Lecture 23. Chemical Kinetics Pt. 2.
Arasasingham, Ramesh D.
 UCI Chem 1C is the third and final quarter of General Chemistry series and covers the following topics: equilibria, aqueous acid-base equilibria, solubility equilibria, oxidation reduction reactions, electrochemistry; kinetics; special topics. 
 Okay, can I have everyone's attention? So let's go ahead and start. But before I do, I wanted to find out, do you guys have any questions? So we have moved on to looking at chemical kinetics. If you guys remember, last time we talked about the fact that when we studied chemical kinetics, we have two important goals. One is we need to figure out how fast reactions proceed. And the second goal of kinetics is to figure out the reaction pathway. When we say the reaction pathway, what we are interested in looking at is the detailed events that take place as reactants convert to products. So we're interested in looking at how fast reactions proceed. And secondly, we're interested in looking at pathways. And we're going to start looking at both of these in detail. But to begin with, we're going to focus on how fast reactions proceed. And we said when we talk about how fast reactions proceed, the term that we use is rate. And last time we talked about the fact that you can talk about rate in terms of average rate, or you can talk about in terms of instantaneous rate. And we said that it is much more useful to us to refer to rate in terms of instantaneous rate. And from now on, when we use the term rate, what we refer to is the instantaneous rate. And then at the end of the class, we talked about the fact that most reactions are reversible. And if a reaction is reversible, that means that there is a rate associated with the forward reaction and there is a rate associated with the reverse reaction. So there are two rates, forward and reverse. And so if you're looking at a reversible reaction, the overall net rate that we measure experimentally is actually the difference between the forward rate and the reverse rate. But to keep things simple, from now on, we're only going to consider forward rate only. And we're going to pretend as if the forward rate is the only one that exists. In reality, you have a forward rate and a reverse rate. But to keep things simple, we're just going to focus on the forward rate. So last time we talked about rate, we talked about the unit for rate. And now you understand what it means when we talk about reaction rate. Now we're going to look at it in a little bit more detail. And we're going to look at the relationship between rate and concentration. So to keep things easy for you, I've just made some short notes. And these notes are available on the class website. And it just kind of highlights the important things that I want you to remember as you study this topic. So today we're going to look at the dependence of rate on concentration. And this looks at the relationship between the rate of a reaction and the concentration. And the concentration that we look at is the concentration of the reactant. So it turns out that the rate of a chemical reaction, and remember we're focusing on the forward reaction, so the rate of a chemical reaction, the forward direction, nearly always depends on the concentration of one or more of the reactants. So if you're looking at a forward reaction, then how fast that reaction proceeds will depend on the concentration of the reactants. If you have only one reactant, it would depend on that concentration, that one reactant. If you have two reactants or more, then it would depend on the concentrations of all of the reactants. So the first thing to remember is the rate depends on the concentration of the reactant. And if you want to mathematically describe this relationship in the form of an equation, we call that the rate expression or the rate law. So I'm going to take the same reaction, I'm just going to write it down. So if we take a reaction where our reactant, we have only one reactant and this reactant goes to products. So if your balanced equation is that one reactant going to products, and our reactant is A, that lowercase A is the coefficient in that balanced equation. So the rate of this reaction is going to depend on the concentration of A. Remember, the rate always depends on the concentration of the reactant. So if I want to mathematically state this, then I would say that the relationship is described as rate equals a constant K. Now remember, I want you to always make sure this is lowercase K, not capital K. When you use capital K, what does capital K represent? Equilibrium constant. So this is lowercase K. And remember, the rate depends on the concentration of the reactant raised to a power n, which is a number n. And the K here is called the rate constant. So the lowercase K is a constant, and that is called the rate constant. This will be the concentration of A, and this n here is called the order with respect to that reactant. So this is order with respect to the reactant. And this relationship, when you write it in this format, where you say rate equals a constant times the concentration raised to the power n, this relationship is what is called the rate law or rate expression. So this is similar to the law of mass action. You guys remember, whenever you write the chemical equilibrium expression, where it's the products, concentration of the products raised to the power of the coefficient is called the law of mass action. Here the relationship between rate and concentration of the reactant is always called the rate law. So that's what I've given you in these notes. And so the same thing is written here. So it's always rate equals K, concentration of the reactant, raised to the power n, where K is a rate constant, and this is a constant. n is the order of the reaction with respect to that reactant. This is the power to which the concentration is raised. The power n in the rate expression has no direct relationship to the coefficients A in the balanced equation. So this number n has no relationship to the coefficient. It's a number that we determine experimentally. So there's no way in advance for anyone to look at a balanced equation and say what that value of n would come out to be. Does that make sense, Sir? So there's no way that you or I can look at a balanced equation and say what that value n is going to turn out to be. The only way to determine the value of n is to experiment. And later on I'll show you how we figure it out. This value of n can be a whole number or a fractional number. So let's see if, and it's called the order. So in some reactions, the value of n comes out to be 1. When the value of n comes out to be 1, we call it first order. That about n means order of the reaction. And therefore when n equals 1, we call it first order. So when n equals 1, the rate law is written as this. Rate equals k, a raised to the power 1. Obviously when something is raised to the power 1, you don't need to state it as 1. If you leave it blank, you know that it is to the power 1. So if you take this example of this reaction where n205 goes to 2n02 plus half n02, it turns out experimentally we know this is first order. So the rate law would be rate equals a constant k, which is the rate constant, times the constant. There's only one reactant, and that one reactant raised to the power 1. So when you write the rate law like that, we call that a first order rate law. Because the order n is 1, and if n comes out to be 1, then the rate law is written where n equals 1. Now there are other reactions where n can be 2. When n is 2, the rate law is written as rate equals k, a to the power 2. And we call that reaction second order. And so if you take an example, this is an example of reaction that has only one reactant, and therefore rate equals k, the concentration of the reactant raised to the power 2, and that would be the rate law for a second order reaction. And this is a reaction where you have only one reactant, and if the order turns out to be 2, it will be second order. Now sometimes this could be 0. When n equals 0, the rate law is written as rate equals k, a raised to the power 0. And you know that if a is raised to the power 0, you know that's going to be 1. And so the rate will be, when n equals to the rate law turns out to be this, and we call this so rate will equal k, because a raised to the power 0 turns out to be 1. And therefore that means rate equals the rate constant. So if the rate equals the rate constant k, then we call that a zero order reaction. Does that make sense to everybody? So it could be zero order. If n equals zero, it's called zero order. If n equals 1, it's first order. If n equals 2, it's second order. And so on and so forth. All right? Usually the common ones are zero, one, and two. All right? And lastly, remember this could be a fractional number. So if we take here, in this reaction, you can see the rate constant comes to 1.5. All right? So here the rate equals k times the concentration raised to the power of a fractional number. This could be 3 halves, or 1.5. And therefore that is the order with respect to that reactor. So the lesson is, to remember is, for you to remember is, that the lesson from all of this is that the reaction order is an experimental property that cannot be predicted from the chemical equation. It has to be deduced from experiment. All right? So every reaction that you look at, the relationship between the concentration of the reactant, all right? And the rate is called the rate law. And the rate law is always written in the same form. Rate equals a constant k, which is the rate constant, times the concentration of the reactant raised to the power of the n. And that n is called the order of the reaction. All right? So what happens if you have more than one reactant? So if you have more than one reactant, and let's take an example of a reaction where you have more than one reactant. So for reactions with two or more reactants, the rate law is given by. So now let's say we have a hypothetical general form of the equation where you have a plus v plus c plus d. You can have any number of reactants going to products. Then the rate law is written as rate equals, once again, the rate constant k. Now we have two reactants. So it'd be a raised to the power of a number m, b raised to the power of a number n. And if you have three reactants, it'll be c raised to the power of y, a number y, all right? And that would describe the rate law. So if you have only two reactants, the rate law would be the concentration of the two reactants, each raised to the power of a number. Okay, where m is the order with respect to a, n is the order with respect to b, and if you add them up, that gives you the overall order, all right? So let's take an example. So if we take this example of this reactant reacting with hydroxide to give you HPO3 2 minus plus H2, let's say we're looking at this reaction. Now if you were asked to write the rate law, the first thing you would do is look at that equation and figure out how many reactants there are. There are two reactants, all right? Therefore the rate law of the rate expression would be rate of that reaction equals a constant k, which is the rate constant, times the concentration of each reactant raised to the power of a number. So it turns out experimentally we figure out that this is first order and hydroxide is second order, all right? So that means we say that this reaction is first order in H2PO2 minus and second order in hydroxide. And the overall order would be 1 plus 2, which is 3. Is that clear to everybody? And that's what we call the rate law, all right? And I want to kind of remind you this is very similar to the law of mass action, where we write down and we say the equilibrium constant k equals the concentrations raised to the power of their coefficients. Likewise, in kinetics when we look at rate law or the rate expression, it's always the rate. This gives the relationship between rate of that reaction and the concentrations of reactants, all right? And the rate of the forward direction, remember we're considering only the rate and the forward direction. And the rate always will depend on the concentrations of reactants and that gives the relationship. So now we want to know how do you remember I said we can't figure this out experimentally. I mean, sorry, we can't look at a balanced equation and just say what this order will turn out to be. This order with respect to reactants vary from reaction to reaction. And as I said before, there's no way for us to look at it in advance and say what the order is going to be. And the only way to figure out the order is to figure it out experimentally, all right? So let me show you how we do this. So it is a mathematical calculation and essentially what you do. So let's take this problem and it's in the worksheet that I have put up on the class website. But let's take a look at how one determines the rate law for a reaction and how one determines the order, okay? So here we're looking at a reaction and the first thing I want you to look at this and if you look at the equation, the balanced equation that's provided to you, can everybody see there's only one reactant, all right? So the rate law is going to only depend on the concentration of that reactant, okay? So now in the lab what you would do is you would do several different trials and at different concentrations of the reactant you can calculate the rate of that reaction. Remember how do we calculate rate? Experimentally what you do is in the lab you would look at concentration versus time, all right? You would make a plot and then you would pick an instantaneous time and you would take the slope at that instant in time, the tangential slope and from that you can figure out what the rate is, all right? So we talked about that last time. So what we have is we're looking at the rates at three different concentrations of the reactant. So here's the first concentration and the corresponding rate. Here's the second concentration and the corresponding rate and this is the third concentration and the corresponding rate, all right? Now we're going to use this data to figure out first the order of the reaction. So the question is determine the order of the reaction and write the rate expression. So we're going to start by writing the general form of the rate law, okay? So without figuring out what the order is, if I'm going to write the general form of the rate law what would that look like? Rate equals, remember what is a rate law tell you? The rate equals what? The rate constant times the concentration of the reactant raised to the power of a number n that we need to establish. So this is the general form of the rate law. Now we have to figure out what that exact rate law is, okay? Now if you look at the experimental measurements can everybody see that we are measuring rate? We know the corresponding concentrations but we don't know what the value of this constant is and we don't know what the value of this n is and our ultimate goal is to figure those out in this problem, all right? So we write the general form of the rate law because we know the rate of the reaction depends on the concentration of the reactor, all right? So if you look at this equation we have two unknowns and mathematically if you have two unknowns all of you recognize that you need two equations to solve for it, all right? So algebraically if you have two unknowns we need two equations. So what we're going to do is we're going to write two equations based on two sets of data. So we'll start with the first set of data. We know that the rate equals k times h raised to the power n. So if one of my rates is 7.5 times 10 to the negative 4 moles per liter per second that will be the rate equals k times, we know the concentration for that will be 0.0050 molar raised to the power of n. So that's one equation. I'm going to go to the second equation and I'm going to say when the rate is 10 to the negative 3 times 10 to the negative 3 moles per liter per second equals the same constant k but now my concentration is 0.010 molar raised to the power n. That gives me 2. So now algebraically I've got to figure out a way to use both of these equations because remember I have two unknowns, I have two equations, now I can solve that and figure out what the value of n is and I can figure out what the value of k is. And you can see that if I divide one equation by the other and usually it's best to keep the math simple, it's always best to take a bigger number and divide the bigger number by a smaller number than vice versa. You should end up with the same answer but it would just keep the math simple. So this is a bigger number. So I'm going to say 3 times 10 to the negative 3 moles per liter per second equals k times 0.010 moles per liter raised to the power n. I'm going to divide that by 7.5 times 10 to the negative 4 moles per liter per second equals k times 0.0050 moles per liter raised to the power n. Can you see now k and k will cancel out? And so what I've got is now an equation with one unknown. And so if you take this and divide this, what I end up with is 4. Over here both of these are raised to the power n. So if I take 0.5, you can see that this comes out to be 2 and that raised to the power n. And so can everybody see I figured out what n is? n is 2. So we know that this reaction is second order. By taking the data, essentially what we're doing is experimentally we're measuring the rate and we're looking at its dependence on concentration. So we're changing the concentration of reactant. By doing this experiment in the lab, we can figure out what the order of the reaction is. So now I've figured out what the rate law is. So my rate law would be that rate equals the constant, rate constant times h i raised to the power 2. So that answers the first part. And this is how in the laboratory we determine what the order of the reaction is. Remember we said we can't figure this out in advance. We can't predetermine or pre-predict what this order n would come out to be. The only way to determine it is to carry out the experiment. You measure the rates at different concentrations and you can figure out what the rate law is. So now that we've figured out what the rate law is, we can move on to looking at the second part of the question. So we determine the order of the reaction and write the rate law. So if I ask you on the exam to write the rate law, this is what you need to give. You cannot give this because this is the general form. This is a generic equation and you have to find out the unique rate law that corresponds to this reaction and we've figured out that it is second order. Now if I go to the second part of that problem, we're asked to calculate the rate constant and give its unit. So if I go to part b, I need to figure out now what k is. Since I've figured out the second, can I ever see I can just take any one of these equations. Now I know that n is 2 and now I have only one unknown I can solve for k. So I can say I know that rate equals k times h i raised to the power 2. Therefore if I take one set of data and I'll take the first set of data which is 7.5 times 10 to the negative 4 moles per liter per second equals k times 0.0050 moles per liter square. And therefore I rearrange this equation. I can say k equals 7.5 times 10 to the negative 4 moles per liter per second divided by 0.0050 moles per let's say this is squared. So the mole squared per liter squared. And if you were to put the numbers into your calculator and work this out, what you would end up with is a two-sig-fig number, so it would be 30. Now I'm dividing moles per liter per second by mole squared per liter squared. So if I take the moles, you can see that one of these moles and one of those will cancel out, you can see between these two, one negative one and this will become negative one. So now what I end up with is liters per mole per second. And so the unit now turns out to be liters per mole per second. And can everybody see that the unit for the rate constant will vary depending on the order of that reaction. So there's no one standard unit for the rate constant. It's like the equilibrium constant. Remember we said the equilibrium constant, the unit varies depending on the law of mass action. But likewise, the unit for the rate constant will change depending on the order. And so we'll say that the unit for the rate constant will vary depending on the rate law or the rate expression. So you need to figure out the appropriate unit. And so that will be the rate constant. And then if we go on to part C, the part C tells us or asks us to calculate the reaction rate for a concentration of HI equal to 0.02 molar. So now we need to figure out the rate. Now that we know what the value of K is, now that we know what the value of N is, if we change the concentration we should be able to figure out the corresponding rate. So once again we go back to the equation, the rate law, which we know is K times HI taken to the power 2. Now we want to figure out the rate, but now we know what the K value is. So that would be 30 times liter per mole per second times the concentration, which is 0.0020 moles per liter squared. And so if you put that into your calculator, it comes out to be 1.2 times 10 to the negative 4. And you can see that the unit now works out to be moles per liter per second. Remember the unit for rate is always concentration per time. So you know that unit has to come out to be the change in concentration, which is moles per liter, and change in time, which is per second. So the moles per liter per second. All right? And you'll see that the units work out. Got it? So this is how, I don't know, have you guys carried out any experiments in the lab with kinetics? So this is how you figure out what the order of the reaction is if you know what the rates are. So we took an example of a problem where we had only one reactant. All right? Now what happens if you have more than one reactant? And so to take an example of how you figure out the order relative to multiple reactants, let's take an example where once again we're going to use an equation that has two reactants. All right? So if you look at this, now we're looking at the reaction where we have 2NO reacting with O2 to give you 2NO2. All right? And so in the laboratory, what we're going to be looking at is now we have two concentrations to look at. So we have two concentrations and we figure out the corresponding rate. All right? And so we have three sets of data. And the way we do this is to keep the math simple. The way we kind of, we construct the experiment in a particular way so that we can figure out what the order of the reaction is. And the way we do that is if you look at this, what we do is we keep one of the concentrations constant. So if you look at NO and look at the first two set of data, we keep those two concentrations the same but only change the other two and we figure out experimentally what the rates are. All right? Then what we do is now for the next set of experiment, we keep these two the same and change these two. You see that? So what we do is we collect data where if you have multiple reactants, we will keep one concentrations of the reactants the same, all right, and change the other two. And that allows us to figure out the order with respect to that. And now the next time we do this, we keep the other two reactants, the concentrations the same, and change the other two. So once again, to solve a problem of this type, we start by writing the general form of the rate law. Okay? So can you help me write that out? So if I'm going to write the general form of this rate law, I would start by saying rate equals what? What do I write down next? The k, the rate constant k, lower case k, times my first reactant is what? NO. It's going to be raised to some arbitrary number m. Now I have another reactant. So my other reactant would be O2 raised to some arbitrary number n. And now ultimately, I have to figure out what m is and what n is if I want to know what the rate law is. Okay? So now we have three sets of data, so I can write three equations. So I can say the first equation would be rate is 2.8 times 10 to the negative 6 moles per liter per second equals the constant k times the concentration of NO would be 1 times 10 to the negative 4 molar raised to the power m times 1 times 10 to the negative 4 molar raised to the power n. That would be my first data. Okay? Finally I have the second data which is 8.4 times 10 to the negative 6 moles per liter per second equals k times 1 times 10 to the negative 4 molar raised to the power m times 3 times 10 to the negative 4 molar raised to the power n. That would be equation 2. The third one would be 3.4 times 10 to the negative 5 moles per liter per second equals k times 2 times 10 to the negative 4 molar raised to the power m times 3 times 10 to the negative 4 molar raised to the power n. Alright? Now I want to go back to this general form of the equation and can you see what we measure experimentally is rate? We know the concentration of n, we know the concentration of NO2 but what we don't know is what m is, we don't know what n is and we don't know what k is. And remember if you have three unknowns, algebraically how many equations do you need to solve for it if you want to figure out the three unknowns? You need three algebraic equations to solve for three unknowns. As I've written down the three equations. Alright? Now we start by, let's take equation 2 which is a bigger number than equation 1 and I'm going to divide equation 2 by equation 1. Alright? So if I take these two you can see what I end up with is 8.4 times 10 to the negative 6 moles per liter per second equals k times 1 times 10 to the negative 4 molar raised to the power m times 3 times 10 to the negative 4 molar raised to the power n. And I divide that by equation 1 which is 2.8 times 10 to the negative 6 moles per liter per second equals k times 1 times 10 to the negative 4 molar raised to the power m times 1 times 10 to the negative 4 molar raised to the power n. And I divide one equation by the other, can you see the 2k's will cancel out. Now by design we're keeping these two concentrations the same and can you see the reason why we do that? If I keep the concentration the same this whole term will cancel out as well. Alright? So essentially I've eliminated one term for that equation. Now if I divide this by this I end up with 3 and over here if I divide this by that I end up with 3 raised to the power of n. So now just very simply I know that n equals 1. So I've quickly figured out what the order is with respect to that by dividing one equation by the other and by design we always keep the concentrations the same so that they cancel out. Okay? So I figured out what n is. Now if I want to figure out the other one can you see by design we're keeping these two concentrations the same. So now all I have to do is to figure out this m I'm going to divide this equation by that. Okay? So now that I know what n is I need to figure out what m is and so for that once again I always divide the bigger number by the smaller number in terms of rate and so I'm going to take this equation and divide this equation by that. So what I end up with is so I'm going to say 3.4 times 10 to the negative 5 moles per liter per second equals k. Alright times 2 times 10 to the negative 4 molar times 3 times 10 to the negative 4 molar. Now I've figured out that this n is 1 so I can put the one there to remind me that is first order. Okay? I'm going to divide that by 8.4 times 10 to the negative 6 moles per liter per second times k times. Now this turns out to be 1 times 10 to the negative 4 molar and this would be 3 times 10 to the negative 4 molar raised to the power 1. So k and k would cancel out. By design now that we take the one where the other two concentrations are the same so that these terms will cancel out. So what I end up with is now this term divided by this term and this term divided by this term and both of these raised to the power of m. Okay? So if we divide this by this what I end up with is 4 and over here I have 2 divided by 1 which is 2 raised to the power of m and therefore I know m comes out to be 2. Alright? So by dividing you know this is just algebra, simple algebra and by taking advantage of designing the experiment so that we always keep one of the concentrations the same and only varying the other we can figure out what the order of the reaction is. So now we figure out what the rate law is. What is my rate law? My rate law that I've determined from experiment turns out to be that rate equals the constant k, lowercase k. My reactant is NO and I know it's second order in NO and first order in O2. So if you want to describe this rate law you'd say it's second order with respect to NO. It's first order with respect to O2 and what is the overall order for this reaction? 2 plus 1? 3. Alright? So ultimately in many of these problems our ultimate goal is to figure out this expression with what the order is and that gives us the rate law or the rate expression. Now what is the second? So now the second part is determine the rate law and now they want us to figure out the value of the rate constant. So to figure out the value of the rate constant you just pick any one of these equations. So let's take the first one where we want to figure out rate constant so we know rate equals k times NO squared times O2. So if I want to know what the rate constant is this would be the rate divided by the concentration of NO squared times the concentration of O2 which if you take the first set of data it would be 2.8 times 10 to the negative 6 moles per liter per second divided by the concentration of NO which would be 1 times 10 to the negative 4 moles per liter the whole thing squared times the concentration of O2 which is 1 times 10 to the negative 4 moles per liter. And so if you work that out the answer comes out to be 2 the answer comes out to be of course we're dividing everything by 1 so the answer comes out to be 2.8 times 10 to the power 6 liter squared per mole squared per second. So what you need to do is look at this will be moles this times this will be moles cubed per liter squared per liter cubed and you cancel the units out and you can see once again that the unit for the rate constant varies depending on the order so there's no one standard unit for the rate constant it's going to vary and you can figure out the appropriate unit. Okay so it's very important because on an exam you're graded not only on your numerical answer but also on the unit so you have to make sure that you check to make sure you cancel the units out and get the appropriate unit for the rate constant. Okay so any questions up to this point so we looked at the fact so the take-home message is what? That when we talk about rate remember last time we introduced you to the idea of what it means when we say the rate of a reaction. Now that we understand what it means when we say rate now we know that the rate actually depends on the concentration of the reactant alright and the relationship between rate and the concentration of the reactant is given by the rate law alright and whenever you refer to a rate law it's got two parts to it's got a rate constant and it's got an order and so we have to figure that out and this is done experimentally there's no way in advance for us to predict what the outcome is going to be so if we can't predict this in advance then you have to deal with experimental data the only way to figure it out is to use experimental data and so the type of problem that you will get will say in an experiment this is the data that we obtained now from this data figure out what the rate law is and we took two examples of how you figure out the rate law alright if you have just one reactant how do you figure out if you have two reactants how do you figure it out ok now while this is interesting and fun we run into a problem when we deal with this is experimentally it's very difficult to measure the rate remember I told you the way you measure the rate of a reaction is to plot concentration versus time and then pick an instantaneous time and then you draw a tangential line that goes to that point and figure out the slope of that line ok now it turns out experimentally when you actually try to do this in the lab it's not it's very difficult to do this precisely alright it's you know getting the right tangential line you know just a slight twist can change it and so it turns out when you are measuring rate as a function of concentration it's very very difficult to precisely measure the rate so while the calculations are pretty straightforward executing this in the laboratory becomes difficult ok and so we need a slightly different approach so it turns out that we are trying to accomplish the same goal alright but now we want to take a slightly different approach alright so instead of looking at remember we said we looked at the dependence of rate on concentration alright and what we find is while it helps us understand the topic actually accomplishing measuring rates in the lab is cumbersome and difficult alright so we need a different approach that makes it a lot easier for us to accomplish this alright so what we do is instead of looking at rate versus concentration what we do is now what is much more easier for us to measure is just concentration versus time alright so we are going to take advantage of concentration versus time so to measure the rate of a chemical reaction we have to determine small changes in concentration occurring during a short time interval sometimes it can be difficult to obtain sufficiently precise experimental data for these small changes because remember you have to pick a point and then you have to get the tangential slope that runs to that point and sometimes that is difficult to do ok and alternative is to fit all the data over a long time interval with an equation that expresses the concentration of the species directly in terms of time so what do I mean by that so remember the way we figure out rate is to plot concentration versus time and usually we look at it in terms of reactants so we look at the concentration of a versus time in seconds and we said it decays like that so if I wanted to figure out instantaneous so if I wanted to carry this out I would make measurements at different time intervals like that so I would measure at different times I would measure the concentration of a so remember what we measure is the amount that is left unreacted so at the beginning we have the initial concentration it reacts a little bit then we figure out how much of a is left over then it would proceed at a different like the first 50 seconds we figure out how much a is unreacted then in 100 seconds we figure out how much a is left unreacted and then at 150 seconds we figure out how much a is left unreacted and you would end up with a plot like that if I wanted to figure out rate if it's at the instant the reaction begins with the initial I would have to draw a tangential line through that and figure out what it is that's hard to do what's more what's much more easier is to take this exponential decay that you see and come up with a mathematical equation so we're going to derive a mathematical equation that actually that describes this whole process alright and we're going to derive a mathematical equation that looks at the dependence of concentration versus time and gives us the whole dependence of that entire plot alright so we want to derive a mathematical equation that describes this plot so in order to do that we have to look at each one separately we want to look at a first order reaction and let's say we're going to derive an equation for a first order reaction where you have only one reactant alright so if I'm going to derive a mathematical expression that describes the whole process and describes the dependence of concentration versus time I will start with it with the rate law and the rate law for this would be rate equals K alright how many reactants do I have only one reactant and my reactant is a and it is first order so you know I put a one there but I don't need to put a one there because it's a first order reaction okay now since I want to derive an equation now I'm going to take advantage of integrating the whole process I'm going to say that if you write the rate remember rate is what delta concentration of A divided by delta T you guys remember so instead of using delta I'm going to say we're looking at small incremental changes so instead of using delta which is change I'm going to use D or dava which represents small incremental changes and so I'm looking at small incremental changes in A as a function of small incremental changes in time so I'm representing rate this way and remember it's a reactant and the reactant is consumed and so we put a negative sign in front of that that will equal K times the concentration of A got it now I'm going to rearrange this equation so that now I'm going to say that the small incremental change divided by A equals negative K dt alright so I'm going to just put all the concentration terms on one side and the time terms on the other side and now remember these are incremental changes and I'm now going to integrate this so that it explains the whole process from when it begins so that so my limits are I start with the initial concentration of A to concentration of A and over here I'm going to integrate time from T equals 0 to T and if you don't understand this process now essentially if I integrate this what I'll end up with is ln A minus ln A0 equals minus KT alright if you don't understand this all I'm interested is that you know what the final outcome is alright so if you haven't done any integration in your math classes all I want you to remember is the outcome of this equation turns out to be this or you can also rewrite this and say this is ln A over ln A0 equals minus KT now what this equation represents is after I integrate it I'm looking at small incremental changes and when I integrate it I'm integrating it to the whole limits starting from initial concentration of A until as A reacts at A concentration of A at time T this is what it would look like got it now if I were to write this so I would expect you to memorize this because it's a first order reaction and for a first order reaction you need to know this is what it looks like alright and this as well now if I want to represent this in the form of a graph can you see if I take ln A minus the initial concentration of ln A0 equals minus KT if I rearrange this I'll end up with ln A equals minus KT plus ln A0 so I'm just taking this over there and now this is in the form of if this were the y-axis and K is the slope it's a negative slope times x-axis that will be the x-axis this would be the slope plus C which is the intercept alright so it's in the form of y equals mx plus C and so you can see if I plot this out what that would tell me is that if I plot ln A versus time I should end up with a straight line graph alright like this where the intercept would be the initial concentration that's obvious and the slope would equal K and you can see it's a negative slope alright so the final take home message that I want you to remember is if it's a first order reaction what you do is you measure you make a graph you look at concentration sorry at different time intervals you figure out the concentration of A so this is A at different time intervals at zero this will be the initial concentration of A at different time intervals you measure the concentration of A alright and now if you take you know in x in the excel you convert A to ln A and if you make a plot of ln A versus time and you end up with a straight line with a negative slope what does that tell you it's first order does that make sense to everybody so now instead of measuring rate what I would do is in the laboratory I would measure concentration versus time and then take the natural log of the concentration and draw a graph and if I end up with a negative slope a straight line with a negative slope what does that tell me it's first order alright so that's a lot more easier to accomplish in the lab.
 Okay, can I have everyone's attention?  So let's go ahead and start.  But before I do, I wanted to find out, do you guys have any questions?  So we have moved on to looking at chemical kinetics.  If you guys remember, last time we talked about the fact that when we studied chemical  kinetics, we have two important goals.  One is we need to figure out how fast reactions proceed.  And the second goal of kinetics is to figure out the reaction pathway.  When we say the reaction pathway, what we are interested in looking at is the detailed  events that take place as reactants convert to products.  So we're interested in looking at how fast reactions proceed.  And secondly, we're interested in looking at pathways.  And we're going to start looking at both of these in detail.  But to begin with, we're going to focus on how fast reactions proceed.  And we said when we talk about how fast reactions proceed, the term that we use is rate.  And last time we talked about the fact that you can talk about rate in terms of average  rate, or you can talk about in terms of instantaneous rate.  And we said that it is much more useful to us to refer to rate in terms of instantaneous  rate.  And from now on, when we use the term rate, what we refer to is the instantaneous rate.  And then at the end of the class, we talked about the fact that most reactions are reversible.  And if a reaction is reversible, that means that there is a rate associated with the forward  reaction and there is a rate associated with the reverse reaction.  So there are two rates, forward and reverse.  And so if you're looking at a reversible reaction, the overall net rate that we measure experimentally  is actually the difference between the forward rate and the reverse rate.  But to keep things simple, from now on, we're only going to consider forward rate only.  And we're going to pretend as if the forward rate is the only one that exists.  In reality, you have a forward rate and a reverse rate.  But to keep things simple, we're just going to focus on the forward rate.  So last time we talked about rate, we talked about the unit for rate.  And now you understand what it means when we talk about reaction rate.  Now we're going to look at it in a little bit more detail.  And we're going to look at the relationship between rate and concentration.  So to keep things easy for you, I've just made some short notes.  And these notes are available on the class website.  And it just kind of highlights the important things that I want you to remember as you  study this topic.  So today we're going to look at the dependence of rate on concentration.  And this looks at the relationship between the rate of a reaction and the concentration.  And the concentration that we look at is the concentration of the reactant.  So it turns out that the rate of a chemical reaction, and remember we're focusing on the  forward reaction, so the rate of a chemical reaction, the forward direction, nearly always  depends on the concentration of one or more of the reactants.  So if you're looking at a forward reaction, then how fast that reaction proceeds will  depend on the concentration of the reactants.  If you have only one reactant, it would depend on that concentration, that one reactant.  If you have two reactants or more, then it would depend on the concentrations of all  of the reactants.  So the first thing to remember is the rate depends on the concentration of the reactant.  And if you want to mathematically describe this relationship in the form of an equation,  we call that the rate expression or the rate law.  So I'm going to take the same reaction, I'm just going to write it down.  So if we take a reaction where our reactant, we have only one reactant and this reactant  goes to products.  So if your balanced equation is that one reactant going to products, and our reactant is A,  that lowercase A is the coefficient in that balanced equation.  So the rate of this reaction is going to depend on the concentration of A. Remember, the rate  always depends on the concentration of the reactant.  So if I want to mathematically state this, then I would say that the relationship is  described as rate equals a constant K. Now remember, I want you to always make sure this  is lowercase K, not capital K. When you use capital K, what does capital K represent?  Equilibrium constant.  So this is lowercase K. And remember, the rate depends on the concentration of the reactant  raised to a power n, which is a number n.  And the K here is called the rate constant.  So the lowercase K is a constant, and that is called the rate constant.  This will be the concentration of A, and this n here is called the order with respect to  that reactant.  So this is order with respect to the reactant.  And this relationship, when you write it in this format, where you say rate equals a constant  times the concentration raised to the power n, this relationship is what is called the  rate law or rate expression.  So this is similar to the law of mass action.  You guys remember, whenever you write the chemical equilibrium expression, where it's  the products, concentration of the products raised to the power of the coefficient is  called the law of mass action.  Here the relationship between rate and concentration of the reactant is always called the rate  law.  So that's what I've given you in these notes.  And so the same thing is written here.  So it's always rate equals K, concentration of the reactant, raised to the power n, where  K is a rate constant, and this is a constant.  n is the order of the reaction with respect to that reactant.  This is the power to which the concentration is raised.  The power n in the rate expression has no direct relationship to the coefficients A in  the balanced equation.  So this number n has no relationship to the coefficient.  It's a number that we determine experimentally.  So there's no way in advance for anyone to look at a balanced equation and say what that  value of n would come out to be.  Does that make sense, Sir?  So there's no way that you or I can look at a balanced equation and say what that value  n is going to turn out to be.  The only way to determine the value of n is to experiment.  And later on I'll show you how we figure it out.  This value of n can be a whole number or a fractional number.  So let's see if, and it's called the order.  So in some reactions, the value of n comes out to be 1.  When the value of n comes out to be 1, we call it first order.  That about n means order of the reaction.  And therefore when n equals 1, we call it first order.  So when n equals 1, the rate law is written as this.  Rate equals k, a raised to the power 1.  Obviously when something is raised to the power 1, you don't need to state it as 1.  If you leave it blank, you know that it is to the power 1.  So if you take this example of this reaction where n205 goes to 2n02 plus half n02, it turns  out experimentally we know this is first order.  So the rate law would be rate equals a constant k, which is the rate constant, times the constant.  There's only one reactant, and that one reactant raised to the power 1.  So when you write the rate law like that, we call that a first order rate law.  Because the order n is 1, and if n comes out to be 1, then the rate law is written where  n equals 1.  Now there are other reactions where n can be 2.  When n is 2, the rate law is written as rate equals k, a to the power 2.  And we call that reaction second order.  And so if you take an example, this is an example of reaction that has only one reactant,  and therefore rate equals k, the concentration of the reactant raised to the power 2, and  that would be the rate law for a second order reaction.  And this is a reaction where you have only one reactant, and if the order turns out to  be 2, it will be second order.  Now sometimes this could be 0.  When n equals 0, the rate law is written as rate equals k, a raised to the power 0.  And you know that if a is raised to the power 0, you know that's going to be 1.  And so the rate will be, when n equals to the rate law turns out to be this, and we call  this so rate will equal k, because a raised to the power 0 turns out to be 1.  And therefore that means rate equals the rate constant.  So if the rate equals the rate constant k, then we call that a zero order reaction.  Does that make sense to everybody?  So it could be zero order.  If n equals zero, it's called zero order.  If n equals 1, it's first order.  If n equals 2, it's second order.  And so on and so forth.  All right?  Usually the common ones are zero, one, and two.  All right?  And lastly, remember this could be a fractional number.  So if we take here, in this reaction, you can see the rate constant comes to 1.5.  All right?  So here the rate equals k times the concentration raised to the power of a fractional number.  This could be 3 halves, or 1.5.  And therefore that is the order with respect to that reactor.  So the lesson is, to remember is, for you to remember is, that the lesson from all of  this is that the reaction order is an experimental property that cannot be predicted from the  chemical equation.  It has to be deduced from experiment.  All right?  So every reaction that you look at, the relationship between the concentration of the reactant,  all right?  And the rate is called the rate law.  And the rate law is always written in the same form.  Rate equals a constant k, which is the rate constant, times the concentration of the reactant  raised to the power of the n.  And that n is called the order of the reaction.  All right?  So what happens if you have more than one reactant?  So if you have more than one reactant, and let's take an example of a reaction where  you have more than one reactant.  So for reactions with two or more reactants, the rate law is given by.  So now let's say we have a hypothetical general form of the equation where you have a plus  v plus c plus d.  You can have any number of reactants going to products.  Then the rate law is written as rate equals, once again, the rate constant k.  Now we have two reactants.  So it'd be a raised to the power of a number m, b raised to the power of a number n.  And if you have three reactants, it'll be c raised to the power of y, a number y, all  right?  And that would describe the rate law.  So if you have only two reactants, the rate law would be the concentration of the two  reactants, each raised to the power of a number.  Okay, where m is the order with respect to a, n is the order with respect to b, and if  you add them up, that gives you the overall order, all right?  So let's take an example.  So if we take this example of this reactant reacting with hydroxide to give you HPO3 2  minus plus H2, let's say we're looking at this reaction.  Now if you were asked to write the rate law, the first thing you would do is look at that  equation and figure out how many reactants there are.  There are two reactants, all right?  Therefore the rate law of the rate expression would be rate of that reaction equals a constant  k, which is the rate constant, times the concentration of each reactant raised to the power of a number.  So it turns out experimentally we figure out that this is first order and hydroxide is second  order, all right?  So that means we say that this reaction is first order in H2PO2 minus and second order  in hydroxide.  And the overall order would be 1 plus 2, which is 3.  Is that clear to everybody?  And that's what we call the rate law, all right?  And I want to kind of remind you this is very similar to the law of mass action, where we  write down and we say the equilibrium constant k equals the concentrations raised to the  power of their coefficients.  Likewise, in kinetics when we look at rate law or the rate expression, it's always the  rate.  This gives the relationship between rate of that reaction and the concentrations of reactants,  all right?  And the rate of the forward direction, remember we're considering only the rate and the forward  direction.  And the rate always will depend on the concentrations of reactants and that gives the relationship.  So now we want to know how do you remember I said we can't figure this out experimentally.  I mean, sorry, we can't look at a balanced equation and just say what this order will  turn out to be.  This order with respect to reactants vary from reaction to reaction.  And as I said before, there's no way for us to look at it in advance and say what the  order is going to be.  And the only way to figure out the order is to figure it out experimentally, all right?  So let me show you how we do this.  So it is a mathematical calculation and essentially what you do.  So let's take this problem and it's in the worksheet that I have put up on the class  website.  But let's take a look at how one determines the rate law for a reaction and how one determines  the order, okay?  So here we're looking at a reaction and the first thing I want you to look at this and  if you look at the equation, the balanced equation that's provided to you, can everybody  see there's only one reactant, all right?  So the rate law is going to only depend on the concentration of that reactant, okay?  So now in the lab what you would do is you would do several different trials and at different  concentrations of the reactant you can calculate the rate of that reaction.  Remember how do we calculate rate?  Experimentally what you do is in the lab you would look at concentration versus time,  all right?  You would make a plot and then you would pick an instantaneous time and you would take the  slope at that instant in time, the tangential slope and from that you can figure out what  the rate is, all right?  So we talked about that last time.  So what we have is we're looking at the rates at three different concentrations of the reactant.  So here's the first concentration and the corresponding rate.  Here's the second concentration and the corresponding rate and this is the third concentration and  the corresponding rate, all right?  Now we're going to use this data to figure out first the order of the reaction.  So the question is determine the order of the reaction and write the rate expression.  So we're going to start by writing the general form of the rate law, okay?  So without figuring out what the order is, if I'm going to write the general form of  the rate law what would that look like?  Rate equals, remember what is a rate law tell you?  The rate equals what?  The rate constant times the concentration of the reactant raised to the power of a number  n that we need to establish.  So this is the general form of the rate law.  Now we have to figure out what that exact rate law is, okay?  Now if you look at the experimental measurements can everybody see that we are measuring rate?  We know the corresponding concentrations but we don't know what the value of this constant  is and we don't know what the value of this n is and our ultimate goal is to figure those  out in this problem, all right?  So we write the general form of the rate law because we know the rate of the reaction depends  on the concentration of the reactor, all right?  So if you look at this equation we have two unknowns and mathematically if you have two  unknowns all of you recognize that you need two equations to solve for it, all right?  So algebraically if you have two unknowns we need two equations.  So what we're going to do is we're going to write two equations based on two sets of data.  So we'll start with the first set of data.  We know that the rate equals k times h raised to the power n.  So if one of my rates is 7.5 times 10 to the negative 4 moles per liter per second that  will be the rate equals k times, we know the concentration for that will be 0.0050 molar  raised to the power of n.  So that's one equation.  I'm going to go to the second equation and I'm going to say when the rate is 10 to the  negative 3 times 10 to the negative 3 moles per liter per second equals the same constant  k but now my concentration is 0.010 molar raised to the power n.  That gives me 2.  So now algebraically I've got to figure out a way to use both of these equations because  remember I have two unknowns, I have two equations, now I can solve that and figure out what the  value of n is and I can figure out what the value of k is.  And you can see that if I divide one equation by the other and usually it's best to keep  the math simple, it's always best to take a bigger number and divide the bigger number  by a smaller number than vice versa.  You should end up with the same answer but it would just keep the math simple.  So this is a bigger number.  So I'm going to say 3 times 10 to the negative 3 moles per liter per second equals k times  0.010 moles per liter raised to the power n.  I'm going to divide that by 7.5 times 10 to the negative 4 moles per liter per second  equals k times 0.0050 moles per liter raised to the power n.  Can you see now k and k will cancel out?  And so what I've got is now an equation with one unknown.  And so if you take this and divide this, what I end up with is 4.  Over here both of these are raised to the power n.  So if I take 0.5, you can see that this comes out to be 2 and that raised to the power n.  And so can everybody see I figured out what n is?  n is 2.  So we know that this reaction is second order.  By taking the data, essentially what we're doing is experimentally we're measuring the  rate and we're looking at its dependence on concentration.  So we're changing the concentration of reactant.  By doing this experiment in the lab, we can figure out what the order of the reaction is.  So now I've figured out what the rate law is.  So my rate law would be that rate equals the constant, rate constant times h i raised to  the power 2.  So that answers the first part.  And this is how in the laboratory we determine what the order of the reaction is.  Remember we said we can't figure this out in advance.  We can't predetermine or pre-predict what this order n would come out to be.  The only way to determine it is to carry out the experiment.  You measure the rates at different concentrations and you can figure out what the rate law is.  So now that we've figured out what the rate law is, we can move on to looking at the second  part of the question.  So we determine the order of the reaction and write the rate law.  So if I ask you on the exam to write the rate law, this is what you need to give.  You cannot give this because this is the general form.  This is a generic equation and you have to find out the unique rate law that corresponds  to this reaction and we've figured out that it is second order.  Now if I go to the second part of that problem, we're asked to calculate the rate constant  and give its unit.  So if I go to part b, I need to figure out now what k is.  Since I've figured out the second, can I ever see I can just take any one of these equations.  Now I know that n is 2 and now I have only one unknown I can solve for k.  So I can say I know that rate equals k times h i raised to the power 2.  Therefore if I take one set of data and I'll take the first set of data which is 7.5 times  10 to the negative 4 moles per liter per second equals k times 0.0050 moles per liter square.  And therefore I rearrange this equation.  I can say k equals 7.5 times 10 to the negative 4 moles per liter per second divided by 0.0050  moles per let's say this is squared.  So the mole squared per liter squared.  And if you were to put the numbers into your calculator and work this out, what you would  end up with is a two-sig-fig number, so it would be 30.  Now I'm dividing moles per liter per second by mole squared per liter squared.  So if I take the moles, you can see that one of these moles and one of those will cancel  out, you can see between these two, one negative one and this will become negative one.  So now what I end up with is liters per mole per second.  And so the unit now turns out to be liters per mole per second.  And can everybody see that the unit for the rate constant will vary depending on the order  of that reaction.  So there's no one standard unit for the rate constant.  It's like the equilibrium constant.  Remember we said the equilibrium constant, the unit varies depending on the law of mass  action.  But likewise, the unit for the rate constant will change depending on the order.  And so we'll say that the unit for the rate constant will vary depending on the rate law  or the rate expression.  So you need to figure out the appropriate unit.  And so that will be the rate constant.  And then if we go on to part C, the part C tells us or asks us to calculate the reaction  rate for a concentration of HI equal to 0.02 molar.  So now we need to figure out the rate.  Now that we know what the value of K is, now that we know what the value of N is, if we  change the concentration we should be able to figure out the corresponding rate.  So once again we go back to the equation, the rate law, which we know is K times HI  taken to the power 2.  Now we want to figure out the rate, but now we know what the K value is.  So that would be 30 times liter per mole per second times the concentration, which is 0.0020  moles per liter squared.  And so if you put that into your calculator, it comes out to be 1.2 times 10 to the negative  4.  And you can see that the unit now works out to be moles per liter per second.  Remember the unit for rate is always concentration per time.  So you know that unit has to come out to be the change in concentration, which is moles  per liter, and change in time, which is per second.  So the moles per liter per second.  All right?  And you'll see that the units work out.  Got it?  So this is how, I don't know, have you guys carried out any experiments in the lab with  kinetics?  So this is how you figure out what the order of the reaction is if you know what the rates  are.  So we took an example of a problem where we had only one reactant.  All right?  Now what happens if you have more than one reactant?  And so to take an example of how you figure out the order relative to multiple reactants,  let's take an example where once again we're going to use an equation that has two reactants.  All right?  So if you look at this, now we're looking at the reaction where we have 2NO reacting  with O2 to give you 2NO2.  All right?  And so in the laboratory, what we're going to be looking at is now we have two concentrations  to look at.  So we have two concentrations and we figure out the corresponding rate.  All right?  And so we have three sets of data.  And the way we do this is to keep the math simple.  The way we kind of, we construct the experiment in a particular way so that we can figure  out what the order of the reaction is.  And the way we do that is if you look at this, what we do is we keep one of the concentrations  constant.  So if you look at NO and look at the first two set of data, we keep those two concentrations  the same but only change the other two and we figure out experimentally what the rates  are.  All right?  Then what we do is now for the next set of experiment, we keep these two the same and  change these two.  You see that?  So what we do is we collect data where if you have multiple reactants, we will keep one  concentrations of the reactants the same, all right, and change the other two.  And that allows us to figure out the order with respect to that.  And now the next time we do this, we keep the other two reactants, the concentrations  the same, and change the other two.  So once again, to solve a problem of this type, we start by writing the general form  of the rate law.  Okay?  So can you help me write that out?  So if I'm going to write the general form of this rate law, I would start by saying  rate equals what?  What do I write down next?  The k, the rate constant k, lower case k, times my first reactant is what?  NO.  It's going to be raised to some arbitrary number m.  Now I have another reactant.  So my other reactant would be O2 raised to some arbitrary number n.  And now ultimately, I have to figure out what m is and what n is if I want to know what  the rate law is.  Okay?  So now we have three sets of data, so I can write three equations.  So I can say the first equation would be rate is 2.8 times 10 to the negative 6 moles per  liter per second equals the constant k times the concentration of NO would be 1 times 10  to the negative 4 molar raised to the power m times 1 times 10 to the negative 4 molar  raised to the power n.  That would be my first data.  Okay?  Finally I have the second data which is 8.4 times 10 to the negative 6 moles per liter  per second equals k times 1 times 10 to the negative 4 molar raised to the power m times  3 times 10 to the negative 4 molar raised to the power n.  That would be equation 2.  The third one would be 3.4 times 10 to the negative 5 moles per liter per second equals  k times 2 times 10 to the negative 4 molar raised to the power m times 3 times 10 to  the negative 4 molar raised to the power n.  Alright?  Now I want to go back to this general form of the equation and can you see what we measure  experimentally is rate?  We know the concentration of n, we know the concentration of NO2 but what we don't know  is what m is, we don't know what n is and we don't know what k is.  And remember if you have three unknowns, algebraically how many equations do you need to solve for  it if you want to figure out the three unknowns?  You need three algebraic equations to solve for three unknowns.  As I've written down the three equations.  Alright?  Now we start by, let's take equation 2 which is a bigger number than equation 1 and I'm  going to divide equation 2 by equation 1.  Alright?  So if I take these two you can see what I end up with is 8.4 times 10 to the negative  6 moles per liter per second equals k times 1 times 10 to the negative 4 molar raised  to the power m times 3 times 10 to the negative 4 molar raised to the power n.  And I divide that by equation 1 which is 2.8 times 10 to the negative 6 moles per liter  per second equals k times 1 times 10 to the negative 4 molar raised to the power m times  1 times 10 to the negative 4 molar raised to the power n.  And I divide one equation by the other, can you see the 2k's will cancel out.  Now by design we're keeping these two concentrations the same and can you see the reason why we  do that?  If I keep the concentration the same this whole term will cancel out as well.  Alright?  So essentially I've eliminated one term for that equation.  Now if I divide this by this I end up with 3 and over here if I divide this by that I  end up with 3 raised to the power of n.  So now just very simply I know that n equals 1.  So I've quickly figured out what the order is with respect to that by dividing one equation  by the other and by design we always keep the concentrations the same so that they cancel  out.  Okay?  So I figured out what n is.  Now if I want to figure out the other one can you see by design we're keeping these  two concentrations the same.  So now all I have to do is to figure out this m I'm going to divide this equation by that.  Okay?  So now that I know what n is I need to figure out what m is and so for that once again I  always divide the bigger number by the smaller number in terms of rate and so I'm going to  take this equation and divide this equation by that.  So what I end up with is so I'm going to say 3.4 times 10 to the negative 5 moles per liter  per second equals k.  Alright times 2 times 10 to the negative 4 molar times 3 times 10 to the negative 4  molar.  Now I've figured out that this n is 1 so I can put the one there to remind me that is  first order.  Okay?  I'm going to divide that by 8.4 times 10 to the negative 6 moles per liter per second  times k times.  Now this turns out to be 1 times 10 to the negative 4 molar and this would be 3 times  10 to the negative 4 molar raised to the power 1.  So k and k would cancel out.  By design now that we take the one where the other two concentrations are the same so that  these terms will cancel out.  So what I end up with is now this term divided by this term and this term divided by this  term and both of these raised to the power of m.  Okay?  So if we divide this by this what I end up with is 4 and over here I have 2 divided by  1 which is 2 raised to the power of m and therefore I know m comes out to be 2.  Alright?  So by dividing you know this is just algebra, simple algebra and by taking advantage of  designing the experiment so that we always keep one of the concentrations the same and  only varying the other we can figure out what the order of the reaction is.  So now we figure out what the rate law is.  What is my rate law?  My rate law that I've determined from experiment turns out to be that rate equals the constant  k, lowercase k.  My reactant is NO and I know it's second order in NO and first order in O2.  So if you want to describe this rate law you'd say it's second order with respect to NO.  It's first order with respect to O2 and what is the overall order for this reaction?  2 plus 1?  3.  Alright?  So ultimately in many of these problems our ultimate goal is to figure out this expression  with what the order is and that gives us the rate law or the rate expression.  Now what is the second?  So now the second part is determine the rate law and now they want us to figure out the  value of the rate constant.  So to figure out the value of the rate constant you just pick any one of these equations.  So let's take the first one where we want to figure out rate constant so we know rate  equals k times NO squared times O2.  So if I want to know what the rate constant is this would be the rate divided by the concentration  of NO squared times the concentration of O2 which if you take the first set of data it  would be 2.8 times 10 to the negative 6 moles per liter per second divided by the concentration  of NO which would be 1 times 10 to the negative 4 moles per liter the whole thing squared  times the concentration of O2 which is 1 times 10 to the negative 4 moles per liter.  And so if you work that out the answer comes out to be 2 the answer comes out to be of  course we're dividing everything by 1 so the answer comes out to be 2.8 times 10 to the  power 6 liter squared per mole squared per second.  So what you need to do is look at this will be moles this times this will be moles cubed  per liter squared per liter cubed and you cancel the units out and you can see once  again that the unit for the rate constant varies depending on the order so there's no  one standard unit for the rate constant it's going to vary and you can figure out the appropriate  unit.  Okay so it's very important because on an exam you're graded not only on your numerical  answer but also on the unit so you have to make sure that you check to make sure you  cancel the units out and get the appropriate unit for the rate constant.  Okay so any questions up to this point so we looked at the fact so the take-home message  is what?  That when we talk about rate remember last time we introduced you to the idea of what  it means when we say the rate of a reaction.  Now that we understand what it means when we say rate now we know that the rate actually  depends on the concentration of the reactant alright and the relationship between rate  and the concentration of the reactant is given by the rate law alright and whenever you refer  to a rate law it's got two parts to it's got a rate constant and it's got an order and  so we have to figure that out and this is done experimentally there's no way in advance  for us to predict what the outcome is going to be so if we can't predict this in advance  then you have to deal with experimental data the only way to figure it out is to use experimental  data and so the type of problem that you will get will say in an experiment this is the  data that we obtained now from this data figure out what the rate law is and we took two examples  of how you figure out the rate law alright if you have just one reactant how do you figure  out if you have two reactants how do you figure it out ok now while this is interesting and  fun we run into a problem when we deal with this is experimentally it's very difficult  to measure the rate remember I told you the way you measure the rate of a reaction is  to plot concentration versus time and then pick an instantaneous time and then you draw  a tangential line that goes to that point and figure out the slope of that line ok now  it turns out experimentally when you actually try to do this in the lab it's not it's very  difficult to do this precisely alright it's you know getting the right tangential line  you know just a slight twist can change it and so it turns out when you are measuring  rate as a function of concentration it's very very difficult to precisely measure the rate  so while the calculations are pretty straightforward executing this in the laboratory becomes difficult  ok and so we need a slightly different approach so it turns out that we are trying to accomplish  the same goal alright but now we want to take a slightly different approach alright so instead  of looking at remember we said we looked at the dependence of rate on concentration alright  and what we find is while it helps us understand the topic actually accomplishing measuring  rates in the lab is cumbersome and difficult alright so we need a different approach that  makes it a lot easier for us to accomplish this alright so what we do is instead of looking  at rate versus concentration what we do is now what is much more easier for us to measure  is just concentration versus time alright so we are going to take advantage of concentration  versus time so to measure the rate of a chemical reaction we have to determine small changes  in concentration occurring during a short time interval sometimes it can be difficult  to obtain sufficiently precise experimental data for these small changes because remember  you have to pick a point and then you have to get the tangential slope that runs to that  point and sometimes that is difficult to do ok and alternative is to fit all the data  over a long time interval with an equation that expresses the concentration of the species  directly in terms of time so what do I mean by that so remember the way we figure out  rate is to plot concentration versus time and usually we look at it in terms of reactants  so we look at the concentration of a versus time in seconds and we said it decays like  that so if I wanted to figure out instantaneous so if I wanted to carry this out I would make  measurements at different time intervals like that so I would measure at different times  I would measure the concentration of a so remember what we measure is the amount that  is left unreacted so at the beginning we have the initial concentration it reacts a little  bit then we figure out how much of a is left over then it would proceed at a different  like the first 50 seconds we figure out how much a is unreacted then in 100 seconds we  figure out how much a is left unreacted and then at 150 seconds we figure out how much  a is left unreacted and you would end up with a plot like that if I wanted to figure out  rate if it's at the instant the reaction begins with the initial I would have to draw a tangential  line through that and figure out what it is that's hard to do what's more what's much  more easier is to take this exponential decay that you see and come up with a mathematical  equation so we're going to derive a mathematical equation that actually that describes this  whole process alright and we're going to derive a mathematical equation that looks at the  dependence of concentration versus time and gives us the whole dependence of that entire  plot alright so we want to derive a mathematical equation that describes this plot so in order  to do that we have to look at each one separately we want to look at a first order reaction  and let's say we're going to derive an equation for a first order reaction where you have  only one reactant alright so if I'm going to derive a mathematical expression that  describes the whole process and describes the dependence of concentration versus time  I will start with it with the rate law and the rate law for this would be rate equals  K alright how many reactants do I have only one reactant and my reactant is a and it is  first order so you know I put a one there but I don't need to put a one there because  it's a first order reaction okay now since I want to derive an equation now I'm going  to take advantage of integrating the whole process I'm going to say that if you write  the rate remember rate is what delta concentration of A divided by delta T you guys remember  so instead of using delta I'm going to say we're looking at small incremental changes  so instead of using delta which is change I'm going to use D or dava which represents  small incremental changes and so I'm looking at small incremental changes in A as a function  of small incremental changes in time so I'm representing rate this way and remember it's  a reactant and the reactant is consumed and so we put a negative sign in front of that  that will equal K times the concentration of A got it now I'm going to rearrange this  equation so that now I'm going to say that the small incremental change divided by A  equals negative K dt alright so I'm going to just put all the concentration terms on  one side and the time terms on the other side and now remember these are incremental changes  and I'm now going to integrate this so that it explains the whole process from when it  begins so that so my limits are I start with the initial concentration of A to concentration  of A and over here I'm going to integrate time from T equals 0 to T and if you don't  understand this process now essentially if I integrate this what I'll end up with is  ln A minus ln A0 equals minus KT alright if you don't understand this all I'm interested  is that you know what the final outcome is alright so if you haven't done any integration  in your math classes all I want you to remember is the outcome of this equation turns out to  be this or you can also rewrite this and say this is ln A over ln A0 equals minus KT  now what this equation represents is after I integrate it I'm looking at small incremental  changes and when I integrate it I'm integrating it to the whole limits starting from initial  concentration of A until as A reacts at A concentration of A at time T this is what it would look  like got it now if I were to write this so I would expect you to memorize this because  it's a first order reaction and for a first order reaction you need to know this is what  it looks like alright and this as well now if I want to represent this in the form of  a graph can you see if I take ln A minus the initial concentration of ln A0 equals minus  KT if I rearrange this I'll end up with ln A equals minus KT plus ln A0 so I'm just taking  this over there and now this is in the form of if this were the y-axis and K is the slope  it's a negative slope times x-axis that will be the x-axis this would be the slope plus  C which is the intercept alright so it's in the form of y equals mx plus C and so you  can see if I plot this out what that would tell me is that if I plot ln A versus time  I should end up with a straight line graph alright like this where the intercept would  be the initial concentration that's obvious and the slope would equal K and you can see  it's a negative slope alright so the final take home message that I want you to remember  is if it's a first order reaction what you do is you measure you make a graph you look  at concentration sorry at different time intervals you figure out the concentration of A so this  is A at different time intervals at zero this will be the initial concentration of A at  different time intervals you measure the concentration of A alright and now if you take you know  in x in the excel you convert A to ln A and if you make a plot of ln A versus time and  you end up with a straight line with a negative slope what does that tell you it's first  order does that make sense to everybody so now instead of measuring rate what I would  do is in the laboratory I would measure concentration versus time and then take the natural log  of the concentration and draw a graph and if I end up with a negative slope a straight  line with a negative slope what does that tell me it's first order alright so that's  a lot more easier to accomplish in the lab.
General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__0_275.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__276_575.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__576_635.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__636_851.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__852_983.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__984_1079.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__1080_1127.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__1128_1175.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__1176_1235.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__1236_1379.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__1380_1439.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__1440_1667.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__1668_1739.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__1740_1787.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__1788_1907.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__1908_1955.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__1956_2027.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__2028_2075.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__2076_2171.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__2172_2219.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__2220_2339.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__2340_2411.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__2412_2471.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__2472_2519.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__2520_2711.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__2712_2759.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__2760_2819.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__2820_2903.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__2904_2951.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__2952_3047.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__3048_3155.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__3156_3203.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__3204_3251.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__3252_3299.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__3300_3347.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__3348_3395.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__3396_3443.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__3444_3491.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__3492_3587.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__3588_3647.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__3648_3815.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__3816_3983.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__3984_4031.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__4032_4079.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__4080_4223.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__4224_4271.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__4272_4343.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__4344_4427.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__4428_4475.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__4476_4523.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__4524_4583.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__4584_4739.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__4740_4787.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__4788_4895.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__4896_4979.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__4980_6491.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__6492_18539.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__18540_39251.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__39252_55559.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__55560_64595.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__64596_77231.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__77232_77327.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__77328_77387.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__77388_77471.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__77472_77567.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__77568_77771.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__77772_77831.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__77832_77879.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__77880_78059.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__78060_78107.png General_Chemistry_1C__Lecture_23__Chemical_Kinetics_Pt__2__78108_78118.png
Okay, can I have everyone's attention? So let's go ahead and start. But before I do, I wanted to find out, do you guys have any questions? So we have moved on to looking at chemical kinetics. If you guys remember, last time we talked about the fact that when we studied chemical kinetics, we have two important goals. One is we need to figure out how fast reactions proceed. And the second goal of kinetics is to figure out the reaction pathway. When we say the reaction pathway, what we are interested in looking at is the detailed events that take place as reactants convert to products. So we're interested in looking at how fast reactions proceed. And secondly, we're interested in looking at pathways. And we're going to start looking at both of these in detail. But to begin with, we're going to focus on how fast reactions proceed. And we said when we talk about how fast reactions proceed, the term that we use is rate. And last time we talked about the fact that you can talk about rate in terms of average rate, or you can talk about in terms of instantaneous rate. And we said that it is much more useful to us to refer to rate in terms of instantaneous rate. And from now on, when we use the term rate, what we refer to is the instantaneous rate. And then at the end of the class, we talked about the fact that most reactions are reversible. And if a reaction is reversible, that means that there is a rate associated with the forward reaction and there is a rate associated with the reverse reaction. So there are two rates, forward and reverse. And so if you're looking at a reversible reaction, the overall net rate that we measure experimentally is actually the difference between the forward rate and the reverse rate. But to keep things simple, from now on, we're only going to consider forward rate only. And we're going to pretend as if the forward rate is the only one that exists. In reality, you have a forward rate and a reverse rate. But to keep things simple, we're just going to focus on the forward rate. So last time we talked about rate, we talked about the unit for rate. And now you understand what it means when we talk about reaction rate. Now we're going to look at it in a little bit more detail. And we're going to look at the relationship between rate and concentration. So to keep things easy for you, I've just made some short notes. And these notes are available on the class website. And it just kind of highlights the important things that I want you to remember as you study this topic. So today we're going to look at the dependence of rate on concentration. And this looks at the relationship between the rate of a reaction and the concentration. And the concentration that we look at is the concentration of the reactant. So it turns out that the rate of a chemical reaction, and remember we're focusing on the forward reaction, so the rate of a chemical reaction, the forward direction, nearly always depends on the concentration of one or more of the reactants. So if you're looking at a forward reaction, then how fast that reaction proceeds will depend on the concentration of the reactants. If you have only one reactant, it would depend on that concentration, that one reactant. If you have two reactants or more, then it would depend on the concentrations of all of the reactants. So the first thing to remember is the rate depends on the concentration of the reactant. And if you want to mathematically describe this relationship in the form of an equation, we call that the rate expression or the rate law. So I'm going to take the same reaction, I'm just going to write it down. So if we take a reaction where our reactant, we have only one reactant and this reactant goes to products. So if your balanced equation is that one reactant going to products, and our reactant is A, that lowercase A is the coefficient in that balanced equation. So the rate of this reaction is going to depend on the concentration of A. Remember, the rate always depends on the concentration of the reactant. So if I want to mathematically state this, then I would say that the relationship is described as rate equals a constant K. Now remember, I want you to always make sure this is lowercase K, not capital K. When you use capital K, what does capital K represent? Equilibrium constant. So this is lowercase K. And remember, the rate depends on the concentration of the reactant raised to a power n, which is a number n. And the K here is called the rate constant. So the lowercase K is a constant, and that is called the rate constant. This will be the concentration of A, and this n here is called the order with respect to that reactant. So this is order with respect to the reactant. And this relationship, when you write it in this format, where you say rate equals a constant times the concentration raised to the power n, this relationship is what is called the rate law or rate expression. So this is similar to the law of mass action. You guys remember, whenever you write the chemical equilibrium expression, where it's the products, concentration of the products raised to the power of the coefficient is called the law of mass action. Here the relationship between rate and concentration of the reactant is always called the rate law. So that's what I've given you in these notes. And so the same thing is written here. So it's always rate equals K, concentration of the reactant, raised to the power n, where K is a rate constant, and this is a constant. n is the order of the reaction with respect to that reactant. This is the power to which the concentration is raised. The power n in the rate expression has no direct relationship to the coefficients A in the balanced equation. So this number n has no relationship to the coefficient. It's a number that we determine experimentally. So there's no way in advance for anyone to look at a balanced equation and say what that value of n would come out to be. Does that make sense, Sir? So there's no way that you or I can look at a balanced equation and say what that value n is going to turn out to be. The only way to determine the value of n is to experiment. And later on I'll show you how we figure it out. This value of n can be a whole number or a fractional number. So let's see if, and it's called the order. So in some reactions, the value of n comes out to be 1. When the value of n comes out to be 1, we call it first order. That about n means order of the reaction. And therefore when n equals 1, we call it first order. So when n equals 1, the rate law is written as this. Rate equals k, a raised to the power 1. Obviously when something is raised to the power 1, you don't need to state it as 1. If you leave it blank, you know that it is to the power 1. So if you take this example of this reaction where n205 goes to 2n02 plus half n02, it turns out experimentally we know this is first order. So the rate law would be rate equals a constant k, which is the rate constant, times the constant. There's only one reactant, and that one reactant raised to the power 1. So when you write the rate law like that, we call that a first order rate law. Because the order n is 1, and if n comes out to be 1, then the rate law is written where n equals 1. Now there are other reactions where n can be 2. When n is 2, the rate law is written as rate equals k, a to the power 2. And we call that reaction second order. And so if you take an example, this is an example of reaction that has only one reactant, and therefore rate equals k, the concentration of the reactant raised to the power 2, and that would be the rate law for a second order reaction. And this is a reaction where you have only one reactant, and if the order turns out to be 2, it will be second order. Now sometimes this could be 0. When n equals 0, the rate law is written as rate equals k, a raised to the power 0. And you know that if a is raised to the power 0, you know that's going to be 1. And so the rate will be, when n equals to the rate law turns out to be this, and we call this so rate will equal k, because a raised to the power 0 turns out to be 1. And therefore that means rate equals the rate constant. So if the rate equals the rate constant k, then we call that a zero order reaction. Does that make sense to everybody? So it could be zero order. If n equals zero, it's called zero order. If n equals 1, it's first order. If n equals 2, it's second order. And so on and so forth. All right? Usually the common ones are zero, one, and two. All right? And lastly, remember this could be a fractional number. So if we take here, in this reaction, you can see the rate constant comes to 1.5. All right? So here the rate equals k times the concentration raised to the power of a fractional number. This could be 3 halves, or 1.5. And therefore that is the order with respect to that reactor. So the lesson is, to remember is, for you to remember is, that the lesson from all of this is that the reaction order is an experimental property that cannot be predicted from the chemical equation. It has to be deduced from experiment. All right? So every reaction that you look at, the relationship between the concentration of the reactant, all right? And the rate is called the rate law. And the rate law is always written in the same form. Rate equals a constant k, which is the rate constant, times the concentration of the reactant raised to the power of the n. And that n is called the order of the reaction. All right? So what happens if you have more than one reactant? So if you have more than one reactant, and let's take an example of a reaction where you have more than one reactant. So for reactions with two or more reactants, the rate law is given by. So now let's say we have a hypothetical general form of the equation where you have a plus v plus c plus d. You can have any number of reactants going to products. Then the rate law is written as rate equals, once again, the rate constant k. Now we have two reactants. So it'd be a raised to the power of a number m, b raised to the power of a number n. And if you have three reactants, it'll be c raised to the power of y, a number y, all right? And that would describe the rate law. So if you have only two reactants, the rate law would be the concentration of the two reactants, each raised to the power of a number. Okay, where m is the order with respect to a, n is the order with respect to b, and if you add them up, that gives you the overall order, all right? So let's take an example. So if we take this example of this reactant reacting with hydroxide to give you HPO3 2 minus plus H2, let's say we're looking at this reaction. Now if you were asked to write the rate law, the first thing you would do is look at that equation and figure out how many reactants there are. There are two reactants, all right? Therefore the rate law of the rate expression would be rate of that reaction equals a constant k, which is the rate constant, times the concentration of each reactant raised to the power of a number. So it turns out experimentally we figure out that this is first order and hydroxide is second order, all right? So that means we say that this reaction is first order in H2PO2 minus and second order in hydroxide. And the overall order would be 1 plus 2, which is 3. Is that clear to everybody? And that's what we call the rate law, all right? And I want to kind of remind you this is very similar to the law of mass action, where we write down and we say the equilibrium constant k equals the concentrations raised to the power of their coefficients. Likewise, in kinetics when we look at rate law or the rate expression, it's always the rate. This gives the relationship between rate of that reaction and the concentrations of reactants, all right? And the rate of the forward direction, remember we're considering only the rate and the forward direction. And the rate always will depend on the concentrations of reactants and that gives the relationship. So now we want to know how do you remember I said we can't figure this out experimentally. I mean, sorry, we can't look at a balanced equation and just say what this order will turn out to be. This order with respect to reactants vary from reaction to reaction. And as I said before, there's no way for us to look at it in advance and say what the order is going to be. And the only way to figure out the order is to figure it out experimentally, all right? So let me show you how we do this. So it is a mathematical calculation and essentially what you do. So let's take this problem and it's in the worksheet that I have put up on the class website. But let's take a look at how one determines the rate law for a reaction and how one determines the order, okay? So here we're looking at a reaction and the first thing I want you to look at this and if you look at the equation, the balanced equation that's provided to you, can everybody see there's only one reactant, all right? So the rate law is going to only depend on the concentration of that reactant, okay? So now in the lab what you would do is you would do several different trials and at different concentrations of the reactant you can calculate the rate of that reaction. Remember how do we calculate rate? Experimentally what you do is in the lab you would look at concentration versus time, all right? You would make a plot and then you would pick an instantaneous time and you would take the slope at that instant in time, the tangential slope and from that you can figure out what the rate is, all right? So we talked about that last time. So what we have is we're looking at the rates at three different concentrations of the reactant. So here's the first concentration and the corresponding rate. Here's the second concentration and the corresponding rate and this is the third concentration and the corresponding rate, all right? Now we're going to use this data to figure out first the order of the reaction. So the question is determine the order of the reaction and write the rate expression. So we're going to start by writing the general form of the rate law, okay? So without figuring out what the order is, if I'm going to write the general form of the rate law what would that look like? Rate equals, remember what is a rate law tell you? The rate equals what? The rate constant times the concentration of the reactant raised to the power of a number n that we need to establish. So this is the general form of the rate law. Now we have to figure out what that exact rate law is, okay? Now if you look at the experimental measurements can everybody see that we are measuring rate? We know the corresponding concentrations but we don't know what the value of this constant is and we don't know what the value of this n is and our ultimate goal is to figure those out in this problem, all right? So we write the general form of the rate law because we know the rate of the reaction depends on the concentration of the reactor, all right? So if you look at this equation we have two unknowns and mathematically if you have two unknowns all of you recognize that you need two equations to solve for it, all right? So algebraically if you have two unknowns we need two equations. So what we're going to do is we're going to write two equations based on two sets of data. So we'll start with the first set of data. We know that the rate equals k times h raised to the power n. So if one of my rates is 7.5 times 10 to the negative 4 moles per liter per second that will be the rate equals k times, we know the concentration for that will be 0.0050 molar raised to the power of n. So that's one equation. I'm going to go to the second equation and I'm going to say when the rate is 10 to the negative 3 times 10 to the negative 3 moles per liter per second equals the same constant k but now my concentration is 0.010 molar raised to the power n. That gives me 2. So now algebraically I've got to figure out a way to use both of these equations because remember I have two unknowns, I have two equations, now I can solve that and figure out what the value of n is and I can figure out what the value of k is. And you can see that if I divide one equation by the other and usually it's best to keep the math simple, it's always best to take a bigger number and divide the bigger number by a smaller number than vice versa. You should end up with the same answer but it would just keep the math simple. So this is a bigger number. So I'm going to say 3 times 10 to the negative 3 moles per liter per second equals k times 0.010 moles per liter raised to the power n. I'm going to divide that by 7.5 times 10 to the negative 4 moles per liter per second equals k times 0.0050 moles per liter raised to the power n. Can you see now k and k will cancel out? And so what I've got is now an equation with one unknown. And so if you take this and divide this, what I end up with is 4. Over here both of these are raised to the power n. So if I take 0.5, you can see that this comes out to be 2 and that raised to the power n. And so can everybody see I figured out what n is? n is 2. So we know that this reaction is second order. By taking the data, essentially what we're doing is experimentally we're measuring the rate and we're looking at its dependence on concentration. So we're changing the concentration of reactant. By doing this experiment in the lab, we can figure out what the order of the reaction is. So now I've figured out what the rate law is. So my rate law would be that rate equals the constant, rate constant times h i raised to the power 2. So that answers the first part. And this is how in the laboratory we determine what the order of the reaction is. Remember we said we can't figure this out in advance. We can't predetermine or pre-predict what this order n would come out to be. The only way to determine it is to carry out the experiment. You measure the rates at different concentrations and you can figure out what the rate law is. So now that we've figured out what the rate law is, we can move on to looking at the second part of the question. So we determine the order of the reaction and write the rate law. So if I ask you on the exam to write the rate law, this is what you need to give. You cannot give this because this is the general form. This is a generic equation and you have to find out the unique rate law that corresponds to this reaction and we've figured out that it is second order. Now if I go to the second part of that problem, we're asked to calculate the rate constant and give its unit. So if I go to part b, I need to figure out now what k is. Since I've figured out the second, can I ever see I can just take any one of these equations. Now I know that n is 2 and now I have only one unknown I can solve for k. So I can say I know that rate equals k times h i raised to the power 2. Therefore if I take one set of data and I'll take the first set of data which is 7.5 times 10 to the negative 4 moles per liter per second equals k times 0.0050 moles per liter square. And therefore I rearrange this equation. I can say k equals 7.5 times 10 to the negative 4 moles per liter per second divided by 0.0050 moles per let's say this is squared. So the mole squared per liter squared. And if you were to put the numbers into your calculator and work this out, what you would end up with is a two-sig-fig number, so it would be 30. Now I'm dividing moles per liter per second by mole squared per liter squared. So if I take the moles, you can see that one of these moles and one of those will cancel out, you can see between these two, one negative one and this will become negative one. So now what I end up with is liters per mole per second. And so the unit now turns out to be liters per mole per second. And can everybody see that the unit for the rate constant will vary depending on the order of that reaction. So there's no one standard unit for the rate constant. It's like the equilibrium constant. Remember we said the equilibrium constant, the unit varies depending on the law of mass action. But likewise, the unit for the rate constant will change depending on the order. And so we'll say that the unit for the rate constant will vary depending on the rate law or the rate expression. So you need to figure out the appropriate unit. And so that will be the rate constant. And then if we go on to part C, the part C tells us or asks us to calculate the reaction rate for a concentration of HI equal to 0.02 molar. So now we need to figure out the rate. Now that we know what the value of K is, now that we know what the value of N is, if we change the concentration we should be able to figure out the corresponding rate. So once again we go back to the equation, the rate law, which we know is K times HI taken to the power 2. Now we want to figure out the rate, but now we know what the K value is. So that would be 30 times liter per mole per second times the concentration, which is 0.0020 moles per liter squared. And so if you put that into your calculator, it comes out to be 1.2 times 10 to the negative 4. And you can see that the unit now works out to be moles per liter per second. Remember the unit for rate is always concentration per time. So you know that unit has to come out to be the change in concentration, which is moles per liter, and change in time, which is per second. So the moles per liter per second. All right? And you'll see that the units work out. Got it? So this is how, I don't know, have you guys carried out any experiments in the lab with kinetics? So this is how you figure out what the order of the reaction is if you know what the rates are. So we took an example of a problem where we had only one reactant. All right? Now what happens if you have more than one reactant? And so to take an example of how you figure out the order relative to multiple reactants, let's take an example where once again we're going to use an equation that has two reactants. All right? So if you look at this, now we're looking at the reaction where we have 2NO reacting with O2 to give you 2NO2. All right? And so in the laboratory, what we're going to be looking at is now we have two concentrations to look at. So we have two concentrations and we figure out the corresponding rate. All right? And so we have three sets of data. And the way we do this is to keep the math simple. The way we kind of, we construct the experiment in a particular way so that we can figure out what the order of the reaction is. And the way we do that is if you look at this, what we do is we keep one of the concentrations constant. So if you look at NO and look at the first two set of data, we keep those two concentrations the same but only change the other two and we figure out experimentally what the rates are. All right? Then what we do is now for the next set of experiment, we keep these two the same and change these two. You see that? So what we do is we collect data where if you have multiple reactants, we will keep one concentrations of the reactants the same, all right, and change the other two. And that allows us to figure out the order with respect to that. And now the next time we do this, we keep the other two reactants, the concentrations the same, and change the other two. So once again, to solve a problem of this type, we start by writing the general form of the rate law. Okay? So can you help me write that out? So if I'm going to write the general form of this rate law, I would start by saying rate equals what? What do I write down next? The k, the rate constant k, lower case k, times my first reactant is what? NO. It's going to be raised to some arbitrary number m. Now I have another reactant. So my other reactant would be O2 raised to some arbitrary number n. And now ultimately, I have to figure out what m is and what n is if I want to know what the rate law is. Okay? So now we have three sets of data, so I can write three equations. So I can say the first equation would be rate is 2.8 times 10 to the negative 6 moles per liter per second equals the constant k times the concentration of NO would be 1 times 10 to the negative 4 molar raised to the power m times 1 times 10 to the negative 4 molar raised to the power n. That would be my first data. Okay? Finally I have the second data which is 8.4 times 10 to the negative 6 moles per liter per second equals k times 1 times 10 to the negative 4 molar raised to the power m times 3 times 10 to the negative 4 molar raised to the power n. That would be equation 2. The third one would be 3.4 times 10 to the negative 5 moles per liter per second equals k times 2 times 10 to the negative 4 molar raised to the power m times 3 times 10 to the negative 4 molar raised to the power n. Alright? Now I want to go back to this general form of the equation and can you see what we measure experimentally is rate? We know the concentration of n, we know the concentration of NO2 but what we don't know is what m is, we don't know what n is and we don't know what k is. And remember if you have three unknowns, algebraically how many equations do you need to solve for it if you want to figure out the three unknowns? You need three algebraic equations to solve for three unknowns. As I've written down the three equations. Alright? Now we start by, let's take equation 2 which is a bigger number than equation 1 and I'm going to divide equation 2 by equation 1. Alright? So if I take these two you can see what I end up with is 8.4 times 10 to the negative 6 moles per liter per second equals k times 1 times 10 to the negative 4 molar raised to the power m times 3 times 10 to the negative 4 molar raised to the power n. And I divide that by equation 1 which is 2.8 times 10 to the negative 6 moles per liter per second equals k times 1 times 10 to the negative 4 molar raised to the power m times 1 times 10 to the negative 4 molar raised to the power n. And I divide one equation by the other, can you see the 2k's will cancel out. Now by design we're keeping these two concentrations the same and can you see the reason why we do that? If I keep the concentration the same this whole term will cancel out as well. Alright? So essentially I've eliminated one term for that equation. Now if I divide this by this I end up with 3 and over here if I divide this by that I end up with 3 raised to the power of n. So now just very simply I know that n equals 1. So I've quickly figured out what the order is with respect to that by dividing one equation by the other and by design we always keep the concentrations the same so that they cancel out. Okay? So I figured out what n is. Now if I want to figure out the other one can you see by design we're keeping these two concentrations the same. So now all I have to do is to figure out this m I'm going to divide this equation by that. Okay? So now that I know what n is I need to figure out what m is and so for that once again I always divide the bigger number by the smaller number in terms of rate and so I'm going to take this equation and divide this equation by that. So what I end up with is so I'm going to say 3.4 times 10 to the negative 5 moles per liter per second equals k. Alright times 2 times 10 to the negative 4 molar times 3 times 10 to the negative 4 molar. Now I've figured out that this n is 1 so I can put the one there to remind me that is first order. Okay? I'm going to divide that by 8.4 times 10 to the negative 6 moles per liter per second times k times. Now this turns out to be 1 times 10 to the negative 4 molar and this would be 3 times 10 to the negative 4 molar raised to the power 1. So k and k would cancel out. By design now that we take the one where the other two concentrations are the same so that these terms will cancel out. So what I end up with is now this term divided by this term and this term divided by this term and both of these raised to the power of m. Okay? So if we divide this by this what I end up with is 4 and over here I have 2 divided by 1 which is 2 raised to the power of m and therefore I know m comes out to be 2. Alright? So by dividing you know this is just algebra, simple algebra and by taking advantage of designing the experiment so that we always keep one of the concentrations the same and only varying the other we can figure out what the order of the reaction is. So now we figure out what the rate law is. What is my rate law? My rate law that I've determined from experiment turns out to be that rate equals the constant k, lowercase k. My reactant is NO and I know it's second order in NO and first order in O2. So if you want to describe this rate law you'd say it's second order with respect to NO. It's first order with respect to O2 and what is the overall order for this reaction? 2 plus 1? 3. Alright? So ultimately in many of these problems our ultimate goal is to figure out this expression with what the order is and that gives us the rate law or the rate expression. Now what is the second? So now the second part is determine the rate law and now they want us to figure out the value of the rate constant. So to figure out the value of the rate constant you just pick any one of these equations. So let's take the first one where we want to figure out rate constant so we know rate equals k times NO squared times O2. So if I want to know what the rate constant is this would be the rate divided by the concentration of NO squared times the concentration of O2 which if you take the first set of data it would be 2.8 times 10 to the negative 6 moles per liter per second divided by the concentration of NO which would be 1 times 10 to the negative 4 moles per liter the whole thing squared times the concentration of O2 which is 1 times 10 to the negative 4 moles per liter. And so if you work that out the answer comes out to be 2 the answer comes out to be of course we're dividing everything by 1 so the answer comes out to be 2.8 times 10 to the power 6 liter squared per mole squared per second. So what you need to do is look at this will be moles this times this will be moles cubed per liter squared per liter cubed and you cancel the units out and you can see once again that the unit for the rate constant varies depending on the order so there's no one standard unit for the rate constant it's going to vary and you can figure out the appropriate unit. Okay so it's very important because on an exam you're graded not only on your numerical answer but also on the unit so you have to make sure that you check to make sure you cancel the units out and get the appropriate unit for the rate constant. Okay so any questions up to this point so we looked at the fact so the take-home message is what? That when we talk about rate remember last time we introduced you to the idea of what it means when we say the rate of a reaction. Now that we understand what it means when we say rate now we know that the rate actually depends on the concentration of the reactant alright and the relationship between rate and the concentration of the reactant is given by the rate law alright and whenever you refer to a rate law it's got two parts to it's got a rate constant and it's got an order and so we have to figure that out and this is done experimentally there's no way in advance for us to predict what the outcome is going to be so if we can't predict this in advance then you have to deal with experimental data the only way to figure it out is to use experimental data and so the type of problem that you will get will say in an experiment this is the data that we obtained now from this data figure out what the rate law is and we took two examples of how you figure out the rate law alright if you have just one reactant how do you figure out if you have two reactants how do you figure it out ok now while this is interesting and fun we run into a problem when we deal with this is experimentally it's very difficult to measure the rate remember I told you the way you measure the rate of a reaction is to plot concentration versus time and then pick an instantaneous time and then you draw a tangential line that goes to that point and figure out the slope of that line ok now it turns out experimentally when you actually try to do this in the lab it's not it's very difficult to do this precisely alright it's you know getting the right tangential line you know just a slight twist can change it and so it turns out when you are measuring rate as a function of concentration it's very very difficult to precisely measure the rate so while the calculations are pretty straightforward executing this in the laboratory becomes difficult ok and so we need a slightly different approach so it turns out that we are trying to accomplish the same goal alright but now we want to take a slightly different approach alright so instead of looking at remember we said we looked at the dependence of rate on concentration alright and what we find is while it helps us understand the topic actually accomplishing measuring rates in the lab is cumbersome and difficult alright so we need a different approach that makes it a lot easier for us to accomplish this alright so what we do is instead of looking at rate versus concentration what we do is now what is much more easier for us to measure is just concentration versus time alright so we are going to take advantage of concentration versus time so to measure the rate of a chemical reaction we have to determine small changes in concentration occurring during a short time interval sometimes it can be difficult to obtain sufficiently precise experimental data for these small changes because remember you have to pick a point and then you have to get the tangential slope that runs to that point and sometimes that is difficult to do ok and alternative is to fit all the data over a long time interval with an equation that expresses the concentration of the species directly in terms of time so what do I mean by that so remember the way we figure out rate is to plot concentration versus time and usually we look at it in terms of reactants so we look at the concentration of a versus time in seconds and we said it decays like that so if I wanted to figure out instantaneous so if I wanted to carry this out I would make measurements at different time intervals like that so I would measure at different times I would measure the concentration of a so remember what we measure is the amount that is left unreacted so at the beginning we have the initial concentration it reacts a little bit then we figure out how much of a is left over then it would proceed at a different like the first 50 seconds we figure out how much a is unreacted then in 100 seconds we figure out how much a is left unreacted and then at 150 seconds we figure out how much a is left unreacted and you would end up with a plot like that if I wanted to figure out rate if it's at the instant the reaction begins with the initial I would have to draw a tangential line through that and figure out what it is that's hard to do what's more what's much more easier is to take this exponential decay that you see and come up with a mathematical equation so we're going to derive a mathematical equation that actually that describes this whole process alright and we're going to derive a mathematical equation that looks at the dependence of concentration versus time and gives us the whole dependence of that entire plot alright so we want to derive a mathematical equation that describes this plot so in order to do that we have to look at each one separately we want to look at a first order reaction and let's say we're going to derive an equation for a first order reaction where you have only one reactant alright so if I'm going to derive a mathematical expression that describes the whole process and describes the dependence of concentration versus time I will start with it with the rate law and the rate law for this would be rate equals K alright how many reactants do I have only one reactant and my reactant is a and it is first order so you know I put a one there but I don't need to put a one there because it's a first order reaction okay now since I want to derive an equation now I'm going to take advantage of integrating the whole process I'm going to say that if you write the rate remember rate is what delta concentration of A divided by delta T you guys remember so instead of using delta I'm going to say we're looking at small incremental changes so instead of using delta which is change I'm going to use D or dava which represents small incremental changes and so I'm looking at small incremental changes in A as a function of small incremental changes in time so I'm representing rate this way and remember it's a reactant and the reactant is consumed and so we put a negative sign in front of that that will equal K times the concentration of A got it now I'm going to rearrange this equation so that now I'm going to say that the small incremental change divided by A equals negative K dt alright so I'm going to just put all the concentration terms on one side and the time terms on the other side and now remember these are incremental changes and I'm now going to integrate this so that it explains the whole process from when it begins so that so my limits are I start with the initial concentration of A to concentration of A and over here I'm going to integrate time from T equals 0 to T and if you don't understand this process now essentially if I integrate this what I'll end up with is ln A minus ln A0 equals minus KT alright if you don't understand this all I'm interested is that you know what the final outcome is alright so if you haven't done any integration in your math classes all I want you to remember is the outcome of this equation turns out to be this or you can also rewrite this and say this is ln A over ln A0 equals minus KT now what this equation represents is after I integrate it I'm looking at small incremental changes and when I integrate it I'm integrating it to the whole limits starting from initial concentration of A until as A reacts at A concentration of A at time T this is what it would look like got it now if I were to write this so I would expect you to memorize this because it's a first order reaction and for a first order reaction you need to know this is what it looks like alright and this as well now if I want to represent this in the form of a graph can you see if I take ln A minus the initial concentration of ln A0 equals minus KT if I rearrange this I'll end up with ln A equals minus KT plus ln A0 so I'm just taking this over there and now this is in the form of if this were the y-axis and K is the slope it's a negative slope times x-axis that will be the x-axis this would be the slope plus C which is the intercept alright so it's in the form of y equals mx plus C and so you can see if I plot this out what that would tell me is that if I plot ln A versus time I should end up with a straight line graph alright like this where the intercept would be the initial concentration that's obvious and the slope would equal K and you can see it's a negative slope alright so the final take home message that I want you to remember is if it's a first order reaction what you do is you measure you make a graph you look at concentration sorry at different time intervals you figure out the concentration of A so this is A at different time intervals at zero this will be the initial concentration of A at different time intervals you measure the concentration of A alright and now if you take you know in x in the excel you convert A to ln A and if you make a plot of ln A versus time and you end up with a straight line with a negative slope what does that tell you it's first order does that make sense to everybody so now instead of measuring rate what I would do is in the laboratory I would measure concentration versus time and then take the natural log of the concentration and draw a graph and if I end up with a negative slope a straight line with a negative slope what does that tell me it's first order alright so that's a lot more easier to accomplish in the lab.
second goal of kinetics is to figure out the is the third concentration and the corresponding