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At a certain temperature, 0.980 mol SO₃ is placed in a 4.50 L container. The reaction is represented by:

2SO₃​(g)⇌2SO₂​(g)+O₂​(g)

At equilibrium, 0.130 mol O₂ is present. Calculate Kc​.

User Dchris
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Final answer:

To calculate the equilibrium constant (Kc) for the reaction 2SO₃(g) ⇌ 2SO₂(g) + O₂(g), the concentrations of the reactants and products at equilibrium are used in the Kc expression.

Step-by-step explanation:

To calculate the equilibrium constant (Kc) for the reaction 2SO₃​(g) ⇌ 2SO₂​(g) + O₂​(g), given the amount of SO₃ and O₂ at equilibrium, we need to use the equilibrium constant expression Kc = [SO₂]^2 [O₂] / [SO₃]^2. We know that initially there was 0.980 mol of SO₃ in a 4.50 L container, which gives us an initial concentration of [SO₃] = 0.980 mol / 4.50 L = 0.218 M.

At equilibrium, 0.130 mol of O₂ is present, which translates to [O₂] = 0.130 mol / 4.50 L = 0.029 M. Since the reaction produces O₂ in half the amount to the SO₂ produced, there will be 0.260 mol of SO₂ formed (twice the amount of O₂), which means [SO₂] = 0.260 mol / 4.50 L = 0.058 M.

The change in concentration for SO3 is equivalent to the amount of O₂ formed, so the equilibrium concentration of [SO₃] is 0.218 M - 0.029 M = 0.189 M.

Plugging these values into the equilibrium expression:

Kc = ([SO₂]^2 [O₂]) / [SO₃]^2 = (0.058 M)^2 x (0.029 M) / (0.189 M)^2

Calculating this gives us:

Kc = 0.00336 / 0.0357 = ≈ 0.094

Therefore, the equilibrium constant for the reaction at this temperature is approximately 0.094.

After calculating the concentrations, Kc is found to be approximately 0.094 at the given temperature.

User Baikho
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