Question

For the equilibrium $N_2{O}_4\rightleftharpoons 2N{O}_2$ in gaseous phase, $N{O}_2$ is $50\%$ of the total volume when equilibrium is set up. Hence percent of dissociation of $N_2{O}_4$ is:

• A. $50\%$
• B. $25\%$
• C. $66.66\%$
• D. $33.33\%$

Answer 1

Answer: (D)

$33.33\%$

At equilibrium, $N{O}_2$ is 50% of total volume. The remaining 50% of total volume will be $N_2{O}_4$

In other words, $V_{NO2}=V_{N_2{O}_4}=\frac{V_{total}}{2}$

Suppose we start with 100 L of $N_2{O}_4$ and x L of $N_2{O}_4$ dissociate to provide 2x L of$N{O}_2$.

$(100-x)$ L of $N_2{O}_4$ will remain at equilibrium.

Total volume at equilibrium $=(100-x)+2x=(100+x)L$.

At equilibrium, $N{O}_2$ is 50% of total volume.

$2xL=\frac{(100+x)L}{2}$

$2x=\frac{(100+x)}{2}$

$4x=(100+x)$

$3x=100$
$x=33.33$

Hence, out of 100 L of $N_2{O}_4$, 33.33 L dissociates to reach equilibrium.

The degree of dissociation $=\frac{x}{100}\times 100=\frac{33.33}{100}\times 100=33.33$.