Chemical And Ionic Equilibrium Result Question 53
53. One mole of $\mathrm{N}_2$ and $3$ moles of $\mathrm{PCl}_5$ are placed in a $100$ L vessel heated to $227^{\circ} \mathrm{C}$. The equilibrium pressure is $2.05$ atm . Assuming ideal behaviour, calculate the degree of dissociation for $\mathrm{PCl}_5$ and $K_p$ for the reaction,
$\mathrm{PCl}_5(g) \rightleftharpoons \mathrm{PCl}_3(g)+\mathrm{Cl}_2(g) \quad$
(1984,6 M)
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Answer:
Correct Answer: 53. $( 0.33 )$
Solution:
- Total moles of 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 $\mathrm{N}_2(\mathrm{~g})$ and remaining $4$ moles for $\mathrm{PCl}_5$ and its dissociation products.
$\begin{aligned} & \mathrm{PCl}_5 \rightleftharpoons \mathrm{PCl}_3+\mathrm{Cl}_2 \\ & 3-x \quad \quad x \quad \quad x \\ & 3+x=4 \Rightarrow x=1 \\ \end{aligned}$
Degree of dissociation $=\frac{1}{3}=0.33$
BETA
