For the reaction $A (g) + 2 B (g) ⇌ 2 C (g)$ , one mole of A and 1.5 mo! of B are taken in a 2.0 litre vessel. At equilibrium, the concentration of C was found to be $0.35 {molL}^{- 1}$ .The equilibrium constant $K_{c}$ of the reaction would be

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$2.36 M^{- 1}$

$0.673 M^{- 1}$

$0.295 M^{- 1}$

$1.178 M^{- 1}$

answer is C.

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Detailed Solution

$\begin{array}{l} A (g) + 2 B (g) ⇌ 2 C (g) \\ 1 mol - x 1.5 mol - 2 x 2 x \end{array}$

Hence,

$\begin{array}{l} \frac{2 x}{2 L} = 0.35 {molL}^{- 1} \Rightarrow x = 0.35 mol \\ K_{c} = \frac{[C]^{2}}{[A] [B]^{2}} = \frac{(0.70 mol / 2 L)^{2}}{(0.65 mol / 2 L) (0.8 mol / 2 L)^{2}} = 2.36 {mol}^{- 1} L \end{array}$

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