2s(s) + 2 o2(g) ⇄ 2 so2(g) k1 = 2 x 10105 2 so2(g) + o2(g) ⇄ 2 so3(g) k2 = 7 x 1024 given the value of the equilibrium constants k1 and k2 for the reactions represented above, what is the value of the equilibrium constant, k3, for the following reaction? 2 s(s) + 3 o2(g) ⇄ 2 so3(g) k3 = ?

The equations and the values of K are:

2S(s) + 2O₂(g) ⇄ 2SO₂(g) K1 = 2x10¹⁰

2SO₂(g) + O₂(g) ⇄ 2SO₃(g) K2 = 7x10²⁴

K3 = 1.4x10³⁵

Explanation:

The equilibrium constant of a global reaction is the multiplication of the constants of the steps reactions. If the equation needed to be multiplied by some constant, the value of the equilibrium constant must be elevated by the same value. If the reaction must be inverted, the value of the equilibrium constant must be inverted (1/K):

2S(s) + 2O₂(g) ⇄ 2SO₂(g) K1 = 2x10¹⁰

2SO₂(g) + O₂(g) ⇄ 2SO₃(g) K2 = 7x10²⁴

2S(s) + 3O₂(g) + 2SO₂(g) ⇄ 2SO₂(g) + 2SO₃(g)

The substances in both sides are simplified:

2S(s) + 3O₂(g) ⇄ 2SO₃(g)

K3 = K1xK2

K3 = 2x10¹⁰ x 7x10²⁴

K3 = 1.4x10³⁵