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Q. One mole of $N_{2}$, and $3$ moles of $P C l_{5}$ are placed in a $100\, L$ vessel heated to $227^{\circ} C$. The equilibrium pressure is $2.05 \,atm$. Assuming ideal behaviour, calculate the degree of dissociation for $PCl _{5}$, and $K_{p}$ for the reaction,
$P C l_{5}(g) \rightleftharpoons P C l_{3}(g)+C l_{2}(g)$

IIT JEEIIT JEE 1984Equilibrium

Solution:

Total moles or gases at equilibrium
$=\frac{p V}{R T}=\frac{2.05 \times 100}{0.082 \times 500}=5.0$
Out of this $5$ moles, $1.0$ mole is for $N_{2}( g )$ and remaining $4$ moles for $PC I_{5} $ and its dissociation products.
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$3+x=4 $
$\Rightarrow x=1$
Degree of dissociation $=\frac{1}{3}=0.33$