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.