Reaction between $N_{2}$ and $O_{2}$ takes place as follows:
$2 N_{2(g)} + O_{2(g)} \longleftrightarrow 2 N_{2}O_{(g)}$
If a mixture of $0.482 \ mol$ of $N_{2}$ and $0.933 \ mol$ of $O_{2}$ is placed in a $10 \ L$ reaction vessel and allowed to form $N_{2}O$ at a temperature for which $K_{c} = 2.0 \times 10^{-37}$,determine the composition of the equilibrium mixture.

The given reaction is: $2 N_{2(g)} + O_{2(g)} \longleftrightarrow 2 N_{2}O_{(g)}$
Initial amounts: $n(N_{2}) = 0.482 \ mol$,$n(O_{2}) = 0.933 \ mol$,$n(N_{2}O) = 0 \ mol$.
Let the amount of $N_{2}O$ formed at equilibrium be $x \ mol$.
At equilibrium,the amounts are: $n(N_{2}) = (0.482 - x) \ mol$,$n(O_{2}) = (0.933 - x/2) \ mol$,$n(N_{2}O) = x \ mol$.
In a $10 \ L$ vessel,the concentrations are: $[N_{2}] = \frac{0.482 - x}{10}$,$[O_{2}] = \frac{0.933 - x/2}{10}$,$[N_{2}O] = \frac{x}{10}$.
Since $K_{c} = 2.0 \times 10^{-37}$ is extremely small,the reaction barely proceeds to the right. Thus,$x$ is negligible compared to initial amounts.
$[N_{2}] \approx \frac{0.482}{10} = 0.0482 \ M$,$[O_{2}] \approx \frac{0.933}{10} = 0.0933 \ M$.
$K_{c} = \frac{[N_{2}O]^{2}}{[N_{2}]^{2}[O_{2}]} = 2.0 \times 10^{-37}$.
$\frac{(x/10)^{2}}{(0.0482)^{2}(0.0933)} = 2.0 \times 10^{-37}$.
$x^{2} = 100 \times 2.0 \times 10^{-37} \times (0.0482)^{2} \times 0.0933 \approx 4.335 \times 10^{-38}$.
$x \approx 6.58 \times 10^{-20} \ mol$.
Equilibrium composition: $[N_{2}] = 0.0482 \ M$,$[O_{2}] = 0.0933 \ M$,$[N_{2}O] = \frac{6.58 \times 10^{-20}}{10} = 6.58 \times 10^{-21} \ M$.