Question

At a certain temperature, 0.680 mol SO3 is placed in a 4.50 L container.

2SO3(g)⇌2SO2(g)+O2(g)

At equilibrium, 0.110 mol O2 is present. 

Answer 1

Initial concentration of SO3 = mol of SO3 / volume in L
= 0.680 mol / 4.50 L
= 0.151 M

Initial concentration of O2 = mol of O2 / volume in L
= 0.110 mol / 4.50 L
= 0.0244 M

ICE Table:

Given at equilibrium,
[O2] = 0.0244
+1x = 0.0244
x = 0.0244
Equilibrium constant expression is
Kc = [SO2]^2[O2]/[SO3]^2
Kc = (+2x)^2(+1x)/(0.151-2x)^2
Kc = (+2*0.0244)^2(+1*0.0244)/(0.151-2*0.0244)^2
Kc = 0.00556
Answer: 5.56*10^-3