Question

The degree of dissociation is $0.4$ at $400K$ & $1.0atm$ for the gaseous reaction
$PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2\left(g\right)$. Assuming ideal behaviour of all gases. Calculate the density of equilibrium mixture at $400K$ & $1.0atm$ pressure.

Answer 1

$PC{l}_5\leftrightharpoons PC{l}_3+C{l}_2$

At $t=0$- $1$ $0$ $0$

At $t=$ equilibrium- $(1-\alpha )$ $\alpha$ $\alpha$

$\alpha =0.4$

$\left[PC{l}_5\right]$ at equilibrium $=0.6$

$\left[PC{l}_3\right]$ at equilibrium $=0.4$

$\left[C{l}_2\right]$ at equilibrium $=0.4$

At equilibrium, Mole fraction of $PC{l}_5=\frac{0.6}{0.6+0.4+0.4}=0.428$

Mole fraction of $PC{l}_3=\frac{0.4}{0.6+0.4+0.4}=0.286$

Mole fraction of $C{l}_2=0.286$

$\therefore$ Average Molecular mass of mixture $\left(M\right)=\sum x_i{M}_i$

$⟹M=(0.428\times 208.5)+(0.286\times 137.5)+(0.286\times 71)$
$⟹M=148.869$ $gmo{l}^{-1}$

$\therefore$ Density of mixture $=\frac{PM}{RT}=\frac{148.869\times 1.0}{0.0821\times 400}=4.533$ $g/L$

$\therefore$ Density of mixture at equilibrium $=4.533$ $g/L$