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2mole of $N_{2}$ is mixed with 6 mol of $H_{2}$ in a closed vessel of one litre capacity. If 50% of $N_{2}$ is converted in to $N H_{3}$ at equilibrium, the value of $K_{c}$ for the reaction $N_{2} (g) + 3 H_{2} (g) ⇌_{}^{} 2 N H_{3} (g)$ is

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1/27

4/27

27/4

answer is A.

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Detailed Solution

$N_{2}$ + $3 H_{2}$ $⇌_{}^{}$ $2 N H_{3}$

Initral 2mol 6mol

At.Eqm. 1mol 3mol 2mol

Molar conc. 1 3 2mol $L^{- 1}$

$k_{c} = \frac{{(2)}^{2}}{1 \times {(3)}^{3}} = \frac{4}{27}$

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