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

Suppose `alpha` is the degree of dissociation, then
`{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),(" Intial conc.",1 " mole",,0,,0),(" At. eqm.",1-alpha,,alpha,,alpha):}`
`:. " Total number of moles at equilibrium " = 1- alpha + alpha + alpha = 1 + alpha `
If P is the total pressure at equilibrium, , then pressures will be
`P_(PCI_(3)) = alpha/(1+alpha)P,p_(Cl_(2)) = alpha/(1+alpha) P, p_(PCl_(5)) = (1-alpha)/(1+alpha)P`
` :. K_(p) = (p_(PCl_(3)) xx 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`
We are given that at P=1 atm, `alpha = 0*2` . Hence, `K_(p) = (0*2)^(2) /(1-(0*2)^(2))xx1 = (0*04)/(0*96)=0*0417`
When dissociation is 50% , i.e., `alpha = 0*5`,suppose total pressure is P'. Then
`0*0417 = (0*5)^(2)/(1-(0*5)^(2)) xx P' or P'= 0*125`