Discussion Forum



1)

A moles of PCl5 is heated in a closed container to equilibrium   $PCl_{5}\rightleftharpoons PCl_{3}(g)+ Cl_{2}(g)$ at a pressure of p  atm. If x moles of PCl5 disscociate at equilibrium  , then

 


A) $\frac{x}{a}=\frac{K_{p}}{K_{p}+P}$

B) $\frac{x}{a}=\left(\frac{K_{p}+P}{K_{p}}\right)^{1/2}$

C) $\frac{x}{a}=\left(\frac{K_{p}}{P}\right)^{1/2}$

D) $\frac{x}{a}=\left(\frac{K_{p}}{K_{p}+P}\right)^{1/2}$

Answer:

Option D

Explanation:

 $PCl_{5}\rightleftharpoons PCl_{3}(g)+ Cl_{2}(g)$ 

 1-$\alpha$           $\alpha$             $\alpha$

 $\alpha$  = x/a  (degree of dissociation )

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

$K_{p}=\frac{P_{PCl_{3}}\times P_{Cl_{2}}}{P_{PCl_{5}}}$

       $=\frac{\left[\frac{\alpha}{1+\alpha}.p\right]\left[\frac{\alpha}{1+\alpha}.p\right]}{\frac{1-\alpha}{1+\alpha}.p}$

 $=\frac{\alpha^{2}p}{1-\alpha^{2}}\Rightarrow\alpha=\left(\frac{K_{p}}{K_{p}+P}\right)^{1/2}$