Question

NO2 dissociates as NO2(g) NO(g) + 1/2O2(g)

If vapour density at equilibrium is found to be 20 when 1 mole of NO2 is taken initially in 1 L flask. The percentage degree of dissociation of NO2 is

Answer 1

Given:

Vapour Density (VD) = 20

Initial amount of $\sf NO_2$ = 1 mole in 1 L flask

To Find:

Percentage degree of dissociation of $\sf NO_2$ $\rm (\alpha)$

Answer:

$\rm \ \ \ \ \ \ \ \ \ \ \ {NO_2} _{(g)} \longrightarrow {NO}_{ (g)} + \dfrac{1}{2}{O_2}_{(g)}$

$\rm t = 0 : 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \ \ \ \ \ \ \ 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \ \ \ 0$

$\rm t = t: 1 - \alpha \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \alpha \: \: \: \: \: \: \: \: \: \: \: \: \: \ \: \: \dfrac{\alpha}{2}$

Average Molecular Mass:

$\rm M_{avg} = 2 \times VD \\ \\ \rm = 2 \times 20 \\ \\ \rm =4 0$

Molecular Mass of $\sf NO_2$ = 96

Molecular Mass of $\sf NO$ = 30

Molecular Mass of $\sf O_2$ = 32

$\rm \leadsto 40 = \dfrac{(1 - \alpha) {M}_{{NO}_{2}} + \alpha \times {M}_{NO} + \dfrac{\alpha}{2} {M}_{{O}_{2}}}{(1 - \alpha) + \alpha + \dfrac{\alpha}{2} } \\ \\ \rm \leadsto 40 = \dfrac{(1 - \alpha) \times 96 + \alpha \times 30 + \dfrac{\alpha}{2} \times 32}{1 + \dfrac{\alpha}{2}} \\ \\ \rm \leadsto 40 \bigg(1 + \dfrac{\alpha}{2} \bigg) = 96 - 96 \alpha + 30 \alpha + 16 \alpha \\ \\ \rm \leadsto 40 + 20 \alpha = 96 - 50 \alpha \\ \\ \rm \leadsto 20 \alpha + 50 \alpha = 96 - 40 \\ \\ \rm \leadsto 70 \alpha = 56 \\ \\ \rm \leadsto \alpha = \dfrac{56}{70} \\ \\ \rm \leadsto \alpha = 0.8 \\ \\ \rm \leadsto \% \alpha = 0.8 \times 100 \% \\ \\ \rm \leadsto \% \alpha = 80 \%$

$\therefore$ Percentage degree of dissociation of $\sf NO_2$ $\rm (\alpha)$ = 80%