Question

$PC{l}_5$ dissociates into $PC{l}_3$ and $C{l}_2$. If the total pressure of the system in equilibrium is $P$ at a density $\rho$ and temperature $T$, show that the degree of dissociation $\alpha =\frac{PM}{\rho RT}-1$, where $M$ is the relative molar mass of $PC{l}_5$. If the vapour density of the gas mixture at equilibrium has the value of 62 when the temperature is ${230}^{o}C$, what is the value of $P/\rho$?

Answer 1

$\alpha =\frac{D-d}{(n-1)d}=\frac{\frac{2085}{2}-62}{(2-1)\times 62}=\frac{42.25}{62}$

$=0.68$

$\alpha =\frac{PM}{\varrho RT}-1$

2) $0.68=\frac{P}{\varrho }\times \frac{208.5}{8.314\times 503}-1$

$\frac{1.68\times 8.314\times 503}{208.5}=\frac{P}{\varrho }$

$=\frac{P}{\varrho }=33.77$

$D=V.D.$ of Reactant