Lecture 24. Chemical Kinetics Pt. 3.
CC Attribution - ShareAlike 3.0 USA:
You are free to use, adapt and copy, distribute and transmit the work or content in adapted or unchanged form for any legal purpose as long as the work is attributed to the author in the manner specified by the author or licensor and the work or content is shared also in adapted form only under the conditions of this
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, let's go ahead and start. If I could have everyone's attention please. I want to make one quick announcement and that is I wanted to remind everybody that there are bonus points for completing the surveys. So if you guys remember at the beginning of the quarter I said that there was one during the second week of class and I hope that you completed the survey for the ninth week of class and the last part that you need to complete to earn those bonus points is to complete the teaching evaluations as well. Okay, so you got to complete three surveys. Alright, if you complete all three then you will get the bonus points. So please take the time to complete it and I would appreciate it if you would comment on the online homework and if you've used sapling before and you're using Wiley Plus or vice versa, if you could comment on both that would be great. Alright, so if you guys remember last time we started looking at what we called the integrated rate law. Okay, as we will see later on it's important for us to figure out the rate law that corresponds to a reaction. We will see later on that we can use that to obtain information about the pathway. Alright, but before we start looking at pathways we want to understand what it means when we say the rate law of a reaction. And we said the only way you can establish the rate law is to carry out the experiment and either look at rate versus concentration or look at concentration versus time. Alright, and we said that even though we looked at how we can use the rate of a reaction and the concentration of the reactant to figure out the rate law or the order of the reaction. We said unfortunately actually executing that in the laboratory becomes hard because measuring rate accurately or precisely sometimes is difficult. And so we have a different approach where we look at concentration versus time and we derive a mathematical equation that describes the concentration versus time dependence. And last time we derived the expression for the first order reaction. So if you have just one reactant and for a first order reaction we said the general form of the reaction turns out to be lnA minus lnA0 equals minus kT. And so if we rearrange that equation into a form that gives us y equals mx plus c we can see that we end up with a straight line with a negative slope. So if you plot the natural log of the concentration of your reactant as a function of time and you have a straight line with a negative slope it means that that reactant is first order. So what you would do is, so this is much easier to do in the lab because essentially what you do is at different time intervals you measure the concentrations of A. Now remember the concentration of A is the amount unreacted. So as the reaction progresses the concentration of A is the amount that's left unreacted. And so different time intervals you can measure the concentration of A. At the beginning when T equals 0 it will be A0 and then you just measure the concentration at different times. And now you take the natural log of A and if you make a graph and if the graph comes out to be a straight line with a negative slope then you know it's first order. If you don't end up with a straight line graph with a negative slope it means it's not first order. If your point is scattered or if it looks something different then you know that it's not first order. Now having said that now we're going to define a new term that we're going to look at and this term is called half life. So the half life is the time it takes for the concentration of a reactant A to fall to half its original value. So if I were just to plot instead of plotting natural log if I were to just plot the concentration of A as a function of time. And so if this was the initial concentration then we know that it decays like that. So instead of plotting natural log A versus time if you just plot concentration versus time this is what it would look like, the plot. Now half life is the time it takes for the original concentration to be halved. So it's exactly half of this and it's exactly half of that this would be half of the original concentration and we call that the time it takes is called t half, t subscript half to indicate the half life. So the half life is the time it takes for the original concentration to be halved. Now if I take this concentration which is half the amount and look at the time it takes for it to be halved that would be, now the concentration would be a quarter of what you started with and this would be two t halfs where this is one t half and then this is the other t half and so t half plus t half gives you two t halfs. Now if you want to take the quarter amount and the time it takes for the quarter amount to become one eighth would be that and that is three half lives or three t halves and that becomes A0 over 8 and so this would be the third. The time will be the same, okay my plot looks like this is a little bit longer but in reality if it is two days for the first half life it will take another two days for the second half life so the second half life would be four days and then the third half life would be six days, do you understand that? So it's the time it takes for the original concentration to be halved, alright and we call that half life. So if you want to derive an equation for half life we can say at half life we know that the concentration of A is half the original concentration and now t becomes t half because we call that the time it takes for that is half life and we call that t half. So if we go back to the original equation that we derived for the first order reaction we know ln A over A0 equals minus kt and we know at half life this becomes ln A is now equal to this and that will be over A0 equals minus kt half where t is now half life so essentially what I'm doing is I'm replacing this term with that and this will become the half life, okay? You can see that the original concentrations will cancel out therefore ln one half equals minus kt half. Now since there's a negative sign here and to get rid of that negative sign I'm going to take the inverse of that so if I take ln half, mathematical is the same as negative ln two equals kt half, alright? And so now I'm going to rearrange that and get a term for t half which is the half life and so if I rearrange that equation I can say t half, oh there's a negative sign there as well so the two negative signs will get rid of each other, t half equals ln two over the rate constant k, ln two is 0.6931 divided by k so the half life is 0.6931 divided by the rate constant. Now that's worth remembering so if you want to calculate, if it's a first order reaction and it has one reactant and you want to calculate what is the half life and remember what is the definition of half life, it's the time it takes for the original concentration that you started with to be halved, okay? So let's apply this in a problem so if you turn to this problem, okay so if I may turn to that, so here is an example of applying what we just learned in a problem. What is the rate constant k for the first order decomposition of N2O5 at 25 degrees Celsius if the half life of N2O5 at that temperature is given, okay? So what they tell us is that it is first order so it's important to check that, what is the order? Remember so far we derived an expression only for first order so this is a first order reaction and we know that the half life for this is t half equals 4.03 times 10 to the 4 seconds, they want to know what is the rate constant for that reaction, alright? So I know that the only relationship between these two that we've looked at thus far is that t half equals 0.6931 divided by k. I know what this is, this is a number, this is what I need to find so you can see k equals 0.6931 over t half and therefore k equals 0.6931 divided by the half life which is 4.03 times 10 to the power 4 seconds and so if you put that in your calculator and work that out it comes out to be 1.72 times 10 to the negative 5 and you can see that the unit is per second. Remember k is rate constant and the unit for rate constant varies depending on what the order of that reaction is so this is first order and you can see that's the unit for the rate constant. So that's pretty straightforward. Now having calculated that, now for the second part they tell us under these same conditions what percent of N2O5 molecules will not have reacted after one day, alright? So what they're asking is when t is one day they want to know what is the percent unreacted. So we want to figure out percent unreacted. So if you want to calculate the percent unreacted now we have to make use of our integrated rate law and we know the integrated rate law is lnA divided by A0 equals minus kT. Now the first part we calculated k, time is one day. They want us to calculate percent unreacted. Can everybody see this would give me the ratio of unreacted. Remember A is always the amount unreacted and since that's the amount unreacted if I want to calculate percent remember what does percent mean? The amount unreacted divided by what you started with of the total which is the original concentration times what? It's a percent times 100. Do you see that? So if I just find the ratio and multiply it by 100 that will give me the percent unreacted. Alright? So what we're going to do is we're going to figure out what this ratio is. So we can say lnA over A0 which is the original concentration or in this case I should put the actual numbers which is lnN2O5 and the original concentration of N2O5 equals minus now I know my k is 1.72 times 10 to the negative 5 per second and times time. Now remember in kinetics what is the unit that we use for time? Seconds. So they give us one day so that means I need to convert that to seconds. So you know that there are 24 hours in a day and 60 minutes in an hour and 60 seconds in a minute. Alright? So hours and hours will cancel out, minutes and minutes will cancel out. This is per second and second will cancel out so it becomes a unitless quantity. And so I know that lnN2O5 over the original concentration of N2O5 equals if you take, figure that term out it comes to 1.49. Okay? Three sig figs, 24 hours is an exact number, 60 minutes is an exact number, 60 seconds is an exact number so we end up with a number that is three significant figures. Alright? Now what I'm going to do is take the inverse log to get rid of this log. So if I take the inverse log on this side I need to take the inverse log on that side so what I end up with is N2O5 divided by the original concentration of N2O5 and if I take the inverse log of that that comes out to 0.23. Now remember I'm taking the inverse log so if I'm keeping track of sig figs I have a three sig fig number when I go from here to here I only count the decimals so it goes to two significant figures. Alright? And then now that I have figured this out now percent unreacted would be N2O5 divided by N2O5 the original concentration times 100 which is 23 percent. Alright so all I've done is just taken the value that I calculated and multiplied by 100 and that gives me a percent so that means in one day 23 percent is left unreacted. Now when you're solving a problem you have to read the question carefully because if they asked you for percent reacted the percent that actually reacted is 77 percent. So when you're carrying out online homework sometimes it will ask you to calculate the percent unreacted sometimes it will ask you to calculate the percent reacted and so if 23 percent is unreacted you know that 77 percent has reacted. Is that clear to everybody? So that was a look at a first order reaction. Now we need to do the same thing that we've done for the first order and we want to look at what happens with a second order reaction. So we're going to derive an expression for the second order reaction and so if you have a case where you have just one reactant and if it is second order we know that the general form of the rate law would be rate equals rate constant times the concentration of A now it is second order it is raised to the power 2. I hope you took time to review the stuff that we covered last Friday because this is going to build on what we've looked at. So all of you know this is the general form of the rate law we know it's second order and we know that rate equals K times A to the power 2. But since we want to derive an integrated rate law we're going to replace this rate law the rate term with change in concentration incremental change in concentration of A as a function of incremental change in time. And we put a negative sign in front because it's a reactant and a reactant is consumed in that reaction. So this equals K A squared. Now I'm going to put all the concentration terms on one side and take everything else to the other side. And so if I keep the negative sign here I can say this over A squared equals K times DT. Alright? Now if I integrate this over the range of initial concentrations the limits are initial concentration to any concentration A and over here that would be at time zero to any time T. And if you integrate that what you end up with is one over A minus one over the initial concentration equals KT. So if it's a second order reaction and the integrated rate law or the dependence of concentration on time comes out to be in this form. So you need to know this equation I won't give you this equation so you need to know whether it's a first order what the equation should be. You need to know what the equation would be for a second order reaction. Now if I want to rearrange this so that I get it in the form of Y equals MX plus C in other words to figure out what form of this equation will give me a straight line graph I can say now that one over A equals KT plus one over A zero so I'm just going to take this term over there. So now this will turn out to be the Y axis. Now you can see that K is the slope time is the X axis plus C gives me the intercept. And if you look at K you can see that it's positive and so this gives me a positive slope. So the general form of this react if I take this and if I plot a graph of one over A versus time this is the Y axis time is the X axis you can see I end up with a straight line graph. So if I were to plot this I would end up with a slope and the slope will give me the rate constant and everybody can see that this is a slope that has a positive slope and the intercept will be one over the initial concentration. So what this means is that if I were to carry out an experiment in the laboratory and so if I were to look at different time intervals I have different concentrations of A and so when at time zero this will be the initial concentration and then let's say at 50 second increments if I measure the concentration of A. Now in your Excel spreadsheet now you take the inverse of A and you can put those values. And if I plot one over A versus time and if I end up with a straight line graph with a positive slope like that then I know that that would be a second order. So you're looking at the variation of concentration versus time and so you're making a plot and if you make a plot of one over A versus time and you end up with a straight line with a positive slope that means it's second order. If you don't get a straight line but if you take these values and take LNA versus time and it gives you a straight line what does that mean? First order. Do you see that? So you look at the shape of the form of the graph and you can figure out which graph gives you a straight line and that will tell you whether it's a first order reaction or a second order reaction. Now we can do the same thing that we did for the first order and we can start looking at half life. So if I take half life for a second order reaction we know at half life we know that the concentration of A is half of the original concentration we started with and we know the time is called t half. And so if I go to 1 over A minus 1 over A0 equals kt and we know if I take this equation that I started with at half life now this A I'm going to replace this A with that term so what I'll have is 2 over A0 minus 1 over A0 equals kt half. So all I've done is taken the equation that we derived and replaced it with these terms because we're looking at half life and you can see that if I subtract this from that what I end up with is 1 over A0 equals kt half and therefore kt half sorry let's we want t half so t half equals 1 over k times the original concentration. So the equation that you would use if you wanted to calculate the half life for a second order reaction is 1 over k times the original. So these are just derivations so what I need you to do is at home go through the derivations and make sure you know the appropriate equation for first order the appropriate equation for second order. Now let's work a problem where we apply this. So if we turn to the next problem. Now we're looking at the application of what we just learned and so the decomposition of C286 to methyl radicals is second order so that's your key. You're told that it's a second order reaction. If the second order in the reactant and second order overall there's only one reactant so you know that it will be overall second order as well. The rate constant is given if the initial concentration is this what is the concentration after 205 seconds. So let's just write down everything that's given to us. We know that the rate constant is 0.0448 liters per mole per second. We know that the initial concentration of C286 is 0.0100 moles per liter. We know that the time is 205 seconds and they want us to calculate what is the amount unreacted at that point. What is the concentration after 205 seconds? So what is the amount of the reactant that's left over after 205 seconds? So what I wanted to keep in mind is there are only two equations that we've learned for second order reactions. One is for half life. They don't mention half life at all. So we know that if I want to figure out this the equation that I'm going to use is 1 over C286 minus the initial concentration equals kT. Can everybody see this is what I need to find? I know this. They've given us the original concentration. They've given us what k is. They've given us what time is. So this is just knowing the appropriate equation and applying it. So I'm going to put the unknown on one side and take everything else to the other side. So this would be 1 over C286 equals kT plus the original concentration. K is 0.0448 liters per mole per second times the time is 205 seconds. So this and this will cancel out plus 1 over 0.0100 moles per liter. Can you see that this moles and that moles would cancel out and this liter and that liter would cancel out because this is the inverse of the inverse. So sorry, they do not cancel out. This will remain moles per liter per moles and this becomes liter per mole as well. So let me just put this number. If you take this and multiply this by this, it's a three-sic thick number and that number comes out to be 9.18 liters per mole plus this is the inverse of liters, moles per liter. So if you take that, this becomes 100 liters per mole as well. The reason the units don't cancel out is because we're not multiplying, we're adding them. And you can only add apples to apples. And you see that you have the unit is liters per mole, the unit for the other is the inverse of moles per liter and therefore it's liters per mole and therefore you can add apples to apples. Does that make sense to everybody? And so you know that you're on the right track because if you end up with two different units, you know you can add the two numbers together. So now that we know that we have the same units and you can add apples to apples, now I know that 1 over C286 equals that and if I add these two together, now keep in track of sig figs. This is a three-sic fig number, this is a three-sic fig number. Now I'm adding these two numbers together. When you're adding numbers, what do you keep track of in terms of significant figures? Decimals. So can you see that that 100 has zero decimals? So when I add these two together, I'm going to give it to the least number of decimals which is 109 liters per mole. I'm going to take the inverse because I want to calculate the concentration. So if I take the inverse of that, this comes out to be 9.17 times 10 to the negative 3. Now the inverse of liters per mole becomes moles per liter and I know that I end up with the right unit and therefore you know that you haven't made any careless mistakes along the way because I'm calculating concentration and I end up with the appropriate unit. So in all of these problems, I want to remind you that keeping the unit there is helpful because it's sort of checking to make sure that you're doing the problem correctly because if you end up with the right unit, then you know that mathematically you've set up your equations correctly. So if you end up with the right unit, you know for sure that you haven't made any careless mistakes along the way. So now we've looked at first order, we've looked at second order, now we're going to look at zero order. These are the common ones. So if we take a zero order reaction, then we're going to do the same thing that we've been doing all along. So if I take a zero order reaction, can everybody see that rate equals k a raised to power 1? If it's a zero order reaction, it should be raised to the power what? Zero. Anything raised to the power zero is what? One. So that means rate equals k. Now I need to derive the integrated rate loss, I'm going to look at small incremental changes of concentrations of a as a function of time giving me k. Now I'm going to rearrange these equations so that I put all the concentration terms on one side and everything else on the other. So this would be concentration of a. On this side I will put negative k changing time, incremental changes in time. So if I integrate this over the range of the initial concentration to any concentration a, and that corresponds to t equals zero to any time t. And if you integrate that, the outcome comes out to be concentration of a minus the initial concentration equals minus k t. So if it is a zero order reaction, then the equation that describes the dependence of concentration versus time is given by this equation. So this is called the integrated rate law. And if I want to look at the form of a straight line graph, I can rearrange this and say concentration a equals minus k t plus a zero. This would be the y axis. Now can everybody see this is m, but this is a negative slope. So I'm going to put the negative sign in front of this to remind me that it has a negative slope. And this is y equals mx plus c. So this would be k t like that. So now if you look at this, this is a graph that has a negative slope. So if I plot this, my y axis will be concentration of a. My, I'm plotting it versus time. This will give me a straight line graph like that, where the intercept is the initial concentration. And the slope of this graph will give me the rate constant k and this is a negative slope. So the rate constant, when you take the slope, the rate constant has no sign. Remember we talked about the fact that rates don't have sign and neither do rate constants. But I want you to pay attention to, if you end up with a negative slope, then you know it's a zero order reaction. And so if it's a zero order reaction, if you take different times, different concentrations of a, when it's zero, this will be the initial concentration. Now a different time interval. As I said, 50 second increments, you measure the concentration of a that's unreacted. And now if I plot a versus time, if I take this data and plot it, and now if I end up with a graph that gives me a straight line with a negative slope, it tells me that it is zero order. And so what we generally do in the lab is, now that we've looked at all three, the way you figure out the order of the reaction is that once you've got this, so essentially in the lab what you do is, like I'm just going to repeat that again. So you have time, and so zero. You measure the concentration of a that's left unreacted. At time zero, that will be the initial concentration. And then at 50 second increments, for example, you can measure the corresponding concentrations. Now you'll make a third column where you'll figure out the natural log a. And then you have another column where you'll have one over a. All right? And what you would do is that you would make three graphs. All right? You make a graph of time t versus a, or a versus t. This is the y-axis. L and a versus t. And one over a versus t. And so it's like this. So let's say you're carrying this out in the laboratory. And essentially what you do is once you complete the experiment, you draw three graphs. All right? And so you make three graphs, and only one of these will give you a straight line. So if you look at this graph, this is one over a. Remember, one over a versus time is what? Second order. Does that give you a straight line? No. And if one over a versus time, it has to be a straight line with a positive slope. We don't get a straight line with a positive. So we know we can rule out second order. Now look at the second plot. It's L and a versus time. L and a versus time gives you a straight line with a negative slope. And you can actually measure the slope, which is negative 1.3 times 10 to the negative 3. All right? So because that gives you a straight line, what does that tell you about this reaction? That is first order. And I can actually, from the slope, I know the rate constant is, you toss out the negative sign, the rate constant is 1.3 times 10 to negative 3. Got it? And then if you take a versus, I don't get a straight line. So it only has to be 1 of the three. Do you guys get it? And so you make three graphs, and from that, you can figure out what the order of the reaction is. So that's what, I hope you carried out this in the laboratory, and you can figure out the order of the reaction based on drawing the three plots. Yes? This is only for one reactant. This is only one reactant. All right? And it can only be one of the three. Do you understand that? So you're only going to end up with one straight line graph. All right? And so that would be the one that tells you the order of the reaction. OK? Now I also want to derive the half-life for a zero-order reaction. So for a zero-order reaction, if I want to derive the half-life, we know that at half-life, we know that the concentration of A equals half the original concentration. We know that time now becomes half-life. OK? So we go to the equation for zero-order, which is 1 over A minus 1 over A zero equals negative kT. We know at half-life, we're going to replace all of this. Now this becomes 2 over A zero. Wait a minute. Sorry. I'm doing the second order. Please bear with me. I'm thinking about something else. So we know that at half-life, concentration of A is A zero divided by two, and we know T equals T half. And we know for a zero-order reaction, we know this is concentration of A minus A zero equals negative kT. That's the one for zero order. Previously I wrote it for the second order. OK? And so if I take this, and now I'm going to say at half-life, I'm going to replace the terms. I know that A equals A zero divided by two minus A zero equals minus kT half. OK? So now we're looking at half-life. And you can see that this is one. This is half. If I subtract one from the other, what I'll end up with is negative A zero divided by two equals negative kT half. So the negative signs will cancel out on either side. And so T half equals the initial concentration divided by two k. All right? So the integrated rate law, because you have to derive this mathematically, we only look at the three common orders. The common ones are zero, first order, and second order. All right? And so we derive the rate laws for the three. Now on the class website, when you're studying at home, I want you to kind of summarize everything this way. So there's a copy of this on the class website. And so what we're looking at is we looked at zero, first, second order. All right? The rate law would be for zero order, rate equals k. For first order, rate equals kA raised to the power one. Rate equals kA raised to the power two for second order. Okay? Now the form of the integrated rate law that we derived is given here. So A equals minus kT plus A zero. L and A equals minus kT plus L and A zero. Or one over A equals kT plus one over A zero. So this is the form of the integrated rate law that we derived. Okay? Now the plot needed to get a straight line. Because remember, we're going to determine whether reaction is first order, second order, zero order, based on which plot gives you a straight line. So for zero order, A versus T will give you a straight line with a negative slope. Okay? So the slope will be negative k. Where for a first order, L and A versus T will give you a straight line graph with a negative slope. And one over A versus T will give you a positive slope for a second order reaction. So that's why we draw three graphs and figure out which one of these will give us a straight line. Okay? Lastly, the half-life equations that we derived is given here. So half-life for zero order is this, half-life for this is this, and half-life for a second order is that. Okay? You need to remember these equations. All right? And if you worked enough examples like the homework as well as the online homework will give you enough practice so that now you should be able to know which equation to use to figure out the half-life for zero order, half-life for second order, and half-life for first order. Is that clear? So what you're doing with, keep this handy with you. Just paste it on the wall. And as you're carrying out homework, just go to that table and you can look at the appropriate equations. And after a while, you'll be so comfortable with it that you can no longer leave that table. All right? Now, the problem is that up to this point, what we've looked at is only one reactant. All right? And if it's only one reactant, then it's pretty straightforward. How do you figure out the order when you have more than one reactant? All right? So if you turn to the next page, we have to come up with something more clever to figure out the integrated rate laws if you have more than one reactant. All right? Now if you guys remember previously when we looked at rate versus concentration, remember we said we designed experiment so that we can cancel out one reactant. Remember when we worked at problems last Friday, we said if you want to figure out what the order of the reaction is, we actually designed the experiment so that we keep the concentrations, you know, we measure rate at different concentrations. But if you have more than one reactant, you always keep the concentrations of one of those reactant at the same sort that they cancel out and change only the other. All right? And likewise, if you want to use the integrated rate law, we have to design our experiments in a special way. All right? So so far we have considered the integrated rate laws for simple reactions with only one reactant. It turns out special approaches are required to deal with more complicated reactions. All right? So let's say we are looking at this. Okay? So if I want to write the rate law, so we're looking at now two reactants, and if I want to figure this out, the rate law, as all of you can see, rate equals k, concentration A, let's say raised to the power m times concentration of B raised to the power n. Okay? So now we have to design this in a special way so that we can look at the rate law of one at a time. The way we do that is we set up the concentrations so that one concentration is really small. This is of the order of 10 to the negative 3. The other is one molar. So can you see that the difference between the two reactants is about thousandfold? All right? So we specially designed the experiment so that the concentration of one reactant is kept really small, like the order of 10 to the negative 3, and the other is of the order of thousand times more, like one molar. And the reason is that when the reaction takes place, the smaller concentration is going to be the limiting reactant. All right? And so what's going to happen is the smaller one is only going to react, and relative to the smaller concentration, the bigger concentration doesn't appear to change. Because remember, the limiting reactant is a thousand times more dilute. So the reaction will take place at the scale of 10 to the negative 3, and the one molar, which is a thousand times more concentrated, stays relatively unchanged for the scale of the reaction. Can you ever see that? And so what happens is that we set it up so that we take A is the one with the smaller concentration, B is the one with the larger concentration. So B concentration is much, much greater than A. Now what happens is because this is much, much greater, it appears relatively unchanged. And so what happens is as the reaction progresses, the concentration of B does not change all that much. So we're going to approximate it and say that as the reaction progresses, the concentration of B will always be the original concentration, which is a constant. All right? So if you go to the rate law where we say rate equals K, A raised to the power m, B raised to the power n, we can say now we can replace the concentration of B with the original concentration of B. And can everybody see that this term does not change? So this is a constant. All right? So because this is a constant, I'm going to include this with the constant. And I come up with K' which I call, this is called the pseudo rate constant. It's not the real rate constant, but because this includes the pseudo rate constant, which is K' is actually the real rate constant, which is this K times this constant, which is the concentration of B. Do you guys see that? So essentially what I'm doing is by keeping one concentration really large, I'm making it look like you have only one reactant. All right? Because one concentration does not change. And now what I would do is I would carry out this experiment and I would take at different time intervals, I would look at concentration of A. B does not change. It's a constant. So at T equals zero, this will be A zero, and I'll measure the concentrations. All right? Now I will take L and A, and I'll also plot one over A. And in this way, I will see whether I end up, which of these will give me a straight line graph. So if this is experiment A over here, this is experiment A, and you can see now which line gives you a straight line, L and A. If it's L and A versus time in straight line, what does that tell me? It's first order in A. All right? And the slope that I calculate, this is where you want to watch out. The slope, so let's say that if I plot this out, and I end up with L and A versus time will give me a negative slope. Okay? So these are the points on that graph. Now the slope gives me K prime, which is the pseudo rate constant. All right? Because we've reduced this to look as if you have only one reactant. All right? The rate constant is the pseudo rate constant, which is K prime. Got it? So now I have to repeat this experiment and do this all over again, but now I have to keep the A concentration one molar, and I'm going to reverse the thing. Now I'm going to make B the one that is smaller in concentration. All right? So now you can see the A concentration is far greater than B, and therefore now the one that will be constant will be this. So now you can see that this term is constant. All right? So now this equation reduces to rate equals K double prime. This is a different one. All right? K double prime times B raised to the power n. And so now once again I have to repeat this experiment where I have T, and then now I'm looking at concentration of B so that this will be zero. This will be the initial concentration of B to begin with. At different time increments, I will measure the different concentrations of B zero. Now I will have to look at L and B versus one over B. And so you have to do two experiments. And so now if you turn to experiment B, now I want to look at this. This is B versus time. Is that a straight line? No. This is one over B versus time. Is that a straight line? No. If I take L and B versus time, it gives me a straight line with a negative slope. So what is the order with respect to B? First order. All right? So now from this plot, and now this slope that I get, so once again it just so happens that you have L and B versus time gives you a straight line. And so these points would be what you would figure out. And now you can see this slope will give me k double prime. All right? And we know that that is first order as well. But I'm not interested in the pseudo rate constant. I'm not interested in k prime. Neither am I interested in k double prime. What do I need to figure out? k. So now that I've figured out its first order, I can go pick any one of these slopes. And I know that k prime equals the real rate constant times B, original concentration of B raised to the power one. Did you guys get that? I know this number I have got from the plot. This is the original concentration which is one molar. I've already figured out its first order. All right? So I can rearrange that and find out what the real rate constant for that reaction. k equals k prime over the initial concentration over the original, raised to the power one because it's first order. Does that make sense everybody? You can use the same equation for k double prime and you should end up with the same answer. It's one rate constant. You see that? You can use a k prime, the pseudo rate constant for experiment A. Or you can use the pseudo rate constant for B. So just hold on. This is on the class website, all right, under your worksheets. All right? And you can see that these have two different slopes. But you will end up with the same rate constant. Okay? We'll stop there for today.
 Okay, let's go ahead and start.  If I could have everyone's attention please.  I want to make one quick announcement and that is I wanted to remind everybody that  there are bonus points for completing the surveys.  So if you guys remember at the beginning of the quarter I said that there was one during  the second week of class and I hope that you completed the survey for the ninth week of  class and the last part that you need to complete to earn those bonus points is to complete  the teaching evaluations as well.  Okay, so you got to complete three surveys.  Alright, if you complete all three then you will get the bonus points.  So please take the time to complete it and I would appreciate it if you would comment  on the online homework and if you've used sapling before and you're using Wiley Plus  or vice versa, if you could comment on both that would be great.  Alright, so if you guys remember last time we started looking at what we called the integrated  rate law.  Okay, as we will see later on it's important for us to figure out the rate law that corresponds  to a reaction.  We will see later on that we can use that to obtain information about the pathway.  Alright, but before we start looking at pathways we want to understand what it means when we  say the rate law of a reaction.  And we said the only way you can establish the rate law is to carry out the experiment  and either look at rate versus concentration or look at concentration versus time.  Alright, and we said that even though we looked at how we can use the rate of a reaction and  the concentration of the reactant to figure out the rate law or the order of the reaction.  We said unfortunately actually executing that in the laboratory becomes hard because  measuring rate accurately or precisely sometimes is difficult.  And so we have a different approach where we look at concentration versus time and we  derive a mathematical equation that describes the concentration versus time dependence.  And last time we derived the expression for the first order reaction.  So if you have just one reactant and for a first order reaction we said the general form  of the reaction turns out to be lnA minus lnA0 equals minus kT.  And so if we rearrange that equation into a form that gives us y equals mx plus c we  can see that we end up with a straight line with a negative slope.  So if you plot the natural log of the concentration of your reactant as a function of time and  you have a straight line with a negative slope it means that that reactant is first order.  So what you would do is, so this is much easier to do in the lab because essentially what  you do is at different time intervals you measure the concentrations of A. Now remember  the concentration of A is the amount unreacted.  So as the reaction progresses the concentration of A is the amount that's left unreacted.  And so different time intervals you can measure the concentration of A. At the beginning when  T equals 0 it will be A0 and then you just measure the concentration at different times.  And now you take the natural log of A and if you make a graph and if the graph comes  out to be a straight line with a negative slope then you know it's first order.  If you don't end up with a straight line graph with a negative slope it means it's not first  order.  If your point is scattered or if it looks something different then you know that it's  not first order.  Now having said that now we're going to define a new term that we're going to look at and  this term is called half life.  So the half life is the time it takes for the concentration of a reactant A to fall to half  its original value.  So if I were just to plot instead of plotting natural log if I were to just plot the concentration  of A as a function of time.  And so if this was the initial concentration then we know that it decays like that.  So instead of plotting natural log A versus time if you just plot concentration versus  time this is what it would look like, the plot.  Now half life is the time it takes for the original concentration to be halved.  So it's exactly half of this and it's exactly half of that this would be half of the original  concentration and we call that the time it takes is called t half, t subscript half to  indicate the half life.  So the half life is the time it takes for the original concentration to be halved.  Now if I take this concentration which is half the amount and look at the time it takes  for it to be halved that would be, now the concentration would be a quarter of what you  started with and this would be two t halfs where this is one t half and then this is  the other t half and so t half plus t half gives you two t halfs.  Now if you want to take the quarter amount and the time it takes for the quarter amount  to become one eighth would be that and that is three half lives or three t halves and  that becomes A0 over 8 and so this would be the third.  The time will be the same, okay my plot looks like this is a little bit longer but in reality  if it is two days for the first half life it will take another two days for the second  half life so the second half life would be four days and then the third half life would  be six days, do you understand that?  So it's the time it takes for the original concentration to be halved, alright and we  call that half life.  So if you want to derive an equation for half life we can say at half life we know that  the concentration of A is half the original concentration and now t becomes t half because  we call that the time it takes for that is half life and we call that t half.  So if we go back to the original equation that we derived for the first order reaction  we know ln A over A0 equals minus kt and we know at half life this becomes ln A is  now equal to this and that will be over A0 equals minus kt half where t is now half  life so essentially what I'm doing is I'm replacing this term with that and this will  become the half life, okay?  You can see that the original concentrations will cancel out therefore ln one half equals  minus kt half.  Now since there's a negative sign here and to get rid of that negative sign I'm going  to take the inverse of that so if I take ln half, mathematical is the same as negative  ln two equals kt half, alright?  And so now I'm going to rearrange that and get a term for t half which is the half life  and so if I rearrange that equation I can say t half, oh there's a negative sign there  as well so the two negative signs will get rid of each other, t half equals ln two over  the rate constant k, ln two is 0.6931 divided by k so the half life is 0.6931 divided by  the rate constant.  Now that's worth remembering so if you want to calculate, if it's a first order reaction  and it has one reactant and you want to calculate what is the half life and remember what is  the definition of half life, it's the time it takes for the original concentration that  you started with to be halved, okay?  So let's apply this in a problem so if you turn to this problem, okay so if I may turn  to that, so here is an example of applying what we just learned in a problem.  What is the rate constant k for the first order decomposition of N2O5 at 25 degrees  Celsius if the half life of N2O5 at that temperature is given, okay?  So what they tell us is that it is first order so it's important to check that, what is the  order?  Remember so far we derived an expression only for first order so this is a first order  reaction and we know that the half life for this is t half equals 4.03 times 10 to the  4 seconds, they want to know what is the rate constant for that reaction, alright?  So I know that the only relationship between these two that we've looked at thus far is  that t half equals 0.6931 divided by k.  I know what this is, this is a number, this is what I need to find so you can see k equals  0.6931 over t half and therefore k equals 0.6931 divided by the half life which is 4.03 times  10 to the power 4 seconds and so if you put that in your calculator and work that out  it comes out to be 1.72 times 10 to the negative 5 and you can see that the unit is per second.  Remember k is rate constant and the unit for rate constant varies depending on what the  order of that reaction is so this is first order and you can see that's the unit for  the rate constant.  So that's pretty straightforward.  Now having calculated that, now for the second part they tell us under these same conditions  what percent of N2O5 molecules will not have reacted after one day, alright?  So what they're asking is when t is one day they want to know what is the percent unreacted.  So we want to figure out percent unreacted.  So if you want to calculate the percent unreacted now we have to make use of our integrated  rate law and we know the integrated rate law is lnA divided by A0 equals minus kT.  Now the first part we calculated k, time is one day.  They want us to calculate percent unreacted.  Can everybody see this would give me the ratio of unreacted.  Remember A is always the amount unreacted and since that's the amount unreacted if I want  to calculate percent remember what does percent mean?  The amount unreacted divided by what you started with of the total which is the original concentration  times what?  It's a percent times 100.  Do you see that?  So if I just find the ratio and multiply it by 100 that will give me the percent unreacted.  Alright?  So what we're going to do is we're going to figure out what this ratio is.  So we can say lnA over A0 which is the original concentration or in this case I should put  the actual numbers which is lnN2O5 and the original concentration of N2O5 equals minus  now I know my k is 1.72 times 10 to the negative 5 per second and times time.  Now remember in kinetics what is the unit that we use for time?  Seconds.  So they give us one day so that means I need to convert that to seconds.  So you know that there are 24 hours in a day and 60 minutes in an hour and 60 seconds in  a minute.  Alright?  So hours and hours will cancel out, minutes and minutes will cancel out.  This is per second and second will cancel out so it becomes a unitless quantity.  And so I know that lnN2O5 over the original concentration of N2O5 equals if you take,  figure that term out it comes to 1.49.  Okay?  Three sig figs, 24 hours is an exact number, 60 minutes is an exact number, 60 seconds  is an exact number so we end up with a number that is three significant figures.  Alright?  Now what I'm going to do is take the inverse log to get rid of this log.  So if I take the inverse log on this side I need to take the inverse log on that side  so what I end up with is N2O5 divided by the original concentration of N2O5 and if I take  the inverse log of that that comes out to 0.23.  Now remember I'm taking the inverse log so if I'm keeping track of sig figs I have a  three sig fig number when I go from here to here I only count the decimals so it goes  to two significant figures.  Alright?  And then now that I have figured this out now percent unreacted would be N2O5 divided  by N2O5 the original concentration times 100 which is 23 percent.  Alright so all I've done is just taken the value that I calculated and multiplied by  100 and that gives me a percent so that means in one day 23 percent is left unreacted.  Now when you're solving a problem you have to read the question carefully because if  they asked you for percent reacted the percent that actually reacted is 77 percent.  So when you're carrying out online homework sometimes it will ask you to calculate the  percent unreacted sometimes it will ask you to calculate the percent reacted and so if  23 percent is unreacted you know that 77 percent has reacted.  Is that clear to everybody?  So that was a look at a first order reaction.  Now we need to do the same thing that we've done for the first order and we want to look  at what happens with a second order reaction.  So we're going to derive an expression for the second order reaction and so if you have  a case where you have just one reactant and if it is second order we know that the general  form of the rate law would be rate equals rate constant times the concentration of A  now it is second order it is raised to the power 2.  I hope you took time to review the stuff that we covered last Friday because this is going  to build on what we've looked at.  So all of you know this is the general form of the rate law we know it's second order  and we know that rate equals K times A to the power 2.  But since we want to derive an integrated rate law we're going to replace this rate  law the rate term with change in concentration incremental change in concentration of A as  a function of incremental change in time.  And we put a negative sign in front because it's a reactant and a reactant is consumed  in that reaction.  So this equals K A squared.  Now I'm going to put all the concentration terms on one side and take everything else  to the other side.  And so if I keep the negative sign here I can say this over A squared equals K times  DT.  Alright?  Now if I integrate this over the range of initial concentrations the limits are initial  concentration to any concentration A and over here that would be at time zero to any  time T.  And if you integrate that what you end up with is one over A minus one over the initial  concentration equals KT.  So if it's a second order reaction and the integrated rate law or the dependence of concentration  on time comes out to be in this form.  So you need to know this equation I won't give you this equation so you need to know  whether it's a first order what the equation should be.  You need to know what the equation would be for a second order reaction.  Now if I want to rearrange this so that I get it in the form of Y equals MX plus C in  other words to figure out what form of this equation will give me a straight line graph  I can say now that one over A equals KT plus one over A zero so I'm just going to take  this term over there.  So now this will turn out to be the Y axis.  Now you can see that K is the slope time is the X axis plus C gives me the intercept.  And if you look at K you can see that it's positive and so this gives me a positive slope.  So the general form of this react if I take this and if I plot a graph of one over A versus  time this is the Y axis time is the X axis you can see I end up with a straight line  graph.  So if I were to plot this I would end up with a slope and the slope will give me the rate  constant and everybody can see that this is a slope that has a positive slope and the  intercept will be one over the initial concentration.  So what this means is that if I were to carry out an experiment in the laboratory and so  if I were to look at different time intervals I have different concentrations of A and so  when at time zero this will be the initial concentration and then let's say at 50 second  increments if I measure the concentration of A. Now in your Excel spreadsheet now you  take the inverse of A and you can put those values.  And if I plot one over A versus time and if I end up with a straight line graph with a  positive slope like that then I know that that would be a second order.  So you're looking at the variation of concentration versus time and so you're making a plot and  if you make a plot of one over A versus time and you end up with a straight line with a  positive slope that means it's second order.  If you don't get a straight line but if you take these values and take LNA versus time  and it gives you a straight line what does that mean?  First order.  Do you see that?  So you look at the shape of the form of the graph and you can figure out which graph gives  you a straight line and that will tell you whether it's a first order reaction or a second  order reaction.  Now we can do the same thing that we did for the first order and we can start looking at  half life.  So if I take half life for a second order reaction we know at half life we know that  the concentration of A is half of the original concentration we started with and we know the  time is called t half.  And so if I go to 1 over A minus 1 over A0 equals kt and we know if I take this equation  that I started with at half life now this A I'm going to replace this A with that term  so what I'll have is 2 over A0 minus 1 over A0 equals kt half.  So all I've done is taken the equation that we derived and replaced it with these terms  because we're looking at half life and you can see that if I subtract this from that  what I end up with is 1 over A0 equals kt half and therefore kt half sorry let's we  want t half so t half equals 1 over k times the original concentration.  So the equation that you would use if you wanted to calculate the half life for a second  order reaction is 1 over k times the original.  So these are just derivations so what I need you to do is at home go through the derivations  and make sure you know the appropriate equation for first order the appropriate equation for  second order.  Now let's work a problem where we apply this.  So if we turn to the next problem.  Now we're looking at the application of what we just learned and so the decomposition of  C286 to methyl radicals is second order so that's your key.  You're told that it's a second order reaction.  If the second order in the reactant and second order overall there's only one reactant so  you know that it will be overall second order as well.  The rate constant is given if the initial concentration is this what is the concentration  after 205 seconds.  So let's just write down everything that's given to us.  We know that the rate constant is 0.0448 liters per mole per second.  We know that the initial concentration of C286 is 0.0100 moles per liter.  We know that the time is 205 seconds and they want us to calculate what is the amount unreacted  at that point.  What is the concentration after 205 seconds?  So what is the amount of the reactant that's left over after 205 seconds?  So what I wanted to keep in mind is there are only two equations that we've learned  for second order reactions.  One is for half life.  They don't mention half life at all.  So we know that if I want to figure out this the equation that I'm going to use is 1 over  C286 minus the initial concentration equals kT.  Can everybody see this is what I need to find?  I know this.  They've given us the original concentration.  They've given us what k is.  They've given us what time is.  So this is just knowing the appropriate equation and applying it.  So I'm going to put the unknown on one side and take everything else to the other side.  So this would be 1 over C286 equals kT plus the original concentration.  K is 0.0448 liters per mole per second times the time is 205 seconds.  So this and this will cancel out plus 1 over 0.0100 moles per liter.  Can you see that this moles and that moles would cancel out and this liter and that  liter would cancel out because this is the inverse of the inverse.  So sorry, they do not cancel out.  This will remain moles per liter per moles and this becomes liter per mole as well.  So let me just put this number.  If you take this and multiply this by this, it's a three-sic thick number and that number  comes out to be 9.18 liters per mole plus this is the inverse of liters, moles per liter.  So if you take that, this becomes 100 liters per mole as well.  The reason the units don't cancel out is because we're not multiplying, we're adding them.  And you can only add apples to apples.  And you see that you have the unit is liters per mole, the unit for the other is the inverse  of moles per liter and therefore it's liters per mole and therefore you can add apples  to apples.  Does that make sense to everybody?  And so you know that you're on the right track because if you end up with two different  units, you know you can add the two numbers together.  So now that we know that we have the same units and you can add apples to apples, now  I know that 1 over C286 equals that and if I add these two together, now keep in track  of sig figs.  This is a three-sic fig number, this is a three-sic fig number.  Now I'm adding these two numbers together.  When you're adding numbers, what do you keep track of in terms of significant figures?  Decimals.  So can you see that that 100 has zero decimals?  So when I add these two together, I'm going to give it to the least number of decimals  which is 109 liters per mole.  I'm going to take the inverse because I want to calculate the concentration.  So if I take the inverse of that, this comes out to be 9.17 times 10 to the negative 3.  Now the inverse of liters per mole becomes moles per liter and I know that I end up with  the right unit and therefore you know that you haven't made any careless mistakes along  the way because I'm calculating concentration and I end up with the appropriate unit.  So in all of these problems, I want to remind you that keeping the unit there is helpful  because it's sort of checking to make sure that you're doing the problem correctly because  if you end up with the right unit, then you know that mathematically you've set up your  equations correctly.  So if you end up with the right unit, you know for sure that you haven't made any careless  mistakes along the way.  So now we've looked at first order, we've looked at second order, now we're going to  look at zero order.  These are the common ones.  So if we take a zero order reaction, then we're going to do the same thing that we've  been doing all along.  So if I take a zero order reaction, can everybody see that rate equals k a raised to power 1?  If it's a zero order reaction, it should be raised to the power what?  Zero.  Anything raised to the power zero is what?  One.  So that means rate equals k.  Now I need to derive the integrated rate loss, I'm going to look at small incremental changes  of concentrations of a as a function of time giving me k.  Now I'm going to rearrange these equations so that I put all the concentration terms  on one side and everything else on the other.  So this would be concentration of a.  On this side I will put negative k changing time, incremental changes in time.  So if I integrate this over the range of the initial concentration to any concentration  a, and that corresponds to t equals zero to any time t.  And if you integrate that, the outcome comes out to be concentration of a minus the initial  concentration equals minus k t.  So if it is a zero order reaction, then the equation that describes the dependence of  concentration versus time is given by this equation.  So this is called the integrated rate law.  And if I want to look at the form of a straight line graph, I can rearrange this and say  concentration a equals minus k t plus a zero.  This would be the y axis.  Now can everybody see this is m, but this is a negative slope.  So I'm going to put the negative sign in front of this to remind me that it has a negative  slope.  And this is y equals mx plus c.  So this would be k t like that.  So now if you look at this, this is a graph that has a negative slope.  So if I plot this, my y axis will be concentration of a.  My, I'm plotting it versus time.  This will give me a straight line graph like that, where the intercept is the initial concentration.  And the slope of this graph will give me the rate constant k and this is a negative slope.  So the rate constant, when you take the slope, the rate constant has no sign.  Remember we talked about the fact that rates don't have sign and neither do rate constants.  But I want you to pay attention to, if you end up with a negative slope, then you know  it's a zero order reaction.  And so if it's a zero order reaction, if you take different times, different concentrations  of a, when it's zero, this will be the initial concentration.  Now a different time interval.  As I said, 50 second increments, you measure the concentration of a that's unreacted.  And now if I plot a versus time, if I take this data and plot it, and now if I end up  with a graph that gives me a straight line with a negative slope, it tells me that it  is zero order.  And so what we generally do in the lab is, now that we've looked at all three, the way  you figure out the order of the reaction is that once you've got this, so essentially  in the lab what you do is, like I'm just going to repeat that again.  So you have time, and so zero.  You measure the concentration of a that's left unreacted.  At time zero, that will be the initial concentration.  And then at 50 second increments, for example, you can measure the corresponding concentrations.  Now you'll make a third column where you'll figure out the natural log a.  And then you have another column where you'll have one over a.  All right?  And what you would do is that you would make three graphs.  All right?  You make a graph of time t versus a, or a versus t.  This is the y-axis.  L and a versus t.  And one over a versus t.  And so it's like this.  So let's say you're carrying this out in the laboratory.  And essentially what you do is once you complete the experiment, you draw three graphs.  All right?  And so you make three graphs, and only one of these will give you a straight line.  So if you look at this graph, this is one over a.  Remember, one over a versus time is what?  Second order.  Does that give you a straight line?  No.  And if one over a versus time, it has to be a straight line with a positive slope.  We don't get a straight line with a positive.  So we know we can rule out second order.  Now look at the second plot.  It's L and a versus time.  L and a versus time gives you a straight line with a negative slope.  And you can actually measure the slope, which is negative 1.3 times 10 to the negative 3.  All right?  So because that gives you a straight line, what does that tell you about this reaction?  That is first order.  And I can actually, from the slope, I know the rate constant is, you toss out the negative  sign, the rate constant is 1.3 times 10 to negative 3.  Got it?  And then if you take a versus, I don't get a straight line.  So it only has to be 1 of the three.  Do you guys get it?  And so you make three graphs, and from that, you can figure out what the order of the reaction  is.  So that's what, I hope you carried out this in the laboratory, and you can figure out  the order of the reaction based on drawing the three plots.  Yes?  This is only for one reactant.  This is only one reactant.  All right?  And it can only be one of the three.  Do you understand that?  So you're only going to end up with one straight line graph.  All right?  And so that would be the one that tells you the order of the reaction.  OK?  Now I also want to derive the half-life for a zero-order reaction.  So for a zero-order reaction, if I want to derive the half-life, we know that at half-life,  we know that the concentration of A equals half the original concentration.  We know that time now becomes half-life.  OK?  So we go to the equation for zero-order, which is 1 over A minus 1 over A zero equals negative  kT.  We know at half-life, we're going to replace all of this.  Now this becomes 2 over A zero.  Wait a minute.  Sorry.  I'm doing the second order.  Please bear with me.  I'm thinking about something else.  So we know that at half-life, concentration of A is A zero divided by two, and we know  T equals T half.  And we know for a zero-order reaction, we know this is concentration of A minus A zero equals  negative kT.  That's the one for zero order.  Previously I wrote it for the second order.  OK?  And so if I take this, and now I'm going to say at half-life, I'm going to replace  the terms.  I know that A equals A zero divided by two minus A zero equals minus kT half.  OK?  So now we're looking at half-life.  And you can see that this is one.  This is half.  If I subtract one from the other, what I'll end up with is negative A zero divided by  two equals negative kT half.  So the negative signs will cancel out on either side.  And so T half equals the initial concentration divided by two k.  All right?  So the integrated rate law, because you have to derive this mathematically, we only look  at the three common orders.  The common ones are zero, first order, and second order.  All right?  And so we derive the rate laws for the three.  Now on the class website, when you're studying at home, I want you to kind of summarize everything  this way.  So there's a copy of this on the class website.  And so what we're looking at is we looked at zero, first, second order.  All right?  The rate law would be for zero order, rate equals k.  For first order, rate equals kA raised to the power one.  Rate equals kA raised to the power two for second order.  Okay?  Now the form of the integrated rate law that we derived is given here.  So A equals minus kT plus A zero.  L and A equals minus kT plus L and A zero.  Or one over A equals kT plus one over A zero.  So this is the form of the integrated rate law that we derived.  Okay?  Now the plot needed to get a straight line.  Because remember, we're going to determine whether reaction is first order, second order,  zero order, based on which plot gives you a straight line.  So for zero order, A versus T will give you a straight line with a negative slope.  Okay?  So the slope will be negative k.  Where for a first order, L and A versus T will give you a straight line graph with a  negative slope.  And one over A versus T will give you a positive slope for a second order reaction.  So that's why we draw three graphs and figure out which one of these will give us a straight  line.  Okay?  Lastly, the half-life equations that we derived is given here.  So half-life for zero order is this, half-life for this is this, and half-life for a second  order is that.  Okay?  You need to remember these equations.  All right?  And if you worked enough examples like the homework as well as the online homework will  give you enough practice so that now you should be able to know which equation to use to figure  out the half-life for zero order, half-life for second order, and half-life for first  order.  Is that clear?  So what you're doing with, keep this handy with you.  Just paste it on the wall.  And as you're carrying out homework, just go to that table and you can look at the  appropriate equations.  And after a while, you'll be so comfortable with it that you can no longer leave that  table.  All right?  Now, the problem is that up to this point, what we've looked at is only one reactant.  All right?  And if it's only one reactant, then it's pretty straightforward.  How do you figure out the order when you have more than one reactant?  All right?  So if you turn to the next page, we have to come up with something more clever to figure  out the integrated rate laws if you have more than one reactant.  All right?  Now if you guys remember previously when we looked at rate versus concentration, remember  we said we designed experiment so that we can cancel out one reactant.  Remember when we worked at problems last Friday, we said if you want to figure out what the  order of the reaction is, we actually designed the experiment so that we keep the concentrations,  you know, we measure rate at different concentrations.  But if you have more than one reactant, you always keep the concentrations of one of those  reactant at the same sort that they cancel out and change only the other.  All right?  And likewise, if you want to use the integrated rate law, we have to design our experiments  in a special way.  All right?  So so far we have considered the integrated rate laws for simple reactions with only one  reactant.  It turns out special approaches are required to deal with more complicated reactions.  All right?  So let's say we are looking at this.  Okay?  So if I want to write the rate law, so we're looking at now two reactants, and if I want  to figure this out, the rate law, as all of you can see, rate equals k, concentration  A, let's say raised to the power m times concentration of B raised to the power n.  Okay?  So now we have to design this in a special way so that we can look at the rate law of  one at a time.  The way we do that is we set up the concentrations so that one concentration is really small.  This is of the order of 10 to the negative 3.  The other is one molar.  So can you see that the difference between the two reactants is about thousandfold?  All right?  So we specially designed the experiment so that the concentration of one reactant is  kept really small, like the order of 10 to the negative 3, and the other is of the order  of thousand times more, like one molar.  And the reason is that when the reaction takes place, the smaller concentration is going  to be the limiting reactant.  All right?  And so what's going to happen is the smaller one is only going to react, and relative to  the smaller concentration, the bigger concentration doesn't appear to change.  Because remember, the limiting reactant is a thousand times more dilute.  So the reaction will take place at the scale of 10 to the negative 3, and the one molar,  which is a thousand times more concentrated, stays relatively unchanged for the scale of  the reaction.  Can you ever see that?  And so what happens is that we set it up so that we take A is the one with the smaller  concentration, B is the one with the larger concentration.  So B concentration is much, much greater than A. Now what happens is because this is much,  much greater, it appears relatively unchanged.  And so what happens is as the reaction progresses, the concentration of B does not change all  that much.  So we're going to approximate it and say that as the reaction progresses, the concentration  of B will always be the original concentration, which is a constant.  All right?  So if you go to the rate law where we say rate equals K, A raised to the power m, B raised  to the power n, we can say now we can replace the concentration of B with the original concentration  of B. And can everybody see that this term does not change?  So this is a constant.  All right?  So because this is a constant, I'm going to include this with the constant.  And I come up with K' which I call, this is called the pseudo rate constant.  It's not the real rate constant, but because this includes the pseudo rate constant, which  is K' is actually the real rate constant, which is this K times this constant, which  is the concentration of B. Do you guys see that?  So essentially what I'm doing is by keeping one concentration really large, I'm making  it look like you have only one reactant.  All right?  Because one concentration does not change.  And now what I would do is I would carry out this experiment and I would take at different  time intervals, I would look at concentration of A. B does not change.  It's a constant.  So at T equals zero, this will be A zero, and I'll measure the concentrations.  All right?  Now I will take L and A, and I'll also plot one over A. And in this way, I will see whether  I end up, which of these will give me a straight line graph.  So if this is experiment A over here, this is experiment A, and you can see now which  line gives you a straight line, L and A. If it's L and A versus time in straight line,  what does that tell me?  It's first order in A. All right?  And the slope that I calculate, this is where you want to watch out.  The slope, so let's say that if I plot this out, and I end up with L and A versus time  will give me a negative slope.  Okay?  So these are the points on that graph.  Now the slope gives me K prime, which is the pseudo rate constant.  All right?  Because we've reduced this to look as if you have only one reactant.  All right?  The rate constant is the pseudo rate constant, which is K prime.  Got it?  So now I have to repeat this experiment and do this all over again, but now I have to  keep the A concentration one molar, and I'm going to reverse the thing.  Now I'm going to make B the one that is smaller in concentration.  All right?  So now you can see the A concentration is far greater than B, and therefore now the one  that will be constant will be this.  So now you can see that this term is constant.  All right?  So now this equation reduces to rate equals K double prime.  This is a different one.  All right?  K double prime times B raised to the power n.  And so now once again I have to repeat this experiment where I have T, and then now I'm  looking at concentration of B so that this will be zero.  This will be the initial concentration of B to begin with.  At different time increments, I will measure the different concentrations of B zero.  Now I will have to look at L and B versus one over B.  And so you have to do two experiments.  And so now if you turn to experiment B, now I want to look at this.  This is B versus time.  Is that a straight line?  No.  This is one over B versus time.  Is that a straight line?  No.  If I take L and B versus time, it gives me a straight line with a negative slope.  So what is the order with respect to B?  First order.  All right?  So now from this plot, and now this slope that I get, so once again it just so happens  that you have L and B versus time gives you a straight line.  And so these points would be what you would figure out.  And now you can see this slope will give me k double prime.  All right?  And we know that that is first order as well.  But I'm not interested in the pseudo rate constant.  I'm not interested in k prime.  Neither am I interested in k double prime.  What do I need to figure out?  k.  So now that I've figured out its first order, I can go pick any one of these slopes.  And I know that k prime equals the real rate constant times B, original concentration of  B raised to the power one.  Did you guys get that?  I know this number I have got from the plot.  This is the original concentration which is one molar.  I've already figured out its first order.  All right?  So I can rearrange that and find out what the real rate constant for that reaction.  k equals k prime over the initial concentration over the original, raised to the power one  because it's first order.  Does that make sense everybody?  You can use the same equation for k double prime and you should end up with the same  answer.  It's one rate constant.  You see that?  You can use a k prime, the pseudo rate constant for experiment A. Or you can use the pseudo  rate constant for B. So just hold on.  This is on the class website, all right, under your worksheets.  All right?  And you can see that these have two different slopes.  But you will end up with the same rate constant.  Okay?  We'll stop there for today.
General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__0_299.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__300_347.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__348_395.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__396_575.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__576_623.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__624_671.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__672_719.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__720_851.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__852_935.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__936_1043.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__1044_1091.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__1092_1175.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__1176_1367.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__1368_1415.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__1416_1511.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__1512_1655.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__1656_1727.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__1728_1787.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__1788_1835.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__1836_1883.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__1884_1943.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__1944_2003.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__2004_2147.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__2148_2195.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__2196_2243.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__2244_2303.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__2304_2435.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__2436_2483.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__2484_2543.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__2544_2591.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__2592_2711.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__2712_2759.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__2760_2807.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__2808_2855.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__2856_2903.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__2904_2951.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__2952_2999.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__3000_3299.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__3300_3347.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__3348_3407.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__3408_3527.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__3528_3623.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__3624_3839.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__3840_3887.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__3888_3935.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__3936_3983.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__3984_4031.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__4032_4079.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__4080_6887.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__6888_25631.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__25632_54227.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__54228_66047.png General_Chemistry_1C__Lecture_24__Chemical_Kinetics_Pt__3__66048_79963.png