In our previous lessons, We were looking at the relationship between temperature activation energy and 'k'. We are going to see this mathematically played out in what is called the Arrhenius equation. Our learning objective for this lesson is to utilize the Arrhenius equation to determine the rate constant, the temperature, or the activation energy of a reaction. We are going to see the way that they interplay and relate off each other. This is the Arrhenius equation. Lets look at the various component of this equation. Talk about each one of them. Many of them we have seen before. This is obviously the rate Talk about each one of them. Many of them we have seen before. This is obviously the rate constant 'k'. We will be able to see how the rate constant changes depending on the activation energy or the temperature. We know that E sub A stand for the activation energy. This is reported and utilized either in kilojoules per mole or joules per mole, we will see as we proceed on we have to make sure our unit cancel properly. R is the ideal gas constant. The ideal gas constant we are going to want to use has units of joules per mole. Notice that I said kilojoules. We cannot put these both in there with these same unit. We either must have them both in joules, so we could convert this one to joules or we could have them both in kilojoules and we convert this one to kilojoules, but those units are going to have to cancel with each other in order for this to work properly. It is very common for students to mess that up. They just plug numbers in and not thinking their units. So I am talking about your units for a reason. It is an easy way to make a mistake here. OK if we are talking about units what will the units of temperature have to be in? As a hint look over at the gas constant and what is the unit of temperature in it? It is Kelvin, so you better put your temperature in Kelvin, or this equation will not work. So we have our 'k' or activation energy our R and T and this E that is the exponent the inverse of the natural log [ln]. our R and T and this E that is the exponent the inverse of the natural log [ln]. What about the A? The A here is called the frequency factor. It is a constant for the reaction system. It takes into account other avenues and things that are unique to that reaction like how do the molecule have to be oriented when they collide. We will talk more about this frequency factor later in our lesson. OK, We have this expression but we want to do first is to think about how this plays out with what we already know is occurring on the molecular level. So over here we have got some conditions. Lets think about changing activation energy. As activation energy gets larger and larger and we are look at this portion of the equation we are raising it to a more and more negative value. E to a larger and larger negative number. How does that affect the value of 'k' If you are raising it to a large negative number it becomes a smaller and smaller fraction. So 'k' is going to go down. Now does that make sense with what we know about activation energy? Lets look and think about the mountain size. If it is a really really high mountain is it going to be a slower reaction? There are going to be few molecules that can get over that is it going to be a slower reaction? There are going to be few molecules that can get over that so it is going to be a slower reaction so you can expect a smaller 'k'. Lets think about temperature's role. Now where is temperature in this equation? Temperature is right here in the denominator. As temperature goes up we are dividing by a larger and larger number. So it is going to decrease the size of this portion of the number. So it is to a smaller negative number. If it is to a smaller negative number then 'k' is going to become bigger and bigger. Well, lets think about how that makes sense we what we know about what is happen on the molecular level. We have a certain mountain to climb. As temperature is going up we have got more molecules We have a certain mountain to climb. As temperature is going up we have got more molecules that have enough energy to get over that mountain. If there are more molecules that have enough energy to get over that mountain wouldn't the reaction take place at a faster speed? Therefore have a larger 'k' value. We are going to do some manipulations. Our goal in the end is going to be is to have a slope intercept form of a line. So we will take the natural log of both sides. Natural log of both sides we are taking the natural log of a product here. So we have the natural log of this product. We are going to have the natural log of A plus the natural log of E to the -E sub A over R. The natural logs will cancel and leave me with th negative E [-E] sub A over RT. Now I have not just rearranging but I have also done regrouping and I want to look at what is happening. I have moved this over here to the left side, but I have also grouped this together a little bit differently. The reason I have done that is so it is in the form y = mx +b. So I have kind of lined it up underneath there so you know form y = mx +b. So I have kind of lined it up underneath there so you know what each thing is. So what will it plot of the natural log of 'k' That is my Y verses 1 over t [1/T] that is my X. It is definitely going to give you a straight line. Well what is the slope going to be equal to? It is this whole portion here, that is m. That is the slope, negative E sub A over R. You could graphically determine the activation energy by running this experiment at various temperatures getting multiple k's and obtaining an activation energy from the graph. Here is an example. and obtaining an activation energy from the graph. Here is an example. If you are give k's and temperatures, you you need at first take those k's and take the natural log of them. You would take those temperatures and take 1 over those temperatures. Then plot out the natural log of k [ln k] along the y-axis. 1 over t [1/t] along the x-axis. Then you can take the best straight line through those points. Once you have the best straight line through those points you could take two values off of that line. You could do a change in Y over a change of X and have a slope. Once you have the slope, we know the slope is equal to negative E sub A over R. Then you could obtain the activation energy. negative E sub A over R. Then you could obtain the activation energy. How many points does it take to make a straight line, only two. It you take this Arrhenius equation, and this is the slope intercept form that equation and you take it for two data points let say for a K2 with a T2 and then do the same thing a natural log of K1 with a 1 over T1 [1/T1] take those two equations and subtract them you can derivate a two point equation for the Arrhenius equation. And then you have got a relationship between two temperatures and two rate constants and it could be a way to obtain an activation energy. But basically in this equation you have got one, two, three, four, five variables and if you know 4 of them you can solve the fifth, and that is very often asked of you within problems in this section. So i have given you one here. I have got a reaction in which I have given you the activation energy. So this is E sub A. I have told you the rate constant k at 35 degrees Celsius . So lets call this K1, and lets call this T1. And we want to know the K2 and we will call this T2. Now remember you have got to have your temperature in Kelvin. So lets figure out T1 in Kelvin it would be 273 plus 35. And T2 would be 273 + 0. That is obviously 273 kelvin. And lets get see that is 308. Just know that blob there is a k. So those are the temperatures put everything we know into our Arrhenius equation. I am going to call K2 what I don't know. I like having it up on the top. It really doesn't matter which you put on top and which you put on bottom as long as your are consistent that whatever k2 is you put it in the spot where t2 needs to go. K1 value is 1.35 x 10 ^ -4 seconds ^ -1. That is going to be equal to E sub A over R. Now here is where to have to watch the energy units. I want to go ahead and convert my kilojoules per mole to joules per mole. That would be 1000 more. So a 102 Thousand Joules per mole is my activation energy. My R is 8.314 Joules per mole times Kelvin both of those in the denominator. Then we have 1 over T2. Now I have assigned the T2 as 273 minus 1 over T1 [1/T1] which was 308. Both of those are in Kelvin. So we look for our units and see how they cancel. Joules are going to cancel. Moles are going to cancel. And now this Kelvin is in the denominator of the denominator. So the kelvins will cancel each other as well. Because if you are in the denominator of the denominator thats takes the reciprocal and puts it up in the numerator. So my kelvins will cancel. I am left with no units, and that is what you should have at this point. OK so lets do a little work here. When I combine, and you want to do all this. Do the mathematics of this entire left side. You will get a negative 5.107. A very common mistake for students is they lose this minus sign, don't do that. That is going to be equation to the natural log of K2 over 1.35 x 10 ^ -4 seconds ^ -1. Now I want to get into that K2 over 1.35 x 10 ^ -4 seconds ^ -1. Now I want to get into that K2 how do I do that? I take E to both sides. When I take E on this side it cancels those out and I am left with K2 divided by 1.35 x 10 ^ -4 seconds ^ -1. And when I take E to divided by 1.35 x 10 ^ -4 seconds ^ -1. And when I take E to negative 5.107 I have 0.00606. Now I am ready to solve for K2 by multiplying both sides by 1.35 x 10 ^ -4. That will give me K2 is equation to 0.00606 times 1.35 x 10 ^ -4 second ^ -1. That will give me a K2 value of 8.18 x 10 ^-7 second ^ -1. That will give me a K2 value of 8.18 x 10 ^-7 second ^ -1. Lets think if this makes sense. What did we do to our temperature? We went from 35 where we knew a K down to zero. So we dropped the temperature. If we drop the temperature, the rate should decrease, and it does so by having a smaller k. if our k smaller? Well certainly it is. So at least we know that this number makes sense for the work we have done. So with out Arrhenius equation we see the relationship between temperature activation energy and k. We need to mention a little bit about that frequency factor A that comes out of that Arrhenius equation. That quantity frequency factor is going to be dependent on a couple of things. Number one it takes into account the collision frequency. Therefore it is called the frequency factor. With that, you take into account the state of matter, for example. Something that is in a gas state With that, you take into account the state of matter, for example. Something that is in a gas state free to move and they move very very rapidly so they can collide with each other but there is great distances between them. But if they are in a liquid state they still can move pretty quickly but they much much closer together. They will collide more frequently so that takes into account the state. If it is a solid, they do not move very freely and that cannot collide very frequently. So that would have a different frequency factor. It also takes into account. orientation of the molecules because sometimes one collision even if it has a lot of energy won't result in a reaction and we see that here in the top image. In the top image if they were to collide no reaction would occur. But if they collided and the green ran into the green then there could be a reaction. If there is an orientation necessary to have a collision that will come out in that Arrhenius equation. In conclusion with lesson and out learning objective here we see that there is mathematical relationship between temperature activation energy and the rate constant k. We saw how to graph it out in order to obtain a linear relationship. How to plug into a two point value equation with a K1, T1 K2, T2 how to be able to obtain an unknown out of that equation. Lastly we see how the frequency factor A plays a role in rates of reactions. Lastly we see how the frequency factor A plays a role in rates of reactions.
