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20 November, 02:41

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) ?

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  1. 20 November, 02:49
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    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
  2. 20 November, 03:02
    0
    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.
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