[SO₃]₀ = 2.00 M, [O₂]₀ = 0.0850 M, [SO₂]₀ = 0 M and [SO₂]_eq = 0.120 M

Calculate K_c for the following reaction:

2 SO₃ (g) $⇌$ 2 SO₂ (g) + O₂ (g)

Question

[SO₃]₀ = 2.00 M, [O₂]₀ = 0.0850 M, [SO₂]₀ = 0 M and [SO₂]_eq = 0.120 M

Calculate K_c for the following reaction:

2 SO₃ (g)

⇌

2 SO₂ (g) + O₂ (g)

| General Chemistry 3

Accepted Answer

K_c =

\frac{{[{SO}_{2}]}_{eq}^{2} {[O_{2}]}_{eq}}{{[{SO}_{3}]}_{eq}^{2}}

From the stoichiometry of the reaction:

[SO₃]_eq = [SO₃]₀ – 2X

[O₂]_eq = [O₂]₀ + X

[SO₂]_eq = [SO₂]₀ + 2X = 2X = 0.120 M

⇒ X = 6.00 x 10^-2 M

At the equilibrium:

[SO₃]_eq = [SO₃]₀ – 2X = 2.00 – 0.120 = 1.88 M

[O₂]_eq = [O₂]₀ + X = 0.0850 + 0.0600 = 0.145 M

[SO₂]_eq = 0.120 M

K_c =

\frac{{[{SO}_{2}]}_{eq}^{2} {[O_{2}]}_{eq}}{{[{SO}_{3}]}_{eq}^{2}}

K_c =

\frac{0 . 120^{2} \times 0.145}{1 . 88^{2}}

K_c = 5.91 x 10^-4

Midterm 1 - Question 2 | General Chemistry 3