Question

$5$ mole of $PC{l}_5$ and $4$ mole of neon are introduced in a vessel of $110litre$ and allowed to attain equilibrium at ${250}^{\circ }C$. At equilibrium, the total pressure of reaction mixture was $4.678atm$. Calculate degree of dissociation of $PC{l}_5$ and equilibrium constant for the reaction.

Answer 1

	$Ne$	$PC{l}_5$	$PC{l}_3$	$C{l}_2$
Initial number of moles	4	5	0	0
Equilibrium number of moles	4	$5-x$	$x$	$x$

The total number of moles is $n=\frac{PV}{RT}=\frac{4.678\times 110}{0.0821\times 523}=11.98$
It is also equal to $n=4+5-x+x+x=9+x$
Hence, $x=2.98\simeq 3$
The degree of dissociation $A=\frac{2.48}{5}=0.596$

	$Ne$	$PC{l}_5$	$PC{l}_3$	$C{l}_2$
Equilibrium number of moles	4	2	3	3
Equilibrium mole fraction	1/3	1/6	1/4	1/4

The equilibrium constant $B=K=\frac{(1/4{)}^2\times 4.678}{1/6}=\frac{6\times 4.678}{16}=1.75$
$100(A+B)=100(0.596+1.75)=235$