For the formation of ammonia from its constituent elements (1 mole of $N_{2}$ and 3 moles of $H_{2}$ ) in a closed vessel of volume $V (l i t),$ the value of $K_{c}$ is [units of $K_{C} = m o l^{- 2} L^{2}$ ]

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$\frac{4 x V^{2}}{9 {(1 - x)}^{3}}$

$\frac{3 x^{2} V^{2}}{9 {(1 - x)}^{4}}$

$\frac{4 x^{2} V^{2}}{27 {(1 - x)}^{4}}$

$\frac{x^{2} V^{2}}{27 {(1 - x)}^{3}}$

answer is C.

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Detailed Solution

$\underset{1 - x}{N_{2}} + \underset{3 - 3 x}{3 H_{2}} ⇌ \underset{2 x}{2 {NH}_{3}} K_{c} = \frac{{(\frac{2 x}{v})}^{2}}{\frac{(1 - x)}{v} \times {(\frac{3 - 3 x}{v})}^{3}} = \frac{4 x^{2} \times v^{2}}{(1 - x) \times [3 (1 - x)]^{3}} = \frac{4 x^{2} \times v^{2}}{27 (1 - x)^{4}}$