Question

Given the following, what is the equilibrium constant for the reaction: 2 A + 3 D â 3 C + 2 B? I: 2 A â 2 B, K = 1x105 II: C â D, K = 2x103 A) 1x10â10 B) 1x10â5 C) 5x101 D) 2x108 E) 8x109 F) 8x1014

Answer 1

K= 1.25*10⁻⁵ (any of the options) , but if the problem's statement is wrong and D → C, K₂ = [C]/[D] = 2*10³ then K= 8*10¹⁴ (option F)

Step-by-step explanation:

for

2 A + 3 D → 3 C + 2 B

K= [C]³*[B]² /([A]²*[D]³) = ([B]/[A])²* ([C][/D])³

where

2 A → 2 B , K₁ = [B]²/[A]² = ([B]/[A])² =1*10⁵

and

C → D, K₂ = [D]/[C] = 2*10³

therefore

K=K₁* (1/K₂)³=K₁/K₂³ = 1*10⁵ / (2*10³)³ = 1.25*10⁻⁵

K= 1.25*10⁻⁵

but if the problem's statement is wrong and

D → C, K₂ = [C]/[D] = 2*10³

then

K=K₁*K₂³=K₁*K₂³ = 1*10⁵ * (2*10³)³ =8*10¹⁴

K= 8*10¹⁴