The vapour density of $N_{2} O_{4}$ at a certain temperature is $30$ . Calculate the percentage dissociation of $N_{2} O_{4}$ at this temperature.

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Solution

$N_{2} O_{4} (g) ⇌ 2 {N O}_{2} (g)$

Mol. mass of $N_{2} O_{4} = (28 + 64) = 92$

Initial Vapour density, $D = \frac{92}{2} = 46$

Let the degree of dissociation be $x$ .

Vapor density at eqm. $d = 30$ ,

Applying the relationship,

$x = \frac{D - d}{d} = \frac{46 - 30}{30} = 0.533$

Degree of dissociation $= 53.3$ %

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The vapour density of N2O4 at a certain temperature is 30. Calculate the percentage dissociation of N2O4 at this temperature.

N2O4(g)⇌2NO2(g)Mol. mass of N2O4=(28+64)=92Initial Vapour density, D=922=46Let the degree of dissociation be x.Vapor density at eqm. d=30,Applying the relationship,x=D−dd=46−3030=0.533Degree of dissociation =53.3%

The vapour density of $N_{2} O_{4}$ at a certain temperature is $30$ . Calculate the percentage dissociation of $N_{2} O_{4}$ at this temperature.