 All right, so let's take one more attempt at this particular equilibrium problem. Dissociation of N2O4 into NO2 at 298 Kelvin where it has this equilibrium constant denominated in terms of pressures. Previously, we saw when we worked that problem at constant volume, we got this particular answer, a certain amount of N2O4, a certain amount of NO2 when we're at equilibrium. When we do things at constant pressure, as we'll see, the answer is a little bit different for reasons that we'll be able to understand. So let's just go ahead first and start tackling this problem the same way we did previously. I can say once I've reached equilibrium, I no longer have the initial amount of NO2 that I had because NO2 is a product. I've generated two additional moles of NO2 for every mole of times the reaction proceeds, likewise for N2O4. So far, this is proceeding exactly like it did at constant volume. N2O4, I consume a molecule of N2O4 every time the reaction goes forward. The next step would be to say, again, for ideal gases, the pressures, if I want to convert molecules to pressures, I just multiply by kT and divide by V, now comes the difficulty. Since V is no longer constant, the V is going to be changing. The volume of the initial system as the reaction proceeds, if one molecule of N2O4 generates two molecules of NO2, I've got more molecules in the gas. If I keep the pressure constant at one atmosphere, that's going to cause the volume to expand. So this volume depends on the extent of reaction. So that's going to make this step not difficult, but one step more complicated than it was previously. So as a first step, I can say, what is the volume, what is the total volume of that system? It's going to be changing as the reaction proceeds, but what is it going to be equal to? It's, as an ideal gas, going to be NKT over P. But the N that I need now, this N is the total number of moles, the total number of molecules of gas in the box. So N total, the total number of molecules, that's just the molecules of reactants and the molecules of product added together, molecules of NO2 plus molecules of N2O4. So if I add these two things together, the total number of moles is however many moles of NO2 we have. That depends on the extent of reaction. Add that to the total number of moles of N2O4 that we have, that depends on the extent of reaction. Two squiggle minus one squiggle, the net result here is that I've got initial NO2 plus initial N2O4. I'll just name the two of those added together. That's my total initial number of moles. And then plus two squiggle minus a squiggle gives me a net plus one squiggle. So that's my total number of moles. I can use that in this expression to say that the volume is going to be N total. So that's the total initial number of moles, plus one factor of squiggle times kT over P, over the constant pressure of one atmosphere at which I'm doing this calculation. One atmosphere because I've got half an atmosphere each of NO2 and N2O4. That's the volume I need to use in this expression when I convert molecules of NO2 to pressure of NO2, molecules of N2O4 to pressure of N2O4. So now let's go ahead and do that. It's taken me a little work to get to this point, but when I multiply by kT over V, I'm going to put the answers down here. My pressure of NO2, molecules, partial pressure of NO2 is molecules of NO2 times kT divided by this volume. So it's going to be molecules of NO2 and NO2 initial plus two squiggle times kT divided by V. So I'm dividing by this fraction divide by N total initial plus squiggle kT and then there's a denominator of the denominator now which I'll move up to the top. So that's an expression for the partial pressure of NO2 and how it depends on the extent of reaction. Notice that's a little more complicated than it was in the previous example where we didn't have to worry about the volume changing. Likewise, the expression for N2O4 is going to be molecules of N2O4 N0 minus a squiggle times kT divided by this expression for the volume. So I'm going to divide by initial moles for the whole system plus a squiggle times kT over P. So that P in the denominator, the denominator again moves up to the top. Okay, that took a little more work, but now we have expressions for partial pressures of the reactants in the products. So I can say at equilibrium what must be true is that the equilibrium constant must be products raised to their stoichiometric coefficients over reactants raised to their own stoichiometric coefficients P NO2 squared over P N2O4 and now I just need to insert these expressions into that relationship. So I've got P NO2 squared. Whole fraction squared. Initial NO2 plus 2 squiggle over initial moles plus 1 squiggle times kT. Yes, actually, you know what I'd rather do? Before I write this expression in terms of moles, I'd rather rewrite these expressions and get rid of moles. So what I'm going to do here is I'm going to rewrite these expressions. There's a kT on the top and a kT on the bottom that would cancel very easily if I wanted them to. What I'm actually going to do is I'm going to take, throw in an extra 1 over V on top, 1 over V on bottom so that I can say molecules times kT over V and kT over V, that's going to give me a pressure. So that's going to convert molecules to pressures and it's going to convert my extent of reaction with units of moles into an extent of reaction in units of pressures. So the result of doing that, kT over V just turns these molecule quantities into pressure quantities. So I'll end up with p0 NO2 plus 2 lambda which is what I'm calling extent of reaction in units of pressure over p total pressure, initial pressure plus a single lambda and then I still have a total pressure, the fixed total pressure of the system outside this fraction. Likewise, pN204 is going to be this expression with n's turned into p's and squiggles turned into lambdas. So I've got initial pressure of N204 minus lambda over total pressure initial plus a lambda all times pressure. All right, so those are the expressions that I'm going to insert into this expression for kP. So insert both of those into here. Now I'll write pN02 squared is pN02 plus 2 lambda over p total initial plus lambda squared and there's also a p that gets squared. Then I'm going to divide it by this expression so I'll just turn this one upside down. pN204 initial is on bottom minus lambda. On top I've got p total plus lambda and this p once I've turned it upside down becomes a 1 over p. So that's my equilibrium expression. There's some simplification that happens. There's a p squared and a 1 over p. So I end up with just one factor of p. There's on the bottom a p total plus lambda that gets squared and on the top p total plus lambda those again partially cancel. So this expression will become kP is equal to pN02 plus 2 lambda quantity squared. That's this term from the numerator. In the denominator I've got one factor of p total plus lambda. From here the other one canceled this one and one factor of pN204 minus lambda and I've got a p total pressure left over outside the fraction as well. So looking at that expression it's not quite simplified as much as we could. If I bring these two terms over I can convert this to a quadratic equation. I'm not going to have any terms bigger than lambda squared from the numerator and bigger than lambda squared from the denominator. So I could do a bit more algebra and write this out as a quadratic equation with enough quadratic equations. I'll leave that as an exercise for you if you're interested in it. But this is going to be a quadratic equation the coefficients of which we know p0 for N204 and N02. This should be a p0 rather than an equilibrium pressure. p0 for N204 and N02 we know those we were given those. kP we know that we were given that. Initial total pressure is just one atmosphere because that's half an atmosphere from N02 to N04. The constant pressure that's also one atmosphere. That's the total pressure that remains constant throughout the entire equilibrium process. So if we solve this expression if I say we've done the work, done the algebras, used the quadratic equation what we get is solutions to this equation. I can either get negative 0.392 atmospheres as a solution or I can get negative 0.109 atmospheres is the result that we'll get. Again, one of these doesn't make sense physically. If I only have half an atmosphere of N02 I can't react the reaction backwards enough to get rid of 0.392 twice that would consume more than all the N02 that I have. So this solution doesn't make any physical sense. The solution that does make sense is the minus 0.109 atmospheres. That tells us that once we reach equilibrium the amounts of Pn02 and N204 can be obtained from these. I've got them up here. This expression and this expression. So, again, I won't work through the arithmetic but I'll write it out the first step. The way I calculate pressure of N02 would be initial pressure of N02.5 atmospheres plus twice lambda so I'm adding twice this value, this solution, minus 0.109 and then I have a denominator. I have initial total pressure of 1 atmospheres plus a single factor of lambda minus 0.109 and multiply that by the total pressure of 1 atmospheres. So that works out to be some number. 0.316 and likewise I won't do as much of the arithmetic on the board. This time Pn024 if I take half an atmosphere minus my value of lambda divide that by a full atmosphere plus my value of lambda multiplied by an atmosphere that works out to be 0.684 atmospheres. So, I guess we should also double check is that or is that not the solution to our equilibrium problem? If I go back to the equilibrium expression and say Pn02 squared over Pn204 N02 squared that's 0.316 atmospheres squared N204 is 0.684 if I work out that ratio that works out to be 0.146 so that's not exactly the answer we came up with the value of Kp that we started with but it's different just in the third sig figs and since we were working this problem with three sig figs there's a little bit of round off error that happened so essentially that's the correct answer the fully correct algebraic answer would involve a little more precision beyond 0.316 and 0.684 so that solves the equilibrium expression we've confirmed that that's correct if I compare those answers to what we got when we did things at constant volume we can learn a thing or two so our N02 pressure is a little different than it was at constant volume N204 pressure is a little different than it was at constant volume both of them are larger than we got at constant volume notice that if I add these two together 0.316 and 0.614 if I add those together I get a total pressure of one atmosphere so that's another useful double check we claimed that we were doing this at constant pressure whatever the N02 and N204 were end up to be at equilibrium they added up to one atmosphere at the start they're adding up to one atmosphere at the end so that means we did indeed maintain constant pressure as we reached equilibrium that doesn't happen over here if I add 0.298 and 0.601 that adds up to 0.899 so as expected when I do something at constant volume the pressure does not remain constant it changes what happened is as this reaction proceeded in the backwards direction every time two molecules of gas turned into one molecule of gas the total number of gas molecules decreased so since I kept the volume the same the pressure dropped alternatively in the constant pressure case since I'm keeping the pressure the same the volume had to increase sorry the volume had to decrease I'm decreasing the number of moles at constant pressure the volume of the system had to decrease in order to keep the pressure at one atmosphere and that's given by this expression here the volume changes as the extent of reaction changes