Question

When heating $PC{l}_5$, it decompose $PC{l}_3$ and $C{l}_2$ in form of gas. The vapour density of gas mixture is $70.2$ and $57.9$ at ${200}^{\circ }C$ and ${250}^{\circ }C$. The degree of dissociation of $PC{l}_5$ at ${200}^{\circ }C$ at ${250}^{\circ }C$ is:

• A. $48.50$% and $80$%
• B. $60$% and $70$%
• C. $70$% and $80$%
• D. $80$% and $90$%

Answer 1

The correct option is A $48.50$% and $80$%

We know that when PCl5 is heated it decomposes to give $PC{l}_3$ and $C{l}_2$

$PC{l}_5\to PC{l}_3+C{l}_2$
Given,
The vapour density at equilibrium at ${200}^{0}C$ and 250 C.The initial Vapour Density at both the temperature MPCl5/2
therefore,initial $V.D.=(31+5\times 35.5)/2=104.5$
V..D. at equilibrium at ${200}^{0}C=70.2\left(given\right)$
$\frac{totalmoleatequilibrium}{totalmolesinitial}=1+\alpha =\frac{initialV.D.}{V.D.atequilibrium}$
$1+\alpha =\frac{104.5}{70.2}$
$=1.485$
$\alpha =0.485$

$\therefore \%\alpha =48.5\%$
At ${250}^{o}C$
$1+\alpha =\frac{104.25}{57.9}$
$=1.8$
$\alpha =0.80$

$\therefore \%\alpha =80\%$