At a certain temperature, 0.660 mol of SO3 is placed in a 3.00-L container

certain-temperature

#1

At a certain temperature, 0.660 mol of SO3 is placed in a 3.00-L container.

2SO3(g) <—> 2SO2 + O2(g)

At equilibrium, 0.160 mol of O2 is present. Calculate Kc.

Answer:

When this equilibrium shifts to the right, forming “X” mol of O2
2SO3(g) —> 2SO2(g) + O2(g)
0.660 - 2X --> --> 2X & X

since X = 0.130
this becomes
0.400 <=> 0.260 & 0.130

find moles / litre:
(0.400) / 4.5L --> --> 0.260 / 4.5L & 0.130 / 4.5 L

which gives these molarities:
2SO3(g) —> 2SO2(g) + 1 O2(g)
[0.08889] --> [0.05778] & [0.02889]

Kc = [SO2]^2 [O2] / [SO3]^2

Kc = [0.05778]^2 [0.02889] / [0.08889]^2

Kc = (0.003338) [0.02889] / [(0.007901)

Kc = 0.0122