Reaction between nitrogen and oxygen takes place as following:
`2N_(2(g))+O_(2)hArr2N_(2)O_((g))`
If a mixture of `0.482 "mole" N_(2)` and `0.933 "mole"` of `O_(2)` is placed in a reaction vessel of volume `10 litre` and allowed to form `N_(2)O` at a temperature for which `K_(c)=2.0xx10^(-37)litre mol^(-1)`. Determine the composition of equilibrium mixture.

Answer

`[N_(2)] = 0.0482 "molL"^(-1)`, `[O_(2)] = 0.0933 "molL"^(-1). [N_(2)O] = 6.6 xx 10^(-21) "molL"^(-1)`

Explanation

Let the concentration of `N_(2)O` at equilibrium be x.
The given reaction is:
`{:(,2N_(2(g)),+,O_(2(g)),harr,2N_(2)O_((g))),("Initial conc.",0.482 "mol",,0.933"mol",,0),("At equilibrium",(0.482-x)"mol",,(1.933-x)"mol",,x"mol"):}`
Therefore, at equilibrium, in the 10 L vessel:
`[N_(2)] = (0.482 -x)/(10), [O_(2)] = (0.933-x//2)/(10), [N_(2)O] = (x)/(10)`
The value of equilibrium constant i.e., `K_(c) = 2.0 xx 10^(-37)` is very small. Therefore, the amount of `N_(2)` and `O_(2)` reacted is also very small. Thus, x can be neglected from the expressions of molar concentrations of `N_(2)` and `O_(2)`.
Then
`[N_(2)] = (0.482)/(10) = 0.482 "mol" L^(-1)` and `[O_(2)] = (0.933)/(10) = 0.0933 "mol" L^(-1)`
Now,
`K_(c) = ([N_(2)O_((g))]^(2))/([N_(2(g))]^(2)[O_(2(g))])`
`rArr 2.0 xx 10^(-37) = ((x)/(10))^(2)/((0.0482)^(2)(0.0933))`
`rArr (x^(2))/(100) = 2.0 xx 10^(-37) xx (0.0482)^(2) xx (0.0933)`
`rArr x^(2) = 43.35 xx 10^(-40)`
`rArr = 6.6 xx 10^(-20)`
`[N_(2)O] = (x)/(10) = (6.6 xx 10^(-20))/(10)`
`= 6.6 xx 10^(-21)`