In part A of this problem, we calculated the delta G in standard state conditions for the reaction to be of 143.7 kilo joules. Now that'll be kilo joules per mole of the balanced reaction as it is balanced up there. And we're asked now to allow this reaction to get to equilibrium and determine the pressure of oxygen. Because of the positive value for the standard delta G, we know that if we started with 1 atmosphere of oxygen, this reaction would proceed to the left. And as it proceeds to the left, the pressure of O2 is certainly going to be dropping as it goes, so we know it's going to be less than 1 atmosphere. The question is, what will its pressure be? Well, if we look at the balanced reaction, we can determine the pressure of the oxygen if we were to calculate Kp. Because Kp would be equal to the pressure of oxygen to the 1/2 power. We leave out the solids, we take our pressure, raised to the power of the coefficients. And so if we could determine Kp, we could determine the pressure of oxygen. So, how will we determine Kp? We will use the equation that the delta G standard is equal to minus RT natural log of K. Now, delta G, we've figured out in part A and we will plug it in here but I'm going to put it in units of joules instead of kilo joules. That would be 143,700 joules per mole. R is a negative 8.314 joules per mole Kelvin, and that's why I chose to put this in joules, so that the joules would cancel. The temperature needs to be in Kelvin, so it's 298 Kelvin, times the natural log of K. And since it's gasses, it would be a Kp value. If we divide both sides by the negative 8.314 and 298, we will be left with the natural log of Kp and this will equal a negative 58.00 and it'll have no units. If we take e to both sides, the natural log will cancel. And we'll be left with Kp. And Kp will be a very tiny number, it is 6.4 times 10 to the negative 26. This is a very small value. And that tells me that the equilibrium lies far to the left, which it should with a positive 143.7 kJ for the delta G standard. Now we're ready to determine the pressure. If Kp is equal to the pressure of O2 to the 1/2 power, and we want to know the pressure of O2, then we will square both sides, and that would give me the pressure of O2. So, Kp of 6.4 times 10 to the minus 26 squared would give me that pressure. And the number is 6.4, no 4.1 times 10 to the minus 51. And now let's add units. The pressures, when we're doing Kp, are always in atmospheres. So, this reaction proceeds far to the left until the pressure of oxygen is only 4.1 times 10 to the minus 51 atmospheres. And then equilibrium will have been established.