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Reaction Quotient

In a 1 lit. c...

Question

In a 1 lit. container following equilibrium is established with equal moles of $N O_{2}$ (g) & $N_{2} O_{4}$ (g).
$N_{2} O_{4} (g) ⇌ 2 N O_{2}$ (g),at equilibrium $M_{a v g .} = \frac{184}{3}, t h e n r a t i o o f$ K_{C}$ & total initial moles is.

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Solution

The correct option is B 3/2
$92 (b - x) + (b + 2 x) 46 = \frac{184}{3} (2 b + x)$ .
Hence, $x = b / 4$ .
The expression for the equilibrium constant becomes $K_{C} = \frac{9 b^{2} / 4}{3 b / 4} = 3 b$ .
Hence, the ratio of $K_{C}$ & total initial moles is $\frac{K_{C}}{2 b} = \frac{3}{2}$ .

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