Question

The vapour density of N2O4 at a certain temperature is 30 . Calculate the percentage dissociation of N2O4 at this temperature N2O4(g)↔2NO2(g)

Answer 1

N2O4 ---------NO2.       
Initially N2O4 has conc.= 1, then at equilibrium N2O4 has conc.1-X. And NO2 has conc. =2x initially we have taken N2O4 =1 Molar mass =92 .V.D at eq.=30 .M.w.=60.
n initial/ n final =M final/M initial 
1/1-X+2X=60/92
1/1+X=60/92
1+X=92/60
X=92/60 -1 
X=32/60
X=0.53           percentage dissociation=0.53 X 100=53

Answer 2

Answer : The percentage dissociation of $N_2O_4$ is, 53.3 %

Explanation :

The balanced reaction will be,

                       $N_2O_4\rightleftharpoons 2NO_2$

initial moles       1               0

final moles      $1-\alpha$        $2\alpha$

First we have to calculate the Van't Hoff factor, $i$.

The Van't Hoff factor in terms of moles will be :

$i=\frac{\text{Final moles}}{\text{Initial moles}}=\frac{1-\alpha+2\alpha}{1}$     ................(1)

The Van't Hoff factor in terms of molecular mass will be :

$i=\frac{\text{Normal molar mass}}{\text{Observed molar mass}}$

Normal molar mass = 92 g/mole

Observed molar mass = $2\times \text{Vapor density}=2\times 30=60$

Now put all the given values in above expression, we get :

$i=\frac{92g/mole}{60g/mole}$      ..................(2)

Now equation equation 1 and 2, we get :

$\frac{1-\alpha+2\alpha}{1}=\frac{92g/mole}{60g/mole}$

$\alpha=0.533$

Now we have to calculate the percentage dissociation.

$\text{percentage dissociation}=0.533\times 100=53.3$

Therefore, the percentage dissociation of $N_2O_4$ is, 53.3 %