[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.00,0:00:01.44,Default,,0000,0000,0000,,- [Instructor] Let's consider a reaction
Dialogue: 0,0:00:01.44,0:00:04.01,Default,,0000,0000,0000,,with the following multi-step mechanism.
Dialogue: 0,0:00:04.01,0:00:09.01,Default,,0000,0000,0000,,In step one, A reacts\Nwith BC to form AC plus B.
Dialogue: 0,0:00:09.67,0:00:14.67,Default,,0000,0000,0000,,And in step two, AC reacts\Nwith D to form A plus CD.
Dialogue: 0,0:00:16.87,0:00:19.05,Default,,0000,0000,0000,,If we add the two steps\Nof our mechanism together
Dialogue: 0,0:00:19.05,0:00:20.64,Default,,0000,0000,0000,,we can find the balanced equation
Dialogue: 0,0:00:20.64,0:00:23.13,Default,,0000,0000,0000,,for this hypothetical reaction.
Dialogue: 0,0:00:23.13,0:00:25.12,Default,,0000,0000,0000,,So we're gonna put all of our reactants
Dialogue: 0,0:00:25.12,0:00:27.56,Default,,0000,0000,0000,,on the left side here,\Nand we're gonna have
Dialogue: 0,0:00:27.56,0:00:30.80,Default,,0000,0000,0000,,all of our products on the right side.
Dialogue: 0,0:00:30.80,0:00:33.13,Default,,0000,0000,0000,,And we can see that AC is on the left
Dialogue: 0,0:00:33.13,0:00:36.45,Default,,0000,0000,0000,,and it's on the right side,\Nso we can cancel that out.
Dialogue: 0,0:00:36.45,0:00:38.60,Default,,0000,0000,0000,,A is also in the left and the right side,
Dialogue: 0,0:00:38.60,0:00:40.53,Default,,0000,0000,0000,,so we can cancel that out.
Dialogue: 0,0:00:40.53,0:00:45.36,Default,,0000,0000,0000,,So the overall equation would be BC plus D
Dialogue: 0,0:00:46.94,0:00:51.67,Default,,0000,0000,0000,,goes to B plus CD.
Dialogue: 0,0:00:56.00,0:00:58.85,Default,,0000,0000,0000,,We've just seen that BC\Nand D are our reactants
Dialogue: 0,0:01:01.22,0:01:05.48,Default,,0000,0000,0000,,and B and CD are the products
Dialogue: 0,0:01:05.48,0:01:07.98,Default,,0000,0000,0000,,for this hypothetical reaction.
Dialogue: 0,0:01:07.98,0:01:11.55,Default,,0000,0000,0000,,If we look at the mechanism,\NA is there in the beginning
Dialogue: 0,0:01:11.55,0:01:13.33,Default,,0000,0000,0000,,and A is there in the end.
Dialogue: 0,0:01:13.33,0:01:16.04,Default,,0000,0000,0000,,But A is not a reactant or a product,
Dialogue: 0,0:01:16.04,0:01:19.36,Default,,0000,0000,0000,,therefore A must be a catalyst.
Dialogue: 0,0:01:20.60,0:01:24.26,Default,,0000,0000,0000,,Something else that's not a\Nreactant or a product is AC.
Dialogue: 0,0:01:24.26,0:01:26.15,Default,,0000,0000,0000,,You notice how AC was generated
Dialogue: 0,0:01:26.15,0:01:27.74,Default,,0000,0000,0000,,in the first step of our mechanism,
Dialogue: 0,0:01:27.74,0:01:31.63,Default,,0000,0000,0000,,and then AC is used up in the\Nsecond step of the mechanism.
Dialogue: 0,0:01:31.63,0:01:36.63,Default,,0000,0000,0000,,Therefore AC must be the\Nintermediate for this reaction.
Dialogue: 0,0:01:41.86,0:01:43.81,Default,,0000,0000,0000,,Next, let's look at the energy profile
Dialogue: 0,0:01:43.81,0:01:45.72,Default,,0000,0000,0000,,for this multi-step reaction.
Dialogue: 0,0:01:45.72,0:01:48.82,Default,,0000,0000,0000,,Energy profiles usually have\Npotential energy in the y-axis
Dialogue: 0,0:01:48.82,0:01:51.64,Default,,0000,0000,0000,,and then reaction progress on the x-axis.
Dialogue: 0,0:01:51.64,0:01:55.27,Default,,0000,0000,0000,,So as we move to the right on the x-axis,
Dialogue: 0,0:01:55.27,0:01:56.96,Default,,0000,0000,0000,,the reaction is occurring.
Dialogue: 0,0:01:58.31,0:02:00.75,Default,,0000,0000,0000,,This first line on our energy profile
Dialogue: 0,0:02:00.75,0:02:03.75,Default,,0000,0000,0000,,represents the energy\Nlevel of our reactants,
Dialogue: 0,0:02:03.75,0:02:06.33,Default,,0000,0000,0000,,which are BC and D.
Dialogue: 0,0:02:06.33,0:02:10.89,Default,,0000,0000,0000,,So let's go ahead and show\Nthe bond between B and C.
Dialogue: 0,0:02:11.75,0:02:14.98,Default,,0000,0000,0000,,And then we also have D present.
Dialogue: 0,0:02:14.98,0:02:17.14,Default,,0000,0000,0000,,Our catalyst is also present
Dialogue: 0,0:02:17.14,0:02:19.21,Default,,0000,0000,0000,,at the very beginning of our reactions.
Dialogue: 0,0:02:19.21,0:02:23.88,Default,,0000,0000,0000,,So I'll go ahead and draw in\NA above our two reactants.
Dialogue: 0,0:02:26.41,0:02:29.87,Default,,0000,0000,0000,,We can see in our energy\Nprofile that we have two hills.
Dialogue: 0,0:02:29.87,0:02:32.44,Default,,0000,0000,0000,,The first Hill corresponds\Nto the first step
Dialogue: 0,0:02:32.44,0:02:34.19,Default,,0000,0000,0000,,of the mechanism and the second hill
Dialogue: 0,0:02:34.19,0:02:37.09,Default,,0000,0000,0000,,corresponds to the second step.
Dialogue: 0,0:02:37.09,0:02:40.95,Default,,0000,0000,0000,,So the peak of the first\Nhill is the transition state
Dialogue: 0,0:02:40.95,0:02:44.25,Default,,0000,0000,0000,,for the first step of the mechanism.
Dialogue: 0,0:02:44.25,0:02:47.33,Default,,0000,0000,0000,,And we can see in the first\Nstep that the catalyst A,
Dialogue: 0,0:02:47.33,0:02:50.51,Default,,0000,0000,0000,,is colliding with BC or reacting with BC
Dialogue: 0,0:02:50.51,0:02:53.49,Default,,0000,0000,0000,,to form our intermediate AC.
Dialogue: 0,0:02:53.49,0:02:58.49,Default,,0000,0000,0000,,So A must collide with BC\Nand at the transition state,
Dialogue: 0,0:02:59.03,0:03:02.37,Default,,0000,0000,0000,,the bond between B and C is breaking,
Dialogue: 0,0:03:02.37,0:03:07.37,Default,,0000,0000,0000,,and at the same time, the bond\Nbetween A and C is forming.
Dialogue: 0,0:03:10.38,0:03:12.76,Default,,0000,0000,0000,,We would still have reactant D present
Dialogue: 0,0:03:12.76,0:03:14.33,Default,,0000,0000,0000,,at the top of this hill too.
Dialogue: 0,0:03:14.33,0:03:17.41,Default,,0000,0000,0000,,So I'll go ahead and draw in D here.
Dialogue: 0,0:03:17.41,0:03:22.41,Default,,0000,0000,0000,,When a collides with BC,\Nthe collision has to have
Dialogue: 0,0:03:22.66,0:03:27.66,Default,,0000,0000,0000,,enough kinetic energy to\Novercome the activation energy
Dialogue: 0,0:03:28.28,0:03:31.27,Default,,0000,0000,0000,,necessary for this reaction to occur.
Dialogue: 0,0:03:31.27,0:03:33.53,Default,,0000,0000,0000,,And on this energy profile,
Dialogue: 0,0:03:33.53,0:03:36.52,Default,,0000,0000,0000,,the activation energy is\Nthe difference in energy
Dialogue: 0,0:03:36.52,0:03:41.52,Default,,0000,0000,0000,,between the reactants\Nand the transition state,
Dialogue: 0,0:03:42.44,0:03:44.00,Default,,0000,0000,0000,,so the very peak of the hill.
Dialogue: 0,0:03:44.00,0:03:46.90,Default,,0000,0000,0000,,So this difference in energy,
Dialogue: 0,0:03:46.90,0:03:49.31,Default,,0000,0000,0000,,corresponds to the activation energy
Dialogue: 0,0:03:49.31,0:03:53.40,Default,,0000,0000,0000,,for the first step of the\Nmechanism which we will call Ea1.
Dialogue: 0,0:03:54.89,0:03:57.76,Default,,0000,0000,0000,,If we assume that the collision\Nhas enough kinetic energy
Dialogue: 0,0:03:57.76,0:04:00.26,Default,,0000,0000,0000,,to overcome the activation energy,
Dialogue: 0,0:04:00.26,0:04:04.61,Default,,0000,0000,0000,,we'll form our intermediate\NAC, and we'd also form B.
Dialogue: 0,0:04:04.61,0:04:06.65,Default,,0000,0000,0000,,So let's go ahead and show the bond
Dialogue: 0,0:04:06.65,0:04:09.96,Default,,0000,0000,0000,,between A and C has now been formed.
Dialogue: 0,0:04:09.96,0:04:12.38,Default,,0000,0000,0000,,So this valley here between our two hills
Dialogue: 0,0:04:12.38,0:04:16.33,Default,,0000,0000,0000,,represents the energy\Nlevel of the intermediate.
Dialogue: 0,0:04:16.33,0:04:18.12,Default,,0000,0000,0000,,We would also have be present,
Dialogue: 0,0:04:18.12,0:04:20.99,Default,,0000,0000,0000,,so I can go ahead and\NI'll just write in B here.
Dialogue: 0,0:04:20.99,0:04:23.13,Default,,0000,0000,0000,,And then we still have some D present,
Dialogue: 0,0:04:23.13,0:04:25.29,Default,,0000,0000,0000,,D still hasn't reacted yet.
Dialogue: 0,0:04:25.29,0:04:27.80,Default,,0000,0000,0000,,So I'll go ahead and draw in D as well.
Dialogue: 0,0:04:30.44,0:04:31.97,Default,,0000,0000,0000,,Next we're ready for the second hill
Dialogue: 0,0:04:31.97,0:04:34.36,Default,,0000,0000,0000,,or the second step of our mechanism.
Dialogue: 0,0:04:34.36,0:04:39.06,Default,,0000,0000,0000,,In the second step, AC the\Nintermediate AC reacts with D
Dialogue: 0,0:04:39.06,0:04:41.84,Default,,0000,0000,0000,,to form A and CD.
Dialogue: 0,0:04:41.84,0:04:45.16,Default,,0000,0000,0000,,So the top of this second hill\Nwould be the transition state
Dialogue: 0,0:04:45.16,0:04:47.24,Default,,0000,0000,0000,,for this second step.
Dialogue: 0,0:04:47.24,0:04:51.97,Default,,0000,0000,0000,,So we can show the bond\Nbetween A and C braking,
Dialogue: 0,0:04:51.97,0:04:56.97,Default,,0000,0000,0000,,and at the same time the bond\Nbetween C and D is forming.
Dialogue: 0,0:04:58.98,0:05:02.12,Default,,0000,0000,0000,,The difference in energy\Nbetween the energy
Dialogue: 0,0:05:02.12,0:05:05.54,Default,,0000,0000,0000,,of the intermediate and the\Nenergy of the transition state
Dialogue: 0,0:05:05.54,0:05:07.66,Default,,0000,0000,0000,,represents the activation energy
Dialogue: 0,0:05:07.66,0:05:09.90,Default,,0000,0000,0000,,for the second step of the mechanism,
Dialogue: 0,0:05:09.90,0:05:12.01,Default,,0000,0000,0000,,which we will call Ea2.
Dialogue: 0,0:05:14.65,0:05:19.39,Default,,0000,0000,0000,,So AC and D must collide\Nwith enough kinetic energy
Dialogue: 0,0:05:19.39,0:05:24.39,Default,,0000,0000,0000,,to overcome the activation\Nenergy for this second step.
Dialogue: 0,0:05:24.45,0:05:28.10,Default,,0000,0000,0000,,If AC and D collide with\Nenough kinetic energy,
Dialogue: 0,0:05:28.10,0:05:31.76,Default,,0000,0000,0000,,we would produce A and CD.
Dialogue: 0,0:05:31.76,0:05:35.13,Default,,0000,0000,0000,,So this line at the end here\Nrepresents the energy level
Dialogue: 0,0:05:35.13,0:05:37.91,Default,,0000,0000,0000,,of our products.
Dialogue: 0,0:05:37.91,0:05:42.91,Default,,0000,0000,0000,,So CD is one of our products,\Nso we'll write that in here.
Dialogue: 0,0:05:43.53,0:05:46.60,Default,,0000,0000,0000,,And remember B is our other products,
Dialogue: 0,0:05:46.60,0:05:51.33,Default,,0000,0000,0000,,which we formed from the\Nfirst step of the mechanisms.
Dialogue: 0,0:05:51.33,0:05:55.64,Default,,0000,0000,0000,,So let's go ahead and\Nwrite in here B plus CD.
Dialogue: 0,0:05:55.64,0:05:58.18,Default,,0000,0000,0000,,And we also reformed our catalyst,
Dialogue: 0,0:05:58.18,0:06:01.54,Default,,0000,0000,0000,,so A would be present here as well.
Dialogue: 0,0:06:01.54,0:06:04.91,Default,,0000,0000,0000,,Next let's compare the\Nfirst activation energy Ea1
Dialogue: 0,0:06:04.91,0:06:08.04,Default,,0000,0000,0000,,with the second activation energy Ea2.
Dialogue: 0,0:06:08.04,0:06:10.49,Default,,0000,0000,0000,,Looking at the energy\Nprofile we can see that Ea1
Dialogue: 0,0:06:11.74,0:06:14.59,Default,,0000,0000,0000,,has a much greater\Nactivation energy than Ea2.
Dialogue: 0,0:06:15.55,0:06:20.55,Default,,0000,0000,0000,,So let's go ahead and write\NEa1 is greater than Ea2.
Dialogue: 0,0:06:20.89,0:06:24.49,Default,,0000,0000,0000,,The smaller the activation\Nenergy, the faster the reaction,
Dialogue: 0,0:06:24.49,0:06:26.91,Default,,0000,0000,0000,,and since there's a\Nsmaller activation energy
Dialogue: 0,0:06:26.91,0:06:28.44,Default,,0000,0000,0000,,for the second step,
Dialogue: 0,0:06:28.44,0:06:32.32,Default,,0000,0000,0000,,the second step must be\Nthe faster of the two.
Dialogue: 0,0:06:32.32,0:06:35.44,Default,,0000,0000,0000,,Since the first step has the\Nhigher activation energy,
Dialogue: 0,0:06:35.44,0:06:40.44,Default,,0000,0000,0000,,the first step must be slow\Ncompared to the second step.
Dialogue: 0,0:06:42.15,0:06:45.07,Default,,0000,0000,0000,,Since the first step of the\Nmechanism is the slow step,
Dialogue: 0,0:06:45.07,0:06:48.08,Default,,0000,0000,0000,,the first step is the\Nrate determining step.
Dialogue: 0,0:06:48.96,0:06:51.62,Default,,0000,0000,0000,,Finally, let's Find the\Noverall change in energy
Dialogue: 0,0:06:51.62,0:06:54.02,Default,,0000,0000,0000,,for our reaction.
Dialogue: 0,0:06:54.02,0:06:57.32,Default,,0000,0000,0000,,So to find the overall change\Nin energy, that's Delta E,
Dialogue: 0,0:06:57.32,0:06:59.64,Default,,0000,0000,0000,,which is final minus initial.
Dialogue: 0,0:06:59.64,0:07:02.22,Default,,0000,0000,0000,,So that would be the\Nenergy of the products
Dialogue: 0,0:07:02.22,0:07:05.87,Default,,0000,0000,0000,,minus the energy of the reactants.
Dialogue: 0,0:07:05.87,0:07:09.32,Default,,0000,0000,0000,,So the energy level of\Nthe products is right here
Dialogue: 0,0:07:09.32,0:07:11.10,Default,,0000,0000,0000,,and then the energy level of the reactants
Dialogue: 0,0:07:11.10,0:07:11.93,Default,,0000,0000,0000,,is at the beginning.
Dialogue: 0,0:07:11.93,0:07:14.21,Default,,0000,0000,0000,,So let me just extend\Nthis dashed line here
Dialogue: 0,0:07:14.21,0:07:16.57,Default,,0000,0000,0000,,so we can better compare the two.
Dialogue: 0,0:07:16.57,0:07:18.54,Default,,0000,0000,0000,,Representing Delta E on a graph,
Dialogue: 0,0:07:18.54,0:07:20.30,Default,,0000,0000,0000,,it would be the difference in energy
Dialogue: 0,0:07:20.30,0:07:22.76,Default,,0000,0000,0000,,between these two lines.
Dialogue: 0,0:07:22.76,0:07:24.75,Default,,0000,0000,0000,,And since the energy of the products
Dialogue: 0,0:07:24.75,0:07:27.65,Default,,0000,0000,0000,,is greater than the\Nenergy of the reactants,
Dialogue: 0,0:07:27.65,0:07:29.93,Default,,0000,0000,0000,,we would be subtracting a smaller number
Dialogue: 0,0:07:29.93,0:07:31.21,Default,,0000,0000,0000,,from a larger number
Dialogue: 0,0:07:31.21,0:07:34.36,Default,,0000,0000,0000,,and therefore Delta E would be positive
Dialogue: 0,0:07:34.36,0:07:37.01,Default,,0000,0000,0000,,for this hypothetical reaction.
Dialogue: 0,0:07:37.01,0:07:39.29,Default,,0000,0000,0000,,And since Delta E is positive,
Dialogue: 0,0:07:39.29,0:07:43.56,Default,,0000,0000,0000,,we know that this reaction\Nis an endothermic reaction.