Reaction between nitrogen and oxygen takes place as following $2 N_{2} (g) + O_{2} (g) ⇌ 2 N_{2} O (g)$
If a mixture of 0.482 mole of $N_{2}$ and 0.933 mole of $O_{2}$ is placed in a reaction vessel of volume 10L and allowed to form $N_{2} O$ at a temperature for which $K_{c} = 2.0 \times 10^{- 37}$ , the equilibrium concentration of $[N_{2} O]$ will be

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$7.06 \times 10^{- 20} m o l L^{- 1}$

$6.58 \times 10^{- 21} m o l L^{- 1}$

$4.82 \times 10^{- 4} m o l L^{- 1}$

$9.36 \times 10^{- 7} m o l L^{- 1}$

answer is B.

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Detailed Solution

$K_{c} = 2.0 \times 10^{- 37}$

Magnitude of $K_{c}$ is very very small such that at equilibrium
$\frac{0.482 - 2 x}{10} \approx \frac{0.482}{10} = 0.0482 {[N_{2}]}_{e q} = 0.0482 m o l L^{- 1} {[O_{2}]}_{e q} = 0.0933 m o l L^{- 1}, K_{c} = \frac{{[N_{2} O]}^{2}}{{[N_{2}]}^{2} [O_{2}]} 2.0 \times 10^{- 37} = \frac{\frac{4 x^{2}}{100}}{{(0.0482)}^{2} (0.0933)} x = 3.29 \times 10^{- 20} [N_{2} O] = \frac{2 x}{10} [N_{2} O] = \frac{2 \times 3.292 \times 10^{- 20}}{10} = 6.58 \times 10^{- 21} m o l L^{- 1}$