 So let's take one final look at how we can obtain this Langmuir isotherm from a slightly different point of view, a kinetic point of view this time. So remember we have this model where we've got a surface that can, to which molecules can adsorb, there's some number of sites, sites on this surface to which these molecules can absorb, there's an equilibrium between molecules adsorbing onto the surface and desorbing off the surface and we expect, because we've seen where this equation comes from, from a couple of points of view so far, surface coverage to depend on pressure in this way. So if we want to think instead about the thermodynamics or the equilibrium of this reaction, if we want to think instead about the kinetics of the reaction, how quickly molecules are desorbing off the surface or adsorbing onto the surface, we can again think of the chemical reaction that's going on. Gas phase molecules are becoming adsorbed molecules. In order to do that, they need to encounter one of these empty surface sites, a site where a molecule could adsorb onto the surface but there's currently not a molecule adsorbed onto the surface. In the forward reaction in the adsorption direction, I can give that reaction a rate constant, lowercase k sub 1 for the forward reaction, for the desorption reaction, just the reverse of this reaction, when an adsorbed species desorbs off the surface becoming a gas phase molecule and leaving an empty site behind as a product of that reaction, I'll write the rate constant for that reaction as k minus 1. It's the reverse of this reaction k sub 1. Now, thinking about that process from a kinetic point of view, I can ask for the adsorbed species, if I watch the number of these adsorbed species over time, how is that quantity of adsorbed species changing? In other words, what's the time derivative of my adsorbed species? Those adsorbed species show up as a product in the adsorption reaction but as a reactant in the desorption reaction. So that net total rate of change of the adsorbed species, I'm gaining adsorbed molecules when the adsorption reaction proceeds. So that rate is going to be this rate constant k 1 times the number of adsorbed species. So remember, sorry, not the adsorbed species. X is the number of free sites, the number of unoccupied sites. So the total number of unoccupied sites, I could write that as, this is the fraction of sites that are not occupied. If I multiply that by the total number of sites on the surface, this is the number of unoccupied sites. So the rate of this reaction is going to be rate constant amount of this reactant, multiplied by the amount of this reactant. So let me go ahead and write that as, no, that's not true at all. Let me go ahead and write that, the amount of this gas phase species, the gas phase species is proportional to the pressure of the gas. So the reaction will proceed proportionally to the pressure of that gas phase species. In the reverse direction, I need to take account of the rate at which these adsorbed molecules are disappearing when they are used as reactants in the desorption reaction and this reaction with the rate constant k minus 1. I'm losing adsorbed species, so that's going to be a negative contribution to the rate of change of the adsorbed species. The rate constant is k minus 1. The only reactant in this reaction is the number of adsorbed species. Out of a total of m sites, some surface coverage theta is occupied. So the total number of adsorbed species is theta times m. So those two terms combine. The rate at which I'm gaining adsorbed species from the forward reaction and the rate minus the rate at which I'm losing adsorbed species from the reverse reaction tells me the rate of change of the adsorbed species. If I reach equilibrium, if I've got as many molecules adsorbing as desorbing in the system as that equilibrium, then that rate of change will be zero. So this is now what's going to allow us to rearrange this equation and solve for theta. So if this term minus this term is equal to zero, I can say the first term k1, 1 minus theta, m times p must be equal to the second term, k negative 1 theta m. There's an m on each side, which I can cancel. If I move the k negative 1 over to this side, so I'll write k1 over k minus 1. I've got a pressure and I'll move this 1 minus theta over to the right side. So on the right side I've got this theta. I'll divide that by 1 minus theta. I don't have a k minus 1 because that's moved over to this side. Now I can see that if I just take this ratio of rate constants, the ratio of the forward rate constant to the backwards rate constant, if I rename that and call it k, capital k, then this equation will become kp is equal to theta over 1 minus theta, or when I do a little bit of algebra and rearrange that equation solving for theta, I'll get the exact same result as the last time I did that. I'll find that theta is equal to kp over 1 plus kp. So that, of course, is the exact same Langmuir isotherm equation that we obtained when we solved this problem using statistical mechanics or with equilibrium thermodynamics. We get the same result when we think about it from a kinetic point of view. Now the only difference is this constant that we're talking about, this capital k, is a ratio of rate constants. We've thought about it as an equilibrium constant, a ratio of rate constants, and a collection of constants that arose from our statistical mechanic definition of the problem. So we've seen three different ways of understanding this Langmuir model of adsorption. We get the same result whichever way we look at it, so you're welcome to think about whichever one of these derivations makes the most sense to you intuitively. And the next question is going to be to see if we can understand a little more quantitatively how to use that equation and what it says about adsorption processes.