Chemistry, asked by hsgmaipcom4408, 1 year ago

20% N2O4 molecules are dissociated in a sample of gas at 27'C and 760 torr as: N2O4 ----> 2NO2 Calculate the density of the equilibrium mixture.

Answers

Answered by antiochus
30

Concept :

For given equilibrium reaction

N_{2}O_{4}   ⇆ 2NO_{2}

Number of moles formed n = 2

We know that the vant hoff factor relation with degree of dissociation

\alpha  = \frac{(i-1)}{(n-1)}

There is another relation between vant hoff factor and molar mass

i = \frac{normal molar mass}{abnormal molar mass}

Explanation :

Given that,

Degree of dissociation (α) = 20% = 0.2

Number of moles formed n = 2

On substituting above values in vant hoff factor expression

0.2 = \frac{(i-1)}{(2-1)}

i = 1.2

We known that,

Normal molar mass of N_{2}O_{4} = 92

Normal molar mass of NO{2} = 46

So, the abnormal molar mass of each would be

N_{2}O_{4} =\frac{92}{1.2} =  76.67

NO_{2} = \frac{46}{1.2} = 38.33

According to ideal gas equation

d_{N_{2}O_{4}  }  = \frac{PM}{RT} = \frac{1*76.67}{0.0821*300} =  3.11 g/cm^3

Similarly,

d_{NO_{2} } = \frac{1*38.33}{0.0821*300}  = 1.56 g/cm^3

Conclusion:

Total density of equilibrium mixture = 3.11+1.56 = 4.67 g/cm^3

Answered by tejal26677
5

AnswerAssume 100 moles. 

After dissociation: 

80 mol N2O4 

40 mol NO2 

PV = nRT ==> V = nRT/P 

V1 = (80)(0.082)(300)/1 => V1 = 1968 L * (m^3/1000 L) = 1.968 m^3 

V2 = (40)(0.082)(300)/1 => V2 = 984 L * (m^3/1000 L) = 0.984 m^3 

m1 = 80 mol * (0.092011 kg/mol) = > m1 = 7.36 kg 

m2 = 40 mol * (0.0460055 kg/mol) => m2 = 1.84 kg 

ρ1 = 7.36 kg / 1.968 m^3 => 3.74 kg/m^3 

ρ2 = 1.84 kg / 0.984 m^3 => 1.87 kg/m^3 

ρm = (ρ1V1 + ρ2V2)/(V1 + V2) 

ρm = (3.74*1.968 + 1.87*0.984)/(1.968+0.984) 

ρm = 3.12 kg/m^3

Explanation:

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