The degree of dissociation of `PCl_(5)` ata certain temperature and atmospheric pressure is `0*2` . Calculate the pressure at which it will be half (50 %) dissociated at the same temperature .

Explanation

`PCl_5` dissociates as :
`PCl_5 hArr PCl_3 + Cl_2`
If `alpha` is the degree of dissociation at certain temperature under atmospheric pressure , then
`{:("Initial conc.",1,0,0),("At equi.",1-alpha,alpha,alpha):}`
Total number of moles at equilibrium = `1-alpha+alpha+alpha`
`=1+alpha`
Partial pressures of `PCl_5, PCl_3` and `Cl_2` will be
`p(PCl_5)=(1-alpha)/(1+alpha).P " " p_(PCl_3)=alpha/(1+alpha).P`
`p_(Cl_2)=alpha/(1+alpha).P`
Now, `K_(p)=(p(PCl_3).p(Cl_2))/(p(PCl_5))`
`=((alpha/(1+alpha).P)xx(alpha/(1+alpha).P))/((1-alpha)/(1+alpha).P)=alpha^2/(1-alpha^2)P`
Putting P=1 atm and `alpha` =0.2
`K_p=(0.2)^2/(1-(0.2)^2)xx1`=0.041
when `alpha=1/2`=0.5 , let pressure is P.
`K_P=alpha^2/(1-alpha^2).P.`
`0.041=((0.5)^2P.)/(1-(0.5)^2)`
`P.=((0.041)[1-(0.5)^2])/(0.5)^2`=0.125 atm