0:01

In part A of this problem, we calculated the delta G in standard

Â state conditions for the reaction to be of 143.7 kilo joules.

Â Now that'll be kilo joules per mole of

Â the balanced reaction as it is balanced up there.

Â And we're asked now to allow this reaction to get to equilibrium and

Â determine the pressure of oxygen.

Â Because of the positive value for the standard delta G, we know that if we

Â started with 1 atmosphere of oxygen, this reaction would proceed to the left.

Â And as it proceeds to the left, the pressure of O2 is certainly going to be

Â dropping as it goes, so we know it's going to be less than 1 atmosphere.

Â The question is, what will its pressure be?

Â Well, if we look at the balanced reaction,

Â we can determine the pressure of the oxygen if we were to calculate Kp.

Â Because Kp would be equal to the pressure of oxygen to the 1/2 power.

Â We leave out the solids, we take our pressure,

Â raised to the power of the coefficients.

Â And so if we could determine Kp, we could determine the pressure of oxygen.

Â So, how will we determine Kp?

Â We will use the equation that the delta G

Â standard is equal to minus RT natural log of K.

Â Now, delta G, we've figured out in part A and we will plug it in here but

Â I'm going to put it in units of joules instead of kilo joules.

Â That would be 143,700 joules per mole.

Â 1:39

R is a negative 8.314 joules per mole Kelvin, and

Â that's why I chose to put this in joules, so that the joules would cancel.

Â The temperature needs to be in Kelvin, so

Â it's 298 Kelvin, times the natural log of K.

Â And since it's gasses, it would be a Kp value.

Â If we divide both sides by the negative 8.314 and

Â 298, we will be left with the natural log of Kp and

Â this will equal a negative 58.00 and it'll have no units.

Â If we take e to both sides, the natural log will cancel.

Â And we'll be left with Kp.

Â And Kp will be a very tiny number,

Â it is 6.4 times 10 to the negative 26.

Â This is a very small value.

Â And that tells me that the equilibrium lies far to the left,

Â which it should with a positive 143.7 kJ for the delta G standard.

Â Now we're ready to determine the pressure.

Â If Kp is equal to the pressure of O2 to the 1/2 power,

Â and we want to know the pressure of O2, then we will square both sides, and

Â that would give me the pressure of O2.

Â So, Kp of 6.4 times 10 to the minus 26 squared would give me that pressure.

Â 3:06

And the number is 6.4,

Â no 4.1 times 10 to the minus 51.

Â And now let's add units.

Â The pressures, when we're doing Kp, are always in atmospheres.

Â So, this reaction proceeds far to the left until the pressure of oxygen is

Â only 4.1 times 10 to the minus 51 atmospheres.

Â