Question

if n2o4 decomposes to 60% no2 then vapour density of equilibrium mixture is​

Answer 1

dissociation reaction is .....

$N_2O_4\Leftrightarrow 2NO_2$

Initial conc. of N2O4 = 1

initial conc. of NO2 = 0

final conc. of N2O4 = 1 - 0.6 = 0.4 [ it is given that N2O4 decompose to 60% NO2 ]

final conc. of NO2 = 2 × 0.6 = 1.2

now, final mole of mixtures = 1.2 + 0.4 = 1.6

initial mole of mixture = 1

so, initial mole/final mole = final molar mass/initial molar mass

initial molar mass of mixture = molar mass of N2O4 = 92 g/mol

so, 1/1.6 = final molar mass /92

final molar mass = 92/1.6 = 230/4 = 57.5 g/mol

now vapor density of mixture = molar mass/2 = 57.5/2 = 28.75

Answer 2

The vapor density of equilibrium mixture is 28.75 g/mol.

Explanation:

N2O4 ----> 2 NO2

• Ratio between initial concentration of N2O4 and NO2 = 1 & 0
• Final concentration of N2O4 = 1 - 0.6 = 0.4  
• Final concentration of NO2 = 2 × 0.6 = 1.2

• Moles of mixture = 1
• Total number of mole of mixtures = 1.2 + 0.4 = 1.6

Ratio between initial and final mole.

Initial  / final  = final /initial

Initial molar mass of mixture = molar mass of N2O4 = 92 g/mol

1 / 1.6 = Final molar mass / 92

Final molar mass = 92 / 1.6 = 57.5 g/mol

Vapor density of equilibrium mixture = molar mass / 2

Vapor density = 57.5  / 2 = 28.75

Hence, vapor density of equilibrium mixture is 28.75 g/mol