0:00:00.000,0:00:01.440
- [Instructor] Let's consider a reaction

0:00:01.440,0:00:04.010
with the following multi-step mechanism.

0:00:04.010,0:00:09.010
In step one, A reacts[br]with BC to form AC plus B.

0:00:09.670,0:00:14.670
And in step two, AC reacts[br]with D to form A plus CD.

0:00:16.870,0:00:19.050
If we add the two steps[br]of our mechanism together

0:00:19.050,0:00:20.640
we can find the balanced equation

0:00:20.640,0:00:23.130
for this hypothetical reaction.

0:00:23.130,0:00:25.120
So we're gonna put all of our reactants

0:00:25.120,0:00:27.560
on the left side here,[br]and we're gonna have

0:00:27.560,0:00:30.800
all of our products on the right side.

0:00:30.800,0:00:33.130
And we can see that AC is on the left

0:00:33.130,0:00:36.450
and it's on the right side,[br]so we can cancel that out.

0:00:36.450,0:00:38.600
A is also in the left and the right side,

0:00:38.600,0:00:40.530
so we can cancel that out.

0:00:40.530,0:00:45.360
So the overall equation would be BC plus D

0:00:46.940,0:00:51.673
goes to B plus CD.

0:00:56.002,0:00:58.850
We've just seen that BC[br]and D are our reactants

0:01:01.220,0:01:05.480
and B and CD are the products

0:01:05.480,0:01:07.980
for this hypothetical reaction.

0:01:07.980,0:01:11.550
If we look at the mechanism,[br]A is there in the beginning

0:01:11.550,0:01:13.330
and A is there in the end.

0:01:13.330,0:01:16.040
But A is not a reactant or a product,

0:01:16.040,0:01:19.363
therefore A must be a catalyst.

0:01:20.600,0:01:24.260
Something else that's not a[br]reactant or a product is AC.

0:01:24.260,0:01:26.150
You notice how AC was generated

0:01:26.150,0:01:27.740
in the first step of our mechanism,

0:01:27.740,0:01:31.630
and then AC is used up in the[br]second step of the mechanism.

0:01:31.630,0:01:36.630
Therefore AC must be the[br]intermediate for this reaction.

0:01:41.860,0:01:43.810
Next, let's look at the energy profile

0:01:43.810,0:01:45.720
for this multi-step reaction.

0:01:45.720,0:01:48.820
Energy profiles usually have[br]potential energy in the y-axis

0:01:48.820,0:01:51.640
and then reaction progress on the x-axis.

0:01:51.640,0:01:55.270
So as we move to the right on the x-axis,

0:01:55.270,0:01:56.963
the reaction is occurring.

0:01:58.310,0:02:00.750
This first line on our energy profile

0:02:00.750,0:02:03.750
represents the energy[br]level of our reactants,

0:02:03.750,0:02:06.330
which are BC and D.

0:02:06.330,0:02:10.890
So let's go ahead and show[br]the bond between B and C.

0:02:11.750,0:02:14.980
And then we also have D present.

0:02:14.980,0:02:17.140
Our catalyst is also present

0:02:17.140,0:02:19.210
at the very beginning of our reactions.

0:02:19.210,0:02:23.883
So I'll go ahead and draw in[br]A above our two reactants.

0:02:26.408,0:02:29.870
We can see in our energy[br]profile that we have two hills.

0:02:29.870,0:02:32.440
The first Hill corresponds[br]to the first step

0:02:32.440,0:02:34.190
of the mechanism and the second hill

0:02:34.190,0:02:37.090
corresponds to the second step.

0:02:37.090,0:02:40.950
So the peak of the first[br]hill is the transition state

0:02:40.950,0:02:44.250
for the first step of the mechanism.

0:02:44.250,0:02:47.330
And we can see in the first[br]step that the catalyst A,

0:02:47.330,0:02:50.510
is colliding with BC or reacting with BC

0:02:50.510,0:02:53.490
to form our intermediate AC.

0:02:53.490,0:02:58.490
So A must collide with BC[br]and at the transition state,

0:02:59.030,0:03:02.370
the bond between B and C is breaking,

0:03:02.370,0:03:07.370
and at the same time, the bond[br]between A and C is forming.

0:03:10.380,0:03:12.760
We would still have reactant D present

0:03:12.760,0:03:14.330
at the top of this hill too.

0:03:14.330,0:03:17.410
So I'll go ahead and draw in D here.

0:03:17.410,0:03:22.410
When a collides with BC,[br]the collision has to have

0:03:22.660,0:03:27.660
enough kinetic energy to[br]overcome the activation energy

0:03:28.280,0:03:31.270
necessary for this reaction to occur.

0:03:31.270,0:03:33.530
And on this energy profile,

0:03:33.530,0:03:36.520
the activation energy is[br]the difference in energy

0:03:36.520,0:03:41.520
between the reactants[br]and the transition state,

0:03:42.440,0:03:44.000
so the very peak of the hill.

0:03:44.000,0:03:46.900
So this difference in energy,

0:03:46.900,0:03:49.310
corresponds to the activation energy

0:03:49.310,0:03:53.403
for the first step of the[br]mechanism which we will call Ea1.

0:03:54.890,0:03:57.760
If we assume that the collision[br]has enough kinetic energy

0:03:57.760,0:04:00.260
to overcome the activation energy,

0:04:00.260,0:04:04.610
we'll form our intermediate[br]AC, and we'd also form B.

0:04:04.610,0:04:06.650
So let's go ahead and show the bond

0:04:06.650,0:04:09.960
between A and C has now been formed.

0:04:09.960,0:04:12.380
So this valley here between our two hills

0:04:12.380,0:04:16.330
represents the energy[br]level of the intermediate.

0:04:16.330,0:04:18.120
We would also have be present,

0:04:18.120,0:04:20.990
so I can go ahead and[br]I'll just write in B here.

0:04:20.990,0:04:23.130
And then we still have some D present,

0:04:23.130,0:04:25.290
D still hasn't reacted yet.

0:04:25.290,0:04:27.803
So I'll go ahead and draw in D as well.

0:04:30.440,0:04:31.970
Next we're ready for the second hill

0:04:31.970,0:04:34.360
or the second step of our mechanism.

0:04:34.360,0:04:39.060
In the second step, AC the[br]intermediate AC reacts with D

0:04:39.060,0:04:41.840
to form A and CD.

0:04:41.840,0:04:45.160
So the top of this second hill[br]would be the transition state

0:04:45.160,0:04:47.240
for this second step.

0:04:47.240,0:04:51.970
So we can show the bond[br]between A and C braking,

0:04:51.970,0:04:56.970
and at the same time the bond[br]between C and D is forming.

0:04:58.980,0:05:02.120
The difference in energy[br]between the energy

0:05:02.120,0:05:05.540
of the intermediate and the[br]energy of the transition state

0:05:05.540,0:05:07.660
represents the activation energy

0:05:07.660,0:05:09.900
for the second step of the mechanism,

0:05:09.900,0:05:12.007
which we will call Ea2.

0:05:14.650,0:05:19.390
So AC and D must collide[br]with enough kinetic energy

0:05:19.390,0:05:24.390
to overcome the activation[br]energy for this second step.

0:05:24.450,0:05:28.100
If AC and D collide with[br]enough kinetic energy,

0:05:28.100,0:05:31.760
we would produce A and CD.

0:05:31.760,0:05:35.130
So this line at the end here[br]represents the energy level

0:05:35.130,0:05:37.910
of our products.

0:05:37.910,0:05:42.910
So CD is one of our products,[br]so we'll write that in here.

0:05:43.530,0:05:46.600
And remember B is our other products,

0:05:46.600,0:05:51.330
which we formed from the[br]first step of the mechanisms.

0:05:51.330,0:05:55.640
So let's go ahead and[br]write in here B plus CD.

0:05:55.640,0:05:58.180
And we also reformed our catalyst,

0:05:58.180,0:06:01.540
so A would be present here as well.

0:06:01.540,0:06:04.910
Next let's compare the[br]first activation energy Ea1

0:06:04.910,0:06:08.040
with the second activation energy Ea2.

0:06:08.040,0:06:10.490
Looking at the energy[br]profile we can see that Ea1

0:06:11.741,0:06:14.587
has a much greater[br]activation energy than Ea2.

0:06:15.550,0:06:20.550
So let's go ahead and write[br]Ea1 is greater than Ea2.

0:06:20.890,0:06:24.490
The smaller the activation[br]energy, the faster the reaction,

0:06:24.490,0:06:26.910
and since there's a[br]smaller activation energy

0:06:26.910,0:06:28.440
for the second step,

0:06:28.440,0:06:32.320
the second step must be[br]the faster of the two.

0:06:32.320,0:06:35.440
Since the first step has the[br]higher activation energy,

0:06:35.440,0:06:40.440
the first step must be slow[br]compared to the second step.

0:06:42.150,0:06:45.070
Since the first step of the[br]mechanism is the slow step,

0:06:45.070,0:06:48.083
the first step is the[br]rate determining step.

0:06:48.960,0:06:51.620
Finally, let's Find the[br]overall change in energy

0:06:51.620,0:06:54.020
for our reaction.

0:06:54.020,0:06:57.320
So to find the overall change[br]in energy, that's Delta E,

0:06:57.320,0:06:59.640
which is final minus initial.

0:06:59.640,0:07:02.220
So that would be the[br]energy of the products

0:07:02.220,0:07:05.870
minus the energy of the reactants.

0:07:05.870,0:07:09.320
So the energy level of[br]the products is right here

0:07:09.320,0:07:11.100
and then the energy level of the reactants

0:07:11.100,0:07:11.933
is at the beginning.

0:07:11.933,0:07:14.210
So let me just extend[br]this dashed line here

0:07:14.210,0:07:16.570
so we can better compare the two.

0:07:16.570,0:07:18.540
Representing Delta E on a graph,

0:07:18.540,0:07:20.300
it would be the difference in energy

0:07:20.300,0:07:22.760
between these two lines.

0:07:22.760,0:07:24.750
And since the energy of the products

0:07:24.750,0:07:27.650
is greater than the[br]energy of the reactants,

0:07:27.650,0:07:29.930
we would be subtracting a smaller number

0:07:29.930,0:07:31.210
from a larger number

0:07:31.210,0:07:34.360
and therefore Delta E would be positive

0:07:34.360,0:07:37.010
for this hypothetical reaction.

0:07:37.010,0:07:39.290
And since Delta E is positive,

0:07:39.290,0:07:43.563
we know that this reaction[br]is an endothermic reaction.