Question

What is K_C for the following equilibrium when the equilibrium concentration of each substances is [SO₂] = 060 M, [O₂] = 0.82 M and [SO₃] = 1.90 M?

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

Answer 1

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

KC = \(\frac{[SO_3]^2}{[SO_2]^2[O_2]}\) = \(\frac{[1.90]^2}{[0.60]^2[0.82]}\)

= \(\frac{3.61}{0.36 \times .82}\) = \(\frac{3.61}{0.295}\) = 12.229 mol/litre