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The equilibrium constant for the reaction, $N_{2} (g) + O_{2} (g) ⇌ 2 NO (g)$ at temperature T is $4 x 10^{- 4}$ . The value of $K_{c}$ for the reaction, $NO (g) ⇌ \frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g)$ at the same temperature is :

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$2.5 \times 10^{2}$

$0.02$

$4 \times 10^{- 4}$

$50$

answer is D.

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Detailed Solution

$N_{2} (g) + O_{2} (g) \to 2 N O (g)$

$K_{c} = \frac{[{NO}^{2}}{[N_{2}] [O_{2}]} = 4 \times 10^{- 4}$

For

$\Rightarrow 2 NO (g) \to N_{2} (g) + O_{2} (g)$

$K_{c}^{'} = \frac{1}{K_{c}} = \frac{[N_{2}] [O_{2}]}{[{NO}_{2}]} = \frac{1}{4 \times 10^{- 4}} = \frac{10^{4}}{4}$

$\Rightarrow$ $N O (g) \to \frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g)$

${K_{c}}^{''} = \frac{{[N_{2}]}^{1 / 2} [O_{2}]}{{[{NO}_{2}]}^{2}}$

$⊨ \sqrt{{K_{c}}^{'}} = \sqrt{\frac{10^{4}}{4}}$

$= \frac{100}{2} = 50$

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