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$A + B \to C + D$ Initially the concentration of A and B are both equal but at equilibrium concentration of D will be twice of that of A then what will be equilibrium constant of reaction is

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$\frac{1}{9}$

$\frac{4}{9}$

$\frac{2}{9}$

$4$

answer is D.

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Detailed Solution

$A + B = C + D$

$\begin{array}{lllll} Initial conc 1 1 0 0 \end{array}$

$Eq. conc 1 - x 1 - x x x$

$i.e., x = 2 \times 1$

$K_{C} = \frac{x \times x}{(1 - x) (1 - x)} = \frac{2 \times 2}{(2 - 1) (2 - 1)} = 4$

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