At a certain temperature, 0.660 mol SO3 is placed in a 2.00 L
container.

2SO3(g)↽−−⇀2SO2(g)+O2(g)
2
SO
3
(
g
)
↽
−
−
⇀
2
SO
2
(
g
)
+
O
2
(
g
)

At equilibrium, 0.150 mol O2
0.150
mol
O
2
is present. Calculate c.
K
c
.

0,15

Explanation:

Kc is the constant of equilibruim of concentration.

[ ] that symbol means concentration.

Kc = Products/Reagents

Kc = [SO2]² . [O2]/[SO3]²

The questions says that in equilibrium we have:

[0,15mol] of O2

Now, looking to the balancing of the equation, we see that SO3 and O2 are in a disposition 2:1. Which means that for every molecule of O2 formed, we need to use 2 molecules of SO3.

Then, we used 0,30 mol of SO3 for that convertion. That also means that we have 0,30mol of SO2, because he is in a disposition of 1:1.

[SO2] = 0,30mol

[SO3] = 0,30mol

Kc = [0,30]². [0,15]/[0,30]²

Kc = 0,15