The correct option is C 41.66
2NH3⇌N2(g)+3H2(g)
t=0a00t=teqa−xx23x2
total moles at equilibrium =a−x+x2+3x2=a+x
V.D of mixture =Mmix2
Mmix=6×2=12
Mmix=n1a1+n2a2+n3a3total moles=(a−x)×(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=12⇒17a=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=D−dd(n−1)=8.5−66(2−1)=0.416
% dissociation= 0.416×100=41.6 %
Theory:
Relation between α and vapour density:
Total moles at equilibrium=a−aα+naα
=a[1+(n−1)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+(n−1)α]Mmix
Mmix=M[1+(n−1)α]..................(1)
Molarmass=2×VapourDensity
Vapour Density=Molarmass2
Since MMmix=[1+(n−1)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+(n−1)α]