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2 moles of $R (g)$ and 4 moles of $S (g)$ are mixed together in a sealed 1L vessel and allowed to attain equilibrium $P (g) + 2 Q (g) ⇌ R (g) + S (g) k_{c}$ for the above equilibrium is $10^{- 8} L m o l^{- 1} .$ The equilibrium concentration of mixture is/are

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a

$R = 16 \times 10^{- 8} M$

b

$Q = 4 M$

c

$P = 2 M$

d

$S = 2 M$

answer is A, B, C, D.

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Detailed Solution

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$\to$ Forward $K c = 10^{- 8}$
Backward $K c' = \frac{1}{K c} = 10^{8}$
$10^{8} > > > > 1 x ≃ 2 4 - x = 4 - 2 ≃ 2 K c' = \frac{1}{K c} = 10^{8} = \frac{[P] {[Q]}^{2}}{[R] [S]} 10^{8} = \frac{x \times {(2 x)}^{2}}{y \times (4 - x)} = \frac{4 x^{3}}{y \times (4 - x)} = \frac{4 \times 2^{3}}{y \times (4 - 2)} 10^{8} = \frac{4 \times 8}{y \times 2} y = 16 \times 10^{- 8}$

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