Question

You start with a flask at 900 K containing only sulfur trioxide at an unknown pressure and allow the equilibrium below to be established. If the final sulfur trioxide pressure is 5.00 atm, what was its initial pressure?

2 SO₂ (g) + O₂ (g)   <-->   2 SO₃ (g)     K_p = 0.345

NOTE: It is possible to choose your variables so that you get a cubic equation that is easily solvable.

Answer 1

2 SO2 (g) + O2 (g)   <-------------> 2 SO3 (g)

   0                0                                     x

2a              a x - 2a

at equilibrium :

pressure of SO3 = 5.00 atm

x - 2a = 5

Kp = P^3SO3 / P^2SO2 x PO2

0.345 = (x - 2a)^2 / (2a)^2 x a

a = 0.3686 , 1.265

x - 2a = 5

x = 5 + 2a = 5 + 2 x 1.265

x = 7.53

initial pressure of SO3 = 7.53 atm