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

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"