For the formation of ammonia from its constituent elements ( 1 mole of $\mathrm{N}_2$ and 3 moles of $\mathrm{H}_2$ ) in a closed vessel of volume $\mathrm{V}(\mathrm{L})$, the value of $\mathrm{K}_{\mathrm{C}}$ is [units of $\mathrm{K}_{\mathrm{C}}=\mathrm{mol}^{-2} \mathrm{~L}^2$ ]

• A

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

  $\frac{4 x \mathrm{~V}^2}{9(1-x)^3}$
• Correct answer
  C

  $\frac{4 x^2 \mathrm{~V}^2}{27(1-x)^4}$
• D

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

Solution

$\begin{aligned} & \therefore \quad \mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3}=\frac{(2 \mathrm{x})^2 / \mathrm{v}^2}{\{(1-\mathrm{x}) / \mathrm{v}\}\left\{(3-3 \mathrm{x})^3 / \mathrm{v}^3\right\}} \\ & \quad=\frac{4 \mathrm{x}^2 \mathrm{v}^2}{(1-\mathrm{x})\{3(1-\mathrm{x})\}^3}=\frac{4 \mathrm{x}^2 \mathrm{v}^2}{27(1-\mathrm{x})^4}\end{aligned}$

Hence, the correct answer is option: 'C'.