TS EAMCET · Chemistry · Chemical Equilibrium
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 \(\frac{4 x \mathrm{~V}^2}{9(1-x)^3}\)
- C \(\frac{4 x^2 \mathrm{~V}^2}{27(1-x)^4}\)
- D \(\frac{x^2 \mathrm{~V}^2}{27(1-x)^3}\)
Answer & Solution
Correct Answer
(C) \(\frac{4 x^2 \mathrm{~V}^2}{27(1-x)^4}\)
Step-by-step Solution
Detailed explanation
\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\}} \\ &…
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