 Alright, so one additional wrinkle that we need to consider in solving equilibrium problems is the fact that they work out a little bit differently when we try to solve them at constant volume rather than at constant pressure. So the easiest way to understand why that happens is to work an example or two. So that's what we'll do. And instead of our gas phase bromine or HBR examples, let's use an example that's a little bit more of an equilibrium. Let's see, we're going to do the decomposition reaction of dinitrogen tetroxide. So as you may be aware, nitrogen dioxide, I guess, you've likely heard of, dimerizes under typical conditions. Two NO2 molecules can dimerize and form an N2O4 molecule, or likewise an N2O4 molecule can decompose into two NO2 molecules. So that reaction is a bona fide equilibrium. It's equilibrium constant under conditions of 298 Kelvin. That equilibrium constant is 0.148 atmospheres. So notice that has units, products over reactants, products, there's two molecules, reactants, there's only one molecule, so the Kp has some units of pressure. If we, if I give you some initial conditions, let's say I take a box and initially, actually let's give the, we're going to start with pressures rather than moles. So I'll say initially, let's say we have half and half NO2 and N2O4. So I fill a box with some NO2 and some N2O4. Total pressure is one atmosphere. So half an atmosphere is the NO2 and half an atmosphere is the N2O4. Those are my initial conditions. The question is, once that reaches equilibrium, what will those, not the initial pressures, but will those pressures equal once this reaction reaches equilibrium? Once either NO2 dimerizes to form N2O4 or vice versa, and this system reaches equilibrium. So the way we'll solve that problem is the way we've considered before. We can start by talking about moles or molecules and the extent of reaction. So however many molecules of NO2 I have at the beginning, when this reaction proceeds in the forward direction, I create two molecules of NO2 every time the reaction happens. So the number of molecules of NO2 is the initial amount plus twice the extent of reaction. Likewise for N2O4, N2O4 is a reactant the way I've written the reaction. So when the reaction proceeds in the forward direction, I lose one molecule of N2O4 every time the reaction proceeds. So I can write down how many moles of NO2 and N2O4 I have as a function of the extent of reaction if I know the initial molecules, number of molecules of each species. I don't know the initial number of molecules, I know the initial pressures. So here's the point at which I'll say, first of all, I'll remind us that for this example, we're going to do this reaction at constant volume. So the pressures may change as the number of molecules change, but we're doing this in a box of fixed volume. So in order to make a connection with the KP, which is what I've been given, I need to convert these molecules into pressures. So I can write down the pressure of NO2 or the pressure of N2O4 based on what I know about the molecules of NO2 or the molecules of N2O4. And the way I do that is if these gases are ideal gases, for an ideal gas, NkT over v is equal to the pressure. So if I just take molecules multiplied by kT over v, I'll get pressures. So molecules times kT over v is molecules. On the right-hand side, I've got initial number of molecules becomes initial pressure. And then when I take twice squiggle times kT over v, I'm going to write that as twice lambda. So in these, if I write the extent of reaction c or squiggle multiplied by kT over v, turning a number of molecules or moles into something with units of pressure, this is just how I represent the extent of reaction in units of pressure. So pressure of N2O4 is its initial pressure minus lambda, extent of reaction as a pressure. All right, that's enough information now for me to plug into the KP. And I guess let me point out that we've now explicitly used the fact that we're doing this calculation a constant volume, because I can say the volume in this expression is just the volume of the container. That's not going to change. If I'm doing it at 298 Kelvin and whatever the volume of my container is, the volume is the same, whether I'm at equilibrium or away from equilibrium. So knowing these equilibrium pressures, or not knowing what they are, but having an expression for these equilibrium pressures allows me to plug them into the equilibrium condition for KP. So KP is equal to, as always, products over reactants. Since it's a KP, I'll write it as a pressure. Pressure of N02 squared divided by pressure of N2O4. I know I have an expression for P of N2O4 and P of N02, so I can write that as initial pressure of N02 plus twice lambda that gets squared in the numerator. And the denominator, pressure of N2O4, is its initial pressure minus lambda. And so everything in this expression, except for lambda, I know a value for. I know my initial pressures of N02 and N2O4. We were given those. I know the KP for this reaction. That's this number. So all I have to do is rearrange this equation now to solve for lambda. So if I rearrange a little bit to get the fraction broken down. So equilibrium constant times the amount of N2O4, pressure of N2O4 on the left is equal to, I'll go ahead and write this, expand the square and write pressure of N02 squared plus four lambda pressure of N02 initial plus four lambda squared. So now we can see pretty clearly that's a quadratic equation. I've got terms that depend on lambda squared, terms that depend on lambda to the first power and some terms that don't depend on lambda at all. So if I rearrange a little bit more to make it look like a quadratic, the standard form of a quadratic equation, the quadratic term, the term with lambda squared is just this four lambda squared term. The linear term, I've got a lambda here and a lambda here. So my linear term is going to look like lambda times four P naught. And I've got a lambda times KP PN2O4 naught. When I bring that over to the left-hand side, that'll be a positive KP P naught for N2O4. And then my constant terms, I have two of those. I have PN02 naught squared and I have KP, oops, I've made an algebra mistake here. This KP, KP times lambda when I bring it over is just KP times lambda. KP times PN2O4 when I bring it over has a negative sign, so I've got a minus KP PN2O4 initial. So I think that should be correct. So I've just rearranged this equation to make it look like a quadratic equation with an A coefficient, a B coefficient, and a C coefficient. We won't work through solving the quadratic equation again. But if we plug those terms into the quadratic equation, we know the values again for P naught N02, for KP, those are the values we were given in the problem. If we solve the quadratic equation now, what we obtain is, of course, two solutions. Lambda, the variable we're solving for, is either equal to negative 0.436 units of atmospheres or negative 0.101 atmospheres. One of those solutions is going to be non-physical. They're both negative numbers. Both of these numbers say that the reaction should shift backwards towards reactants. The one that's okay, the one that's allowed, is negative 0.101, that's the one we're interested in. Negative 0.436, that one's not possible because since I started with only half an atmosphere worth of N02, if the reaction goes backwards, negative, backwards meaning negative 0.436 times, I'm going to lose twice that many atmospheres of N02. So if I double that number, it's larger than 0.5. So this first option would consume more than all of my N02 when the reaction goes backwards. So that's the version that's not physically reasonable. So this is the value for lambda that we are going to use. So we solve for lambda, what we're interested in is not the value of lambda, not how much the reaction shifts forwards and backwards, but how much of each species there is. So the pressure of N02 and the pressure of N204, once we've reached equilibrium. So we have expressions for those. Pressure of N02 is its initial pressure plus 2 lambda, or initial pressure of N02 is half an atmosphere. I'm going to add twice the value of lambda, which is negative a tenth of an atmosphere roughly. So 0.5 minus 0.202, that leaves me with only 0.298 atmospheres of N02. So that's the partial pressure of N02 that we have once we've reached equilibrium. The pressure of N204, likewise, is its initial pressure minus lambda, 0.5 atmospheres minus a negative 0.101 atmospheres, that gives me 0.601 atmospheres. So these are the answers to the question I posed at the beginning. How much, what is the partial pressure of N02 and N204? It's these two numbers. So we've solved the problem. As usual, when I've done this much algebra, I've caught some of my own mistakes, but I might not have caught them all. So we can double check and ask, is this in fact the correct answer? Are these the pressures that obey our equilibrium condition? So we can plug those back into the equilibrium condition just to double check. It's always a good idea. So pressure of N02 squared over pressure of N204 should equal the numerical value of the equilibrium constant. In other words, 0.298 atmospheres squared, if I divide that by 0.601 atmospheres, again if we get a calculator out and do that calculation, that number works out to, as it turns out, 0.148 with units of atmospheres, atmospheres squared on top, atmospheres in the denominator, and that is in fact exactly what we expected to see, exactly the value of the equilibrium constant that we were trying to solve for. So what that means is that is indeed the solution to the problem. The double check worked. This particular example that we worked with constant volume worked out in this way. What we'll do next is work that same example not at constant volume, but at constant pressure to see how things work out differently.