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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

Respuesta :

Edufirst

Answer:

[tex]\large\boxed{\large\boxed{2.0\times 10^{-50}M^-1}}}[/tex]

Explanation:

The reaction is:

        [tex]3O_2(g)\rightleftharpoons 2O_3(g)[/tex]

The equlibrium constant equation is:

       [tex]K_c=\dfrac{[O_3g)]^2}{[O_2(g)]^3}[/tex]

Substitute:

      [tex]K_c=\dfrac{(2.86\times 10^{-28}M)^2}{(1.6\times 10^{-2}M)^3}=2.0\times 10^{-50}M^{-1}[/tex]

harryjuice

Answer:

A.) 2.0 x 10*50

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