Question

The vapor density of ${\mathrm{N}}_2{\mathrm{O}}_4$ at certain temperature is $30$. Calculate the percentage dissociation of ${\mathrm{N}}_2{\mathrm{O}}_4$ at this temperature.

Answer 1

Step 1 - Given Data

• Vapor Density of ${\mathrm{N}}_2{\mathrm{O}}_4$ $=30$
• $\mathrm{Molar}\mathrm{Mass}=\mathrm{Vapour}\mathrm{Density}\times 2=30\times 2=60\mathrm{g}/\mathrm{mol}$

Step 2 - Concentration of ${\mathrm{N}}_2{\mathrm{O}}_4$

• The reaction for dissociation of ${\mathrm{N}}_2{\mathrm{O}}_4$ into ${\mathrm{NO}}_2$ is written as,

${\mathrm{N}}_2{\mathrm{O}}_4\rightleftarrows 2{\mathrm{NO}}_2$

• Consider the initial concentration of ${\mathrm{N}}_2{\mathrm{O}}_4$ $=1$
• Hence, at equilibrium, the concentration of ${\mathrm{N}}_2{\mathrm{O}}_4$ $=1-\mathrm{x}$
• Initial concentration of ${\mathrm{NO}}_2$$=0$
• Hence, at equilibrium, the concentration of ${\mathrm{NO}}_2$ $=2\mathrm{x}$

	${\mathrm{N}}_2{\mathrm{O}}_4$	${\mathrm{NO}}_2$
Initial moles	$1$	$0$
Moles at equilibrium	$1-\mathrm{x}$	$2\mathrm{x}$
Mole fraction	$\frac{\left(1-\mathrm{x}\right)}{\left(1+\mathrm{x}\right)}$	$\frac{2\mathrm{x}}{\left(1+\mathrm{x}\right)}$

• Now, the molar mass of ${\mathrm{N}}_2{\mathrm{O}}_4$ $=92\mathrm{g}/\mathrm{mol}$
• Molar mass of ${\mathrm{NO}}_2$ $=46\mathrm{g}/\mathrm{mol}$
• Therefore, $92\times \left[\frac{1-\mathrm{x}}{1+\mathrm{x}}\right]+46\times \left[\frac{2\mathrm{x}}{1+\mathrm{x}}\right]=60$.
• $$\begin{gathered} \mathrm{x}=0.533 \\ \%\mathrm{x}=53.3\% \end{gathered}$$
• Therefore the percentage dissociation of ${\mathrm{N}}_2{\mathrm{O}}_4$ = $53.3\%$.