Calculate the volume percent of chlorine gas at equilibrium in the dissociation of `PCl_(5)(g)` under a total pressure of `1.5` atm. The `K_(p)` for its dissociation `=0.3`.

Answer

B, C

Explanation

`{:(,PCl_(5)(g),hArr,Pcl_(3)(g),+,Cl_(2)(g)),(underset(("moles"))(t=0),a,,0,,0),(t=t_("eq"),a-x,,x,,x):}`
`K_(p)=(P_(PCl_(3)).P_(Cl_(2)))/(P_(PCl_(5)))=([x/(a+x)P][x/(a+x)P])/((a-x)/(a+x)P)`
`rArr K_(p)=(x^(2)P)/(a^(2)-x^(2))=(alpha^(2)P)/(1-alpha^(2))=0.3[alpha=x/a]rArr alpha=sqrt(1/6)`
`%Cl_(2)=x/(a+x)=alpha/(1+alpha)xx100%=28.6%`