Question

The vapour density of ${N_2}{O_4}$ at a certain temperature is 30. Calculate the percentage dissociation of ${N_2}{O_4}$ at this temperature.

Answer 1

At equilibrium ${N_2}{O_4}$ dissociates into 2 molecules of $N{O_2}$. The formula for molar mass is: Molar mass = 2 × Vapour density
Van’t hoff factor ‘i’ can be calculated using:
$i = \dfrac{{{n_f}}}{{{n_i}}}$
Where, ${n_f}$ = final moles; and ${n_i}$ = initial moles
Also, $i = \dfrac{{{M_{Actual}}}}{{{M_{Observed}}}}$
Where, ${M_{Actual}}$ = Actual molar mass; and ${M_{Observed}}$ = Observed molar mass
Vapour density = 30

Complete step by step answer:
-First, we should know that $N{O_2}$ is an odd electron species. So, it dimerise to form ${N_2}{O_4}$. Both $N{O_2}$ and ${N_2}{O_4}$ are found at equilibrium and are inter-convertible into the two forms.
-So, the chemical reaction for the dissociation of ${N_2}{O_4}$ will be:

Initially in the reacting vessel only ${N_2}{O_4}$ is present and the concentration of $N{O_2}$ will be 0. At equilibrium some of the ${N_2}{O_4}$ will react and its concentration will decrease and the same amount of $N{O_2}$ will be formed.
\left( {\begin{array}{*{20}{c}} {{N_2}{O_4}}& \rightleftharpoons &{2N{O_2}} \\\ 1&{}&0 \\\ {1 - \alpha }&{}&{2\alpha } \end{array}} \right)
Therefore total moles = (1 – α) + 2α = (1 + α) ; where α is the degree of dissociation.

-The van't hoff factor ‘i’ is equal to the ratio of total moles at equilibrium to the initial moles present in the reaction vessel.
Now, we will calculate van’t hoff factor:
$i = \dfrac{{{n_f}}}{{{n_i}}}$ = $\dfrac{{1 + \alpha }}{1}$ (1)
-Now, we will find the actual molar mass of ${N_2}{O_4}$
The atomic mass of N and O are 14 and 16 respectively.
Actual molar mass of ${N_2}{O_4}$ = (2 × 14) + (4 × 16) = 28 + 64 = 92
-Now, we will find the observed molar mass using vapour density.
The vapour density of ${N_2}{O_4}$ is given in the question above = 30
The formula for molar mass using vapour density is:
Molar mass = 2 × Vapour density
Observed molar mass = 2 × 30 = 60

-Now, we know that:
$i = \dfrac{{{M_{Actual}}}}{{{M_{Observed}}}}$ = $\dfrac{{92}}{{60}}$ (2)
From equation (1) and equation (2), we get:
$\dfrac{{1 + \alpha }}{1} = \dfrac{{92}}{{60}}$
Cross multiplication : 60 (1 + α) = 92 × 1
60 + 60 α = 92
60α = 92 – 60
60α = 32
α = $\dfrac{{32}}{{60}}$ = 0.5333
Therefore, α = 0.5333
Hence, the percentage dissociation will be = 0.5333 × 100 = 53.33%
So, the answer of the given question is 53.33%

Additional Information:
-Vapour density is defined as the density of a gas or vapours relative to the density of hydrogen gas at the same temperature and pressure.
-Van't hoff factor is the ratio of actual molar mass to the observed molar mass, it is represented by ‘i’.

Note: Vapour density is a unit less quantity. ${N_2}{O_4}$ exists as a dimer. $N{O_2}$ is an odd electron species. $N{O_2}$ is unstable so it forms a dimer as ${N_2}{O_4}$. Both $N{O_2}$ and ${N_2}{O_4}$ are found at equilibrium and are inter-convertible into the two forms.