Lecture 26. Chemical Kinetics Pt. 5.
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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 please? So we're down to the last stretch and so I'm just going to finish up the last topic in kinetics that we will cover and then I will work, you know, finish up that worksheet that we started looking at last week and then I'd be happy to answer any questions, alright, and I'll talk a little bit about the final exam as well. Okay, so the last topic that we're going to look at is, remember we're looking at kinetics, last time we talked about, you know, the rate law and how we used the experimentally measured rate law to figure out the reaction mechanism, alright. Today we're going to look at the factors that affect the rate of a reaction. So what affects how fast a reaction takes place? And there are many factors that affect the rate of a reaction and in the answer I want to kind of tell you that all of these factors are related to each other, okay. So let's start, I've listed them out here and let's start with the first one and that is that the temperature affects how fast the reaction takes place. So chemical reactions speed up when the temperature is raised, okay, and the general rule of thumb is that for every 10 degree increment in temperature the rate actually doubles, alright, approximately it doubles, so that's a good rule of thumb to keep in mind. The reason that the reaction speeds up when you raise the temperature is because remember when we looked at reaction mechanisms we said for a reaction to take place molecules must collide, alright. That's what a bimolecular reaction means, that's what a termolecular reaction means, in a reaction mechanism each step results from molecules colliding with each other. So what happens is when you raise the temperature you raise the thermal energy and the thermal energy increases how frequently the molecules collide with each other because now the molecules have greater kinetic energy, they're moving around, bouncing around much faster and the chances that they collide with each other increases as you raise the temperature and so the collisions increase and as the molecules collide and have the right kinetic energy the reaction speeds up. Okay, so that the first factor that affects the rate of a reaction is the temperature, alright, and if you want to make a plot of how temperature affects rate is that if you plot the rate as a function of time kind of increases like that, okay. Now if we move on to the second property or second effect that is the frequency of collisions, okay. So remember we said that the rate depends on molecules colliding and so the more often the molecules collide with each other the faster the reaction would be. So the rate depends on the frequency of the collisions and the more frequent the collisions are then the faster the reaction. So the first criterion for a reaction to take place is that molecules must collide, alright. If they collide then you'll have a reaction taking place and as the frequency of the collision increases the rate of that reaction increases, alright. So for reactions to occur molecules must collide as concentrations of reactants are increased or if the temperature is increased the frequency of the collision increases. So one way to increase the frequency of collision is to increase the number of molecules that are there. If you have more molecules it's going to get more crowded and therefore you're going to have more collisions and therefore the reaction is going to be faster and that's why reaction rate depends on concentration because as the reactants get more concentrated you're going to have more molecules in there. The more crowded the molecules are the chances that they'll collide with each other increases, alright. So the frequency of collision increases with concentration. Also if you provide thermal energy they're going to move a lot faster and if they're moving a lot faster they can collide even faster. Remember all of these are interrelated with each other. The third factor that influences the rate is the kinetic energy of the collisions. So we look to the fact that if you want the reaction to take place the molecules must collide and the more frequent you have the collisions the faster the reaction. But it turns out just having collisions is not enough, alright. When you have molecules colliding they also have to have just the right amount of energy, alright. And why do they need the right amount of energy or the minimum amount of energy? The reason is that when they collide chemical bonds must be broken and the new bonds are being formed. And as all of you know to break a bond does it cost energy or does it release energy? Breaking bonds always cost energy. Bond formation when a new bond is formed energy is released, alright. So if you want to break bonds then can everybody see that you have a minimum amount of kinetic energy that's required to break those bonds. So let's take an example. If we take this reaction where we're looking at 2 NOBr giving you 2 NO plus BR2. If I draw the lowest structure can you see that the central atom will be nitrogen and this is the lowest structure of your reactant. The two molecules of these collide with each other. You end up with the products which is 2 NO. So this is what the lowest structure of your reactants and products are. So you know as you can see is an odd electron molecule so it's got an odd number of electrons and so you see there's one odd electron. And if you look at this you can see that in this reaction bonds are being broken. So which bond is being broken? Can you see that this bond is broken and this bond is broken. So these are broken and this bond is formed. So in this reaction you can see chemical bonds are broken and new bonds are being formed. So the act of the collision when the two molecules collide they must have sufficient energy. The kinetic energy. Remember the molecules are moving around and the energy that the molecules have by virtue of their motion is the kinetic energy. The kinetic energy of the molecules that are moving when they collide has to be sufficient to overcome the potential energy required to break these bonds. Can you ever see that? And so it's something like this. You can think about it as if I'm going to use an analogy and the same you know if you want to look at the minimum amount of energy that you need it's like if you wanted to hit a golf ball over a hill. So say you have a golf ball right here. And let's say you hit this golf ball. Can everybody see that if I hit this ball I have to hit it with sufficient energy to get it to the top. That's the tipping point. If it is sufficient if I hit the ball with just the right amount of energy to exceed the tipping point then the ball will roll over onto the other side. If I don't hit it with sufficient energy let's say the ball ends up here you know that's going to roll back here. Likewise when two molecules collide you have to have sufficient energy to get beyond that tipping point. The tipping point is having sufficient energy to break those bonds. If you hit that tipping point then everything else is downhill and therefore the second point is that the molecules when they collide not every collision actually leads to products. The collision has to have sufficient kinetic energy to overcome that energy barrier and that the energy barrier is what? The breaking of the bonds. So we describe this in the form of what we call a reaction profile. And the reaction profile is a plot of potential energy versus progress of reaction. And so over here this is the reactant and so the reactants this is the potential energy of the reactants and in this case the reactants are 2 NOBr. Over here are the products and we have 2 NO plus Br2 as our products. And so when reactants go to products this reaction is an exothermic reaction so this is delta H is negative. As reactants go you start with a high potential energy and you go into a product with a low potential energy and the difference in remember you can't create or destroy energy and so the energy is released as heat and that's delta H. Now it turns out that this tipping point is shown here. There's a certain amount of minimum amount of energy that needs to be provided. This is the little hill in that golf ball example. This is the minimum amount of energy. This is the tipping point. The minimum amount of energy that needs to be put in for to break the potential energy associated with the bonds. Okay? Break bonds. And we call that energy. So this difference in energy between here and here is called the activation energy. And the activation energy is given the symbol E subscript 8. Alright? So the second factor we said was that you know we said molecules must collide. If you want a reaction to take place molecules must collide and the more often they collide the better. So the higher the frequency of collisions the better. But the second point is it's just not enough for them to collide. When the collision occurs they have to have sufficient energy. Alright? And they have to have sufficient energy because bonds are being broken. And so the kinetic energy of the molecules when they collide the kinetic energy has to be sufficient to overcome the potential energy required to break bonds. Alright? And that that energy is called the activation energy. Alright? So overall the collision has to provide sufficient kinetic energy to get above this hump. Now this highest point here, the highest energy point is called the transition state. Alright? So this plot of potential energy versus the reaction progress is called a reaction profile. And in the reaction profile you always indicate what the activation energy is. And so if you provide the kinetic energy of the collision provides that amount of energy, that's the minimum amount, that's the tipping point. If it has energy greater than that, great. It gets over the hump and then once it gets over the hump now it's all downhill and it gets to the products. And so this is the amount of thermal energy that will be released. So this is the amount of energy that's released but this is the amount of energy that needs to be put in at the beginning to get the process going. You see that? And that's called the activation energy. So that's the second factor. So you can see here that when two molecules collide the kinetic energy is changed to potential energy as the molecules undergo bond breaking and bond formation leading to product molecules. However, not all collisions lead to products. This is because the molecules need a minimum amount of energy to react. This energy requirement is called the activation energy of the reaction and the symbol that we use is E sub A. Unless the kinetic energy of the collision is equal or greater than the activation energy, no reaction will occur. So we're going to go back again. We said if you want a reaction to take place they have to collide. And the more often they collide the better. So the more frequent the collisions the better. So if you make it concentrated, if you increase the temperature you increase the frequency of collisions. Second is, but that's not good enough. Just because they collide doesn't mean it's going to lead to products. They also have to have sufficient kinetic energy. So when you heat the reactants now you increase the thermal energy and therefore the molecules are moving much faster. They're going to have greater kinetic energy. So if you raise the temperature they're going to react faster because you're going to have some, it's not only are you increasing the frequency of collision but also you're increasing the kinetic energy of the molecules and therefore the rate will go up. But the third is, so we said just colliding is not good enough. They have to have sufficient kinetic energy. Second is, even if they collide and even if they have the right amount of kinetic energy that's not good enough. They also have to collide in just the right way. So they have to have what we say the right orientation. So not every collision with the right amount of kinetic energy actually leads to products. The reason is, as I said before, when the molecules hit each other they have to hit each other in just the right way. And what do I mean by just the right way? To do that I'm going to show the space filling model for the same reaction. So let's say I have oxygen is smaller, nitrogen is a little bigger and then bromine is the biggest. So if I take NOBr this is what the space filling model would look like. So this is oxygen, nitrogen, bromine. Now let's say that these molecules hit each other this way. So when they hit each other the point of contact is between oxygen and bromine. So if they hit each other in this orientation, we call this the orientation of the directionality, the way, the orientation in which they hit, if they hit each other this way there is no reaction. So they're hitting each other in the wrong orientation. However, if they hit each other in this way, so this is oxygen, nitrogen is a little bigger, bromine is the biggest one, this is the space filling model. But now, if they hit each other now this way and the point of contact is like this, now remember in this reaction there's a new bond being formed. Can everybody see that? The new bond is being formed between two bromine atoms. So if they hit, so the point of contact is between the two bromines, can you see now this is where the new bond is being formed? So what will happen is this is the new bond to be formed and you can see when that happens, when the new bond is being formed you can see that now this can break up like that. So can everybody see that? Kind of everything is set up so that instant they come into contact you have a new bond being formed and the other two will break off. And so now they're hitting each other in just the right orientation so that you end up with the BR2 molecule being formed and then the two oxygens being formed as well. So it turns out it's much more complicated than just molecules colliding. So what you have to keep in mind is they have to collide the first and the more often they collide the better. Two they have the right amount of kinetic energy so the more kinetic energy, sufficient kinetic energy for the bonds to be broken. And thirdly, even if they have lots of collisions and even if they have the right amount of kinetic energy you won't get the reaction taking place unless the molecules hit in just the right way. And so when they collide they have to collide in the right orientation. So all these factors influence how fast a reaction takes place. Now this orientation is called the steric factor. Steric refers to three dimensional orientation. And so the orientation depends on the three dimensions and how the molecules hit each other in three dimensions. And so we call this the steric factor. The orientation and molecules hit each other. Now so if I ask you what are the factors that influence how fast a reaction takes place then you should know that not only temperature affects it but also the frequency of the collisions affected, the kinetic energy of the collisions affected and also the orientation affects how fast the reaction proceeds and whether a reaction will take place or not. Now Arrhenius came up with a mathematical equation that takes all of these factors into account. So if you want to come up with a mathematical equation that describes the relationship of all of these factors combined then it's given by this equation which is called the Arrhenius equation. One form of the equation is written as k which is the rate constant for the reaction. Now here we're using rate constant instead of rate because a rate of a reaction, remember, depends on concentration. We've seen that the rate varies on concentration. The more concentrated it is the faster the reaction would be. Now rate constant is independent. It's a constant. And so it's a better factor. If you want to compare different reactions it's better to look at rate constant because rate constant is independent of the concentration because it's a concentration. So we're looking at the dependence of rate constant, k, equals what we call a constant A. This is called the Arrhenius factor and this accounts for collision frequency and the steric factor. Remember the steric factor is the orientation. Collision frequency is how often they collide. So the Arrhenius factor A takes into account how often they collide as well as the steric factor and every reaction has a unique value for A. A is a constant and it's available in tables and every unique reaction will have a unique constant that corresponds to A. Ea is the activation energy and the activation energy usually is in units of kilojoules per mole. Arr is the gas constant. Now remember, I want you to keep in mind that in this term the activation energy is divided by the gas constant and the unit for activation energy is kilojoules per mole. That means if you're using this Arr, remember Arr is the gas constant and the Si units, the unit for gas constant is joules per Kelvin per mole. So I want to convert that to kilojoules and to do that all of you recognize that 1 kilojoules is 10 to the 3 joules. So when I use this gas constant in this equation it's important that I divide it by a thousand so that I have the unit kilojoules per mole rather than joules per Kelvin per mole. So we have that and of course T is the Kelvin temperature. So this form of the reaction can also be written like this where lnK equals, now if I take the natural log lnK equals lnA minus Ea over RT and if I rearrange this equation so that I have lnK equals negative Ea over R, 1 over T plus lnA. I'm just putting this in the form of y equals mx plus c. So if I take, the rate constant is the y-axis then this term will be the slope and that will be the x-axis plus c would be the intercept. So if I can measure the rate constant at different temperatures, so if I make a plot of ln, lower case k, okay, not capital K because that's equilibrium constant, this is rate constant. Experimentally we can measure rate constants at different temperatures and if I plot that versus 1 over T and remember this is in units of Kelvin and if I end up with, then I would end up with a graph that has a negative slope where the slope will give me the activation energy divided by the gas constant and keep in mind it's a negative slope. So this is how experimentally we figure out the activation energy for any reaction. If you want to figure out the activation energy for a reaction what you do is you measure the rate constants at different temperatures and you plot it lnK versus 1 over T and you figure out what the slope is and from the slope you can divide, take the slope and multiply it by the gas constant and kilojoules per mole and that will give you the activation energy. You may have done this in the lab when you were looking at the kinetics experiment. Now lastly if you want to look at the dependence of lnK on temperature and let's say we're looking at, so let's take this equation where we say lnK equals, so this is lower case K minus EA over R, 1 over T plus lnA and let's say we're looking at this at 1 temperature so if this is K1 this would be T1, alright? Activation energy does not change, R does not change and lnA remember is a constant that's unique for each reaction. So if I take another reaction where it's lnK2 this would be negative EA over R, 1 over T2 plus lnA. Now since these two are lnAs if I rearrange this can everybody see that if I take this term over here I can say lnK1 plus EA over R, 1 over T1 equals lnK2 plus EA over R, 1 over T2. Can everybody see what I'm doing? Because lnA is a constant, remember A is unique for a particular reaction so as long as we're looking at the same reaction, alright? What we're looking at is the dependence of rate constant on temperature, this term lnA would be this plus that, this lnA would be this plus this if I take it over here and since lnA is the same for both I know these two are equal to each other and therefore if I take lnK1 minus lnK2 now this would become EA over R, 1 over T2 minus 1 over T1 and lnK this would become lnK1 over K2 would equal EA over R, 1 over T2 minus 1 over T1. So if you're, if ever you want to figure out the rate constant at one temperature and you want to figure out the rate constant at another temperature you would use that equation, alright? And that would help you figure it out. Now this same form of the equation can be written like in some textbooks you'll see this written out like this. lnK1 over K2 equals negative EA over R but now it's T1 minus T2. Either one we, okay? So all I'm doing is switching the sign here, flipping this around and putting a negative sign in front both those equations are the same, alright? And some textbooks give the latter equation, some give the former equation, alright? So that kind of completes what we're going to look at and so the last thing that we looked at is the factors that influence how fast a reaction proceeds and we looked at a chemical equation that relates all the factors that influence how fast a reaction takes place. Now within the last few minutes I want to kind of take an example because ultimately it's about applying these in problems. So let's stop there for today and for the one last thing that I just want to talk briefly about the exam, okay? The last minute. And I want to remind everybody that the final exam is on Wednesday. It's at 8 a.m. So your exam time is different from the class time, okay? The exam will be cumulative but greater burden will be on the new material. So almost half the exam will be on kinetics, alright? The other half will include everything that we covered this quarter. So please review the intro to equilibria. You should be able to draw equilibrium problems. You should be able to do equilibrium problems. You should know the relationship between equilibrium and free energy. Remember we said delta G equals minus RTL and K, you know, the relationship between free energy and equilibrium constant, alright? Which goes back to the beginning of equilibria right up to what we've covered to electrochemistry. But I understand that you will not be able to study everything. So for the material that's cumulative other than kinetics, I'm looking at the big picture. I want to see whether you see relationships. So it's more big picture, more conceptual rather than, you know, sort of very detailed problems where you have to go through a lot of detail, okay? So I want to see whether you've mastered just the big picture items, alright? So this is the last quarter. I'm going to see you. So good luck. I wish you all the very best. Alright? Thank you. Thank you. Thank you.
 Okay, can I have everyone's attention please?  So we're down to the last stretch and so I'm just going to finish up the last topic in  kinetics that we will cover and then I will work, you know, finish up that worksheet that  we started looking at last week and then I'd be happy to answer any questions, alright,  and I'll talk a little bit about the final exam as well.  Okay, so the last topic that we're going to look at is, remember we're looking at kinetics,  last time we talked about, you know, the rate law and how we used the experimentally measured  rate law to figure out the reaction mechanism, alright.  Today we're going to look at the factors that affect the rate of a reaction.  So what affects how fast a reaction takes place?  And there are many factors that affect the rate of a reaction and in the answer I want  to kind of tell you that all of these factors are related to each other, okay.  So let's start, I've listed them out here and let's start with the first one and that  is that the temperature affects how fast the reaction takes place.  So chemical reactions speed up when the temperature is raised, okay, and the general rule of thumb  is that for every 10 degree increment in temperature the rate actually doubles, alright, approximately  it doubles, so that's a good rule of thumb to keep in mind.  The reason that the reaction speeds up when you raise the temperature is because remember  when we looked at reaction mechanisms we said for a reaction to take place molecules must  collide, alright.  That's what a bimolecular reaction means, that's what a termolecular reaction means,  in a reaction mechanism each step results from molecules colliding with each other.  So what happens is when you raise the temperature you raise the thermal energy and the thermal  energy increases how frequently the molecules collide with each other because now the molecules  have greater kinetic energy, they're moving around, bouncing around much faster and the  chances that they collide with each other increases as you raise the temperature and  so the collisions increase and as the molecules collide and have the right kinetic energy  the reaction speeds up.  Okay, so that the first factor that affects the rate of a reaction is the temperature,  alright, and if you want to make a plot of how temperature affects rate is that if you  plot the rate as a function of time kind of increases like that, okay.  Now if we move on to the second property or second effect that is the frequency of collisions,  okay.  So remember we said that the rate depends on molecules colliding and so the more often  the molecules collide with each other the faster the reaction would be.  So the rate depends on the frequency of the collisions and the more frequent the collisions  are then the faster the reaction.  So the first criterion for a reaction to take place is that molecules must collide,  alright.  If they collide then you'll have a reaction taking place and as the frequency of the collision  increases the rate of that reaction increases, alright.  So for reactions to occur molecules must collide as concentrations of reactants are increased  or if the temperature is increased the frequency of the collision increases.  So one way to increase the frequency of collision is to increase the number of molecules that  are there.  If you have more molecules it's going to get more crowded and therefore you're going to  have more collisions and therefore the reaction is going to be faster and that's why reaction  rate depends on concentration because as the reactants get more concentrated you're going  to have more molecules in there.  The more crowded the molecules are the chances that they'll collide with each other increases,  alright.  So the frequency of collision increases with concentration.  Also if you provide thermal energy they're going to move a lot faster and if they're  moving a lot faster they can collide even faster.  Remember all of these are interrelated with each other.  The third factor that influences the rate is the kinetic energy of the collisions.  So we look to the fact that if you want the reaction to take place the molecules must  collide and the more frequent you have the collisions the faster the reaction.  But it turns out just having collisions is not enough, alright.  When you have molecules colliding they also have to have just the right amount of energy,  alright.  And why do they need the right amount of energy or the minimum amount of energy?  The reason is that when they collide chemical bonds must be broken and the new bonds are  being formed.  And as all of you know to break a bond does it cost energy or does it release energy?  Breaking bonds always cost energy.  Bond formation when a new bond is formed energy is released, alright.  So if you want to break bonds then can everybody see that you have a minimum amount of kinetic  energy that's required to break those bonds.  So let's take an example.  If we take this reaction where we're looking at 2 NOBr giving you 2 NO plus BR2.  If I draw the lowest structure can you see that the central atom will be nitrogen and  this is the lowest structure of your reactant.  The two molecules of these collide with each other.  You end up with the products which is 2 NO.  So this is what the lowest structure of your reactants and products are.  So you know as you can see is an odd electron molecule so it's got an odd number of electrons  and so you see there's one odd electron.  And if you look at this you can see that in this reaction bonds are being broken.  So which bond is being broken?  Can you see that this bond is broken and this bond is broken.  So these are broken and this bond is formed.  So in this reaction you can see chemical bonds are broken and new bonds are being formed.  So the act of the collision when the two molecules collide they must have sufficient energy.  The kinetic energy.  Remember the molecules are moving around and the energy that the molecules have by virtue  of their motion is the kinetic energy.  The kinetic energy of the molecules that are moving when they collide has to be sufficient  to overcome the potential energy required to break these bonds.  Can you ever see that?  And so it's something like this.  You can think about it as if I'm going to use an analogy and the same you know if you  want to look at the minimum amount of energy that you need it's like if you wanted to hit  a golf ball over a hill.  So say you have a golf ball right here.  And let's say you hit this golf ball.  Can everybody see that if I hit this ball I have to hit it with sufficient energy to  get it to the top.  That's the tipping point.  If it is sufficient if I hit the ball with just the right amount of energy to exceed  the tipping point then the ball will roll over onto the other side.  If I don't hit it with sufficient energy let's say the ball ends up here you know that's  going to roll back here.  Likewise when two molecules collide you have to have sufficient energy to get beyond that  tipping point.  The tipping point is having sufficient energy to break those bonds.  If you hit that tipping point then everything else is downhill and therefore the second  point is that the molecules when they collide not every collision actually leads to products.  The collision has to have sufficient kinetic energy to overcome that energy barrier and  that the energy barrier is what?  The breaking of the bonds.  So we describe this in the form of what we call a reaction profile.  And the reaction profile is a plot of potential energy versus progress of reaction.  And so over here this is the reactant and so the reactants this is the potential energy  of the reactants and in this case the reactants are 2 NOBr.  Over here are the products and we have 2 NO plus Br2 as our products.  And so when reactants go to products this reaction is an exothermic reaction so this  is delta H is negative.  As reactants go you start with a high potential energy and you go into a product with a low  potential energy and the difference in remember you can't create or destroy energy and so  the energy is released as heat and that's delta H.  Now it turns out that this tipping point is shown here.  There's a certain amount of minimum amount of energy that needs to be provided.  This is the little hill in that golf ball example.  This is the minimum amount of energy.  This is the tipping point.  The minimum amount of energy that needs to be put in for to break the potential energy  associated with the bonds.  Okay?  Break bonds.  And we call that energy.  So this difference in energy between here and here is called the activation energy.  And the activation energy is given the symbol E subscript 8.  Alright?  So the second factor we said was that you know we said molecules must collide.  If you want a reaction to take place molecules must collide and the more often they collide  the better.  So the higher the frequency of collisions the better.  But the second point is it's just not enough for them to collide.  When the collision occurs they have to have sufficient energy.  Alright?  And they have to have sufficient energy because bonds are being broken.  And so the kinetic energy of the molecules when they collide the kinetic energy has to  be sufficient to overcome the potential energy required to break bonds.  Alright?  And that that energy is called the activation energy.  Alright?  So overall the collision has to provide sufficient kinetic energy to get above this hump.  Now this highest point here, the highest energy point is called the transition state.  Alright?  So this plot of potential energy versus the reaction progress is called a reaction profile.  And in the reaction profile you always indicate what the activation energy is.  And so if you provide the kinetic energy of the collision provides that amount of energy,  that's the minimum amount, that's the tipping point.  If it has energy greater than that, great.  It gets over the hump and then once it gets over the hump now it's all downhill and it  gets to the products.  And so this is the amount of thermal energy that will be released.  So this is the amount of energy that's released but this is the amount of energy that needs  to be put in at the beginning to get the process going.  You see that?  And that's called the activation energy.  So that's the second factor.  So you can see here that when two molecules collide the kinetic energy is changed to potential  energy as the molecules undergo bond breaking and bond formation leading to product molecules.  However, not all collisions lead to products.  This is because the molecules need a minimum amount of energy to react.  This energy requirement is called the activation energy of the reaction and the symbol that  we use is E sub A. Unless the kinetic energy of the collision is equal or greater than  the activation energy, no reaction will occur.  So we're going to go back again.  We said if you want a reaction to take place they have to collide.  And the more often they collide the better.  So the more frequent the collisions the better.  So if you make it concentrated, if you increase the temperature you increase the frequency  of collisions.  Second is, but that's not good enough.  Just because they collide doesn't mean it's going to lead to products.  They also have to have sufficient kinetic energy.  So when you heat the reactants now you increase the thermal energy and therefore the molecules  are moving much faster.  They're going to have greater kinetic energy.  So if you raise the temperature they're going to react faster because you're going to have  some, it's not only are you increasing the frequency of collision but also you're increasing  the kinetic energy of the molecules and therefore the rate will go up.  But the third is, so we said just colliding is not good enough.  They have to have sufficient kinetic energy.  Second is, even if they collide and even if they have the right amount of kinetic energy  that's not good enough.  They also have to collide in just the right way.  So they have to have what we say the right orientation.  So not every collision with the right amount of kinetic energy actually leads to products.  The reason is, as I said before, when the molecules hit each other they have to hit each other  in just the right way.  And what do I mean by just the right way?  To do that I'm going to show the space filling model for the same reaction.  So let's say I have oxygen is smaller, nitrogen is a little bigger and then bromine is the  biggest.  So if I take NOBr this is what the space filling model would look like.  So this is oxygen, nitrogen, bromine.  Now let's say that these molecules hit each other this way.  So when they hit each other the point of contact is between oxygen and bromine.  So if they hit each other in this orientation, we call this the orientation of the directionality,  the way, the orientation in which they hit, if they hit each other this way there is no  reaction.  So they're hitting each other in the wrong orientation.  However, if they hit each other in this way, so this is oxygen, nitrogen is a little bigger,  bromine is the biggest one, this is the space filling model.  But now, if they hit each other now this way and the point of contact is like this, now  remember in this reaction there's a new bond being formed.  Can everybody see that?  The new bond is being formed between two bromine atoms.  So if they hit, so the point of contact is between the two bromines, can you see now  this is where the new bond is being formed?  So what will happen is this is the new bond to be formed and you can see when that happens,  when the new bond is being formed you can see that now this can break up like that.  So can everybody see that?  Kind of everything is set up so that instant they come into contact you have a new bond  being formed and the other two will break off.  And so now they're hitting each other in just the right orientation so that you end up with  the BR2 molecule being formed and then the two oxygens being formed as well.  So it turns out it's much more complicated than just molecules colliding.  So what you have to keep in mind is they have to collide the first and the more often they  collide the better.  Two they have the right amount of kinetic energy so the more kinetic energy, sufficient  kinetic energy for the bonds to be broken.  And thirdly, even if they have lots of collisions and even if they have the right amount of  kinetic energy you won't get the reaction taking place unless the molecules hit in just  the right way.  And so when they collide they have to collide in the right orientation.  So all these factors influence how fast a reaction takes place.  Now this orientation is called the steric factor.  Steric refers to three dimensional orientation.  And so the orientation depends on the three dimensions and how the molecules hit each  other in three dimensions.  And so we call this the steric factor.  The orientation and molecules hit each other.  Now so if I ask you what are the factors that influence how fast a reaction takes place  then you should know that not only temperature affects it but also the frequency of the collisions  affected, the kinetic energy of the collisions affected and also the orientation affects  how fast the reaction proceeds and whether a reaction will take place or not.  Now Arrhenius came up with a mathematical equation that takes all of these factors into  account.  So if you want to come up with a mathematical equation that describes the relationship of  all of these factors combined then it's given by this equation which is called the Arrhenius  equation.  One form of the equation is written as k which is the rate constant for the reaction.  Now here we're using rate constant instead of rate because a rate of a reaction, remember,  depends on concentration.  We've seen that the rate varies on concentration.  The more concentrated it is the faster the reaction would be.  Now rate constant is independent.  It's a constant.  And so it's a better factor.  If you want to compare different reactions it's better to look at rate constant because  rate constant is independent of the concentration because it's a concentration.  So we're looking at the dependence of rate constant, k, equals what we call a constant  A. This is called the Arrhenius factor and this accounts for collision frequency and  the steric factor.  Remember the steric factor is the orientation.  Collision frequency is how often they collide.  So the Arrhenius factor A takes into account how often they collide as well as the steric  factor and every reaction has a unique value for A. A is a constant and it's available  in tables and every unique reaction will have a unique constant that corresponds to A.  Ea is the activation energy and the activation energy usually is in units of kilojoules per  mole.  Arr is the gas constant.  Now remember, I want you to keep in mind that in this term the activation energy is divided  by the gas constant and the unit for activation energy is kilojoules per mole.  That means if you're using this Arr, remember Arr is the gas constant and the Si units,  the unit for gas constant is joules per Kelvin per mole.  So I want to convert that to kilojoules and to do that all of you recognize that 1 kilojoules  is 10 to the 3 joules.  So when I use this gas constant in this equation it's important that I divide it by a thousand  so that I have the unit kilojoules per mole rather than joules per Kelvin per mole.  So we have that and of course T is the Kelvin temperature.  So this form of the reaction can also be written like this where lnK equals, now if I take  the natural log lnK equals lnA minus Ea over RT and if I rearrange this equation so that  I have lnK equals negative Ea over R, 1 over T plus lnA.  I'm just putting this in the form of y equals mx plus c.  So if I take, the rate constant is the y-axis then this term will be the slope and that  will be the x-axis plus c would be the intercept.  So if I can measure the rate constant at different temperatures, so if I make a plot of ln, lower  case k, okay, not capital K because that's equilibrium constant, this is rate constant.  Experimentally we can measure rate constants at different temperatures and if I plot that  versus 1 over T and remember this is in units of Kelvin and if I end up with, then I would  end up with a graph that has a negative slope where the slope will give me the activation  energy divided by the gas constant and keep in mind it's a negative slope.  So this is how experimentally we figure out the activation energy for any reaction.  If you want to figure out the activation energy for a reaction what you do is you measure  the rate constants at different temperatures and you plot it lnK versus 1 over T and you  figure out what the slope is and from the slope you can divide, take the slope and multiply  it by the gas constant and kilojoules per mole and that will give you the activation  energy.  You may have done this in the lab when you were looking at the kinetics experiment.  Now lastly if you want to look at the dependence of lnK on temperature and let's say we're  looking at, so let's take this equation where we say lnK equals, so this is lower case K  minus EA over R, 1 over T plus lnA and let's say we're looking at this at 1 temperature  so if this is K1 this would be T1, alright? Activation energy does not change, R does  not change and lnA remember is a constant that's unique for each reaction.  So if I take another reaction where it's lnK2 this would be negative EA over R, 1 over T2  plus lnA.  Now since these two are lnAs if I rearrange this can everybody see that if I take this  term over here I can say lnK1 plus EA over R, 1 over T1 equals lnK2 plus EA over R, 1  over T2.  Can everybody see what I'm doing?  Because lnA is a constant, remember A is unique for a particular reaction so as long  as we're looking at the same reaction, alright?  What we're looking at is the dependence of rate constant on temperature, this term lnA  would be this plus that, this lnA would be this plus this if I take it over here and  since lnA is the same for both I know these two are equal to each other and therefore  if I take lnK1 minus lnK2 now this would become EA over R, 1 over T2 minus 1 over T1  and lnK this would become lnK1 over K2 would equal EA over R, 1 over T2 minus 1 over T1.  So if you're, if ever you want to figure out the rate constant at one temperature and you  want to figure out the rate constant at another temperature you would use that equation, alright?  And that would help you figure it out.  Now this same form of the equation can be written like in some textbooks you'll see  this written out like this.  lnK1 over K2 equals negative EA over R but now it's T1 minus T2.  Either one we, okay?  So all I'm doing is switching the sign here, flipping this around and putting a negative  sign in front both those equations are the same, alright?  And some textbooks give the latter equation, some give the former equation, alright?  So that kind of completes what we're going to look at and so the last thing that we looked  at is the factors that influence how fast a reaction proceeds and we looked at a chemical  equation that relates all the factors that influence how fast a reaction takes place.  Now within the last few minutes I want to kind of take an example because ultimately  it's about applying these in problems.  So let's stop there for today and for the one last thing that I just want to talk briefly  about the exam, okay?  The last minute.  And I want to remind everybody that the final exam is on Wednesday.  It's at 8 a.m.  So your exam time is different from the class time, okay?  The exam will be cumulative but greater burden will be on the new material.  So almost half the exam will be on kinetics, alright?  The other half will include everything that we covered this quarter.  So please review the intro to equilibria.  You should be able to draw equilibrium problems.  You should be able to do equilibrium problems.  You should know the relationship between equilibrium and free energy.  Remember we said delta G equals minus RTL and K, you know, the relationship between  free energy and equilibrium constant, alright?  Which goes back to the beginning of equilibria right up to what we've covered to electrochemistry.  But I understand that you will not be able to study everything.  So for the material that's cumulative other than kinetics, I'm looking at the big picture.  I want to see whether you see relationships.  So it's more big picture, more conceptual rather than, you know, sort of very detailed  problems where you have to go through a lot of detail, okay?  So I want to see whether you've mastered just the big picture items, alright?  So this is the last quarter.  I'm going to see you.  So good luck.  I wish you all the very best.  Alright?  Thank you.  Thank you.  Thank you.
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