:44.620 a very different equation. 00:38:44.620 --> 00:38:47.420 A lot of things cancel out. 00:38:47.420 --> 00:38:51.240 So we can also write that expression as K 00:38:51.240 --> 00:38:56.050 obs times NO times Br2. 00:38:56.050 --> 00:39:02.000 And the overall order of that reaction would be what? 00:39:02.000 --> 00:39:04.226 Yell it out. 00:39:04.226 --> 00:39:05.170 AUDIENCE: Two. 00:39:05.170 --> 00:39:07.090 CATHERINE DRENNAN: Yes. 00:39:07.090 --> 00:39:10.030 So this is what we would get for a first step that's slow; 00:39:10.030 --> 00:39:11.860 second step that's fast. 00:39:11.860 --> 00:39:16.450 Now let's consider if the first step is fast 00:39:16.450 --> 00:39:19.300 and the second step is slow. 00:39:19.300 --> 00:39:22.510 So if the first step is fast that means K minus 1, 00:39:22.510 --> 00:39:25.000 the rate constant for the reverse step, 00:39:25.000 --> 00:39:29.910 is going to be a lot bigger than K2 times NO. 00:39:29.910 --> 00:39:32.530 And so now we can look up at this expression 00:39:32.530 --> 00:39:35.890 and say, OK, if this is much bigger than this 00:39:35.890 --> 00:39:38.710 then that cancels out. 00:39:38.710 --> 00:39:41.890 And then we're left with this expression, 00:39:41.890 --> 00:39:44.410 which I can put down here. 00:39:44.410 --> 00:39:48.190 So that leaves us-- we can't cancel any more at this point. 00:39:48.190 --> 00:39:51.190 So that leaves us with 2 times K1 times 00:39:51.190 --> 00:39:57.950 K2-- these-- NO squared Br2 over K minus 1. 00:39:57.950 --> 00:40:01.390 So assuming different things about how fast and slow 00:40:01.390 --> 00:40:06.070 the steps are gives you very different rate laws. 00:40:06.070 --> 00:40:08.890 We can also write that to make it look a little simpler 00:40:08.890 --> 00:40:14.530 as K obs, but you'll note that the overall order is 00:40:14.530 --> 00:40:15.560 very different. 00:40:15.560 --> 00:40:18.330 So what's the overall order here? 00:40:18.330 --> 00:40:19.000 Three. 00:40:19.000 --> 00:40:20.080 Right. 00:40:20.080 --> 00:40:25.060 So let's remind ourselves what the experimental rate law was. 00:40:25.060 --> 00:40:29.380 And it was NO first order Br2 first order. 00:40:29.380 --> 00:40:35.170 So that means that this one would be consistent. 00:40:35.170 --> 00:40:40.210 So the mechanism is likely to involve a slow first step 00:40:40.210 --> 00:40:42.670 and a fast second step. 00:40:42.670 --> 00:40:45.400 And so that's how you do a lot of these problems. 00:40:45.400 --> 00:40:47.830 You think about what is going to change when you have 00:40:47.830 --> 00:40:50.350 different fast and slow steps. 00:40:50.350 --> 00:40:52.150 One of them will be more consistent 00:40:52.150 --> 00:40:54.310 with the experimental data and one of them 00:40:54.310 --> 00:40:56.210 will not be consistent. 00:40:56.210 --> 00:41:01.690 OK Let's do one more fast example. 00:41:01.690 --> 00:41:07.720 Here we have rate law for two molecules of ozone O3 going 00:41:07.720 --> 00:41:11.010 to three molecules of O2. 00:41:11.010 --> 00:41:14.290 And ozone has been in the news a lot recently. 00:41:14.290 --> 00:41:17.020 So we want to keep our ozone layer. 00:41:17.020 --> 00:41:21.070 We don't want it to go away. 00:41:21.070 --> 00:41:25.690 So we have O3 going to O2 plus O, 00:41:25.690 --> 00:41:28.450 and you're forming an intermediate O. 00:41:28.450 --> 00:41:33.010 That intermediate is reacting with O3, forming 00:41:33.010 --> 00:41:35.510 our two molecules of O2. 00:41:35.510 --> 00:41:39.130 So let's just write out what our rate 00:41:39.130 --> 00:41:41.270 is for the forward reaction. 00:41:41.270 --> 00:41:47.430 So we have K1 times the concentration of O3. 00:41:47.430 --> 00:41:53.770 For the reverse we have K minus 1 times concentration 00:41:53.770 --> 00:41:59.120 of O2 times the concentration of our intermediate O. 00:41:59.120 --> 00:42:02.860 For the next one we have K2 times our intermediate O 00:42:02.860 --> 00:42:06.640 times the concentration of O3. 00:42:06.640 --> 00:42:09.400 So now we're told that there's a fast reversible step 00:42:09.400 --> 00:42:11.300 and a slow step. 00:42:11.300 --> 00:42:15.650 So the rate will be determined by the slow step. 00:42:15.650 --> 00:42:19.420 So we can write out the rate of formation of O2 00:42:19.420 --> 00:42:21.160 based on the slow step, which happens 00:42:21.160 --> 00:42:23.770 to be the second step, which is what we've done all along. 00:42:23.770 --> 00:42:26.560 So there's not really a huge change right now. 00:42:26.560 --> 00:42:29.230 So the formation-- again, two molecules of O2 00:42:29.230 --> 00:42:31.090 were formed so we have a 2. 00:42:31.090 --> 00:42:33.100 We have K2 times the concentration 00:42:33.100 --> 00:42:37.120 of our intermediate O times the concentration of O3. 00:42:37.120 --> 00:42:39.910 But again, O is an intermediate so we 00:42:39.910 --> 00:42:42.640 need to solve for it in terms of our products 00:42:42.640 --> 00:42:45.310 reactants of rate constants. 00:42:45.310 --> 00:42:48.730 But now we're told something about fast and slow steps 00:42:48.730 --> 00:42:51.250 right up front. 00:42:51.250 --> 00:42:53.560 And so if we have a fast reversible step 00:42:53.560 --> 00:42:56.260 followed by a slow step, how can we 00:42:56.260 --> 00:42:58.990 solve for our concentration of our intermediate 00:42:58.990 --> 00:43:01.660 in a simpler way than we've been doing? 00:43:01.660 --> 00:43:04.000 What do we use? 00:43:04.000 --> 00:43:05.874 Or what can we use? 00:43:05.874 --> 00:43:07.310 AUDIENCE: [INAUDIBLE]. 00:43:07.310 --> 00:43:10.900 CATHERINE DRENNAN: We can use the equilibrium expression. 00:43:10.900 --> 00:43:12.130 So we can put that in. 00:43:12.130 --> 00:43:15.070 We can say our equilibrium expression products 00:43:15.070 --> 00:43:20.020 over reactants equals little K1 over K minus 1, 00:43:20.020 --> 00:43:23.260 or equilibrium constant K1. 00:43:23.260 --> 00:43:29.380 Solve for O and get either big equilibrium constant K1 00:43:29.380 --> 00:43:33.070 or our little rate constant K1 over K minus 1, 00:43:33.070 --> 00:43:36.130 and we have O3 over O2 here. 00:43:36.130 --> 00:43:39.820 So this was a lot simpler than doing all of that. 00:43:39.820 --> 00:43:42.760 So again, if you have a fast reversible step followed 00:43:42.760 --> 00:43:45.460 by a slow step you can solve for the concentration 00:43:45.460 --> 00:43:48.160 of your intermediate using an equilibrium expression which 00:43:48.160 --> 00:43:49.660 you all know how to write. 00:43:49.660 --> 00:43:51.610 So that makes your life easier. 00:43:51.610 --> 00:43:54.610 Then we can substitute that back in 00:43:54.610 --> 00:43:58.810 and we are able to put this back. 00:43:58.810 --> 00:44:02.200 And we'll solve it for O and we'll plug in our K1 00:44:02.200 --> 00:44:03.390 over K minus 1. 00:44:03.390 --> 00:44:04.330 O3. 00:44:04.330 --> 00:44:05.300 We had an O3. 00:44:05.300 --> 00:44:06.680 So that's squared over O2. 00:44:09.520 --> 00:44:14.650 Or we could write that in terms of K obs O3 concentration 00:44:14.650 --> 00:44:17.710 to the 2 over O2. 00:44:17.710 --> 00:44:21.550 So let's end with some fun, thinking about what 00:44:21.550 --> 00:44:23.440 we would observe here. 00:44:23.440 --> 00:44:24.940 First, the order and then what would 00:44:24.940 --> 00:44:26.980 happen if we double things. 00:44:26.980 --> 00:44:29.800 So what is the order with respect to O3? 00:44:29.800 --> 00:44:31.196 You can yell that out. 00:44:31.196 --> 00:44:32.340 AUDIENCE: [INAUDIBLE]. 00:44:32.340 --> 00:44:33.298 CATHERINE DRENNAN: Yup. 00:44:33.298 --> 00:44:36.256 What is the order for O2? 00:44:36.256 --> 00:44:38.852 AUDIENCE: [INAUDIBLE]. 00:44:38.852 --> 00:44:40.060 CATHERINE DRENNAN: Oh, sorry. 00:44:40.060 --> 00:44:40.990 I had something here. 00:44:40.990 --> 00:44:42.700 Double this what happens? 00:44:42.700 --> 00:44:44.024 AUDIENCE: [INAUDIBLE]. 00:44:44.024 --> 00:44:45.190 CATHERINE DRENNAN: The rate. 00:44:45.190 --> 00:44:46.930 Well, four times. 00:44:46.930 --> 00:44:47.720 Order here. 00:44:47.720 --> 00:44:49.750 Some people yelled it out. 00:44:49.750 --> 00:44:50.260 Oh, no. 00:44:50.260 --> 00:44:50.760 Oh, man. 00:44:50.760 --> 00:44:51.760 It's a clicker question. 00:44:51.760 --> 00:44:53.009 I forgot about that. 00:44:53.009 --> 00:44:53.550 Don't listen. 00:44:53.550 --> 00:44:55.675 But luckily, everyone yelled out different answers. 00:45:06.071 --> 00:45:06.570 All right. 00:45:06.570 --> 00:45:07.433 10 more seconds. 00:45:22.210 --> 00:45:23.440 OK. 00:45:23.440 --> 00:45:24.730 Yup. 00:45:24.730 --> 00:45:27.760 So it's minus 1. 00:45:27.760 --> 00:45:31.090 And so if you double it, it will half. 00:45:31.090 --> 00:45:36.910 And then finally, the overall order would be 1 because again, 00:45:36.910 --> 00:45:39.010 the overall order is the sum. 00:45:39.010 --> 00:45:41.950 So 2 minus 1 is 1. 00:45:41.950 --> 00:45:45.415 And last clicker question. 00:45:56.070 --> 00:45:56.570 All right. 00:45:56.570 --> 00:45:58.760 10 more seconds. 00:45:58.760 --> 00:46:01.370 I know you're really in a hurry to do those extra problems 00:46:01.370 --> 00:46:03.480 for exam four. 00:46:03.480 --> 00:46:05.300 I'm the rate determining step. 00:46:05.300 --> 00:46:07.430 I am with my clicker question. 00:46:07.430 --> 00:46:09.495 I admit it. 00:46:09.495 --> 00:46:11.037 AUDIENCE: Yay. 00:46:11.037 --> 00:46:12.245 CATHERINE DRENNAN: All right. 00:46:14.790 --> 00:46:15.290 All right. 00:46:15.290 --> 00:46:16.100 See you Wednesday. 00:46:16.100 --> 00:46:18.890 Remember, final clicker competition 00:46:18.890 --> 00:46:22.430 of the year before the finals.