Question

For the following 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

The correct option is D 33.33%

$N_2{O}_4\rightleftharpoons 2N{O}_2$

$N{O}_2$ is 50% of the total volume when equilibrium is set up.

Thus, the volume fraction (at equilibrium) of $N{O}_2=\frac{50}{100}=0.5$

Volume fraction (at equilibrium) is proportional to mole fraction. The mole fraction (at equilibrium) of $N{O}_2=0.5$

Let initially n moles of $N_2{O}_4$ are present and a is the degree of dissociation.

The equilibrium number of moles of $N_2{O}_4=n(1-\alpha )$

The equilibrium number of moles of $N{O}_2=2n\alpha$

The mole fraction of $N{O}_2=\frac{2n\alpha }{n(1-\alpha )+2n\alpha }=\frac{2n\alpha }{n(1+\alpha )}$. But it is equal to 0.5

$\frac{2n\alpha }{n(1+\alpha )}=0.5$

$4n\alpha =n+n\alpha$

$3n\alpha =n$

$\alpha =\frac{1}{3}$
The percent dissociation $=100\alpha =\frac{100}{3}=33.33$ %.
Thus, the option (D) is the answer.