Question

20% $N_2{O}_4$ molecules are dissociated in a sample of gas at ${27}^{o}C$ and 760 torr as:
$N_2{O}_4\rightleftharpoons 2N{O}_2$
Calculate the density of the equilibrium mixture.

Answer 1

As $\alpha =20$% $=0.2$

moles $N_2{O}_4\rightleftharpoons 2{NO}_2$ $n_T$

initial $1$ $-$

after time $t$ $1-0.2$ $0.2\times 2$ $1+0.2$

$x_{N_2{O}_4}=\frac{0.8}{1.2}$ $x_{{NO}_2}=\frac{0.4}{1.2}$

effective molecular mass $M_{eff}=x_{N_2{O}_4}{M}_{N_2{O}_4}+x_{{NO}_2}.M_{{NO}_2}$

$M_{eff}=\frac{0.8}{1.2}\times 92+\frac{0.4}{1.2}\times 46$

$M_{eff}=76.67$

as $dRT=P{M}_{eff}⟶\left(1\right)$

$d\to$ density of mixture.

$T={27}^{0}C=300K$

$P=760$ torr $=1$ atm

$R=0.0821$ atm L $K^{-1}{mol}^{-1}$

Thus eqn (1) become

$d\times 0.0821\times 300=1\times 76.67$

$d=3.11$ $g/L$

This is the density of mixture.