> #chemical-reaction , #chemistry > If 50% of SO3 os decomposed to SO2c and O2 what is the kc value considering initial concentration is 0.2mole?
Anonymous
Jun 23, 2023

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

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Anonymous
Jun 23, 2023
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.
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