A sample of SO3 is introduced into an evacuated sealed container and heated to 600 K. The following equilibrium is established:

2SO3 (g) ⇌2SO2 (g) + O2 (g).

The total pressure in the system is found to be 3.0 atm and the mole fraction of O2 is 0.12. Find Kp ...

Kp = 0.0506

Explanation:

0.12 mole fraction of O2 x 3.0 atm = 0.36 atm

The mole fraction of SO2 will be double the mole fraction of O2 since the balanced equation states that 2 moles of SO2 will be obtained for every one mole of O2.

So the mole fraction of SO2 is 2 (0.12) = 0.24

Pressure due to SO2 = 0.24 x 3.0 atm = 0.72 atm

The pressure due to SO3 = total pressure minus pressure of SO2 and O2

3.0 atm - (0.36 + 0.72) = 1.92 atm

Kp = (pSO2) ^2 (pO2) / (pSO3) ^2

Kp = (0.72) ^2 (0.36) / (1.92) ^2

Kp = 0.0506