Answer:
\[\underset{{{t}_{\begin{smallmatrix}
0 \\
_{\begin{smallmatrix}
{{t}_{eq}} \\
Equilibrium
\\
concentration
\end{smallmatrix}}
\end{smallmatrix}}}}{\overset{{}}{\mathop{{}}}}\,\underset{\underset{\begin{smallmatrix}
(0.482-x) \\
\frac{(0.482-x)}{10}
\end{smallmatrix}}{\mathop{0.482}}\,}{\mathop{2{{N}_{2}}(g)}}\,\,\,\,+\,\,\,\,\,\,\,\underset{\underset{\begin{smallmatrix}
(0.933-x/2) \\
\frac{(0.933-x/2)}{10}
\end{smallmatrix}}{\mathop{0.933}}\,}{\mathop{{{O}_{2}}(g)}}\,\rightleftharpoons
\underset{\begin{smallmatrix}
0 \\
x
\\
\frac{x}{10}
\end{smallmatrix}}{\mathop{2{{N}_{2}}O(g)}}\,\]
The value of
is very small, i.e., extent of
reaction
will also be very small.
\[\therefore
\] \[[{{N}_{2}}]=\left[ \frac{0.482-x}{10} \right]=0.0482M\]
\[[{{O}_{2}}]=\left[
\frac{0.933-x/2}{10} \right]=0.0933M\]
\[[{{N}_{2}}O]=0.1x\]
\[{{K}_{c}}=\frac{{{[{{N}_{2}}O]}^{2}}}{{{[{{N}_{2}}]}^{2}}[{{O}_{2}}]}=\frac{{{(0.1x)}^{2}}}{{{(0.0482)}^{2}}(0.0933)}\]
\[2\times
{{10}^{-37}}=\frac{0.01{{x}^{2}}}{{{(0.0482)}^{2}}\times (0.0933)}\]
On solving \[x=6.6\times
{{10}^{-20}}\]
\[\therefore \] \[[{{N}_{2}}O]=0.1x\]
\[=0.1\times
6.6\times {{10}^{-20}}\]
\[=6.6\times
{{10}^{-21}}M.\]
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