question_answer

In the dissociation of \[PC{{l}_{5}}\] as \[PC{{l}_{5}}(g)\rightleftharpoons PC{{l}_{3}}(g)+C{{l}_{2}}(g)\] if the degree of dissociation is a 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) \[{{K}_{P}}=\frac{{{P}^{2}}}{1-{{\alpha }^{2}}}\]

D) \[{{K}_{P}}=\frac{{{\alpha }^{2}}P}{1-{{\alpha }^{2}}}\]

Correct Answer: D

Solution :

[d] Total mole\[=1-\alpha +\alpha +\alpha =1+\alpha \] \[{{K}_{p}}=\frac{\frac{\alpha }{1+\alpha }P.\frac{\alpha }{1+\alpha }.P}{\frac{1-\alpha }{1+\alpha }P}=\frac{{{\alpha }^{2}}P}{1-{{\alpha }^{2}}}\]