chemistry pls help

Question

At a certain temperature, 0.620 mol SO3 is placed in a 2.50 L container.

2SO3(g)↽−−⇀2SO2(g)+O2(g)

At equilibrium, 0.150 mol O2 is present. Calculate 𝐾c

Answer 1

2SO₃(g) <===> 2SO₂(g) + O₂(g)

0.620 mol...........0.................0...........Initial

-2x.....................+2x.............+x.........Change

0.620-2x............2x................x..........Equilibrium

Since we are told that at equilibrium there are 0.150 mol O₂, we now know that x = 0.150.

So, equilibrium concentrations are as follows:

[SO₂] = 2x = 0.300 mol / 2.50 L = 0.12 M

[O₂] = x = 0.150 mol / 2.50 L = 0.06 M

[SO₃] = 0.620 - 0.12 = 0.50 mol / 2.50 L = 0.20 M

Kc = [SO₂]²[O₂] / [SO₃]²

Kc = (0.12)²(0.06) / (0.20)²

Kc = 0.0216