WEBVTT 00:00:00.000 --> 00:00:01.440 - [Instructor] Let's consider a reaction 00:00:01.440 --> 00:00:04.010 with the following multi-step mechanism. 00:00:04.010 --> 00:00:09.010 In step one, A reacts with BC to form AC plus B. 00:00:09.670 --> 00:00:14.670 And in step two, AC reacts with D to form A plus CD. 00:00:16.870 --> 00:00:19.050 If we add the two steps of our mechanism together 00:00:19.050 --> 00:00:20.640 we can find the balanced equation 00:00:20.640 --> 00:00:23.130 for this hypothetical reaction. 00:00:23.130 --> 00:00:25.120 So we're gonna put all of our reactants 00:00:25.120 --> 00:00:27.560 on the left side here, and we're gonna have 00:00:27.560 --> 00:00:30.800 all of our products on the right side. 00:00:30.800 --> 00:00:33.130 And we can see that AC is on the left 00:00:33.130 --> 00:00:36.450 and it's on the right side, so we can cancel that out. 00:00:36.450 --> 00:00:38.600 A is also in the left and the right side, 00:00:38.600 --> 00:00:40.530 so we can cancel that out. 00:00:40.530 --> 00:00:45.360 So the overall equation would be BC plus D 00:00:46.940 --> 00:00:51.673 goes to B plus CD. 00:00:56.002 --> 00:00:58.850 We've just seen that BC and D are our reactants 00:01:01.220 --> 00:01:05.480 and B and CD are the products 00:01:05.480 --> 00:01:07.980 for this hypothetical reaction. 00:01:07.980 --> 00:01:11.550 If we look at the mechanism, A is there in the beginning 00:01:11.550 --> 00:01:13.330 and A is there in the end. 00:01:13.330 --> 00:01:16.040 But A is not a reactant or a product, 00:01:16.040 --> 00:01:19.363 therefore A must be a catalyst. 00:01:20.600 --> 00:01:24.260 Something else that's not a reactant or a product is AC. 00:01:24.260 --> 00:01:26.150 You notice how AC was generated 00:01:26.150 --> 00:01:27.740 in the first step of our mechanism, 00:01:27.740 --> 00:01:31.630 and then AC is used up in the second step of the mechanism. 00:01:31.630 --> 00:01:36.630 Therefore AC must be the intermediate for this reaction. 00:01:41.860 --> 00:01:43.810 Next, let's look at the energy profile 00:01:43.810 --> 00:01:45.720 for this multi-step reaction. 00:01:45.720 --> 00:01:48.820 Energy profiles usually have potential energy in the y-axis 00:01:48.820 --> 00:01:51.640 and then reaction progress on the x-axis. 00:01:51.640 --> 00:01:55.270 So as we move to the right on the x-axis, 00:01:55.270 --> 00:01:56.963 the reaction is occurring. 00:01:58.310 --> 00:02:00.750 This first line on our energy profile 00:02:00.750 --> 00:02:03.750 represents the energy level of our reactants, 00:02:03.750 --> 00:02:06.330 which are BC and D. 00:02:06.330 --> 00:02:10.890 So let's go ahead and show the bond between B and C. 00:02:11.750 --> 00:02:14.980 And then we also have D present. 00:02:14.980 --> 00:02:17.140 Our catalyst is also present 00:02:17.140 --> 00:02:19.210 at the very beginning of our reactions. 00:02:19.210 --> 00:02:23.883 So I'll go ahead and draw in A above our two reactants. 00:02:26.408 --> 00:02:29.870 We can see in our energy profile that we have two hills. 00:02:29.870 --> 00:02:32.440 The first Hill corresponds to the first step 00:02:32.440 --> 00:02:34.190 of the mechanism and the second hill 00:02:34.190 --> 00:02:37.090 corresponds to the second step. 00:02:37.090 --> 00:02:40.950 So the peak of the first hill is the transition state 00:02:40.950 --> 00:02:44.250 for the first step of the mechanism. 00:02:44.250 --> 00:02:47.330 And we can see in the first step that the catalyst A, 00:02:47.330 --> 00:02:50.510 is colliding with BC or reacting with BC 00:02:50.510 --> 00:02:53.490 to form our intermediate AC. 00:02:53.490 --> 00:02:58.490 So A must collide with BC and at the transition state, 00:02:59.030 --> 00:03:02.370 the bond between B and C is breaking, 00:03:02.370 --> 00:03:07.370 and at the same time, the bond between A and C is forming. 00:03:10.380 --> 00:03:12.760 We would still have reactant D present 00:03:12.760 --> 00:03:14.330 at the top of this hill too. 00:03:14.330 --> 00:03:17.410 So I'll go ahead and draw in D here. 00:03:17.410 --> 00:03:22.410 When a collides with BC, the collision has to have 00:03:22.660 --> 00:03:27.660 enough kinetic energy to overcome the activation energy 00:03:28.280 --> 00:03:31.270 necessary for this reaction to occur. 00:03:31.270 --> 00:03:33.530 And on this energy profile, 00:03:33.530 --> 00:03:36.520 the activation energy is the difference in energy 00:03:36.520 --> 00:03:41.520 between the reactants and the transition state, 00:03:42.440 --> 00:03:44.000 so the very peak of the hill. 00:03:44.000 --> 00:03:46.900 So this difference in energy, 00:03:46.900 --> 00:03:49.310 corresponds to the activation energy 00:03:49.310 --> 00:03:53.403 for the first step of the mechanism which we will call Ea1. 00:03:54.890 --> 00:03:57.760 If we assume that the collision has enough kinetic energy 00:03:57.760 --> 00:04:00.260 to overcome the activation energy, 00:04:00.260 --> 00:04:04.610 we'll form our intermediate AC, and we'd also form B. 00:04:04.610 --> 00:04:06.650 So let's go ahead and show the bond 00:04:06.650 --> 00:04:09.960 between A and C has now been formed. 00:04:09.960 --> 00:04:12.380 So this valley here between our two hills 00:04:12.380 --> 00:04:16.330 represents the energy level of the intermediate. 00:04:16.330 --> 00:04:18.120 We would also have be present, 00:04:18.120 --> 00:04:20.990 so I can go ahead and I'll just write in B here. 00:04:20.990 --> 00:04:23.130 And then we still have some D present, 00:04:23.130 --> 00:04:25.290 D still hasn't reacted yet. 00:04:25.290 --> 00:04:27.803 So I'll go ahead and draw in D as well. 00:04:30.440 --> 00:04:31.970 Next we're ready for the second hill 00:04:31.970 --> 00:04:34.360 or the second step of our mechanism. 00:04:34.360 --> 00:04:39.060 In the second step, AC the intermediate AC reacts with D 00:04:39.060 --> 00:04:41.840 to form A and CD. 00:04:41.840 --> 00:04:45.160 So the top of this second hill would be the transition state 00:04:45.160 --> 00:04:47.240 for this second step. 00:04:47.240 --> 00:04:51.970 So we can show the bond between A and C braking, 00:04:51.970 --> 00:04:56.970 and at the same time the bond between C and D is forming. 00:04:58.980 --> 00:05:02.120 The difference in energy between the energy 00:05:02.120 --> 00:05:05.540 of the intermediate and the energy of the transition state 00:05:05.540 --> 00:05:07.660 represents the activation energy 00:05:07.660 --> 00:05:09.900 for the second step of the mechanism, 00:05:09.900 --> 00:05:12.007 which we will call Ea2. 00:05:14.650 --> 00:05:19.390 So AC and D must collide with enough kinetic energy 00:05:19.390 --> 00:05:24.390 to overcome the activation energy for this second step. 00:05:24.450 --> 00:05:28.100 If AC and D collide with enough kinetic energy, 00:05:28.100 --> 00:05:31.760 we would produce A and CD. 00:05:31.760 --> 00:05:35.130 So this line at the end here represents the energy level 00:05:35.130 --> 00:05:37.910 of our products. 00:05:37.910 --> 00:05:42.910 So CD is one of our products, so we'll write that in here. 00:05:43.530 --> 00:05:46.600 And remember B is our other products, 00:05:46.600 --> 00:05:51.330 which we formed from the first step of the mechanisms. 00:05:51.330 --> 00:05:55.640 So let's go ahead and write in here B plus CD. 00:05:55.640 --> 00:05:58.180 And we also reformed our catalyst, 00:05:58.180 --> 00:06:01.540 so A would be present here as well. 00:06:01.540 --> 00:06:04.910 Next let's compare the first activation energy Ea1 00:06:04.910 --> 00:06:08.040 with the second activation energy Ea2. 00:06:08.040 --> 00:06:10.490 Looking at the energy profile we can see that Ea1 00:06:11.741 --> 00:06:14.587 has a much greater activation energy than Ea2. 00:06:15.550 --> 00:06:20.550 So let's go ahead and write Ea1 is greater than Ea2. 00:06:20.890 --> 00:06:24.490 The smaller the activation energy, the faster the reaction, 00:06:24.490 --> 00:06:26.910 and since there's a smaller activation energy 00:06:26.910 --> 00:06:28.440 for the second step, 00:06:28.440 --> 00:06:32.320 the second step must be the faster of the two. 00:06:32.320 --> 00:06:35.440 Since the first step has the higher activation energy, 00:06:35.440 --> 00:06:40.440 the first step must be slow compared to the second step. 00:06:42.150 --> 00:06:45.070 Since the first step of the mechanism is the slow step, 00:06:45.070 --> 00:06:48.083 the first step is the rate determining step. 00:06:48.960 --> 00:06:51.620 Finally, let's Find the overall change in energy 00:06:51.620 --> 00:06:54.020 for our reaction. 00:06:54.020 --> 00:06:57.320 So to find the overall change in energy, that's Delta E, 00:06:57.320 --> 00:06:59.640 which is final minus initial. 00:06:59.640 --> 00:07:02.220 So that would be the energy of the products 00:07:02.220 --> 00:07:05.870 minus the energy of the reactants. 00:07:05.870 --> 00:07:09.320 So the energy level of the products is right here 00:07:09.320 --> 00:07:11.100 and then the energy level of the reactants 00:07:11.100 --> 00:07:11.933 is at the beginning. 00:07:11.933 --> 00:07:14.210 So let me just extend this dashed line here 00:07:14.210 --> 00:07:16.570 so we can better compare the two. 00:07:16.570 --> 00:07:18.540 Representing Delta E on a graph, 00:07:18.540 --> 00:07:20.300 it would be the difference in energy 00:07:20.300 --> 00:07:22.760 between these two lines. 00:07:22.760 --> 00:07:24.750 And since the energy of the products 00:07:24.750 --> 00:07:27.650 is greater than the energy of the reactants, 00:07:27.650 --> 00:07:29.930 we would be subtracting a smaller number 00:07:29.930 --> 00:07:31.210 from a larger number 00:07:31.210 --> 00:07:34.360 and therefore Delta E would be positive 00:07:34.360 --> 00:07:37.010 for this hypothetical reaction. 00:07:37.010 --> 00:07:39.290 And since Delta E is positive, 00:07:39.290 --> 00:07:43.563 we know that this reaction is an endothermic reaction.