in this video, we're going to begin our discussion on reaction orders. So recalling our previous lesson videos, we talked about the rate law and we said that the rate law is just another way to express the reaction velocity or V here. And so the rate law says that the reaction velocity is equal to the rate constant k times each initial reactant concentration raised to the power of the order. And so this reaction order here is exactly what we're gonna be focusing on in this video. And so the reaction order is a proportionality relationship between the reaction rate V and each initial substrate concentration or each initial reactant concentration. Remember that substrates are the reactant of enzyme catalyzed reactions now recall from our previous lesson videos, we said that the reaction orders are frequently going to be equal to the coefficients of the substrates, especially in our course moving forward. However, we said that the coefficients of the substrates do not always equal the reaction orders, and so the reaction orders actually need to be experimentally determined. Now it turns out that the substrate coefficients will Onley equal the reaction order for elementary reactions and elementary reactions air just those reactions with Onley one transition state. And so if the reaction on Lee has one transition state, then it's considered an elementary reaction, and we know that the substrate coefficients will equal the reaction order. However, if it's a non elementary reaction, meaning that it's not elementary, then we know that this will not be true and that the substrate coefficients will not equal the reaction order now. What's important to note is that the overall reaction order, regardless of if the reaction is non elementary or elementary, will always be equal to the sum of the individual reaction orders for all of the substrates. And we'll be able to see how that works better down below in our example. Now, in this example, we're going to determine the overall reaction order for the following reactions. And the reaction over here on the left notice of what we have is ozone gas plus oxygen gas being converted into two moles of oxygen dioxide gas, and so notice that it's telling us that this is an elementary reaction. So we know that this reaction here on Lee has one transition state and that one transition state is being represented by this one arrow here in this reaction. Now, if we were to draw the rate law for this particular reaction, notice that the rate law is expressed right here, and the rate law is just the reaction velocity, which is equal to the rate constant k times the concentration of the initial, uh, substrates or reacting. So we have one substrate here, ozone gas, and we have another substrate here just oxygen. And then because this is an elementary reaction, we know that the substrate coefficients are going to be equal to the substrate reaction order. And so we noticed that the reaction orders that air replaced here do match the substrate coefficients of one. And so what's important to note is that the overall reaction order is always going to be the sum of the individual reaction orders. And so because we have the first reaction order as one and the second reaction orders one we know that one plus one will give us the overall reaction order. So we know that the overall reaction order is second order for this reaction. Now, if we take a look at this non elementary reaction over here on the right notice. What we have is nitrogen dioxide gas plus carbon monoxide gas will give us nitrogen monoxide gas and carbon dioxide gas. And so notice that this is a non elementary reaction, which means that it has more than one transition state. And in this reaction, we can see that's true because we have more than one reaction arrow, which means that there's some kind of intermediate that forms that's not being shown in this overall reaction here. And just because there are two reaction arrows here, uh, doesn't mean that this reaction will always be expressed with two reaction arrow. Sometimes we'll see that this will be expressed with one reaction arrow, and it will make it look like it's an elementary reaction when in reality it's actually not in elementary reaction. And it's really broken up into two smaller reactions. And so what that means is we can't really rely on the arrows toe. Let us know if this is a non elementary reaction or not. However, moving forward in our course, we know that all of the enzyme catalyzed reactions we're going to assume our elementary reactions with Onley one transition state. However, taking a look at this non elementary reaction if we were to draw out the rate law for this reaction, which is shown here, notice that the rate law is equal to the reaction velocity, which is equal to the rate constant K times, the concentration, the initial concentration of the reactant. So we have nitrogen dioxide here and carbon monoxide here, and then notice that because this is a non elementary reaction, the substrate coefficients do not necessarily equal the substrate reaction order, and we can see that here as well. So notice that the reaction order for this nitrogen dioxide gas is, too. But the coefficient here is not to the coefficient is actually one, so they don't match each other. And that's okay, because this is a non elementary reaction, and the same applies for the carbon monoxide gas. Notice that it's reaction order zero. Even though it's coefficient is one and so because they don't match, we know that's OK because that's a non elementary reaction. But still, even though this is a non non elementary reaction, the overall reaction order is always going to be the sum of the individual reaction orders for all the substrates and So we have a reaction order of two plus a reaction order of zero. So to play zero is still too. So that means that the overall reaction order is still going to be second border. And so notice that the reaction order the overall reaction order for this elementary reactions, Second order. And so is the overall reaction order for this non elementary reaction order. So the overall reaction order does not necessarily tell us information about the individual reaction orders for each reactant. And, uh, that is why we say that the individual reaction orders must be experimentally determined. And so again, uh, our course, our biochemistry course. Most of the professors are not going to expect us to use data to determine the reaction orders. And instead, what we're going to do is on Lee, look at reactions and assume that reactions are elementary reactions for the particular enzyme catalyzed reactions that we'll be looking at. And so what we're going to talk about next are the three common overall reaction orders, and those are zero order reactions, first order reactions and second order reactions. And so moving forward in our course, we're gonna talk about each of these common overall reaction orders and their own separate videos. And so we'll start off by talking about zero order reaction, so I'll see you guys in that video.

All right, So now that we know a little bit about reaction orders in this video, we're going to talk about reactions that have an overall reaction order of zero or zero order reactions. And so zero order reactions are reactions where the substrate concentration has zero effect on the reaction rate or the reaction velocity. And so if the substrate concentration has zero effect on the reaction rate, that means that even when we change the substrate concentration, the reaction rate will not change. And so enzyme catalyzed reactions will actually exhibit zero order kinetics when an enzyme is completely saturated with substrate and we'll be able to see an example of that down below in our image. Now, the zero order rate constant units, our, uh, specifically molar ity times inverse seconds, which is express right here. But I really don't want you guys to focus on the units of the rate constant just yet, because later in our course and a different video, we're gonna focus specifically on all of the rate constant units. And so if these units are a bit confusing to you right now, that's okay, because again, we're gonna talk about the units later in a different video now down below. In our image notice here we have a specific reaction where we have reacting a being converted into product B, and we're told that this is a zero order reaction, which means that the overall reaction order is zero. And so if we take a look at the rate law for this reaction, we know that the rate law is just another way to express the reaction rate or the reaction velocity. And it's equal to the rate constant k times the concentration of the reactant or the substrate on. We only have one substrate here it's substrate s O. It's the rate constant k times the concentration of substrate. And then it's gonna be raised to the reaction order. And because we're told that this is a zero order reaction, we know that the reaction order must be zero. And so any number raised to the power of zero is going to be one. Try it in your calculators, take any number and raise it to the power of zero. You'll get an answer of one. And so essentially, what we have here is one times K and one times K is going to be K. And so essentially what we're saying is that 40 order reactions, the reaction rate or the reaction velocity will be equal to the rate constant K. And so, if we take a look at this graph that we have over here, notice that we have the reaction rate on the Y axis and the substrate concentration on the X axis, and this is specifically a zero order reaction. So that means that the substrate concentration has zero effect on the reaction rate. So even when we change the substrate, concentration toe, low substrate concentrations and high substrate concentrations, it has zero effect on the reaction rate. So notice that the reaction rate remains constant and the value of this reaction right here, we know is going to be exactly equal to the rate constant value. And so, essentially, what we have over here is another graph that we're gonna be analyzing Ah, lot as we move forward throughout our course. And this is just showing a typical curve for a typical enzyme catalyzed reaction. And so on the Y axis, what we have is the initial reaction rate and on the X axis, what we have is the substrate concentration. And again, this is a typical curve that we see for any kind of enzyme catalyzed reaction when we plot this data here. And so what you'll notice is that there comes a point in the reaction where the reaction is actually displaying zero order kinetics. And we can kind of see that here, because notice we have this horizontal line just like we have the horizontal line over here. And so we can see that by changing the substrate concentrations in this area where it's displaying zero or two kinetics. The reaction velocity does not change. And so what this means is that within this region where zero ordered kinetics are being displayed, the reaction rate does not depend on the substrate concentration. And again, the reason that an enzyme catalyzed reaction reaches ah, point where it displays zero order kinetics is because the enzyme is being saturated with substrate. And so at this point, uh, here on our graph coming down, the enzyme becomes saturated with substrate. And so even by increasing the substrate here, uh, substrate concentrations were not able to affect the reaction rate and so that's important to note that, uh, enzymes. When they're saturated with substrate, they display zero order kinetics. And so that concludes our lesson here on zero order reactions. And we'll be able to take a look at an example of this in our next video, so I'll see you guys there.

All right. So now that we've covered zero order reactions in this video, we're going to talk about first order reactions. And so first order reactions are just reactions where the rates of the reaction are directly proportional to Onley one substrate concentration. And this includes uni molecular reactions or reactions that only have one molecule uni meaning one. And so these thes reactions involve Onley a single substrate, for example, Converting substrate a into product be now the first order rate. Constant units are specifically inverse seconds. But again, I don't want you guys to focus on the units of the rate constant just yet because later in our course, we're gonna have ah specific video dedicated to helping you guys understand rate constant units. So if this is confusing to you guys right now, that's okay, because again, later we'll talk more about it to clear it up so down below. In our example, what we have is an example of first order kinetics, and so notice that we're looking at a uni molecular reaction that involves only one molecule, and we're converting substrate a into product B. And we're told that this is a first order reaction And so when we, uh, consider the rate law, which is another way to express the reaction rate or the reaction velocity, we know that it's gonna be equal to the rate constant k times the concentration of the Onley Substrate year, which is substrate A. And of course, because we're told this is a first order reaction, we know that the reaction order must be one for this substrate. And so, if we take a look at this graph over here, where we have the reaction rate on the Y axis and the substrate concentration on the X axis, notice that instead of getting a horizontal line like what we saw for zero order reactions instead, this time what we're seeing is a diagonal line that's going up like this. And that's because the rate of the reaction is directly proportional to the substrate concentration. So we can say here that the rate of the reaction here is directly proportional to the substrate concentration. This is a symbol that means directly proportional. And all that means is that as the substrate concentration, uh, increases, so does the reaction rate. And so if we take a look at this other graph over here of a typical enzyme catalyzed reaction where we have the initial reaction rate on the Y axis and the substrate concentration on the X axis notice again from our last lesson video. We already know that if we increase the concentration toe a point where it's saturating the, uh, enzyme, then the enzyme will be displaying zero order kinetics where the rate of the reaction does not respond depend on the substrate concentration on. We already know this again from our last lesson video. And so here in this video we can see that early on where we have ah, lower substrate concentration and the reaction notice that the reaction the enzyme catalyzed reaction is displaying first order kinetics, where the rate of the reaction actually is directly proportional and depends on the substrate concentration. So you can kind of see that it's going up in this diagonally in this fashion, similar to how we can see. It's going diagonally over here in this fashion. And so we'll be able to apply these concepts here on first order reactions as we move along through our course. But for now, this concludes our lesson here on first order reactions, and I'll see you guys in our next video

So now that we've refreshed our memories on first order reactions in our last lesson video in this video, we're going to focus on second order reactions. And so second order reactions, our reactions where the rates of those reactions are directly proportional to two substrate concentrations instead of just one substrate concentration like we saw with first order reactions. And so what's also important to note is that second order reaction rates could also be proportional to the square of Onley one substrate concentration. And so, essentially, what we're saying is that really there's two common ways to express the rate law. For a second order reaction, It could be expressed as being directly proportional to two substrate concentrations, or the rate law could be expressed as directly proportional to the square of only one substrate concentration. And we'll be able to see how that works better down below when we talk about the two common rate laws for a second order reaction. Now, what's important to note before we actually get to our example is that second order reactions include by molecular reactions that include two substrates, and so the by here means to, and molecules means molecular means molecules. And so second order reactions include these by molecular reactions. For example, this reaction right here where we have substrate a interacting with substrate B to form products, See. And so the second order rate constant units for K are actually in units of inverse similarity in inverse seconds. But again, we're going to focus on the rate constant units later in our course in a different video. So if you're a bit confused about this, hang on tight because again, we'll talk more about the units of the rate constant later on in our course and so down below, we're showing you guys an example of second order kinetics. So we're looking at this by molecular reaction here, where we have substrate a interacting with substrate, be toe form products, See? And so with when it comes to second order reaction rates again, as we mentioned up above, there are really two different ways to common ways to express the rate laws for a second order reaction and so recall that the rate law is just an alternative way to express the reaction rate or the reaction velocity V for reaction. And so the rate law says that the reaction Velocity V is going to be equal to the rate constant k times, the concentration of the substrates or the reactant. And we have to reactant or substrates a and B. So it's gonna be multiplied by those substrate concentrations and raised to the reaction order. And so when it's a second order reaction, that is a simple reaction. Then we can assume that the coefficients are going to be the reaction order, So that would be assuming that the coefficients of one would be the reaction order. And so this would mean that the total of these reaction orders individual reaction orders is one plus one, which is equal to two, meaning that the overall reaction is second order reaction. And so this is one way. As we set up above that, the rate of AH second order reaction can be expressed as directly proportional to two substrate concentrations, uh, directly proportional to substrate A and directly proportional to substrate. Be now, as we also mentioned up above. The second order reaction rate can also be expressed in a different way, where it's proportional to the square of only one substrate concentration so down below notice that we have the second common rate law for a second order reaction. And so this rate law would say that the reaction rate V is equal to the rate constant k times the concentration of the substrates. But notice that the coefficients do not necessarily match the, um, reaction orders that we have here. And so that would be indicating that this here is, uh, indicating that this reaction would not be a simple reaction on dso. Essentially here notice that it's on Lee uh, the concentration of a the square of the concentration of only one substrate concentration A that affects the reaction rate and the concentration of B is zero order with respect to the reaction rate. And so that means that altering the concentration of B would not affect the rate of the reaction. So these are the two possibilities for the rate laws when it comes to second order reactions, and so notice over here. What we have is our graph that we've seen before in our previous lesson videos. But this is specifically for second order reactions, and so notice that instead of getting a horizontal line like we got for zero order reactions And instead of getting a straight line with a positive slope like we got for first order reactions and a second order reaction, what we get is this exponential curve here when we plot the reaction rate V on the Y axis and the substrate concentration on the X axis. And so notice that over here on the right, what we're showing you guys is an example of a second order reactions, uh, that occurs within our cells. And so this reaction here is actually actually showing the breakdown of glycogen. And so glycogen recall is just a carbohydrate that's made up of a bunch of individual glucose molecules linked together. And so this little end here on the glycogen represents the number of glucose molecules that are in a particular glycogen molecule. And so, uh, up above, what we have is, um, or visual way to see the structures of the reaction and down below. What we have is a different format for the same exact reaction. And so notice that here what we have is a glycogen molecule, uh, that's interacting with a inorganic phosphate molecule. And so when the glycogen interacts with the inorganic phosphate it produces this glucose one phosphate molecule right here. And it also produces this glycogen molecule that we see here that has one less glucose. And so essentially, what's happening in this reaction is that this inorganic phosphate is being added to this molecule here and in the process. The bonds here are being cleaved to separate this green structure from the blue structure here. And so now we can see that the green structure is separated from the blue structure, a zoo we see on the right. And so if we were to write the rate law for this reaction, we know that the rate law is just another way to express the reaction rate or the reaction velocity, which is just express with a V. And the rate law always says that the reaction velocity is equal to the rate constant k times the concentration of the substrates. So we have to substrates. Here we have glycogen and we have inorganic phosphate. And so notice that the inorganic phosphate is already in here. So we just need to include the glycogen here so we can write in glycogen. And of course, it's going to be with a superscript here of n subscript here of n and so essentially, what we need to recall is that these concentrations are going to be raised to their reaction orders. And so we're always going to assume with the reactions that we're going to cover, moving forward in this course that these were going to be simple reactions. And so the coefficients are going to be the reaction orders. And so the coefficients of both of these is just ah one. And so that means that they're both going to be raised to the power of one. And so when we add up these individual reaction orders one plus one, that shows us that this is a second order reaction. And so this here concludes our lesson on second order reactions, and we'll be able to get more practice utilizing these concepts as we move forward in our course. So I'll see you guys in our next lesson. Video

all right. So now that we've covered both 1st and 2nd order reactions in this video, we're going to briefly talk about pseudo first order reactions. And so sudo is just a word that pretty much just means fake. And so pseudo first order reactions are just fake first order reactions. And really, all they are are reactions that appear to be first order reactions. But in reality, they're actually second order reactions, and really other than that, there's nothing special about them. And so pseudo first order reactions can occur when the concentration of one substrate is much greater than the concentration of the other. And so, for instance, uh, if the concentration of B is much, much, much greater than the concentration of a, then we could have ourselves a pseudo first order reaction. And so remember that when the concentrations of a substrate are really, really, really high, like the concentration of be here, then they're going to be at saturating concentrations. And when the enzyme is saturated with substrate, it's going to have zero order kinetics, and it's going to appear that the concentration of B does not affect the reaction rate when indeed it might actually do that in a second order reaction. And when the concentration of A is so small in comparison to be, this makes the substrate a the limiting re agent and as the limiting re agent, it's going to appear that Onley the concentration of a alone will dictate the rate of the reaction when in reality, it's not just the concentration of a it's also the concentration of B. And so if we take a look up above at the graph from our second order kinetics, uh, reaction that we talked about in our previous lesson video. What we're saying is that a pseudo first order reaction is really just it's actually just a second order reaction, and it's nothing special about that. However, when we graph a pseudo first order reaction, it appears toe look just like a first order reaction. And that's because, again, the concentration of one substrate is much greater than the other. And so, even though with a pseudo first order reaction, the reaction appears to be first order. We know that it's actually a second order reaction, and so this can sometimes, uh, cause a little bit of complications and challenges for a biochemist when they're trying to study a reaction. So they have to vary the concentrations to make sure that these conditions here don't exist, because otherwise it could fake out a biochemist. And so that concludes our brief lesson on pseudo first order reactions, and we'll be able to get some practice as we move along with our practice video, so I'll see you guys there.