Question
A reaction between N2 and O2 takes place as follows:
2N2 (g) + O2 (g) ⇌ 2N2O (g)
If a mixture of 0.482 mol of N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and allowed to form N2O at a temperature for which Kc = 2.0 × 10–37, determine the composition of equilibrium mixture.
Solution
Let x moles of N2(g) take part in the reaction. According to the equation, x/2 moles of O2 (g) will react to form x moles of N2O(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)`
