The value of Kcfor the reaction 3O2 (g) ⇌ 2O3 (g) is 2.0 ×1050 at 25°C. If the equilibrium concentration of O2 in the air at 25°C is 1.6 ×10–2, what is the concentration of O3?

Given:

The reaction is 3O2 (g) ⇌ 2O3 (g)


The equilibrium constant of the given reaction = 2.0×10–50


Temperature = 25° C


Equilibrium concentration of O2 ([O2]) = 1.6 ×10–2 M


To find out the equilibrium concentration of O3, we will apply the


Law of Chemical Equilibrium, i.e., Kc =


Where X and Y are the products and A and B are the reactants.


Hence, equilibrium constant (Kc) of the given reaction,


3O2 (g) ⇌ 2O3 (g)


Kc =


As Kc = 2.0×10–50 (given) and [O2] = 1.6 ×10–2 M (given)


∴ 2.0×10–50 =


⇒[O3]2 = 2.0×10–50 × (1.6 × 10-2 M)3


⇒[O3]2 = 8.192 × 10-56


⇒[O3] =


⇒[O3] = 2.86 × 10-28 M


Thus, the equilibrium concentration of O3 is 2.86 × 10-28M


34