A reaction between N_{2} and O_{2} takes place as follows:

2N_{2} (g) + O_{2} (g) ⇌ 2N_{2}O (g)

If a mixture of 0.482 mol of 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 × 10^{–37}, determine the composition of equilibrium mixture.

#### Solution

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 and at the equilibrium point is:

`2N_2(g) + O_2(g) ⇌ 2N_2O(g)`

Initial mole/litre 0.482/10 0.933/10 Zero

Mole/litre at equation point `(0.482 - x)/10` `(0.933 - x/2)/10` x/10

The value of equilibrium constant (2.0 x 10^{-37}) is extremely small. This means that only small amounts of reactants have reacted. Therefore, is extremely small and can be omitted as far as the reactants are concerned.

Applying Law of chemical Equilibrium `K_c = [N_2O(g)]^2/([N_2(g)]^2[O_2(g)])`

`2.0 xx 10^(-37) = (x/10)^2/((0.482/10)^2 xx(0.933/10)) = (0.01 x^2)/(2.1676 xx 10^(-4))`

`x^2 = 43.352 xx 10^(-40)` or `x = 6.6 xx 10^(-20)`

As x is extermely 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_2O` = `0.1 xx x`

`= 0.1 xx 6.6 xx 10^(-20)` mol `L^(-1)`

`= 6.6 xx 10^(-21) mol L^(-1)`