An 800 g sample of SO3 was placed an evacuated container whe

An 8.00 g sample of SO3 was placed an evacuated container where it is decomposed at 600 degrees C according to the following reaction. SO3(g) <---> SO(g) + 1/2O2(g) At equilibrium the total pressure and density of the gaseous mixture were 1.80 atm and 1.60 g/L respectively. Calculate Kp for the reaction.

Solution

You start with pure SO3. It has a mole wt of 80. For each mole of this that decomps to SO2 and O2, you wind up with 1 mole of something with a 64 mole wt and 1/2 mole of something with a 32 mole wt. So your density for any combination of these 3 is 1.6 g/L. So this tells us that the VOLUME is 5 Liters. With T,V and n known, we can compute the initial pressure. Do this.

HOWEVER, the pressure does not follow the same arrangement as the rxn moves towards equilibrium. This is because at constant T and V, P is proportional to moles. For each mole of SO3 decomposed, we now have 1.5 moles, so the pressure in the container will RISE. But we can compute the pressure of each component.
If we call the initial pressure we computed
(Pso3)o, and the amount lost (x), then at equilibium,
Pso3+Pso2+Po2 = 1.8
(Pso3)o-x + x + x/2 = 1.8.
Solve for all 3 of these and find Kp as
[Pso2] sqrt[Po2] /[Pso3] = Kp