At a certain temperature, 0.680 mol of SO3 is placed in a 3.50-L container. At equilibrium,...

Question

Question

At a certain temperature, 0.680 mol of SO3 is placed in a 3.50-L container. At equilibrium, 0.170 mol of O2 is present. Calcu

Answer 1

Answer #1

For this reaction,

K_c = ( [SO₂]² x [O₂] ) / [SO₃]²

No. of moles of O₂ = 0.170 moles

Volume of container = 3.50 L

Concentration of O₂ = no . of moles / volume in L = 0.170 / 3.50 = 0.0486 mols/L

As per the chemical equation,

2 moles of SO₃ gives 1 mole of O₂ .

No. of moles of SO₃ = 0.170 x 2 = 0.340 moles

Concentration of SO₃ = no . of moles / volume in L = 0.340 / 3.50 = 0.0971 mols/L

Since 2 moles of SO₃ gives 2 moles of SO₂ ,

Concentration of SO₂ = Concentration of SO₃ = 0.0971 mols/L

Substituting in equation,

K_c = ( [SO₂]² x [O₂] ) / [SO₃]² = (0.0971 mols/L)² x (0.0486 mols/L)² / (0.0971 mols/L)²

= 0.0486² mol² L² = 0.002361 mol² L² = 2.36 x 10^-3 mol² L²

Best Wishes.

Answer 2

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