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
Concept :
For given equilibrium reaction
Number of moles formed n = 2
We know that the vant hoff factor relation with degree of dissociation
There is another relation between vant hoff factor and 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
i = 1.2
We known that,
Normal molar mass of = 92
Normal molar mass of = 46
So, the abnormal molar mass of each would be
=
According to ideal gas equation
Similarly,
Conclusion:
Total density of equilibrium mixture = 3.11+1.56 = 4.67 g/cm^3
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: