Skip to main content
Loading Explore guides…
Reaction Mechanisms · Acedra
Home
Explore
Chemistry
Reaction Mechanisms
Sign in to save
Chat
If YouTube asks you to sign in, open the video on YouTube.
Watch on YouTube
Reload player
Transcript
75 segments
Copy
Hide video
1:27
Aloha and welcome back. In this lecture, which is our final one on the subject of chemical kinetics, we discuss reaction mechanisms. By now you probably know that chemical reactions often occur in multiple steps. In other words, certain reactant molecules can form products in one reaction, and those products can go on and react further to form different products in a second reaction, which can even go on in a third reaction, an so on.
1:57
And all of those different steps make up what's called the reaction mechanism. A reaction mechanism is when a chemical reaction occurs through a sequence of elementary steps. So each of those elementary reactions is called an elementary reaction, or a simple reaction. Here's an example of a generic two-step mechanism, In the first step a reactant molecule A forms two molecules B, and these react in the second step to form the final product C.
2:37
The overall effect is for the molecule A to form the product C. These molecules B that appear in the first step then react again in the second one are called intermediates. Intermediates are produced in an earlier step then consumed in a later step. So they're not there at the beginning nor at the end, but they pop up along the way. You can visualize this two-step mechanism in the diagram here. The reactant molecule A forms the two intermediate molecules B
3:13
in the first step. You can imagine this molecule A breaking apart into two equivalent pieces. And these two intermediates need to rearrange before they attach together to form the final product C. You should realize that all of the atoms that were present in the reactant molecule A are still there in product molecule C, they're just attached differently. And so we call these two isomers of one another Isomers are molecules that have the same chemical formula but different structures.
3:56
More information regarding this mechanism involves the number of molecules required to react in each of the elementary reactions. In the first step one molecule of A is required to react so we say that it's a unimolecular step. The second step is bimolecular because two molecules of B are needed to react here. Termolecular reactions do exist but they are kind of rare, because to have a termolecular reaction three molecules need to be at the same place and at the same time
4:33
in order for them to react. And that doesn't happen too often. If you think about asteroids out there in space or something, how often do you think it is that three asteroids collide at the same place and at the same time. Probably not too often. And it's the same situation for molecules to react like that, for three particles to be at the same place and at the same time. So although termolecular reactions do exist they're not as common as unimolecular or bimolecular reaction are.
5:10
Even more information regarding this mechanism involves the speed of each of the elementary reactions. The first step is slower than the second step, and that ends up slowing the overall reaction we say that it's the rate-determining-step. You can visualize the rate-determining-step down here. I've drawn a bunch of boxes for you here, and I always enjoy drawing boxes because I find it challenging to get all the vertical and horizontal lines just right,
5:42
and I hope it looks okay to you, but I see all kinds of imperfections I still have fun doing it, though. But imagine these boxes are atoms or something, and we're trying to get them from situation A over to situation C, where two steps are involved. The first step the boxes are carried up the ladder, then passed to this guy who carries them over the platform and drops them off at the final location C. Now the second step is much faster than the first step,
6:14
In fact, as soon as this guy carries them up the ladder, and passes the box to the second guy the second guy quickly comes over drops it off and returns again, and he's still waiting on the first guy to make the round trip. And it looks like he's waiting. kind-of impatiently. So, when you have a two-step process, where the second step is fast it cannot really proceed because it's waiting on the first step. The rate at which product C is formed is limited by the rate of the slowest step.
6:45
So, in chemical reaction, the rate of the overall reaction is equal to the rate of the rate-determining-step. Let's take a look at a specific example, which is very much like our model here. This is the isomerization of difluorocyclobutane. Starting with the reactant molecule, 1,2-difluorocyclobutane. You see it's chemical formula and the structure. Cyclobutane is a 4-carbon ring, and 1,2-difluoro means there are fluorines attached to the first and second carbon.
7:25
So the way this reaction works, in the first step, the molecule breaks apart into two equivalent pieces, these two vinyl fluoride intermediates, and these intermediates, one of them needs to flip over before they can reattach to form the final product, which is the 1,3 isomer because now fluorines are attached to the first and third carbon. So, in this two-step isomerization mechanism the overall effect is for the one fluorine to move over to the other position.
8:07
This mechanism can be organized a little bit differently, in stages, You see the overall reaction, in which the 1,2 isomer forms the final product 1,3 isomer, and the two elementary reactions, where the first one is the rate-determining-step, because it was the slow step. All of this information over to the right involves the kinetics of the overall reaction as well as the two elementary reactions. The rate of the overall reaction can be described
8:41
by it's differential rate law, which we discussed back in Lecture 12. The differential rate law tells us how the rate of this reaction depends on the concentration of the reactants. And so the rate equals a constant, called the rate constant, times the concentration of the reactant raised to some exponent n. The differential rate law takes that form, but in order to know the rate law completely we need to know the rate constant and especially the order -
9:17
the order is a little more important than the rate constant is. Now, the order of the reaction is unknown without more information. And back in Lecture 12 we discussed how to use experimental data to get the order of the reaction as well the rate constant. Perhaps you'd like to go back and see some of those examples, but I'll briefly outline the steps that we did there. The way you solve for the rate law is you study this reaction in the laboratory and you do it multiple times,
9:53
each time changing the concentration of the reactant, and each time measuring how fast it goes. And with all of that data that you collect you're able to then do some calculations to figure out the order with respect to the reactant as well as the rate constant. So it takes some lab work to get the rate law. But, there's an alternative way to get that information, and that's by knowledge of the mechanism. If you know what the mechanism is for the reaction,
10:25
and you know which is the rate-determining-step, then you can also figure out the order as well as gain knowledge about the rate constant. So let's see how this is done. The first step in the reaction is a unimolecular step. One molecule is required to react. And so the rate of this first step, we'll call it R1, equals a rate constant, k1, times the concentration of the reactant raised to the 1 power, since it's unimolecular. The nice thing about elementary reactions is
11:01
we automatically know the exponent - it's simply the number of molecules required to react. Now, we talked about first order reactions back in Lecture 12, when discussing nuclear decay, and a first order reaction occurs when doubling the concentration of your reactant doubles the rate. So that's a linear relation between concentration and rate, and that's why you have an exponent 1, because a linear relation. This is y equals m x, not y equals m x squared,
11:42
it's y equals m x, it's a linear relation, because doubling the concentration doubles the rate. Now you can make sense of this, if you think about a unimolecular reaction - Suppose you have this reaction occurring in a beaker, and this beaker is at a certain temperature, say room temperature. Now, suppose there are 100 molecules of the reactant in that beaker, and you're watching the reaction go over the course of one minute. Now, in those 100 molecules at room temperature,
12:20
there's a distribution of energies, that Boltzmann distribution, some of the molecules are more energetic, some of them are less energetic, and it's those more energetic molecules that are going to be the ones that react. So suppose at room temperature, 50 of those molecules have enough energy to react. Now, if we add another 100 molecules to the beaker and we do this again, so now there are 200 molecules, well in the first 100, 50 of those would react,
12:56
in the second 100, 50 of those would react as well, cause they're the same type of molecule, everyone's at the same temperature, and whether or not a molecule reacts depends just on that one molecule, nothing around, nothing in it's environment, it's a unimolecular reaction. And doubling the amount doubles the rate at which you form product, and so that's this situation, which means the exponent is 1. Now, the second step is a bimolecular step, two molecules of these vinyl fluoride intermediates
13:34
are required to react, and so the exponent is 2 in the rate law of this second step. Now, we discussed second order reactions in Lecture 12 as well, and that's the collision model. So, the way this works is these two molecules need to be at the same place and at the same time to react. In other words, they need to collide together. And this time when you double the amount of molecules, if you did that collisions would occur four times as often. We explained exactly why that was in Lecture 12,
14:12
but when you increase the number of molecules there's a lot more collisions that can occur, it ends up being four times the number of collisions that occur, and so doubling the concentration this time quadruples the rate, and this is an exponent 2 situation, that's y = x squared situation. Now, before we use the mechanism to get the overall rate law, let's first take a look, a quick look, at the reaction profile, for this two-step mechanism.
14:44
The reaction profile explains how the energy of the molecules changes as the reaction progresses. So, the reactant molecule, the 1,2 isomer has a certain energy, and the product molecule has a different energy, it looks like a little bit lower than the reactant molecule's energy, and so this is the profile for an exothermic reaction, the energy that you end up losing overall is given off as heat. Now, remember the way I want you to think about energy is in terms of structure.
15:21
This reactant molecule has a certain structure, and when something happens to that structure, such as it's deformed from it's normal position, bonds are bent or bonds are broken - that will cost energy. And when bond angles are returned back to place, and new bonds are formed, energy is released again. And so when a chemical reaction proceeds, where bonds are broken and new bonds are formed, and then more bonds are broken, and new bonds are formed - that costs energy, energy is released,
16:02
it costs more energy, and more energy is released. And you can see, along the way, there's a stable energetic structure here, and those are the intermediate molecules that are formed along the way. So, you can see that this is a two-step mechanism. Now the first step, this amount of energy that it costs, before it gets up here to that transition point, before sliding back down, that's called the activation energy of this first step. And down here,
16:39
that's the activation energy of the second step - it's a little bit smaller than that of the first step. And in Lecture 14 we discussed reaction profiles and saw how the activation energy represents a barrier for the reaction to proceed. Larger activation energy means a slower reaction. And we talked about the Arrhenius equation, which tells us how the rate constant of a reaction depends on the activation energy. A larger activation energy up here in the exponent
17:17
means a smaller rate constant, which means a slower rate. And so this first step, with a larger activation barrier is the slow step, and the second step is the fast step. You can imagine what a three-step mechanism would look like, and you should be able to identify which one is the slow step - it would be the one with the largest barrier. So, pretty interesting. Now, let's go ahead and figure out the overall rate law - this is very easy to do at this point.
17:54
We know that the overall rate of the reaction equals the rate of the rate-determining step. So the rate equals R1. And we know how to express R1 - it's simply it's rate constant, k1, times the concentration raised to the 1 power. And so we can see that the overall rate law does, in fact, take that form that it's supposed to - it's rate constant is k1 and the order is first order. Before we move to the next slide, let's take a look at one more thing. This is Dr. Claw here, who is our lab helper.
18:41
And if you want to know more about Dr. Claw, who he is or where he came from, take a look at my previous lecture, that I posted previous to this one, where I was practicing in front of the camera. But for now, just understand that Dr. Claw is here to help us, and he has something to say, a Technical Note: Other reaction mechanisms may be proposed, but the true mechanism must be verified by experiment. What Dr. Claw's trying to say is that this two-step mechanism
19:15
is one possible mechanism for the isomerization reaction. Suppose someone else proposes another one, maybe a three-step or even a four-step mechanism, how do we know which one is the correct one? Well, one way to support a mechanism, or provide support, is to see if you can detect the presence of any intermediate molecules in your mechanism. So, if you study this reaction in the laboratory, and you're able to detect the presence of these vinyl fluoride intermediates,
19:51
that would actually support this two-step mechanism. It would not prove it because there may be an alternative mechanism that also has vinyl fluoride intermediates, and, in fact, that is the true mechanism. There is another four-step mechanism, where vinyl fluorides are also intermediates, and the way the true mechanism works, chemists have figure this out, that that one is the true one, and this one is not quite the true one. The four-step mechanism is a little bit more complicated.
20:30
The way it works, is this ring, before detaching into the two intermediates, it does so in stages. First, one of these carbon-carbon bonds breaks to form an intermediate, and then the second one breaks to form the two vinyl fluoride intermediates. Then after one flips around, one of the carbon-carbons attach to form a third intermediate, and then the second one attaches to form the final product 1,3-isomer. So alternative mechanisms - the four-step one is actually the true one,
21:06
and it also includes vinyl fluoride intermediates. Pretty interesting. In the previous slide we examined a reaction whose first step was the slow step. This time let's look at a case where the second step is slow, and as you'll see things are a bit different. This reaction occurs in three elementary steps - this time it's the second step, which is the rate-determining-step. And since the rate of the overall reaction is limited by that of the rate-determining-step, it's also slow.
21:43
You can obtain this overall chemical equation if you add all of the reactants in the three steps as well as the products, and then cancel chemicals appearing on both sides - those are your intermediates. So you can see the N2O2 is produced in the first step then consumed again in the second one, so it doesn't appear in the overall equation. And also the N2O is produced then consumed again. So the overall equation is for the two hydrogens
22:16
to react with the two NOs to produce the two waters and the nitrogen. And it's also known that this reaction is exothermic, the overall reaction is. Since the first step is faster than the second step it reaches a point of chemical equilibrium, which is represented by this 2-way arrow. And what chemical equilibrium is, is it's sort-of like the finishing point of a reaction. By now you've probably heard that chemical reactions do not go all the way to completion.
22:55
You see, once a certain amount of reactant has reacted and a certain amount of product has been generated, it seems like the reaction no longer proceeds - those concentrations will no longer change after a certain amount of time has passed. And that's called chemical equilibrium, it's like the reaction has stopped. So, we'll examine equilibrium in just a few moments, but first a few more things regarding this reaction mechanism. First of all, we know that the overall reaction
23:29
has a differential rate law, which takes this form. And, if you recall back in Lecture 12, we have actually studied this overall reaction before. And also back in that lecture we determined the differential rate law, and we saw that it was a third order reaction. That means that the two orders - the 'n' plus the 'm' should equal 3. And, what we're going to do this time is use our knowledge of the mechanism to get the differential rate law. But it'll be interesting if you want to compare
24:13
the differential rate law that we get this time with the rate law that we got back in Lecture 12. So, two alternative ways to get at the differential rate law - Lecture 12 we used experimental data - this time we're using the mechanism. Now, you can visualize the rate-determining-step in this three-step mechanism in the diagram here, which has workers carrying boxes again. So the boxes are being carried from the initial location to the final location,
24:47
but there are three steps involved this time, and it's the second step which is the slow step. You see, when the second step is the slow step, the third step cannot proceed because he's waiting on this worker to carry that box up. And also, the first worker cannot really proceed - and you can imagine why if you think of this little drop-off point as only being able to contain a certain number of boxes - so he cannot fit anymore boxes in this location - it's reached it's maximum.
25:26
So this first step has reached a point at which it cannot proceed. And so, both of these workers are now irritated. And I...I guess I really feel sorry for this guy [laughing] this time. But again, that's life buddy. Get used to it. Now, the mechanism can also be visualized in the reaction profile, which plots the energy of the chemicals versus reaction progress. The overall reactants have a certain energy, and the products have a different energy.
26:06
And again, since this is an exothermic reaction, the products have less energy than the reactants do, and that energy which is lost is released as heat. And so the enthalpy change for the reaction is negative - it loses a certain amount of heat. And you can see in the profile that this is a three-step mechanism. And also, by looking at the energy barriers, the second barrier is the largest, so this is your slow step. Now, before we get the differential rate law from the mechanism,
26:47
let's first examine chemical equilibrium a little bit more closely. Let's take this first elementary reaction and see how equilibrium is achieved. So here's that first elementary reaction. Now the two-way arrow is indicated because this reaction, like all reactions do, can proceed in both directions. You've probably heard that before, but chemical reactions can always go in both directions - the reactants can convert into product or the product can convert back into reactant.
27:30
Up till now in our course we've just been examining forward reactions but we're going to start looking at equilibrium now, and so you're going to be seeing a lot more two-way arrows. And so, if you imagine this reaction being able to occur in both directions, we can ask about the rate of the forward reaction and also the rate of the reverse reaction. The forward reaction rate, since it's a bimolecular step, we write it as R1 equals a constant k1 times the concentration of the NO squared
28:08
and the reverse reaction, since it's unimolecular, it's rate can be written as R-negative-1 equals k-negative-1 times the concentration of the N2O2 raised to the 1-power. The way this reaction reaches equilibrium can be imagined here. Suppose you're studying this elementary reaction, and you're doing it inside of some reaction container, and at the very beginning, at 0 seconds, you put a certain amount of reactant molecule, NO, inside that container.
28:46
And since there's no product in that container yet the reaction has no choice but to proceed in the forward direction. And it proceeds rather quickly since there's a lot of reactant in that container, because the rate depends on the concentration of the reactant, and a lot of reactant means a fast rate. And so, the forward arrow is rather long, meaning it's going rather quickly. But after a certain amount of time passes, say 20 seconds, some of the reactant concentration has decreased,
29:24
which will slow down the forward rate. As the reactant concentration decreases that makes the forward rate slow down, and that also causes some product to appear, so the product concentration is increasing, which means the reverse rate is beginning to speed up. So as the N2O2 concentration increases, R-minus-1 speeds up. But you still have a lot more reactant than product in this container, and so the forward arrow is still longer than the reverse arrow. But after a certain amount of time passes
30:04
your reactant concentration will decrease to a certain point, and the product concentration will increase to a certain point, in which the forward rate and the reverse rate are now matching each other. You see eventually, R1 will match R-minus-1, and when that happens, that's chemical equilibrium. You see, when you're reacting in the forward direction as fast as the reverse direction, that means as fast as you produce products they're being consumed again,
30:41
and so it looks like concentrations are no longer changing, because the reaction is just proceeding in both directions at the same speed. And so that's chemical equilibrium - it looks like nothing's happening, but that's kind of an illusion - the reaction is still proceeding, again, but it's just the concentrations aren't changing. So next let's now use our knowledge of the mechanism to figure out the overall differential rate law. We know that the overall rate law takes this form,
31:17
I've just copied it down again for you. And since the overall rate matches the rate of the rate-determining-step, the rate equals R2, which is the rate of the second step. And we know what the rate of the second step looks like, because it's a bimolecular step. One of these reacts with one of the those so it's bimolecular, and so the rate of the second step equals a constant, k2, times the concentration of hydrogen raised to the 1-power times the concentration of N2O2 raised to the 1-power.
31:54
Now, once we get here we're not done yet, and that's because our overall rate, which looks like this, does not look like the form that it's supposed to look like. You see, the rate of the overall reaction cannot depend on the concentration of an intermediate. It's supposed to depend on hydrogen and NO concentrations. And so we need to get rid of this intermediate concentration some how, and the way we'll do that is by using the chemical equilibrium that occurs in the first step.
32:36
In the first step chemical equilibrium has been established, which means the forward rate matches the reverse rate. And so we can write the expression for the forward rate and set it equal to the expression for the reverse rate. And if we solve for the concentration of N2O2, that's what we get, and we can substitute that in right here, and if we simplify we'll get this expression right here. And this expression does have the form that it's supposed to have.
33:14
You see, the overall reaction rate equals a constant, which is a bunch of little constants all put together in one (all of this is your constant 'k'), times the concentration of hydrogen raised to the 1-power times the concentration of NO squared. So it's first-order in hydrogen, second-order in NO, and it's third-order overall. But it's not because of a 3-particle collision. You see, 3-particle collisions do give you third-order kinetics, but since we know about the mechanism this time
33:56
we know that's not what's going on. This is actually a sequence of bimolecular reactions. But the fast initial step and the slow second step is giving us a third order reaction, which if we didn't know about the mechanism we might think it's due to a 3-body collision, but now since we know the mechanism we know that's not the case. You see, the mechanism is kind-of a useful thing to know about. It tells you how the reaction actually happens. So, pretty interesting.
34:32
Reaction mechanisms help us understand many chemical phenomena. One application, which is important in the field of biochemistry, is that of enzyme catalysis. Enzymes are very large molecules that are really important for living organisms. You see, living organisms, which are composed of many different cells - sometimes trillions of different cells, inside each one of those cells, thousands of chemical reactions are always occurring.
35:04
And each one of those chemical reactions is managed by a corresponding enzyme. So, you can think of a cell, a very large cell, and inside each cell there are thousands of different enzymes, which are also very large molecules. And an enzyme is responsible for managing a certain chemical reaction. So the way it does it is the chemicals, the reactant molecules, attach to the surface of the enzyme. And once the reactant molecules turn into product, the product molecules detach.
35:47
So the enzyme provides a place for the reaction to proceed. Let's take a look at, uh, how they work a little more closely now. So a catalyst increases the reaction rate without being consumed itself. And an enzyme is a biological catalyst, or one that's important for biological reactions. We discussed enzymes back in Lectures 2 and 12. In Lecture 2 we talked about the structure of an enzyme and how the forces between one part of the enzyme molecule and another part
36:30
help pull the enzyme molecule together. So rather than being a very long molecule the enzyme actually folds up into a blobby type of molecule. And it's due to the forces - the hydrogen bonds and the dipole-dipoles. So go back to Lecture 2, to the 7th slide - I think it was the last slide - if you want a little closer look on that. And then Lecture 12 on the 2nd slide we discussed the reaction rate and how an enzyme system is a zero-order reaction.
37:07
In other words the rate of the reaction does not depend on the reactant concentration. So, larger reactant concentrations - the reaction rate remains constant. And so that's an enzyme system - zero-order system. Let's take a look at a generic reaction here. An uncatalyzed reaction versus a catalyzed reaction. So, with no catalyst, the reactant molecule A turns into the product molecule B. And this reaction is slow. With a catalyst - we'll represent the catalyst by the letter E,
37:50
standing for enzyme, the reactant molecule A combines with the catalyst to produce a reactant molecule/catalyst complex. You can think of this as the reactant molecule being attached to the surface of the enzyme. So that little complex - is an intermediate. And the next step is for the reactant molecule to turn into the product molecule. So the second intermediate - now you have the product attached to the enzyme. And the last step is for the product to be released, and you get your enzyme back.
38:31
So the enzyme is produced in the last step here. Catalysts are consumed in an early step then produced in a later step. So it's consumed at the beginning then produced at the end. A little bit different than intermediates. Intermediates are produced in an early step then consumed in a later step. Catalysts are the exact opposite - consumed then produced. We can visualize this difference of mechanisms here in the reaction profile.
39:09
With no catalyst the reactant molecule A turns into the final product B. And supposing it's a one-step mechanism, and it's slow, that means we would expect the barrier to be a large barrier. So, a much larger barrier versus the catalyzed version. The catalyzed version is a three-step mechanism represented in the green curve here - one, two, three steps - but being much faster, the barriers are slower versus that of the uncatalyzed one. So even though it's three steps,
39:49
each of these three steps has a smaller barrier than the uncatalyzed barrier. So the first step produces this reactant enzyme complex. The second step you get the enzyme-product complex. And then the last step you get your product and the enzyme as well. So the enzyme becomes available to catalyze another reaction. And that's the way the enzyme works. After catalyzing one reaction then it'll do it again. And again. And the whole thing just continues.
40:25
Let's take a look at a specific enzyme-catalyzed reaction, one that's very dear to my heart, which is the hydrolysis of sucrose. Sucrose is table sugar. The formula is C12 H22 O11. And I love table sugar. In fact I have a weakness for sweets, especially chocolate cake. So, what happens is the sucrose molecule is hydrolyzed. Hydrolysis, or hydrolyzing something, means you're using water to break it apart. So you're sucrose molecule is a large molecule here.
41:08
And what happens is this molecule breaks apart into two smaller molecules with the help of water. So that's hydrolysis. The bond that breaks is highlighted here, and what happens is one of these hydrogens becomes attached right there, and that's your glucose, and then the O-H becomes attached right there, and that's the fructose. Glucose and fructose, the two products, have the same chemical formula but different structures. So these are two isomers of one another.
41:48
Now this molecule, or this reaction, proceeds with the help of the enzyme sucrase. Sucrase greatly speeds this reaction up within the cell. But let's take the enzyme away for a second. Suppose this reaction needs to occur without the help of the enzyme - then it would be extremely slow - hydrolysis would occur at a very slow pace. And that's because in order for this reaction to occur the water molecule needs to react right there at that carbon atom. But there's no room.
42:27
A lot of other atoms are crowding the reaction center. And so there's no room for the water to fit, so this would very rarely occur. Now what the enzyme does is it basically distorts the structure here and it makes a little bit of room for the water to come in and react. So let's observe how this happens. Here's the enzyme sucrase, a very large enzyme. In fact, it is composed of 509 amino acids. So 509 amino acids all connected together in a chain,
43:12
which are folded up into one very large blobby molecule, sucrase. Now, the sucrose molecule, the reactant, attaches to the surface of the enzyme right here at the active site, which is highlighted in brown. In fact the shape of the sucrose molecule seems to match the corresponding shape of the surface of the enzyme. And on top of that, not only do their shapes match up, but the electrostatic character of the reactant molecule matches the electrostatic character of the enzyme.
43:55
And so the enzyme not only provides a surface for the reactant molecule to attach to, but it kind of pulls it in there, because of the forces of interaction between the enzyme and the reactant molecule. Intermolecular forces pull the sucrose into the active site. And that's how all enzymes usually work, they're kind-of designed for a certain reactant molecule to attach. Now, once the reactant molecule attaches to the enzyme the structure of the molecule becomes distorted.
44:37
You see, now this carbon atom is exposed because the molecule has become distorted after attaching to the enzyme. And since it's exposed water can come in here and react right there. and the reaction can occur. You see, the enzyme provides energy to the substrate - the substrate is another name for the reactant molecule. So sucrose here is called the substrate. So, the enzyme provides energy to the substrate, which helps the reaction occur.
45:17
You see, there's a lot of energy trapped in this enzyme. Lots of atoms in here, the enzyme has got lots of vibrations occuring, and it imparts some of that energy to the substrate sucrose molecule. And when it does that it distorts the structure of that sucrose, gives it...gives a little room right there at that carbon atom, which allows water to attack. And once that happens, the chemical reaction occurs, and you generate the products.
45:53
So you can think of incoming substrate attaching at the active site right here. And after it attaches, that's when the reaction occurs. Now you have your two products, which are the glucose and the fructose. And since the identity of the molecules are changed now, the enzyme senses that, the forces of interaction between the enzyme and the products are a little bit different than the forces of interaction between the enzyme and the substrate.
46:29
And so the enzyme sensing that pushes those products out, allowing another substrate to attach. So, the enzyme pulls the substrate in, the reaction occurs, then it pushes the products out. And it keeps on going. So enzymes can do thousands of chemical reactions per second, they're very quick. So we can compare the uncatalyzed versus catalyzed reaction rates for this reaction. If you have 0.3 molarity sucrose at room temperature (298 K), with no catalyst the reaction rate is
47:13
1.4 times 10 to the minus 11 molarity per second. Okay, so what does that mean? Well, let's look at the half-life. The half-life is 440 years. In other words, it would take 440 years for this reaction to occur by itself, for half of the concentration to react by itself with no catalyst. Now with a catalyst, if you add 10 to the minus 6 molarity sucrose, then the reaction rate becomes 1.6 times 10 to the minus 4 molarity per second.
47:56
That's seven orders of magnitute greater, and let's go to the half-life again. The half-life is now 15.4 minutes. So it greatly speeds up the rate of this reaction. Enzymes are a very rich source of study in the field of chemistry and biochemistry. So, pretty interesting. And the way they work is they provide an alternative mechanism for the chemical reaction to occur, which allows the reaction to occur much more quickly. So, I hope you have enjoyed this lecture.
48:39
This concludes the chemical kinetics portion of the course. The next lecture will begin a new portion, chemical equilibrium. So, we're getting ready to change gears here. In chemical equilibrium, we're not interested in as much how fast chemical reactions go, but how far do they go. So chemical equilibrium, if you remember, is like the finishing point of the reaction.
Here to help you learn
Sign in to keep a private conversation with your course tutor.
Ask me anything about the material…
Sign in to chat