Question

Consider the reaction.

3 upper O subscript 2 (g) double-headed arrow 2 upper O subscript 3 (g).

At 298 K, the equilibrium concentration of O2 is 1.6 x 10-2 M, and the equilibrium concentration of O3 is 2.86 x 10-28 M. What is the equilibrium constant of the reaction at this temperature?
A) 2.0 x 10^-10
B) 2.0 x 10^10
C) 1.8 x 10^-10
D) 1.8 x 10^10

Answer 1

A.) 2.0 x 10*50

Answer 2

`\large\boxed{\large\boxed{2.0* 10^(-50)M^-1}}}`

Step-by-step explanation:

The reaction is:

`3O_2(g)\rightleftharpoons 2O_3(g)`

The equlibrium constant equation is:

`K_c=([O_3g)]^2)/([O_2(g)]^3)`

Substitute:

`K_c=((2.86* 10^(-28)M)^2)/((1.6* 10^(-2)M)^3)=2.0* 10^(-50)M^(-1)`