Question

In the dissociation of $PC{l}_5$ as:
$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$ if the degree of dissociation is $\alpha$ at equilibrium pressure P, then the the equilibrium constant for the reaction is:

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

Answer 1

The correct option is C $K_p=\frac{\alpha P^2}{1-{\alpha }^2}$

$PC{l}_5$ ${}_{\left(g\right)}$ $\rightleftharpoons$ $PC{l}_3$ ${}_{\left(g\right)}$ +$C{l}_2$ ${}_{\left(g\right)}$

Initial $t=0$	$1$	$0$	$0$
At equilibrium	$1-\alpha$	$\alpha$	$\alpha$

Total number of molus $=1-\alpha +\alpha +\alpha$

$=1+\alpha$

$K_P$ will known that partial pressure is the product of mole fraction and the total pressure role fraction is the number of moles of that component divided by the number of mole in mixture

$P_{PC{l}_5}=\frac{(1-\alpha )p}{1+\alpha }$,$P_{PC{l}_3}=\frac{\alpha P}{1+\alpha }$ , $P_{PC{l}_2}=\frac{\alpha P}{1+\alpha }$

$Kp=\frac{P_{PC{l}_3}.P_{C{l}_2}}{P_{PC{l}_5}}$

Put all these value:

$kp=\frac{\frac{\alpha P}{1+\alpha }.\frac{\alpha P}{1+\alpha }}{\frac{(1-\alpha )P}{(1+\alpha )}}\Rightarrow \frac{\frac{{\alpha }^{2}p}{(1+\alpha {)}^2}}{\frac{(1-\alpha )}{(1+\alpha )}}=\frac{{\alpha }^{2}P}{1-{\alpha }^2}$

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