At a certain temperature, 0.860 mol SO3 is placed in a 2.50 L container.

Question

2SO3(g)↽−−⇀2SO2(g)+O2(g)

At equilibrium, 0.100 mol O2 is present. Calculate 𝐾c.

𝐾c=_____

Answer 1

2SO3(g) <==> 2SO2(g) + O2(g)

0.860 mol.........0 mol........0 mol ... Initial

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

0.860-2x..........2x.............x.............Equilibrium

x = 0.100 mols

mols SO₃ = 0.860 - 0.1 = 0.760 mols

mols SO₂ = 2x = 0.200

mols O₂ = 0.100

In a volume of 2.50 L, the following concentrations are present at equilibrium:

[SO₃] = 0.760 mol / 2.50 L = 0.304 M

[SO₂] = 0.200 mol / 2.50 L = 0.0800 M

[O₂] = 0.0400 M

Kc = [SO₂]²[O₂] / [SO₃]²

Kc = (0.0800)2(0.0400) / (0.304)²

Kc = 0.00277