# The value of <spa

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

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :

Bromine monochlorNCERT - Chemistry Part-I

The equilibrium cNCERT - Chemistry Part-I