The equilibrium constant for the reaction; 5 O_ 3 (g) 5 O_ 2 (g)+ 1 2 O_ 2 (g) is K c = 4.9 × 10 -2 . The value of K c for the reaction; 25 O_ 2 (g)+O_ 2 (g) 25 O_ 3 (g) will be:

The equilibrium constant for the reaction; 5 O_ 3 (g) 5 O_ 2 (g)+ 1 2 O_ 2 (g) is K c =…

Chemistry · JEE Main · NTA Exams — Equilibrium

The equilibrium constant for the reaction;

\(5 O_{3}(g) \rightleftharpoons 5 O_{2}(g)+\frac{1}{2} O_{2}(g)\)

is K_c = 4.9 × 10^-2. The value of K_c for the reaction;

\(25 O_{2}(g)+O_{2}(g) \rightleftharpoons 25 O_{3}(g)\)

will be:

(B) 2.40 × 10 -3

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