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In the reaction $A + 2 B \underset{}{⇌} 2 C + D$ , the initial concentration of B is 3/2 times to that of A. If the equilibrium concentrations of A & C are equal. The value of Kc for the reaction is

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0.04

0.32

0.16

2.4

answer is B.

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Detailed Solution

$A + 2 B \underset{}{⇌} 2 C + D$

$x$ 3/2x 0 0

$(x - α) (3 / 2 x - 2 α) 2 α α$

$x - α = 2 α$

$α = \frac{x}{3}$

$k_{c} = \frac{{(\frac{2 x}{3})}^{2} \times (\frac{x}{3})}{(\frac{2 x}{3}) {(\frac{5 x}{6})}^{2}} = 0.32$

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