Question

For gaseous decomposition of ${PCI}_5$ in a closed vessel the degree of dissociation '$\alpha$', equilibrium pressure 'P' & ${{}^{\prime }K}_p^{\prime }$ are related as

• A. $\alpha =\sqrt{\frac{K_p}{P}}$
• B. $\alpha =\frac{1}{\sqrt{K_p+P}}$
• C. $\alpha =\sqrt{\frac{K_p+P}{K_p}}$
• D. $\alpha =\sqrt{K_p+P}$

Answer 1

The correct option is A $\alpha =\sqrt{\frac{K_p}{P}}$

${PCl}_5\rightleftharpoons {PCl}_{3\left(9\right)}+{Cl}_{2\left(9\right)}Initialmole100Aftermole1-\alpha \alpha \alpha decomposition$

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

Total pressure$=P$

Partial pressure of $PC{l}_5=P\left(\frac{1-\alpha }{1+\alpha }\right)$

PArtial pressure of $PC{l}_3=P\left(\frac{\alpha }{1+\alpha }\right)$

Partial pressure of $PC{l}_2=P\left(\frac{\alpha }{1+\alpha }\right)$

Then $K_p=\frac{\left(PC{l}_3\right)\left(C{l}_2\right)}{\left(PC{l}_5\right)}=\frac{P\left(\frac{\alpha }{1+\alpha }\right)P\left(\frac{\alpha }{1+\alpha }\right)}{P\left(\frac{1-\alpha }{1+\alpha }\right)}=\frac{P^2{\alpha }^2}{(1+\alpha {)}^2}\times \frac{(1+\alpha )}{P(1-\alpha )}{K}_p=\frac{P{\alpha }^2}{1-{\alpha }^2}$

now, $1-{\alpha }^2<<1$

so that $K_p=P{\alpha }^2{\alpha }^2=\frac{K_p}{P}\alpha =\sqrt{\frac{K_p}{P}}$