Question

Vapour density of the equilibrium mixture of the reaction $2N{H}_3\left(g\right)\rightleftharpoons N_2\left(g\right)+3H_2\left(g\right)$
is $6.0$. The Percent disssociation of ammonia gas is:

• A. 13.88
• B. 58.82
• C. 41.66
• D. None of these

Answer 1

The correct option is C 41.66

$2N{H}_3\rightleftharpoons N_2\left(g\right)+3H_2\left(g\right)$
$\begin{array}{cccc}t=0& a& 0& 0\\ t=t_{eq}& a-x& \frac{x}{2}& \frac{3x}{2}\end{array}$
total moles at equilibrium $=a-x+\frac{x}{2}+\frac{3x}{2}=a+x$
V.D of mixture $=\frac{M_{mix}}{2}$
$M_{mix}=6\times 2=12$

$M_{mix}=\frac{n_1{a}_1+n_2{a}_2+n_3{a}_3}{totalmoles}=\frac{(a-x)\times \left(17\right)+\left(\frac{x}{2}\right)\times \left(28\right)+\left(\frac{3x}{2}\right)\times \left(2\right)}{a+x}=12$

where $a_1$, $a_2$, $a_3$ are molar masses of $N{H}_3$, $N_2$ and $H_2$ resepctively & $n_1,n_2,n_3$ are number of moles of $N{H}_3,N_2$ and $H_2$ at equilibrium respectively.

$\Rightarrow \frac{17a}{a+x}=12\Rightarrow 17a=12a+12x$
$5a=12x$
$\alpha =\frac{\text{moles dissociated}}{\text{initial moles}}$
$\alpha =\frac{x}{a}=\frac{5a}{12a}\times 100=41.667\%$
hence, Percent disssociation of ammonia gas is 41.66%.


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

$\alpha$= 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

$D=\frac{MolecularweightN{H}_3}{2}=\frac{17}{2}=8.5$
$d=6$
$n=\frac{4}{2}=2$
$\alpha =\frac{D-d}{d(n-1)}=\frac{8.5-6}{6(2-1)}=0.416$
% dissociation= $0.416\times 100=41.6$ %


Theory:

Relation between $\alpha$ and vapour density:
Total moles at equilibrium$=a-a\alpha +na\alpha$
$=a[1+(n-1\left)\alpha \right]$

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

Let the initial molar mass be $Mand{M}_{mix}$ be the molar mass of the mixture:

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

Mass$=moles\times molarmass$

$Ma=a[1+(n-1\left)\alpha \right]M_{mix}$

$M_{mix}=\frac{M}{[1+(n-1\left)\alpha \right]}$..................(1)

$Molarmass=2\times VapourDensity$

Vapour Density$=\frac{Molarmass}{2}$
Since $\frac{M}{M_{mix}}=[1+(n-1\left)\alpha \right]$

Where$\frac{M}{M_{mix}}=\frac{\left(\frac{D}{2}\right)}{\left(\frac{d}{2}\right)}$

Let $D=\text{Initial vapour density (i.e. of pure form)}$
$d=\text{Final vapour density (i.e. of mixture at equil.)}$
so, $\frac{D}{d}=[1+(n-1\left)\alpha \right]$