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For the reaction,
$N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)$
The equilibrium constant is $K_{1},$
The equilibrium constant is $K_{2}$ for the reaction
$2 N O (g) + O_{2} (g) ⇌ 2 N O_{2} (g)$
The equilibrium constant“ K” for the following reaction would be
$N O_{2} (g) ⇌ \frac{1}{2} N_{2} (g) + O_{2} (g) ?$

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$\frac{1}{(K_{1} K_{2})}$

$\frac{1}{(2 K_{1} K_{2})}$

$\frac{1}{(4 K_{1} K_{2})}$

${[\frac{1}{(K_{1} K_{2})}]}^{\frac{1}{2}}$

answer is D.

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Detailed Solution

$N_{2} (g) + O_{2} (g) ⇌ 2 N O (g), K_{1}$

$2 N O (g) + O_{2} (g) ⇌ 2 N O_{2} (g), K_{2}$

$N_{2} (g) + 2 O_{2} (g) ⇌ 2 N O_{2} (g), K = K_{1} \times K_{2}$

$∴ f o r N O_{2} (g) ⇌ 1 / 2 N_{2} (g) + O_{2} (g); K = {[\frac{1}{K_{1} \times K_{2}}]}^{1 / 2}$

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