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The reaction $N_{2} + 3 H_{2} \Leftrightarrow 2 {NH}_{3}$ takes place at $450^{0} C$ . 1 mole of $N_{2}$ and 2 mole of $H_{2}$ are mixed in a 1 litre vessel and 1 mole of ${NH}_{3}$ is formed at equilibrium. Then $K_{c}$ for the above reaction is

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answer is C.

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Detailed Solution

$N_{2} + 3 H_{2} \Leftrightarrow 2 {NH}_{3} 1 2 0 1 - x 2 - 3 x 2 x 2 x = 1, x = \frac{1}{2} K_{c} = \frac{{({NH}_{3})}^{2}}{(N_{2}) {(H_{2})}^{3}} = \frac{1^{2}}{(\frac{1}{2}) {(\frac{1}{2})}^{3}} = 16 .$

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