Question

The degree of dissociation of $PC{l}_5$ at one atmosphere is $\mathrm{0.3.}$ The pressure at which $PC{l}_5$ is dissociated to $50\%$ is

• A. 0.273 atm
• B. 0.3 atm
• C. 1.34 atm
• D. 2.73 atm

Answer 1

Answer: (B)

$K_P={\left(\frac{\alpha }{1+\alpha }\right)}^2{P}^2\times \frac{(1+\alpha )}{(1-\alpha )P}=\frac{{\alpha }^{2}P}{\left(1-{\alpha }^2\right)}$
Since $\alpha =0.3$ at $1atm$ pressure
$K_p=\frac{0.3\times 0.3\times 1}{1-0.09}=\frac{0.09}{0.91}$
$=\frac{9\times 10^{-2}}{91\times 10^{-2}}\simeq \frac{9}{91}$
$K_P=\frac{9}{91}$
When $\alpha =50\%=0.5$, the pressure can be calculated as follows
$K_p=\frac{{\alpha }^{2}P}{\left(1-{\alpha }^2\right)}$
$\frac{9}{91}=\frac{0.5\times 0.5\times P}{\left(1-0.5^2\right)}=\frac{0.5\times 0.5P}{0.75}$
Or $P=\frac{9\times 0.75}{91\times 0.5\times 0.5}=0.297=0.3atm$