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Q.
$PCl _{5}$ is $50 \%$ dissociated at $20^{\circ} C$ and $1\, atm$ pressure. The value of $K_{p}$ is
Equilibrium
Solution:
Let there be $1$ mole of $PCl _{5}$ initially
Number of moles dissociated $\frac{1 \times 50}{100}=0.5$
$P _{ PCl _{5}}=\frac{0.5}{1.5} \times 1 ; P _{ PCl _{3}}=\frac{0.5}{1.5} \times 1= P _{ Cl _{2}}$
$K _{ P }=\frac{ P _{ pcl _{3}} P _{ cl _{2}}}{ P _{ pcl _{5}}}$
$=\frac{(0.5 / 1.5)(0.5 / 1.5)}{(0.5 / 1.5)}$
$=0.333$
