The reaction a (g) ⇌2b (g) has an equilibrium constant of k = 0.030. what is the equilibrium constant for the reaction b (g) ⇌12a (g) ?

Chemical reaction 1: A ⇄ 2B. K₁ = 0,030.

Chemical reaction 2: B ⇄ 1/2A. K₂ = ?.

K₁ = [B]² / [A]. [B] = √K₁·[A].

Second chemical reaction is reversed first chemical reaction and diveded by two.

K₂ = √[A]/[B].

K ₂ = √1/K₁.

K₂ = √1/0,03.

K₂ = 5,77; the equilibrium constant for the second chemical reaction.

First, we have to correct the equation in the question to b (g) ⇆ 1/2 A (g)

at the first equation A (g) ⇆ 2 B (g) so,

Kc = [B]^2 [ A] = 0.03

by reverse the equation 2B⇆ A

∴ Kc (original) = [A] / [B]^2

= 1/0.03 = 33 M^-1

and the new equation B⇆ (1/2) A

So, the new Kc = √Kc (original = √33

∴ KC = 5.7