At a certain temperature, 0.680 mol of SO3 is placed in a 5.00 L container. 2SO3(g)↽−−⇀2SO2(g)+O2(g) At equilibrium, 0.0200 M O2 is present. Calculate 𝐾c.

Answer 1

2SO₃(g) <==> 2SO₂(g) + O₂(g) ... balanced equation

Kc = [SO₂]²[O₂] / [SO₃]²

Converting 0.680 mols SO₃ to moles/L (M), and setting up an ICE table, we have

0.680 mols SO₃ / 5.00 L = 0.136 M

2SO₃(g) <==> 2SO₂(g) + O₂(g)

0.136..............0..............0............Initial

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

0.136-2x........2x.............x...........Equilibrium

@ equilibrium O₂ = 0.02 M = x

[SO₃] = 0.136 - 2x = 0.136 - 0.04 = 0.096 M

[SO₂] = 2x = 0.04 M

[O₂] = 0.02 M

Kc = [SO₂]²[O₂] / [SO₃]²

Kc = (0.04)²(0.02) / (0.096)²

Kc = 0.00347

(Be sure to check all math)