If 50% of SO3 os decomposed to SO2c and O2 what is the kc value considering initial concentration is 0.2mole?

To determine the equilibrium constant (Kc) for the given reaction, we need the balanced equation for the reaction between SO₃, SO₂, and O₂. Assuming the reaction is as follows:

2 SO₃(g) ⟶ 2 SO₂(g) + O₂(g)

The balanced equation shows that two moles of SO₃ react to form two moles of SO₂ and one mole of O₂. Therefore, the stoichiometric ratio between SO₃ and SO₂ is 1:1.

Let's denote the initial concentration of SO₃ as [SO₃]₀ = 0.2 M. As the reaction proceeds, let's assume x moles of SO₃ decompose to form x moles of SO₂ and x moles of O₂.

At equilibrium, the concentration of SO₃ will be [SO₃] = [SO₃]₀ - x. The concentrations of SO₂ and O₂ will be [SO₂] = [SO₂]₀ + x and [O₂] = [O₂]₀ + x, respectively.

Now, let's write the expression for the equilibrium constant (Kc) in terms of these concentrations:

Kc = ([SO₂] * [O₂]) / ([SO₃])²

Substituting the expressions for the concentrations:

Kc = (([SO₂]₀ + x) * ([O₂]₀ + x)) / ([SO₃]₀ - x)²

Since the stoichiometric ratio between SO₃ and SO₂ is 1:1, [SO₂]₀ = [SO₃]₀ = 0.2 M.

Substituting the values:

Kc = ((0.2 + x) * ([O₂]₀ + x)) / (0.2 - x)²

Note that the value of [O₂]₀ is not provided in the question. Without the initial concentration of O₂, it is not possible to determine the exact value of Kc.