 Welcome to module 12 of Chemical Kinetics and Transition State Theory. In last several modules, we have developed a theory for of a collision theory to be able to calculate a rate constant, a number out for a given reaction. In the last module, we looked at a few examples of how to calculate this number. Today, we will continue on and solve a few more problems so that you become comfortable. Today, not only we will look at numerical problems, but a little bit of derivations as well. So, let us continue. So, our first problem of today is that let us assume our air has only nitrogen gas. A nitrogen can be assumed to have some particular radii given you diameter of 395 picometer and the question is how many collisions each molecule of N2 will be having per second in regular conditions of pressure of 1 atmosphere and temperature of 300 Kelvin. So, that is our question. So, let us see how do we solve it. And so, the answer is very much related to collision theory. In collision theory, we are finding a rate constant as the rate of collisions. And the rate of collisions is of course, directly related to the number of collisions that you are having. So, let us start. Let us say this is N2 with some radii. And essentially, we construct the same idea. We construct a cylinder. This is U del T. This cylinder's length, you can go back to this proof will be 2 R N2. It was RA plus RB if A and B were different. A and B are same. So, you get 2 R N2. And all that we have to do is to find the volume of the cylinder. So, number of collisions is equal to, so this is in time delta T will be equal to volume of cylinder, the density of nitrogen. So, the density of nitrogen will tell me what is the probability that another nitrogen molecule is in here. So, that is the idea. So, this is equal to then pi 2 R N2 square U delta T rho N2. Is this correct or have I missed something? There is a little factor that is missing, a factor of half. So, whenever I have A equal to B, I get a factor of half. Why? Because I am double counting the collisions. The N2, this N2 let us me call just A arbitrarily. If it collides with another N2 which is B, well N2 B also collided with N2 A. So, I have double counted the collision. So, that is why the factor of 2. Again, we discussed this is in some detail of few modules back. So, please revise that if this is not clear. So, collisions per unit time will simply be half 4 pi R N2 square. And this U in collision theory, we replace by the average speed, average thermal speed which is. So, we are going to calculate all of these quantities one by one and multiply them together. So, first thing was sigma equal to 4 pi R N2 square. We have to be just very, very careful of units. Never mess on that 395 picometer. 395 picometer is 10 to the power of minus 12 meter square. I have calculated this out. This thing is equal to 196 meter square. Just a quick note on how many digits you want to write the answer to. If you will plug it on a calculator, you will of course get many, many more decimal places. But since my question have was having 3 significant figures, I have written the answer also in 3 significant figures. So, that is a general rule of thumb. So, the next quantity I have to calculate is the average speed. But to calculate average speed, first let me get the mass. This is equal to, you can just quickly check, this is equal to M of N2 divided by 2. And the mass of nitrogen is what? Nitrogen is atomic mass is 14 gram per mole. So, this is 28 grams per mole divided by 2. I want the answer in kilograms in SI units. So, I have to convert my units. If you use grams per mole, you will get a completely wrong answer, units, careful. So, this I convert in 1 kilogram over 1000 gram into 1 mole divided by the Avogadro number. So, the gram cancels with gram, moles cancels with mole and I will get the answer in kilogram. And I have done the math, this comes out to be 2.3 into 10 to the power of minus 26 kilogram. So, once you start doing these kind of calculations, you will actually this thing is wrong, this should be minus 20 here. So, when you keep on doing these calculations, you will also get a sense of what are the numbers you expect to get. Mass is generally in the order of 10 to the power of minus 26 or 10 to the power of minus 27. This reaction cross section is generally in this order of 190 like 100 angstrom square. So, get a sense of these numbers. So, even if you make a mistake, you will be able to quickly see, you will be able to sense, your sixth sense will tell you that something is not right. Finally, I have to not finally, but I have to calculate the thermal speed as 8 kB 1.38 into 10 to the power of minus 23, what is the unit of kB kilogram meter square, second square Kelvin, 300 Kelvin divided by 8 kT divided by pi into the mass and the mass I have found is 2.3 into 10 to the power of minus 26 kilogram. So, Kelvin cancels with Kelvin, kilogram cancels with kilogram and I will get square root of meter square per second square, which is same as meters per second. So, this I can again plug it on a calculator and I get it as 677 meters per second. Finally, I have to calculate the rho. I have to first get the pen rho. So, to calculate rho, I am going to use ideal gas law. In the question to calculate something, you are making an assumption that is fine to do. So, if it is a reasonable assumption, but you should mention what assumption you are making, N2 is ideal gas. So, my answer will be valid under this assumption. So, rho is nothing but N over V, which is nothing but P over RT, which is nothing but 1 atmosphere or I have given already in appropriate units into temperature is 300 Kelvin. So, what is the unit of rho I am looking at? Rho is density and density should be, it is a number density. So, I want to cancel liters with meter cube. So, I will have 1000 liters in 1 meter cube and I will have 6.02 into 10 to the power of 23 per mole. Units, pay attention. If you do not practice, you will not get these right. Liter cancels with liter, mole cancels with mole, Kelvin cancels with Kelvin and everything else is just plugging it on a calculator, which gives me 24.5 into 10 to the power of 24, something atmosphere also cancels, correct, meter minus 3. So, make sure that my numbers are right. I am punching them myself on a calculator and there is certainly a chance that I have made a mistake. So, it is upon you to make sure that these numbers are correct. Sorry, I still have to find the final answer, which is now easy, again my bad. So, finally, the collision was half into sigma, which we found to be 196 into 10 to the power of minus 20 meter square into thermal speed, which we found to be equal to 677 meter per second into rho, which we have found equal to 24.5 into 10 to the power of 24, 1 over meter cube, okay. So, if you do all this, first thing is that units will cancel meter square, meter square with meter cube and I will get 1.63 into 10 to the power of 10 collisions per second per molecule of N2, okay. So, that was already built into my equation. So, let us look at the next question. This is about transforming between different variables. It is extremely important and collision theory is really built upon it. You should be comfortable in converting between different variables, okay. So, this is something we have done in one of in while deriving, but let us do explicitly once more. I am doing this again because it is very important. You should practice this a lot, okay. So, I have given the rho equilibrium in speed and the question asked to find the distribution in energy, in kinetic energy, this one. So, how do we do it? Remember the volume element. Again, I can promise you if you do not practice, you are going to forget volume elements in your exam, okay. So, we are going to do it carefully. First thing is we write this rho with a volume element du and this is given to me. So, what do I do? E is half Mu square. First note is u equal to 2. This is simple manipulation. This is not hard and d epsilon is Mu du. And I will just play around with it. You can do in any fashion that you like. I have a u square here. I will take this. I will write as u here. 1u I will take and write here. So, I get u into du which I will use here. So, this I will write as m over 2 pi kBt to the power of 3 half 4 pi u. u is root 2 epsilon t over m e to the power of minus beta. Half Mu square is epsilon t. u into du is d over m. So, I can simplify this a little bit. Again, more you practice, more you will become comfortable on how to do these kind of tricks. This comes only and only with practice. No alternate. You cannot just watch these lectures serious and become comfortable with these. 4 pi over m root 2 over m root epsilon t e to the power of minus beta epsilon t d epsilon t. So, let me just write this since to be correct. This I have converted now. So, I will see several factors are going to cancel. m cancels with m. 2 pi cancels here to give me 2. Anything else cancels? Yes, m cancels with m. This 2 cancels with this 2. So, the net result I get is I have written it here. You can simplify this 2 over kt into 1 over root pi kbt root epsilon t e to the power of minus beta epsilon t. This is equal to rho of epsilon t. So, you can go back one slide, make sure that this equation is correct. I have cancelled all the factors correctly. So, this is about transformation of variables, very important concept. So, our second problem is to calculate average speed and most probable speed in 2D. So, we have actually solved a more complex problem which is of 3D. Here we will solve this problem now in 2D. So, we will have to go back to our basics, how we derived it in 3D and follow that proof in 2D. We have to be just very, very careful. So, what are basics? Well, our basics says the Boltzmann distribution that will always be true. I have only 2D. So, I am writing only px and py minus beta h and the h here is only px square over 2m plus py square over 2m and never forget your partition function. So, earlier we have done the same thing, but with pz as well. So, it was more complex. So, I had a p plus pz square as well as integral over pz in the denominator. Why I have to be converting units between different variables and when I do that, what do I do? I always keep track of volume element. First let us find out the denominator. So, let us calculate dpx e to the power of minus beta half mpx square. This integral you can find here and in this if I have to convert a will be equal to m over 2 kbt. So, this is equal to m pi divided by a which is m divided by 2 kbt. So, this thing then becomes equal to e to the power of minus beta over 2m px square plus py square divided by 2 pi kt dpx dpy. So, the next thing is we have to move to polar coordinates. In 3D we went to spherical polar coordinates of r theta and phi. Here we have only 2D. So, we are going to move to polar. So, we will take the row that we had in the last slide and we want to move to p comma phi. So, this will be equal to now we will convert. So, what is our basis? p is root of px square plus py square and importantly we will see this thing here dpx dpy is mod p dp d phi. So, this thing will be equal to 1 over 2 pi kt over m. So, that is simply this factor. I think I have forgotten or divided by m here. So, this thing to keep on going back to pen e to the power of minus beta I had px square plus py square which becomes mod p square over 2m into the volume element. So, this thing. Now, the point is I do not care about phi I only care about the overall speed overall momentum. So, what do we do? We integrate over phi just like we integrated over theta and phi and 3D. So, this thing I want is integral over d phi sorry d phi. So, this is easy we note that most of the terms here are independent of phi actually everything is independent of phi and I simply multiply this by 2 pi. So, I get an integral here from 0 to 2 pi everything here is independent of phi. So, it is an easy integral. So, I will just cancel 2 pi's and I get let me go back m over kbt e to the power of minus beta over 2 p over 2m mod p square mod p. Now, we are going to convert to speed this is momentum. So, p is equal to mu dp equal to m du. So, we will just replace these things carefully. So, if I do this thing here, you can quickly verify this is mu square mod p becomes mu and dp becomes m du. I have made a mistake here this is m in the numerator this m is in the numerator this thing will carry on m is in the numerator here. So, this m is here my apologies. So, 1m is supposed to cancel and I get mu over kbt e to the power of minus beta half m u square. So, once we have found the equilibrium density we will then find the average speed and most probable speed. In summary, today we have solved many few more examples, two important points to note when you are transforming in between units remember your volume element when you are doing a numerical calculation remember your units. Thank you.