 So let's take this new equation for a spin, the one that relates our pressure based equilibrium constant to an equilibrium constant determined in terms of moles that we might calculate from partition functions. So if we stick with our favorite reaction for the moment to bromine atoms dimerizing to form a bromine molecule, if that's happening in the gas phase then we might want to use that reaction, we might want to describe the amounts of those products in terms of the pressures of bromine atoms and pressures of Br2 gas. So we'd like to be able to use Kp. We do know that for the specific case of this reaction when we're at 298 Kelvin the equilibrium constant is equal to 1.6 times 10 to the 7th liters times 1 over the volume. So if we were doing that in a volume of 1 liter then the 1 over 1 liter would cancel these units of liter and that would be our equilibrium constant. But for now I've written this in general so as long as we're at 298 Kelvin doesn't matter what the volume is we can compute what the equilibrium constant is with this expression. So if we want to know what Kp is equal to this expression tells us that's going to be Kt over v raised to this difference in stoichiometric coefficients. When I turn two atoms of bromine into one molecule of Br2 I've lost, I've turned two gas species into one gas species so I've lost one species in the gas phase. So my stoichiometric coefficient is negative one, I've lost one molecule. So Kp is Kt over v to the negative one times Kn or v over Kt times Kn. But Kn is this quantity 1.6 times 10 to the 7th liters times 1 over v so now very conveniently the volume in this conversion ratio cancels the volume in the Kn and there's no more volume dependence left in Kp. So that's actually very convenient Kp is not a volume dependent quantity it will depend on temperature both because this number depends on temperature as well as the extra factor of temperature that arises here. For this specific example of doing improvement 298 Kelvin I can calculate Kp as this Kn divided by Boltzmann's constant and also divided by the temperature. So the Kelvin units will cancel Kelvin cancels Kelvin but I've got a liters on top and a joules in the denominator those don't cancel and we won't really understand what that's talking about until I convert units of joules into units of liter atmospheres so I'll at least get some partial cancellation. So now these joules will cancel these joules these liters will cancel these liters I can do the arithmetic and calculate the value of this equilibrium constant is 3.9 times 10 to the 29th the units that I have left over is per atmosphere and that makes sense as we'll see in just a second because the equilibrium constant written in terms of pressures will be a ratio of pressures over pressures that has more pressures in the denominator than in the numerator so that's how we obtain a value for the equilibrium constant written in terms of pressures how do we use that equilibrium constant we can work an example and just to see how it works we'll use the same example that we've used previously for this Br forming Br2 reaction let's suppose that initially again at 298 Kelvin initially we have a mole of Br2 and no Br atoms and I'll give you those initial conditions just so we're sure that it matches the problem we've solved previously what we're interested in to solve a problem with a Kp is what is the initial pressure of Br2 so if the bromine is acting like an ideal gas we would say that's n Kt over V so we'll skip the details of the ideal gas calculation but one mole times gas constant times 298 Kelvin divided by volume and let's go ahead and say now we'll say we're doing this in a volume of one liter then the pressure is going to work out to 24 and a half atmosphere so our initial pressure I've got a one liter box into which I put 24.5 atmospheres worth of compressed Br2 gas and no Br atoms initially and the question will be how much Br does we form in equilibrium after this reaction shifts backwards a little bit so the equilibrium expression that we're going to need to plug into is going to involve pressure of Br2 and pressure of Br at equilibrium pressure of Br2 is going to be I can think of that as moles of Br2 times RT over V at equilibrium just like we've seen previously the moles of Br2 is going to be the initial amount of Br2 plus the extent of reaction if I start out with some moles of Br2 and the reaction proceeds some number of times I've gained that many moles of Br2 let me let me keep writing these as Kt over V just for consistency so I could think of that as n not Br2 Kt over V plus extent of reaction Kt over V where remember extent of reaction is the number of times or the number of moles of times the reaction is preceded this quantity and not Br2 Kt over V that's exactly what I calculated here that's the initial pressure of Br2 squiggle Kt over V that's some number of moles of times the reaction is preceded times Kt over V that's going to be in units of pressure let me go ahead and label that quantity extent of reaction times Kt over V that's just converting extent of reaction and moles into extent of reaction as a pressure so I'll call I'll give that a different variable lambda lambda is the amount by which the pressure has increased due to this much extent of reaction so instead of saying moles is initial moles plus extent of reaction now I've said pressure is initial pressure plus extent of reaction just written in terms of pressures instead of moles likewise the pressure of Br is going to be the initial amount minus twice the extent of reaction sorry Kt on top V on the bottom so since squiggle Kt over V is equal to lambda I'll write that as minus 2 lambda so the rest of the problem is going to seem very similar to the way it worked when we did it in terms of moles but now we're just working in terms of pressures instead so I'll say the equilibrium constant is the amount of products divided by the amount of reactants so pressure of Br2 divided by pressure of Br2 products divided by reactants raised to stoichiometric coefficients with the work I've done in the previous lines I can write these pressure of Br2 as initial pressure plus lambda pressure of Br atoms is negative 2 lambda when I square it that becomes 4 lambda squared and now that expression looks familiar we had something very much like that written in terms of moles and squiggle rather than pressures and lambdas so the solution to that problem is again going to be a quadratic equation the exact same quadratic equation we saw previously lambda is going to turn out to be negative 1 plus or minus the square root of 1 plus 4 kp initial pressure over 8k kp in this case so algebra is exactly the same as previously the difference is the numbers we're plugging in for kp we're plugging in 3.9 times then the 29th per atmosphere notice the per atmosphere in kp is going to cancel the p0 the initial pressure 24 and a half atmospheres but if I plug in 3.9 times 10 to the 29th per atmosphere 24 and a half atmospheres and also plug in kp down here I get two solutions the two solutions I obtain are either positive or negative 10 to the 15th 10 to the minus 15th atmospheres and I haven't written the positive solution because that's the one that is non-physical that the lambda the extent of reaction written as a pressure is negative 4 times 10 to the minus 15th atmospheres so what that means is the amount of Br2 initial pressure plus that amount so 24 and a half atmospheres plus this ridiculously small number of atmospheres that's still 24 and a half atmospheres the pressure of Br pressure of Br is negative 2 times this extent of reaction in units of atmospheres so negative 2 times that is equal to 8 times 10 to the minus 15 atmospheres just to make sure we like that number that's the same result as we've gotten previously if I convert that to moles using for example moles is PV divided by RT if I do the ideal gas calculation and convert this many atmospheres at 298 Kelvin in a volume of 1 liter into moles what I find is not surprisingly perhaps bromine atoms with this pressure in 1 liter volume at 298 Kelvin is that amount of pressure is generated by 3.2 times 10 to the minus 16th moles of Br atoms so that is exactly the same answer that we got when we solved the equilibrium problem in terms of moles actually when we solved it in terms of molecules and converted to moles so that shows us that using Kp works just fine if we would rather solve an equilibrium problem in terms of pressures rather than moles we can do that and now we know how to convert back and forth between equilibrium constants in terms of molecules in terms of moles and in terms of pressures there's one more complication to consider the problem we've just done has assumed constant volume I used one liter of volume both at the start of the problem and at equilibrium so we assumed the volume was not changing things work out a little bit differently when the volume is constant than when the pressure is constant so that's a complication we need to consider also