Reaction between `N_(2)` and `O_(2-)` takes place as follows :
`2N_(2)(g) + O_(2)(g) hArr 2N_(2)O (g)`
If a mixture of `0.482` mol `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 xx 10^(-37)`, determine the composition of equilibrium mixutre.

Explanation

`2N_(2)(g)+O_(2)(g) hArr 2N_(2)O(g)`
`{:(0.482 mol,0.933 mol,,,"Initial"),(0.482-x,0.933-x//2,,x,"At equilibrium"),((0.482-x)/10,(0.933-x//2)/10,,x/10,"Molar conc."):}`
As `K=2.0xx10^(-37)` is very small, this means that the amount of `N_(2)` and `O_(2)` reacted `(x)` is very very small. Hence, at equilibrium, we have
`[N_(2)]=0.0482 mol L^(-1)`,
`[O_(2)]=0.0933 mol L^(-1), [N_(2)O]=0.1x`
`:. K_(c )=((0.1 x)^(2))/((0.0482)^(2)(0.0933))=2.0xx10^(-37) ("given")`
On solving, this gives `x=6.6xx10^(-20)`
`:. [N_(2)O]=0.1x=6.6xx10^(-21) mol L^(-1)`