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Question

Vapour density of the equilibrium mixture of the reaction 2NH3(g)N2(g)+3H2(g)
is 6.0. The Percent disssociation of ammonia gas is:

A
13.88
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B
58.82
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C
41.66
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D
None of these
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Solution

The correct option is C 41.66
2NH3N2(g)+3H2(g)
t=0a00t=teqaxx23x2
total moles at equilibrium =ax+x2+3x2=a+x
V.D of mixture =Mmix2
Mmix=6×2=12

Mmix=n1a1+n2a2+n3a3total moles=(ax)×(17)+(x2)×(28)+(3x2)×(2)a+x=12

where a1, a2, a3 are molar masses of NH3, N2 and H2 resepctively & n1,n2,n3 are number of moles of NH3,N2 and H2 at equilibrium respectively.

17aa+x=1217a=12a+12x
5a=12x
α=moles dissociatedinitial moles
α=xa=5a12a×100=41.667%
hence, Percent disssociation of ammonia gas is 41.66%.


This can also be solved easily using direct formula.
Alternative Solution:

α= degree of dissociation
D = vapour Density of pure substance
d = Experimetal Vapour Density (of equilibrium mixture)
n = No. of moles of product formed from one mole of reactant

D=Molecular weight NH32=172=8.5
d=6
n=42=2
α=Ddd(n1)=8.566(21)=0.416
% dissociation= 0.416×100=41.6 %


Theory:

Relation between α and vapour density:
Total moles at equilibrium=aaα+naα
=a[1+(n1)α]

Molar mass of the mixture is the mass of 1 mole of the mixture.

Let the initial molar mass be MandMmix be the molar mass of the mixture:

From laws of conservation of Mass :
Initial mass of the solution = Final mass of the solution

Mass=moles×molarmass

Ma=a[1+(n1)α]Mmix

Mmix=M[1+(n1)α]..................(1)

Molarmass=2×VapourDensity

Vapour Density=Molarmass2
Since MMmix=[1+(n1)α]

WhereMMmix=(D2)(d2)

Let D=Initial vapour density (i.e. of pure form)
d=Final vapour density (i.e. of mixture at equil.)
so, Dd=[1+(n1)α]

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