Question

In the dissociation of N2O4 into NO2, (1 + $\alpha$) values with the vapour densities ratio $\left( \frac{D}{d} \right)$ is as given by :

[$\alpha$ -degree of dissociation, D-vapour density before dissociation, d-vapour density after dissociation]

• A.
• B.
• C.
• D.

Answer 1

Answer: (A)

Answer 2

N2O4 $\rightleftharpoons$2NO2

$M_{mix} = \frac{(1 - \alpha)92 + 2\alpha \times 46}{1 + \alpha}$

1 – $\alpha$ 2$\alpha$

$M_{mix} = \frac{92}{1 + \alpha}$

$\frac{D}{d} = \frac{46}{46/(1 + \alpha)} = (1 + \alpha).$

Therefore, (1) option is correct.