WEBVTT 00:00:00.000 --> 00:00:02.400 The following content is provided under a Creative 00:00:02.400 --> 00:00:03.820 Commons license. 00:00:03.820 --> 00:00:06.840 Your support will help MIT OpenCourseWare continue to 00:00:06.840 --> 00:00:10.520 offer high quality educational resources for free. 00:00:10.520 --> 00:00:13.390 To make a donation, or view additional materials from 00:00:13.390 --> 00:00:17.590 hundreds of MIT courses, visit MIT OpenCourseWare at 00:00:17.590 --> 00:00:20.600 ocw.mit.edu. 00:00:20.600 --> 00:00:21.180 PROFESSOR: OK. 00:00:21.180 --> 00:00:24.410 So last time you started kinetics which is a completely 00:00:24.410 --> 00:00:26.640 different topic from thermodynamics. 00:00:26.640 --> 00:00:30.570 They're related and we'll see a relationship at some point. 00:00:30.570 --> 00:00:32.980 And you did first order kinetics. 00:00:32.980 --> 00:00:37.550 And today we're going to move on and go ahead and do more 00:00:37.550 --> 00:00:41.190 complicated kinetics and hopefully get to some 00:00:41.190 --> 00:00:44.890 interesting stuff in a couple lectures. 00:00:44.890 --> 00:00:48.080 Right now we just have to do the review of stuff that 00:00:48.080 --> 00:00:51.100 you've probably seen before. 00:00:51.100 --> 00:00:53.680 And so we're going to go reasonably fast. 00:00:53.680 --> 00:00:55.140 So you saw first order reactions. 00:00:55.140 --> 00:00:56.870 Today we're going to do second order reactions. 00:00:56.870 --> 00:01:00.440 At least begin with second order reactions. 00:01:00.440 --> 00:01:05.980 Second order kinetics. 00:01:05.980 --> 00:01:09.720 Of the form, and then there are two kinds. 00:01:09.720 --> 00:01:21.410 There's first order, rather, second order, in one reactant 00:01:21.410 --> 00:01:26.900 of the form A goes to products. 00:01:26.900 --> 00:01:32.700 And then you have first order in two reactants first order 00:01:32.700 --> 00:01:35.980 in two reactants. 00:01:35.980 --> 00:01:44.850 So I'll have the form A plus B goes of products. 00:01:44.850 --> 00:01:46.960 Where that's first order in A and first order in B, and this 00:01:46.960 --> 00:01:49.640 is second order in A. So you can think of this as A plus A 00:01:49.640 --> 00:01:53.210 goes to products if you want. 00:01:53.210 --> 00:01:56.070 And there's some rate constant k, 00:01:56.070 --> 00:01:57.860 associated with this reaction. 00:01:57.860 --> 00:01:59.380 And we do the rate analysis. 00:01:59.380 --> 00:02:02.330 We write the rate of this process. 00:02:02.330 --> 00:02:03.580 Minus dA/dt. 00:02:07.120 --> 00:02:10.440 And for the purpose of writing on the board, and you might 00:02:10.440 --> 00:02:12.030 want to do this also in your homework, when 00:02:12.030 --> 00:02:15.320 you're tired of writing. 00:02:15.320 --> 00:02:18.130 I'm going to skip the brackets. 00:02:18.130 --> 00:02:23.040 The little brackets that we usually put for concentration. 00:02:23.040 --> 00:02:24.780 I'm going to skip those, because it's just too much 00:02:24.780 --> 00:02:26.125 work to write them. 00:02:26.125 --> 00:02:30.510 And you can understand that A is a concentration of A. So 00:02:30.510 --> 00:02:34.250 this is equal to k A squared. 00:02:34.250 --> 00:02:39.490 Second order on A. And the units for k, it's important to 00:02:39.490 --> 00:02:43.280 keep track of your units, at the end of the calculations, 00:02:43.280 --> 00:02:46.030 often you want to make sure your units work out. 00:02:46.030 --> 00:02:48.730 So the units for k are going to be of this, A is 00:02:48.730 --> 00:02:50.400 in moles per liter. 00:02:50.400 --> 00:02:53.600 The units for k, you're going to have to be able to match 00:02:53.600 --> 00:02:55.200 the units on this side here. 00:02:55.200 --> 00:02:59.070 So the units for k are going to be liters squared per mole 00:02:59.070 --> 00:03:04.960 squared per second. 00:03:04.960 --> 00:03:06.830 To make the units match. 00:03:06.830 --> 00:03:11.140 Then you integrate this, on both sides, from zero to t, or 00:03:11.140 --> 00:03:18.630 from A0, the initial rate, to A. Or from zero to t and this 00:03:18.630 --> 00:03:22.180 doesn't go like this. 00:03:22.180 --> 00:03:32.470 So integrate with A0 to A. You've got dA over A squared, 00:03:32.470 --> 00:03:35.520 you put all the A's on one side, all the t's on the other 00:03:35.520 --> 00:03:43.160 side, from zero to t dt with a minus k on this side here. 00:03:43.160 --> 00:03:50.420 And then you get your rate equation, integrated rate 00:03:50.420 --> 00:03:52.630 equation for this. 00:03:52.630 --> 00:04:01.240 Which gives you one over A is equal to kt plus one over A0. 00:04:01.240 --> 00:04:03.600 So this gives you A as a function of time. 00:04:03.600 --> 00:04:06.610 And this is a convenient way to write it, because it's 00:04:06.610 --> 00:04:07.410 linear in time. 00:04:07.410 --> 00:04:09.650 So you always try to get things to be linear in time. 00:04:09.650 --> 00:04:15.360 Because then you can plot them as a straight line. 00:04:15.360 --> 00:04:20.460 Plot on this axis here you plot one over A. on this axis 00:04:20.460 --> 00:04:25.650 here you plot t as a function of time. 00:04:25.650 --> 00:04:34.420 And then you get a straight line where the slope is k. 00:04:34.420 --> 00:04:39.460 Gives you the rate, and the intercept is one over A0. 00:04:39.460 --> 00:04:39.720 Yes. 00:04:39.720 --> 00:04:45.450 STUDENT: [INAUDIBLE] 00:04:45.450 --> 00:04:51.030 PROFESSOR: k is moles per liter squared per second. 00:04:51.030 --> 00:04:53.760 You're right, liters per mole per second, yes. 00:04:53.760 --> 00:05:03.880 That is correct. because it has to work. 00:05:03.880 --> 00:05:05.000 Otherwise it doesn't work. 00:05:05.000 --> 00:05:08.860 So I need to turn on my brain. 00:05:08.860 --> 00:05:10.530 OK, think. 00:05:10.530 --> 00:05:15.250 Liters per mole per second. 00:05:15.250 --> 00:05:24.630 Thank you. 00:05:24.630 --> 00:05:28.210 And the other thing that you want to know is the half-life. 00:05:28.210 --> 00:05:30.050 What is the half-life. 00:05:30.050 --> 00:05:34.030 So you set 2 A0. 00:05:34.030 --> 00:05:39.600 So A0 over two, you look for the time where you get to A0 00:05:39.600 --> 00:05:43.310 over two, so you put A0 over two in here. 00:05:43.310 --> 00:05:47.110 And that's going to be equal to k times t 1/2 00:05:47.110 --> 00:05:48.270 plus one over A0. 00:05:48.270 --> 00:05:52.490 So you solve for the half-life and you get a half-life 00:05:52.490 --> 00:05:57.590 of one over k A0. 00:05:57.590 --> 00:06:00.520 So the half-life is inversely proportional to the amount of 00:06:00.520 --> 00:06:02.820 stuff you started out with. 00:06:02.820 --> 00:06:06.820 Unlike the first order reaction, where the half-life 00:06:06.820 --> 00:06:14.010 was independent of the amount that you started out with. 00:06:14.010 --> 00:06:15.680 So this was the easy one. 00:06:15.680 --> 00:06:18.760 The next one is a little bit more complicated. 00:06:18.760 --> 00:06:27.270 Which is when your first order in each of two reactants. 00:06:27.270 --> 00:06:30.500 Then your rate, your differential rate equation, 00:06:30.500 --> 00:06:32.320 looks like this. 00:06:32.320 --> 00:06:36.240 Because the k times A times B, and now you don't know what to 00:06:36.240 --> 00:06:38.960 do with B, a priori. 00:06:38.960 --> 00:06:43.740 So you want to rewrite this equation a little bit 00:06:43.740 --> 00:06:47.350 differently in terms of the amount of A that's used up. 00:06:47.350 --> 00:06:51.740 So you define A, x is equal to A0 minus A, this is the amount 00:06:51.740 --> 00:06:54.070 of A that's used up. 00:06:54.070 --> 00:06:56.120 This is what you started out with, this is 00:06:56.120 --> 00:06:56.990 what you're left with. 00:06:56.990 --> 00:07:01.440 And so the difference is what's being used up. 00:07:01.440 --> 00:07:09.050 And dx/dt is minus dA/dt. 00:07:09.050 --> 00:07:15.180 And by stoichiometry, what you've used up, of A, is also 00:07:15.180 --> 00:07:19.310 what you've used up of B. Because for every A that 00:07:19.310 --> 00:07:25.575 reacts, you have to use up one mole of B. For every mole of A 00:07:25.575 --> 00:07:28.510 that reacts, you use up one mole of B. And so you also 00:07:28.510 --> 00:07:34.780 have x to B0 minus B. So, you can plug this in here. 00:07:34.780 --> 00:07:37.420 And get a differential equation which is purely in 00:07:37.420 --> 00:07:39.450 terms of one variable, which is x. 00:07:39.450 --> 00:07:41.480 Right here, it looks like it's in terms of two variables, 00:07:41.480 --> 00:07:42.730 which makes it complicated. 00:07:42.730 --> 00:07:46.160 By doing this change of variables, you see that A and 00:07:46.160 --> 00:07:47.370 B actually related. 00:07:47.370 --> 00:07:51.150 Because of the reaction stoichiometry. 00:07:51.150 --> 00:07:59.010 So you can rewrite that as dx/dt is equal to k times A0 00:07:59.010 --> 00:08:02.230 minus x times B0 minus x. 00:08:02.230 --> 00:08:05.210 And now you have a differential equation in one 00:08:05.210 --> 00:08:09.170 variable, which with some tricks you can solve. 00:08:09.170 --> 00:08:13.350 So we want to have the integrated equation. 00:08:13.350 --> 00:08:15.540 So we take an integral of both sides. 00:08:15.540 --> 00:08:17.560 We put in all the x's on one side. 00:08:17.560 --> 00:08:19.500 All the times on the other side. 00:08:19.500 --> 00:08:28.670 We'll go from x equals zero to x, dx, A0 minus x times B0 00:08:28.670 --> 00:08:34.920 minus x is equal to k from zero to t dt. 00:08:34.920 --> 00:08:37.990 And now you have to dig back into the last time you took 00:08:37.990 --> 00:08:42.830 integration calculus, which for me was about 00:08:42.830 --> 00:08:44.600 two centuries ago. 00:08:44.600 --> 00:08:49.450 And figure out how to do this integral here. 00:08:49.450 --> 00:08:51.430 And the trick for doing this integral is 00:08:51.430 --> 00:09:01.640 to use partial fractions. 00:09:01.640 --> 00:09:04.510 So you use partial fractions to do this integral here. 00:09:04.510 --> 00:09:13.890 Which means that if you take this ratio, one over A0 minus 00:09:13.890 --> 00:09:20.570 x times B0 minus x and rewrite it as some number, n1 divided 00:09:20.570 --> 00:09:29.200 by A0 minus x, plus some number n2 divided by B0 minus 00:09:29.200 --> 00:09:35.210 x, you solve for n1 and n2, and you find that n1 here is 00:09:35.210 --> 00:09:43.600 equal to one over B0 minus A0 and n2 is minus one 00:09:43.600 --> 00:09:45.930 over B0 minus A0. 00:09:45.930 --> 00:09:48.840 And so you plug, now, this in here. 00:09:48.840 --> 00:09:52.250 And instead of having this complicated denominator, you 00:09:52.250 --> 00:09:55.560 have a sum of two integrals that you know how to do. 00:09:55.560 --> 00:10:00.690 Because they're basically of the form one over x. 00:10:00.690 --> 00:10:03.410 And in doing this, you also realize that you have to be 00:10:03.410 --> 00:10:06.870 careful because when A0 is equal to B0, when you have the 00:10:06.870 --> 00:10:09.930 same amount of A and B, then things blow up 00:10:09.930 --> 00:10:10.770 and you're in trouble. 00:10:10.770 --> 00:10:13.980 So that's going to be a special case. 00:10:13.980 --> 00:10:16.790 So always look out for special cases for these things. 00:10:16.790 --> 00:10:23.450 So you assume, then, that B0 is not the same as A0. 00:10:23.450 --> 00:10:25.800 And then you can go forward with solving it. 00:10:25.800 --> 00:10:29.510 So you integrate, and at the end of the process, I'm not 00:10:29.510 --> 00:10:33.620 going to go through it, it's really complicated, you get 00:10:33.620 --> 00:10:34.520 something that looks like this. 00:10:34.520 --> 00:10:46.420 A0 minus B0 log of A B0 over A0 B. And you have your 00:10:46.420 --> 00:10:46.980 equation here. 00:10:46.980 --> 00:10:49.660 Now, we don't really have a good way to plot it against 00:10:49.660 --> 00:10:51.350 one variable. 00:10:51.350 --> 00:10:54.970 And the usual thing is look at specific cases. 00:10:54.970 --> 00:10:56.340 And limiting cases. 00:10:56.340 --> 00:11:00.160 And there's one limiting case we already brought up. 00:11:00.160 --> 00:11:02.340 Which is when you started with the same amount of material. 00:11:02.340 --> 00:11:04.650 What does it look like if you have the same amount of 00:11:04.650 --> 00:11:06.830 material to begin with? 00:11:06.830 --> 00:11:11.200 So if you have A0 equal to B0, you start out with the same 00:11:11.200 --> 00:11:15.680 amount of stuff, but if you start out with the same amount 00:11:15.680 --> 00:11:18.310 of stuff, then this doesn't look so different from, at 00:11:18.310 --> 00:11:21.260 least mathematically, from this one right here. 00:11:21.260 --> 00:11:24.570 If you start out with A0 is equal to B0, and for every 00:11:24.570 --> 00:11:27.345 mole of A that you use up you use a mole of B, then 00:11:27.345 --> 00:11:30.560 throughout the whole reaction, the concentration of A and the 00:11:30.560 --> 00:11:33.480 concentration of B are going to be the same. 00:11:33.480 --> 00:11:38.010 So for the whole reaction, if you started with this, A is 00:11:38.010 --> 00:11:41.290 equal to B for all times. 00:11:41.290 --> 00:11:42.000 That makes it easy. 00:11:42.000 --> 00:11:45.030 Because now you can go back and instead of writing A times 00:11:45.030 --> 00:11:47.320 B, you can write A times A, which is A squared, which is 00:11:47.320 --> 00:11:47.890 what we had here. 00:11:47.890 --> 00:11:51.870 And then you have the whole thing solved. 00:11:51.870 --> 00:11:57.150 So in that case here you just have minus dA/dt is equal to k 00:11:57.150 --> 00:12:06.610 A squared, and one over A is equal to kt plus one over A0. 00:12:06.610 --> 00:12:09.430 You don't even have to do any math, you just look at it. 00:12:09.430 --> 00:12:10.790 So that's one case. 00:12:10.790 --> 00:12:14.340 Another case is if one of the reactants is in much higher 00:12:14.340 --> 00:12:16.380 concentration than the other reactant. 00:12:16.380 --> 00:12:18.180 And that's called flooding. 00:12:18.180 --> 00:12:21.960 You basically flood the system with one reactant. 00:12:21.960 --> 00:12:27.670 And that's something that we'll use again, hopefully by 00:12:27.670 --> 00:12:29.710 the end of class today. 00:12:29.710 --> 00:12:33.740 Let's say that we take, so this is another limiting case. 00:12:33.740 --> 00:12:35.860 Let's say we take A0 to be much bigger than B0. 00:12:35.860 --> 00:12:39.010 So we flood the system with A0. 00:12:39.010 --> 00:12:45.550 As a result, the concentration of A doesn't change 00:12:45.550 --> 00:12:47.470 very much in my pot. 00:12:47.470 --> 00:12:50.260 It's hugely concentrated in A, there's a 00:12:50.260 --> 00:12:53.190 little bit of B around. 00:12:53.190 --> 00:12:54.150 Again, with the process. 00:12:54.150 --> 00:12:57.600 If I use all of the B up, the difference in A is going to be 00:12:57.600 --> 00:12:58.850 very small. 00:12:58.850 --> 00:13:00.840 So at the end of the process, I'm basically still going to 00:13:00.840 --> 00:13:04.240 have A0 left in the pot. 00:13:04.240 --> 00:13:07.230 So during the whole process, during the whole time period, 00:13:07.230 --> 00:13:13.280 I might as well assume that A is equal to A0. 00:13:13.280 --> 00:13:14.650 And that makes my life much easier. 00:13:14.650 --> 00:13:17.850 Because now if I write my differential equation in terms 00:13:17.850 --> 00:13:24.950 of B instead of A, so the rate of destruction of B, k A times 00:13:24.950 --> 00:13:30.160 B, instead of writing A here, it's pretty much constant for 00:13:30.160 --> 00:13:30.730 the whole time. 00:13:30.730 --> 00:13:34.310 I'm just going to write A0. 00:13:34.310 --> 00:13:37.650 So now, if k times A0 is a constant, and this looks 00:13:37.650 --> 00:13:42.160 awfully like a first order reaction. 00:13:42.160 --> 00:13:44.430 So I can solve for it. 00:13:44.430 --> 00:13:48.800 And I get that, so I can just write the answer because I've 00:13:48.800 --> 00:13:49.590 done this already. 00:13:49.590 --> 00:13:50.940 I don't have to do it again. 00:13:50.940 --> 00:13:56.760 The concentration of B then goes like B0 e to the minus k 00:13:56.760 --> 00:13:59.080 prime t, that's the first order reaction. 00:13:59.080 --> 00:14:03.670 Where k prime here is this new rate constant, this new 00:14:03.670 --> 00:14:10.860 number, which is k, times A0. 00:14:10.860 --> 00:14:14.260 So that's easy to solve also. 00:14:14.260 --> 00:14:15.910 So always go to the limiting cases, because 00:14:15.910 --> 00:14:18.150 they tend to easy. 00:14:18.150 --> 00:14:25.120 And if you were to go to the full solution and put in this 00:14:25.120 --> 00:14:29.220 limiting case, then you'd find that you get the right answer 00:14:29.220 --> 00:14:30.840 this way as well. 00:14:30.840 --> 00:14:32.890 You can directly go to the easy way of doing it, or you 00:14:32.890 --> 00:14:35.170 can go through the whole process of solving it and 00:14:35.170 --> 00:14:39.140 putting the approximation in there. 00:14:39.140 --> 00:14:41.530 And do the cancellations and get this. 00:14:41.530 --> 00:14:43.710 But this is much easier. 00:14:43.710 --> 00:14:46.590 Just writing the answer down is always much easier. 00:14:46.590 --> 00:14:48.540 So it's a pseudo first order reaction. 00:14:48.540 --> 00:14:53.320 We call this a pseudo first order reaction. 00:14:53.320 --> 00:14:56.460 So we're done with the simple stuff now. 00:14:56.460 --> 00:15:04.960 Any questions on first order and second order reactions? 00:15:04.960 --> 00:15:07.990 So the next step is, you've got a reaction. 00:15:07.990 --> 00:15:11.130 It could be a gas phase reaction, it could be a 00:15:11.130 --> 00:15:13.080 solution phase reaction. 00:15:13.080 --> 00:15:17.000 There's some quantity, some property, of the solution 00:15:17.000 --> 00:15:17.760 that's going to change. 00:15:17.760 --> 00:15:20.000 That's going to allow you to follow it as 00:15:20.000 --> 00:15:21.420 a function of time. 00:15:21.420 --> 00:15:22.920 And that property could be many things. 00:15:22.920 --> 00:15:24.730 It could be spectroscopic. 00:15:24.730 --> 00:15:27.770 It could be that there's an absorption in the visible that 00:15:27.770 --> 00:15:30.980 changes as the concentration of one of 00:15:30.980 --> 00:15:33.760 your reactants changes. 00:15:33.760 --> 00:15:37.210 Or one of the products could have an absorption band that 00:15:37.210 --> 00:15:38.600 you could follow in time. 00:15:38.600 --> 00:15:39.640 Or you could using infrared spectroscopy to 00:15:39.640 --> 00:15:41.090 follow it in time. 00:15:41.090 --> 00:15:44.360 Or if you have a reaction in the gas phase, and you have 00:15:44.360 --> 00:15:46.730 more products or less products than the reactants, and the 00:15:46.730 --> 00:15:48.224 pressure is going to change in time if you 00:15:48.224 --> 00:15:49.730 have a finite volume. 00:15:49.730 --> 00:15:52.840 So there's usually some quantity that you can use to 00:15:52.840 --> 00:15:59.150 follow the reaction in time to extract out data. 00:15:59.150 --> 00:16:03.140 And then from that data, that you want to know what are the 00:16:03.140 --> 00:16:04.790 kinetics of this reaction. 00:16:04.790 --> 00:16:05.790 Because eventually you're going to 00:16:05.790 --> 00:16:07.770 try to find a mechanism. 00:16:07.770 --> 00:16:10.440 You're going to try to find a mechanism that's consistent 00:16:10.440 --> 00:16:12.520 with the data. 00:16:12.520 --> 00:16:13.780 So you get data. 00:16:13.780 --> 00:16:18.980 Then you want extract out of the rate constants and orders. 00:16:18.980 --> 00:16:21.790 So let's assume that you've found a way to get data. 00:16:21.790 --> 00:16:24.270 And now you've got to analyze your data. 00:16:24.270 --> 00:16:30.980 And suppose that you've found a way to measure the reactant 00:16:30.980 --> 00:16:33.150 concentration as a function of time. 00:16:33.150 --> 00:16:38.210 And let's say that in the first case, the simplest case 00:16:38.210 --> 00:16:43.100 is that you have one reactant. 00:16:43.100 --> 00:16:50.140 So you have one reactant, A. So A goes to products. 00:16:50.140 --> 00:16:53.540 And you've managed to extract A as a function of time. 00:16:53.540 --> 00:16:57.670 Well, the obvious thing to do is to plot A versus time and 00:16:57.670 --> 00:16:59.670 see what it fits like. 00:16:59.670 --> 00:17:03.520 So you take A versus time, and you plot it. 00:17:03.520 --> 00:17:07.620 And you know that if you plot log A versus time and it's a 00:17:07.620 --> 00:17:10.515 straight line, well, that's going to be 00:17:10.515 --> 00:17:12.730 a first order process. 00:17:12.730 --> 00:17:16.320 Plot log A versus time in a straight line, you know that's 00:17:16.320 --> 00:17:18.290 going to be first order. 00:17:18.290 --> 00:17:20.040 If it doesn't go to a straight line you know 00:17:20.040 --> 00:17:21.100 it's not first order. 00:17:21.100 --> 00:17:26.810 So then you go ahead and plot one over A versus time. 00:17:26.810 --> 00:17:32.630 And if it's a straight line, you know it's second order. 00:17:32.630 --> 00:17:35.230 And if it's not second order, it's not a straight line. 00:17:35.230 --> 00:17:38.140 It's not a straight line, you look for some other order. 00:17:38.140 --> 00:17:39.590 So that's one way to do it. 00:17:39.590 --> 00:17:44.060 And you've got to have enough points on your graph, because 00:17:44.060 --> 00:17:48.780 if I were to plot a, let's say this is my data point, if I 00:17:48.780 --> 00:17:52.700 have a second order process, at the beginning it's going to 00:17:52.700 --> 00:17:54.570 look an awful lot like a first order process. 00:17:54.570 --> 00:17:56.460 It's not until after a while that it's 00:17:56.460 --> 00:17:59.130 going to start to deviate. 00:17:59.130 --> 00:18:03.460 So you've got to have enough points down in time to make 00:18:03.460 --> 00:18:06.810 sure that you can differentiate between a 00:18:06.810 --> 00:18:10.300 straight line and a line that's not straight. 00:18:10.300 --> 00:18:15.170 And usually, that's often a mistake that 00:18:15.170 --> 00:18:16.330 experimentalists make. 00:18:16.330 --> 00:18:17.500 They look at the beginning. 00:18:17.500 --> 00:18:21.050 They say, oh, it's a straight line, work is done. 00:18:21.050 --> 00:18:22.730 Go home. 00:18:22.730 --> 00:18:30.930 But usually you need to have a good amount of the reactant 00:18:30.930 --> 00:18:33.180 consumed before you can tell the difference between first 00:18:33.180 --> 00:18:37.700 and second order. 00:18:37.700 --> 00:18:41.000 So you'll have an opportunity to do this on the homework. 00:18:41.000 --> 00:18:46.470 This kind of exercise of extracting the order of a 00:18:46.470 --> 00:18:48.340 simple reaction. 00:18:48.340 --> 00:18:51.520 Another way to do it if you have a simple reaction is to 00:18:51.520 --> 00:18:53.340 look at half-lives. 00:18:53.340 --> 00:19:00.230 That would be the half-life method. 00:19:00.230 --> 00:19:02.700 If you can measure A as a function of time, then you 00:19:02.700 --> 00:19:07.060 know when you've gotten A over two. 00:19:07.060 --> 00:19:13.500 So, you know that if I look at the half-life versus the 00:19:13.500 --> 00:19:16.870 concentration, the initial concentration, of my reactant, 00:19:16.870 --> 00:19:18.370 that tells me something about the order. 00:19:18.370 --> 00:19:22.720 Because we saw that for our first order, t 1/2 was 00:19:22.720 --> 00:19:30.530 independent of the initial concentration. 00:19:30.530 --> 00:19:34.640 And that for a second order, t 1/2 was proportional to one 00:19:34.640 --> 00:19:39.620 over the inverse of the initial concentration. 00:19:39.620 --> 00:19:44.580 So if you plot t 1/2 versus A0, or have a few A0 versus t 00:19:44.580 --> 00:19:47.640 1/2s, then you can tell the difference between first order 00:19:47.640 --> 00:19:51.770 and a second order reaction, and see which one fits. 00:19:51.770 --> 00:19:59.710 Sometimes to get even more solid numbers, because from 00:19:59.710 --> 00:20:01.450 here you can also extract k. 00:20:01.450 --> 00:20:04.670 If you have a bunch of points, of t 1/2 versus A0, you can 00:20:04.670 --> 00:20:06.470 extract k, the rate constant. 00:20:06.470 --> 00:20:08.040 Not just the order, but also the rate 00:20:08.040 --> 00:20:10.040 constant out of this data. 00:20:10.040 --> 00:20:14.650 You can use multiple lifetimes if you have enough data. 00:20:14.650 --> 00:20:15.910 Multiple lifetimes. 00:20:15.910 --> 00:20:21.540 So you can define, you can define a t 3/4. 00:20:21.540 --> 00:20:26.470 Which is the amount of time it takes for the concentration of 00:20:26.470 --> 00:20:30.800 A to be 1/4 of what you started out with. 00:20:30.800 --> 00:20:34.400 So 3/4 is gone. 00:20:34.400 --> 00:20:39.010 And then you can put that into your first order rate law. 00:20:39.010 --> 00:20:46.530 So when you have log of A over A0 is equal to minus kt, you 00:20:46.530 --> 00:20:51.765 get that t 3/4, we can solve for t 3/4, and you get that 00:20:51.765 --> 00:20:57.670 that's equal to two log two, over k. 00:20:57.670 --> 00:20:58.970 And then you can do the same thing for a 00:20:58.970 --> 00:21:01.440 second order process. 00:21:01.440 --> 00:21:05.390 So this is first order. 00:21:05.390 --> 00:21:10.190 You plug in t 3/4 and A is equal to 1/4 A0, and you solve 00:21:10.190 --> 00:21:16.240 for t 3/4 for a second order process. 00:21:16.240 --> 00:21:21.880 And you get that this is equal to three over A0 times k. 00:21:21.880 --> 00:21:27.030 So it's the same functional form as these two. 00:21:27.030 --> 00:21:28.960 But there's a pre-factor that's different here. 00:21:28.960 --> 00:21:31.420 So here there's a two that comes in there. 00:21:31.420 --> 00:21:34.500 And here there's there's a three that comes in there. 00:21:34.500 --> 00:21:37.910 And so there's an obvious way to tell, then, if you have the 00:21:37.910 --> 00:21:39.630 t 1/2 and the t 3/4 signs. 00:21:39.630 --> 00:21:43.640 Basically, you follow, instead of having many reactions, 00:21:43.640 --> 00:21:46.130 instead of having many reactions to do, with 00:21:46.130 --> 00:21:49.050 different A0's here you can do one reaction. 00:21:49.050 --> 00:21:52.450 If you do one reaction and you watch the reactant go away, 00:21:52.450 --> 00:21:53.670 and you time it. 00:21:53.670 --> 00:21:57.430 When 1/2 of it is gone, that's one time, then you keep going, 00:21:57.430 --> 00:22:00.060 like 3/4 is gone, that's another time. 00:22:00.060 --> 00:22:03.210 Then you can take the ratio of those two times. 00:22:03.210 --> 00:22:10.800 Of t 3/4 versus t 1/2. t 3/4 versus t 1/2. 00:22:10.800 --> 00:22:13.050 And if it's a first order process, the ratio 00:22:13.050 --> 00:22:19.090 here is just two. 00:22:19.090 --> 00:22:24.840 And if you take t 3/4 over t 1/2 and it's a second order 00:22:24.840 --> 00:22:32.990 process, the ratio is three. 00:22:32.990 --> 00:22:38.010 So with one experiment, then, you can extract out the order. 00:22:38.010 --> 00:22:39.990 You can't extract out, well, you can extract 00:22:39.990 --> 00:22:41.630 out the rate constant. 00:22:41.630 --> 00:22:44.150 If you know the order then you know which equation fits, and 00:22:44.150 --> 00:22:46.230 you can extract out the the rate constant, 00:22:46.230 --> 00:22:46.980 with a big error bar. 00:22:46.980 --> 00:22:50.590 You're always better off doing many multiple lifetimes of 00:22:50.590 --> 00:22:55.540 different A0's or many of these just to get more 00:22:55.540 --> 00:22:58.470 statistics in the result. 00:22:58.470 --> 00:23:00.780 So this is the simple process. 00:23:00.780 --> 00:23:02.640 And you always try, if you have something complicated, 00:23:02.640 --> 00:23:05.920 you always try to bring it back to a one component 00:23:05.920 --> 00:23:08.580 process by doing something like flooding. 00:23:08.580 --> 00:23:14.470 So if you have five different reactants, if you make four of 00:23:14.470 --> 00:23:19.290 them in very large quantities and keep one of them in very 00:23:19.290 --> 00:23:22.170 small quantities, then all the four that are in large 00:23:22.170 --> 00:23:25.140 quantities are basically constant over the process. 00:23:25.140 --> 00:23:28.410 And you basically are looking at only one 00:23:28.410 --> 00:23:31.090 reactant going away. 00:23:31.090 --> 00:23:34.270 And then you can use these methods to figure out what the 00:23:34.270 --> 00:23:39.070 order is for that one reactant. 00:23:39.070 --> 00:23:45.890 So, questions about simple one reactant sort of processes. 00:23:45.890 --> 00:23:48.250 It's pretty straightforward. 00:23:48.250 --> 00:24:02.480 So now, let's say we have more complicated reactions. 00:24:02.480 --> 00:24:05.210 There are two ways that we can deal with that. 00:24:05.210 --> 00:24:07.360 The first, I already mentioned, which is the 00:24:07.360 --> 00:24:09.850 flooding which we'll get back to. 00:24:09.850 --> 00:24:13.880 And another way is called, so we have 00:24:13.880 --> 00:24:18.110 some complicated reactions. 00:24:18.110 --> 00:24:20.400 Complex reactions, with multiple reactants, that's A 00:24:20.400 --> 00:24:24.100 plus B plus C goes to products. 00:24:24.100 --> 00:24:24.600 And there could be some 00:24:24.600 --> 00:24:28.270 stoichiometry in front of there. 00:24:28.270 --> 00:24:29.970 So one of the ways to deal with that is called the 00:24:29.970 --> 00:24:33.160 initial rate method. 00:24:33.160 --> 00:24:38.630 We want to find out orders and rate constants. 00:24:38.630 --> 00:24:46.630 Initial rate method. 00:24:46.630 --> 00:24:53.270 So if I look at minus dA/dt, one of the reactants or the 00:24:53.270 --> 00:24:57.300 rate of the reaction near time t equals zero. 00:24:57.300 --> 00:24:59.310 That's the initial rate. 00:24:59.310 --> 00:25:00.820 Right as the reaction starts. 00:25:00.820 --> 00:25:03.450 I mix everything together and I, just as after I mix it 00:25:03.450 --> 00:25:06.640 together, I watch the process of A disappearing or B 00:25:06.640 --> 00:25:08.530 disappearing or C disappearing. 00:25:08.530 --> 00:25:13.260 And so in reality what I'm doing is minus delta A / delta 00:25:13.260 --> 00:25:18.090 t near t equals zero, where delta t is a small interval. 00:25:18.090 --> 00:25:20.910 So there's not much change in delta A. Experimentally 00:25:20.910 --> 00:25:21.460 that's what I do. 00:25:21.460 --> 00:25:24.410 And that's pretty much, essentially getting this 00:25:24.410 --> 00:25:26.550 number out. 00:25:26.550 --> 00:25:28.590 And we're going to call that the initial rate. 00:25:28.590 --> 00:25:32.990 And the initial rate is k, and at the beginning I haven't 00:25:32.990 --> 00:25:33.790 used up anything. 00:25:33.790 --> 00:25:37.200 So all the initial concentrations are there A0 to 00:25:37.200 --> 00:25:41.280 the alpha, B0 to the beta, C0 to the gamma. 00:25:41.280 --> 00:25:46.480 Et cetera if you have more reactants. 00:25:46.480 --> 00:25:47.310 So you measure this. 00:25:47.310 --> 00:25:55.320 You measure this R0, and then you repeat the same process 00:25:55.320 --> 00:25:59.880 with a new concentration of one of those three reactants. 00:25:59.880 --> 00:26:03.690 So with A0 prime, let's say. 00:26:03.690 --> 00:26:09.260 And then you get a new R0, R0 prime. 00:26:09.260 --> 00:26:14.340 Then you take the ratios of these R0 and R0 primes, R0 00:26:14.340 --> 00:26:16.640 divided by R0 prime. 00:26:16.640 --> 00:26:24.640 So R0 is k A0 to the alpha, B0 to the beta, C0 to the gamma, 00:26:24.640 --> 00:26:29.990 then you have k A0 prime to the alpha. 00:26:29.990 --> 00:26:34.210 B0 to the beta, C0 to the gamma, 00:26:34.210 --> 00:26:40.240 and the k's you disappear. 00:26:40.240 --> 00:26:42.250 The B0's disappear. 00:26:42.250 --> 00:26:43.150 The C0's disappear. 00:26:43.150 --> 00:26:45.210 The only thing that we've changed is the concentration 00:26:45.210 --> 00:26:46.450 of A0 to begin with. 00:26:46.450 --> 00:26:50.230 So the ratios of these initial rates is the ratios of the 00:26:50.230 --> 00:26:56.310 A0's to the alpha power, to the order in terms of A0. 00:26:56.310 --> 00:27:00.330 So if you're clever about your choice of ratios, then you can 00:27:00.330 --> 00:27:02.690 get alpha pretty easily. 00:27:02.690 --> 00:27:16.280 So if you choose, so if you now choose A0 prime to be 00:27:16.280 --> 00:27:27.770 equal to 1/2 A0, then you measure R0 over R0 prime. 00:27:27.770 --> 00:27:35.560 And if you get one, then you know that that alpha's zero. 00:27:35.560 --> 00:27:38.250 Alpha has to be zero here. 00:27:38.250 --> 00:27:39.610 Then you know alpha is zero, that gives you the order. 00:27:39.610 --> 00:27:41.100 It's a zero order reaction. 00:27:41.100 --> 00:27:44.740 If you get that R0 over R0 prime is square root of two, 00:27:44.740 --> 00:27:51.460 or rather 1/2, then a square root of two, square root of 2, 00:27:51.460 --> 00:27:54.950 then you know that alpha is 1/2. 00:27:54.950 --> 00:27:56.490 And that's a half order. 00:27:56.490 --> 00:28:00.060 We haven't seen any half order reactions yet, but we will. 00:28:00.060 --> 00:28:04.320 Those are indicative of a complicated mechanisms. 00:28:04.320 --> 00:28:07.940 But that's what you would get out of this experiment. 00:28:07.940 --> 00:28:15.230 If you get that it's equal to two, if you get that this 00:28:15.230 --> 00:28:16.880 ratio is equal to two, then you know that alpha is equal 00:28:16.880 --> 00:28:19.500 to one, et cetera. 00:28:19.500 --> 00:28:24.880 So it's a pretty easy way to get the order. 00:28:24.880 --> 00:28:26.410 Then you repeat the experiment. 00:28:26.410 --> 00:28:30.535 Now instead of changing A0, you keep A0 constant 00:28:30.535 --> 00:28:31.600 and you change B0. 00:28:31.600 --> 00:28:38.710 Or, you can use flooding or isolation, which is the next 00:28:38.710 --> 00:28:40.760 process to get the other orders. 00:28:40.760 --> 00:28:43.400 And eventually you get the rate constant. 00:28:43.400 --> 00:28:46.720 Once you have all the orders, and you have R0, then you have 00:28:46.720 --> 00:28:50.070 the rate constant. 00:28:50.070 --> 00:28:52.540 OK, so that's one way of doing it. 00:28:52.540 --> 00:28:55.260 And a second way of doing it is the 00:28:55.260 --> 00:28:57.360 way we already mentioned. 00:28:57.360 --> 00:29:02.500 To solve the second order reaction in two components. 00:29:02.500 --> 00:29:06.300 Which is to flood the reaction with 00:29:06.300 --> 00:29:08.110 everything except for one. 00:29:08.110 --> 00:29:14.430 So this is called flooding or isolation. 00:29:14.430 --> 00:29:20.920 Basically, you isolate one reactant and watch it. 00:29:20.920 --> 00:29:22.460 You're trying to get back to a system, 00:29:22.460 --> 00:29:23.840 which is a simple system. 00:29:23.840 --> 00:29:25.990 Which is a system of one reaction. 00:29:25.990 --> 00:29:31.090 So let's say you take A0 to be much smaller than all the 00:29:31.090 --> 00:29:31.910 other species. 00:29:31.910 --> 00:29:35.150 You flood with B and C, and you isolate A0. 00:29:35.150 --> 00:29:41.510 And then your rate minus dA/dt, it's going 00:29:41.510 --> 00:29:43.250 to be A to the alpha. 00:29:43.250 --> 00:29:47.840 And instead of B, well, during that process B's going to say 00:29:47.840 --> 00:29:49.030 pretty much constant. 00:29:49.030 --> 00:29:51.110 Because it's hugely concentrated in B. You can 00:29:51.110 --> 00:29:51.930 replace B with B0. 00:29:51.930 --> 00:29:54.810 You can replace C with C0. 00:29:54.810 --> 00:29:58.720 So now you have an effective constant, an effective rate 00:29:58.720 --> 00:30:06.890 constant, and then you have a process which is effectively, 00:30:06.890 --> 00:30:09.820 or pseudo, one reactant. 00:30:09.820 --> 00:30:14.380 Then you can use these methods here, you can plot A versus 00:30:14.380 --> 00:30:17.410 time, you can find path lines, et cetera, to gather 00:30:17.410 --> 00:30:19.440 the order for it. 00:30:19.440 --> 00:30:28.240 Then you can get alpha and k prime. 00:30:28.240 --> 00:30:30.100 And if you can change B0 and C0, then you 00:30:30.100 --> 00:30:33.220 get k out of this. 00:30:33.220 --> 00:30:36.340 So this is basically all fairly straightforward, just 00:30:36.340 --> 00:30:42.440 tedious experimentation to get all these numbers out. 00:30:42.440 --> 00:30:44.240 OK, any questions on this? 00:30:44.240 --> 00:30:46.080 You'll get experience on the homework. 00:30:46.080 --> 00:30:49.470 There's likely to be a question on the final where 00:30:49.470 --> 00:30:52.680 you're given data and asked to extract out 00:30:52.680 --> 00:30:59.030 orders and rate constants. 00:30:59.030 --> 00:31:00.730 So let's move on now. 00:31:00.730 --> 00:31:07.260 So, so far we've looked at first and second order 00:31:07.260 --> 00:31:10.480 elementary processes, and we've looked at taking data 00:31:10.480 --> 00:31:13.390 and extracting out rate and rate constant. 00:31:13.390 --> 00:31:19.370 And the next step is to build mechanisms. 00:31:19.370 --> 00:31:31.300 So a mechanism is when you take a complicated reaction, 00:31:31.300 --> 00:31:37.820 like A plus B plus C goes to D plus E. And you break it up 00:31:37.820 --> 00:31:39.050 into elementary steps. 00:31:39.050 --> 00:31:40.060 What's an elementary step? 00:31:40.060 --> 00:31:43.400 An elementary step is a step which 00:31:43.400 --> 00:31:45.440 happens in a single reaction. 00:31:45.440 --> 00:31:51.690 So I could hypothesize that this complicated reaction 00:31:51.690 --> 00:31:55.590 happens in three steps, where I need to have a molecule of A 00:31:55.590 --> 00:32:00.220 and a molecule of B collide with each other to first form 00:32:00.220 --> 00:32:03.630 an intermediate F. Then I want a molecule of F plus a 00:32:03.630 --> 00:32:07.850 molecule of B to collide to form intermediate G plus a 00:32:07.850 --> 00:32:12.250 product D. Then have the intermediate G plus reactant C 00:32:12.250 --> 00:32:17.390 collide together to form the product E. So this set of 00:32:17.390 --> 00:32:20.020 elementary steps, where at each step you have a collision 00:32:20.020 --> 00:32:25.390 of two or three molecules together, three is not so 00:32:25.390 --> 00:32:29.520 common but two is very common, those elementary steps are 00:32:29.520 --> 00:32:33.660 called the steps of the mechanisms. 00:32:33.660 --> 00:32:44.440 And these elementary steps you can define something called 00:32:44.440 --> 00:32:53.130 molecularity, which is the number of species that you 00:32:53.130 --> 00:32:56.460 need to collide with each other in one of these 00:32:56.460 --> 00:32:57.510 elementary steps. 00:32:57.510 --> 00:33:00.210 So the molecularity here would be two, you need two 00:33:00.210 --> 00:33:01.160 molecules to react. 00:33:01.160 --> 00:33:03.240 Here it's two, here it's two. 00:33:03.240 --> 00:33:06.200 If I have an elementary step which is a zero order in one 00:33:06.200 --> 00:33:09.490 reactant, then the molecularity would be one. 00:33:09.490 --> 00:33:12.690 Or I could have A plus A, the same molecules have to collide 00:33:12.690 --> 00:33:13.400 with each other. 00:33:13.400 --> 00:33:15.330 Molecularity would be two. 00:33:15.330 --> 00:33:18.800 And the molecularity and the order of the 00:33:18.800 --> 00:33:20.600 reaction are connected. 00:33:20.600 --> 00:33:23.790 So if you have something which is a molecularity of one, then 00:33:23.790 --> 00:33:25.730 it's going to be a first order reaction. 00:33:25.730 --> 00:33:29.570 One reactant is just sitting by itself, falls apart, like 00:33:29.570 --> 00:33:31.350 in radioactive decay. 00:33:31.350 --> 00:33:33.270 Molecularity of one, that's a first order of process. 00:33:33.270 --> 00:33:36.510 If I need to have two molecules come together, then 00:33:36.510 --> 00:33:37.790 it's a second order process. 00:33:37.790 --> 00:33:40.900 If I have to have three molecules collide at the same 00:33:40.900 --> 00:33:44.520 time together, molecularity of three, then it's going to 00:33:44.520 --> 00:33:49.650 depend on the concentration of all three at the same time. 00:33:49.650 --> 00:33:52.600 That's called a ternary reaction, and those are really 00:33:52.600 --> 00:33:54.140 quite rare. 00:33:54.140 --> 00:33:57.300 Termolecular reactions, you need to have your 00:33:57.300 --> 00:34:01.820 concentrations very, very high to statistically get an event 00:34:01.820 --> 00:34:07.370 happening where all three molecules collide together. 00:34:07.370 --> 00:34:12.750 So three body reaction is hard and anything higher than three 00:34:12.750 --> 00:34:15.050 body is essentially impossible. 00:34:15.050 --> 00:34:19.880 So that limits your choices, which is nice. 00:34:19.880 --> 00:34:21.940 So that's the mechanism. 00:34:21.940 --> 00:34:23.700 And so what we're going to do next is go through some 00:34:23.700 --> 00:34:24.200 mechanisms. 00:34:24.200 --> 00:34:28.730 Some simple mechanisms and build up the complexity. 00:34:28.730 --> 00:34:33.910 Any questions about mechanisms here? 00:34:33.910 --> 00:34:37.430 What we're doing here is, we're formulating a framework. 00:34:37.430 --> 00:34:39.715 Where we can go back and look at things that are more 00:34:39.715 --> 00:34:42.530 complicated, like chain reactions or explosions or 00:34:42.530 --> 00:34:47.680 enzymatic reactions, and know when to apply approximations 00:34:47.680 --> 00:34:49.530 and et cetera. 00:34:49.530 --> 00:34:52.750 So we basically, here, are just laying 00:34:52.750 --> 00:34:58.210 down the ground rules. 00:34:58.210 --> 00:35:03.820 So let's go to our first example of a mechanism, a more 00:35:03.820 --> 00:35:05.320 complicated reaction. 00:35:05.320 --> 00:35:08.050 And what we're going to do is we're going to extract out 00:35:08.050 --> 00:35:12.340 integrated rate laws out of all these mechanisms. 00:35:12.340 --> 00:35:15.710 And see what it looks like, as a function of time. 00:35:15.710 --> 00:35:20.610 So the first one we're going to do is 00:35:20.610 --> 00:35:22.340 called parallel reactions. 00:35:22.340 --> 00:35:23.280 Simple mechanism. 00:35:23.280 --> 00:35:31.030 In this case here I have one reactant, and that reactant 00:35:31.030 --> 00:35:34.040 has a choice. 00:35:34.040 --> 00:35:37.030 You can think of it as a radioactive decay, an atom 00:35:37.030 --> 00:35:41.050 decaying in two different channels. 00:35:41.050 --> 00:35:44.670 So it can decay into B, or it can decay into C. There are 00:35:44.670 --> 00:35:50.470 two rate constants, k1 and k2. 00:35:50.470 --> 00:35:54.360 So you can write it like this, or you can write it as A goes 00:35:54.360 --> 00:35:58.740 to B plus C. This is how you would write a reaction. 00:35:58.740 --> 00:36:01.290 And this is how you would write your mechanism. 00:36:01.290 --> 00:36:10.090 A goes to B and A goes to C. Each elementary step, these 00:36:10.090 --> 00:36:12.550 are the elementary steps and this is the complex reaction, 00:36:12.550 --> 00:36:15.440 each elementary step is unimolecular. 00:36:15.440 --> 00:36:18.480 It's a first order process. 00:36:18.480 --> 00:36:20.890 So in all of these examples, the first thing you do is you 00:36:20.890 --> 00:36:24.330 write your rate law. 00:36:24.330 --> 00:36:28.190 The rate at which A gets created or destroyed. 00:36:28.190 --> 00:36:31.240 And there are two paths. 00:36:31.240 --> 00:36:34.310 It gets destroyed and into B, with a rate which is 00:36:34.310 --> 00:36:37.540 proportional to the concentration of A, and it 00:36:37.540 --> 00:36:43.040 gets destroyed into C, proportional to the 00:36:43.040 --> 00:36:49.670 concentration of A. So you write down all the ways that A 00:36:49.670 --> 00:36:51.100 can get destroyed. 00:36:51.100 --> 00:36:52.160 There are two ways here. 00:36:52.160 --> 00:36:54.660 Two channels. 00:36:54.660 --> 00:36:59.270 This one happens to be fairly easy to solve. k1 plus k2 00:36:59.270 --> 00:37:00.970 times A, and you've seen this before. 00:37:00.970 --> 00:37:03.590 It's minus dA/dt as a constant times A. That's 00:37:03.590 --> 00:37:04.720 a first order process. 00:37:04.720 --> 00:37:06.530 So you can just write down the answer. 00:37:06.530 --> 00:37:08.420 You don't need to do any math here. 00:37:08.420 --> 00:37:11.680 You recognize that we just call this one k prime. 00:37:11.680 --> 00:37:15.010 And that the rate is A as a function of time. 00:37:15.010 --> 00:37:22.185 This is A0 e to the minus k1 plus k2 times t. 00:37:22.185 --> 00:37:24.520 And everything you've learned about plotting first order 00:37:24.520 --> 00:37:28.370 processes et cetera, is applicable here, where the 00:37:28.370 --> 00:37:32.920 rate constant is the sum of these two. 00:37:32.920 --> 00:37:34.580 So that's for the reactant. 00:37:34.580 --> 00:37:36.730 The products are also interesting to plot as a 00:37:36.730 --> 00:37:40.260 function of time, to see how they are related to each other 00:37:40.260 --> 00:37:41.650 in terms of their concentrations. 00:37:41.650 --> 00:37:52.400 So let me go through this also. 00:37:52.400 --> 00:37:54.420 When things get more complicated we'll quickly go 00:37:54.420 --> 00:37:56.110 and make approximations. 00:37:56.110 --> 00:37:59.380 But for now, we can still do everything exactly. 00:37:59.380 --> 00:38:04.960 So you write down your rate law for the product. dB/dt is 00:38:04.960 --> 00:38:14.270 equal to k1 A. dC/dt is equal to k2 times A. The formation 00:38:14.270 --> 00:38:17.420 of B depends linearly on A. The formation of C depends 00:38:17.420 --> 00:38:21.050 linearly on A, because they're both first order processes. 00:38:21.050 --> 00:38:25.800 To make B and C. And you integrate. 00:38:25.800 --> 00:38:31.180 You integrate here from zero to B, dB. 00:38:31.180 --> 00:38:38.430 Is equal from zero to t, A dt, k1. 00:38:38.430 --> 00:38:39.650 A is a function of time. 00:38:39.650 --> 00:38:41.570 And we've already solved for that. 00:38:41.570 --> 00:38:43.700 It's this exponential up there. 00:38:43.700 --> 00:38:51.370 So you plug in here A of time. 00:38:51.370 --> 00:38:53.430 And you turn the crank and you integrate, and it's an 00:38:53.430 --> 00:38:53.990 exponential. 00:38:53.990 --> 00:38:56.020 So it's not so hard to integrate. 00:38:56.020 --> 00:39:00.280 And you get that B is a function of time is k1 times 00:39:00.280 --> 00:39:10.480 A0 over k1 plus k2 times one minus e to the minus k1 plus 00:39:10.480 --> 00:39:13.680 k2 times the time. 00:39:13.680 --> 00:39:15.730 Things are already starting to get a little bit more 00:39:15.730 --> 00:39:18.190 messy in the math. 00:39:18.190 --> 00:39:19.700 And then to get C, you actually 00:39:19.700 --> 00:39:20.470 don't need to do anything. 00:39:20.470 --> 00:39:23.280 Because you notice that the only difference between B and 00:39:23.280 --> 00:39:26.670 C here is replacing k2 with k1. 00:39:26.670 --> 00:39:28.360 So don't worry about doing any math. 00:39:28.360 --> 00:39:34.100 Just write down the answer. k2 A0, you interchange k1 and k2 00:39:34.100 --> 00:39:36.890 at every step. 00:39:36.890 --> 00:39:42.000 One minus e to the minus k1 plus k2 times the time. 00:39:42.000 --> 00:39:45.480 The only difference is up here in the k2 term. 00:39:45.480 --> 00:39:48.090 And those are your equations for k1 and k2. 00:39:48.090 --> 00:39:53.610 And what you find, is the ratio of B to C is a constant. 00:39:53.610 --> 00:39:56.980 If I divide B by C, everything cancels out except for the k1 00:39:56.980 --> 00:39:58.690 and the k2 here. 00:39:58.690 --> 00:40:02.960 Is equal to k1 over k2. 00:40:02.960 --> 00:40:11.280 And that is called the branching ratio. 00:40:11.280 --> 00:40:13.390 The branching ratio, because there are two branches out of 00:40:13.390 --> 00:40:15.210 the reactions. 00:40:15.210 --> 00:40:18.900 And this gives you the ratio of which one is more likely to 00:40:18.900 --> 00:40:20.030 happen than the other one. 00:40:20.030 --> 00:40:22.510 And so if k1 is much bigger than k2, the 00:40:22.510 --> 00:40:25.400 rate is per unit time. 00:40:25.400 --> 00:40:30.420 The units of k1 are per second or per minute or per hour. 00:40:30.420 --> 00:40:33.730 So if this is big, if k1 is big, then mostly you're going 00:40:33.730 --> 00:40:38.830 from A to B, and only a little bit of C is formed. 00:40:38.830 --> 00:40:40.830 And the ratio of B and C is always constant. 00:40:40.830 --> 00:40:47.500 And so you can plot, then, you can plot the result. 00:40:47.500 --> 00:40:57.900 You can sketch out the result. 00:40:57.900 --> 00:41:03.520 So you know that A is going to come down exponentially. 00:41:03.520 --> 00:41:06.250 Time on this axis here, concentrations 00:41:06.250 --> 00:41:10.010 on this axis here. 00:41:10.010 --> 00:41:13.080 So this is A as a function of time. 00:41:13.080 --> 00:41:15.510 That's that equation up here, in exponential decay. 00:41:15.510 --> 00:41:21.490 And the quantity of A. And B and C are going to come up in 00:41:21.490 --> 00:41:27.120 time, also, with this exponential format here. 00:41:27.120 --> 00:41:30.370 B is going to saturate at this ratio right here, k1 A0 00:41:30.370 --> 00:41:31.650 divided by k1 plus k2. 00:41:31.650 --> 00:41:35.560 C is going to saturate at this quantity here. 00:41:35.560 --> 00:41:38.070 So they're going to start both at zero. 00:41:38.070 --> 00:41:52.410 And B is eventually going to go to k1 A0 over k1, plus k2. 00:41:52.410 --> 00:42:08.550 And C, eventually, will go to k2 A0 over k1 plus k2. 00:42:08.550 --> 00:42:14.140 And this, and the ratio of these two lines at every point 00:42:14.140 --> 00:42:22.760 is k1 over k2. 00:42:22.760 --> 00:42:25.960 So, a simple question, for instance, that you might be of 00:42:25.960 --> 00:42:32.520 the type that you might be asked to look at is, suppose 00:42:32.520 --> 00:42:39.220 that k1 is 1/10 of k2. 00:42:39.220 --> 00:42:41.630 Which would you expect? 00:42:41.630 --> 00:42:48.340 Would you expect to have, let's see. 00:42:48.340 --> 00:42:51.610 So C is in green here. 00:42:51.610 --> 00:43:05.810 C, B. Or do you expect, so A comes down. 00:43:05.810 --> 00:43:10.030 B comes up. 00:43:10.030 --> 00:43:17.280 C comes up like this, or do you expect the last choice, 00:43:17.280 --> 00:43:39.380 I'm going to put the last choice here, so this is, let's 00:43:39.380 --> 00:43:41.710 call this choice number one. 00:43:41.710 --> 00:43:44.450 Choice number two. 00:43:44.450 --> 00:43:49.240 Choice number three. 00:43:49.240 --> 00:43:54.670 So k1, the rate k1 is 1/10 of the rate k2. 00:43:54.670 --> 00:43:57.560 That tells you something about the branching ratio. 00:43:57.560 --> 00:44:00.960 So do you expect this one here to be the right one? 00:44:00.960 --> 00:44:05.880 How many people think this is the right one? 00:44:05.880 --> 00:44:06.870 What about this one here? 00:44:06.870 --> 00:44:09.350 How many people think this is the right one? 00:44:09.350 --> 00:44:11.130 One person. 00:44:11.130 --> 00:44:13.500 What about this one here? 00:44:13.500 --> 00:44:15.760 So the branching ratio is the ratio of the two. 00:44:15.760 --> 00:44:16.750 It's 1/10. 00:44:16.750 --> 00:44:20.670 So this is approximately, in my sketch, poor sketch, 00:44:20.670 --> 00:44:22.350 granted, but this is approximately 10 00:44:22.350 --> 00:44:24.840 times bigger than this. 00:44:24.840 --> 00:44:26.550 So that's the ratio that you'd expect. 00:44:26.550 --> 00:44:31.490 And it's the right, here k1 is the rate into B. It's slower 00:44:31.490 --> 00:44:38.420 than the rate into C. So, you got it right. 00:44:38.420 --> 00:44:41.120 So for more complicated, we have a more complicated 00:44:41.120 --> 00:44:43.740 process, we're going to ask you the same sort of stuff. 00:44:43.740 --> 00:44:48.070 And it won't be as straightforward. 00:44:48.070 --> 00:44:52.140 Any questions on this beginning here? 00:44:52.140 --> 00:44:59.370 Next time we're going to finish with the parallel first 00:44:59.370 --> 00:45:01.120 and second order processes. 00:45:01.120 --> 00:45:05.280 And hopefully we'll get done with the complex reactions and 00:45:05.280 --> 00:45:10.740 mechanisms and move on to, I forgot 00:45:10.740 --> 00:45:11.740 what's next on the list. 00:45:11.740 --> 00:45:17.150 But some explosions or chain reactions.