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Q.
The degree of dissociation of $PCl _{5}$ at one atmosphere is $0.3 .$ The pressure at which $PCl _{5}$ is dissociated to $50 \%$ is
Equilibrium
Solution:
$K _{ P }=\left(\frac{\alpha}{1+\alpha}\right)^{2} P ^{2} \times \frac{(1+\alpha)}{(1-\alpha) P }=\frac{\alpha^{2} P }{\left(1-\alpha^{2}\right)}$
Since $\alpha=0.3$ at $1 atm$ 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
$K _{ p }=\frac{\alpha^{2} P }{\left(1-\alpha^{2}\right)}$
$\frac{9}{91}=\frac{0.5 \times 0.5 \times P }{\left(1-0.5^{2}\right)}=\frac{0.5 \times 0.5 P }{0.75}$
Or $P =\frac{9 \times 0.75}{91 \times 0.5 \times 0.5}=0.297=0.3 \,atm$
