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

Let x moles of `N_(2) (g)` take part in the reaction. According to the equation, `x//2` moles of `O_(2) (g)` will react to form x moles of `N_(2)O(g)`. The molar concentration per litre of different species before the reaction nad at the equilibrium point is :
`2N_(2)(g) +O_(2)(g) hArr 2N_(2)O(g)`
`" Initial moles"//"litre : "" " (0.482)/(10) " " (0.933)/(10)" " "Zero"`
`" Moles"//"litre at eqm. point : "" " (0.482 -x)/(10) " " (0.933 -(x)/(2))/(10) " " (x)/(10)`
The value of equilibrium constant `(2.0xx 10^(-37))` is extremely small. This means that only small amounts of reactants have reacted. Therefore , x is extremely small and can be omitted as far as the reactants are concerned.
Applying Law of chemical equilibrium ,`K_(c) =[[N_(2)O(g)]^(2)]/[[N_(2)(g)]^(2)[O_(2)(g)]^(2)]`
`2.0 xx 10^(-37) =((x)/(10))^(2)/(((0.482)/(10))^(2) xx((0.933)/(10)))=(0.01 x^(2))/(2.1676xx10^(-4))`
`x^(2) =43. 352 xx 10^(-40) or x =6.6 xx 10^(-20)`
As x is extremely small , it can be neglected.
Thus in the equilibrium mixture
Molar conc. of `N_(2) = 0.0482 mol L^(-1)`
Molar conc. of `O_(2) =0.0933 mol L^(-1)`
Molar conc. of `N_(2)O= 0.1 xx X = 0.1 xx 6.6 xx 10^(-20) mol L^(-1)`
`=6.6 xx 10^(-21) mol L^(-1)`