Question

At ${250}^{0}C$ and 1 atm pressure, the v.d. of $PC{l}_5$ is 57.9 Calculate

(i) $K_p$ for the reaction (ii) % dissociation when P is doubled.

Find $X=i+ii$

The answer X must be an integer.

Answer 1

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

$(V.d.{)}_{obs.}$ = $57.9$

$\Rightarrow$ $M_{mixture}$ = $2\times 57.9$ = $115.8$

$\Rightarrow$ Observed mol.wt. = $\frac{Theoreticalmol.wt.}{Totalno.ofmoles}$

$M_{mixture}$ = $\frac{M_{th}}{[1+(n-1\left)\alpha \right]}$ = $\frac{208.5}{1+\alpha }$ $\Rightarrow$ $\alpha$ = $0.8$

$\Rightarrow$ $K_p$ = $\frac{{\alpha }^{2}P}{1-{\alpha }^2}$ = $\frac{16}{9}=1.7$

When P is doubled,
$\frac{16}{9}$ = $K_p$ = $\frac{{\left(a^{\prime }\right)}^2}{1-({a^{\prime })}^2}{P}^{\prime }$ $[\because$ $P^{{}^{\prime }}$ = $2atm]$

$\therefore$ $a$ = $0.69$ (ii) $Ans.$

Therefore, $X=i+ii$

$X=1.7+0.69=2.39\simeq 2$