Question

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)?

Answer 1

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

Answer 2

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].
K₂ = √1/K₁.
K₂ = √1/0,03.
K₂ = 5,77; the equilibrium constant for the second chemical reaction.