Question

For gaseous decomposition of $PC{l}_5$ in a closed vessel the degree of dissociation $\alpha$, equilibrium pressure $P$ and $K_p$ are related as:

• A. $\alpha =\sqrt{\frac{K_p}{K_p+P}}$
• 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 $\alpha =\sqrt{\frac{K_p}{K_p+P}}$

$PC{l}_5\to PC{l}_3+C{l}_2$

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

$\frac{(1-\alpha )P}{1+\alpha }\frac{\alpha P}{1+\alpha }\frac{\alpha }{1+\alpha }P$

$k_P=\frac{{\alpha }^2{P}^2}{(1+\alpha {)}^2}\frac{(1+\alpha }{(1-\alpha )P}$

$=\frac{{\alpha }^{2}P}{1-{\alpha }^2}$

$\frac{k_P}{P}=\frac{{\alpha }^2}{1-{\alpha }^2}$

$\frac{P}{k_P}=\frac{1-{\alpha }^2}{{\alpha }^2}$

$\frac{P+k_P}{k_P}=\frac{1}{{\alpha }^2}$

${\alpha }^2=\frac{k_P}{k_P+P}$

$\alpha =\sqrt{\frac{k_P}{k_P+P}}$