Question

In the dissociation of $PCl _{5}$ as:
$PCl _{5}( g ) \to PCl _{3}( g )+ Cl _{2}( g )$, If the degree of dissociation is $\alpha$ at equilibrium pressure "P", then the equilibrium constant for the reaction is:

• A. $K_{p}=\frac{\alpha^{2}}{1+\alpha^{2} p}$
• B. $K_{p}=\frac{\alpha^{2} p^{2}}{1-\alpha^{2}}$
• C. $K_{p}=\frac{\alpha p^{2}}{1-\alpha^{2}}$
• D. $K_{p}=\frac{\alpha^{2} p}{1-\alpha^{2}}$

Answer 1

Answer: (D)

$\underset{\text{equilibrium; }}{\text{Initial :}} \underset{1 - \alpha}{PCl _{5}} \rightleftharpoons \underset{\alpha}{PCl _{3}}+ \underset{\alpha}{Cl _{2}}$

Total moles: $1-\alpha+\alpha+\alpha=1+\alpha$

Total pressure $= P$

$\therefore P_{PCl_{5}}=\frac{(1-\alpha) P}{1+\alpha}, P_{PCl_{3}}=\frac{\alpha P}{1+\alpha}, P_{C l_{2}}=\frac{\alpha P}{1+\alpha}$

$K_{p}=\frac{P_{P Cl_{3}}, \times P_{C l_{2}}}{P_{P C l_5}}=\frac{\alpha^{2} P^{2}}{(1+\alpha)(1-\alpha)(P)}=\frac{\alpha^{2} P}{1-\alpha^{2}}$