Question

For equilibrium $N_2{O}_4\left(g\right)\rightleftharpoons {2NO}_2\left(g\right)$ the observed vapour density of $N_2{O}_4$ is 40 at 350 K. Calculate percentage dissociation of $N_2{O}_4\left(g\right)$ at 350 K?

• A. $13.1$%
• B. $25.25$%
• C. $53.0$%
• D. None of these

Answer 1

Answer: (C)

$53.0$%

$N_2{O}_4\left(g\right)\rightleftharpoons 2{NO}_2$

Initially $N_2{O}_4$ has concentration $=1$

Then at equilibrium $N_2{O}_4$ has concentration $=(1-x)$ and ${NO}_2$ has concentration $=2x$

Initially we have taken $N_2{O}_4=1$

Molar mass $=92$

Vapour density $=40$

$M.W=60$

So,

$\frac{n_{initial}}{n_{final}}=\frac{m_{final}}{m_{initial}}$

$\Rightarrow \frac{1}{1-x}+2x=\frac{60}{92}$

$\Rightarrow \frac{1+2x(1-x)}{1-x}=\frac{60}{92}$

$\Rightarrow \frac{1}{1+x}=\frac{60}{92}$

$\Rightarrow 1+x=\frac{92}{60}$

$\Rightarrow x=\frac{92}{60}-1$

$\Rightarrow x=\frac{32}{60}=0.53$

Percentage dissociation $=0.53\times 100=53$%