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.

Explanation

` {:(,2 N_(2) (g) ,+,O_(2)(g),hArr,2 N_(2)O(g)),("Intial",0*482 "mol",,0*933"mol",,),("At eqm.",0*482 "mol",,0.933-x//2,,x),("Molar conc." ,(0*482-x)/10,,(0*933-x//2)/10,,x/10):}`,
As K ` = 2*0 xx 10 ^(-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*1 x `
` :. K_(c) = (0*1 x)^(2) /((0*0482)^(2) (0*0933)) = 2* 0 xx 10^(-37) ` (Given)
On solving , this gives `x = 6*6 xx 10^(20)`
` :. [N_(2)O ] = 0*1 x 6*6 xx 10^(-21) "mol"L^(-1)`