Question

The equilibrium constant K, for the oxidation of sulfur dioxide into sulfur trioxide is 5.6x10¹ at 350°C. If the equilibrium partial pressures of SO, and SO, are 0.64 atm and 0.58 atm, respectively, what is the partial pressure of O, in the system?

2SO2(g) + O2(g) 250,(g)

Answer 1

2SO₂ + O₂ = 2SO₃

K_c = 5,6 * 10^-1 = 0,56

P (SO₂) = 0,64

P (SO₃) = 0,58

P (O₂) = ?

$K_c = \frac {[SO_3]²}{[SO_2]²[O_2]}$

$[O_2]= \cfrac {[SO_3]²}{\frac{[SO_2]²}{K_c}}$

$[O_2]= \cfrac {[0,58]²}{\frac{[0,64]²}{0,56}}= 0,46$

ANSWER: 0,46