2.69 g of a sample of PCl_(5) was placed in a 1 litre flask and completely vaporised to a temperature of 250^(@)C. The pressure observed at this temperature was 1 atm. The possibility exists that some of PCl_(5) may have dissociated according to the equation. PCl_(5)(g) hArr PCl_(3)(g) + Cl_(2)(g). What are partial pressure of PCl_(5), PCl_(3) and Cl_(2) under these experimental conditions ?

2.69 g of a sample of PCl_(5) was placed in a 1 litre flask and comple

1.	2.69 g of a sample of PCl_(5) was placed in a 1 litre flask and completely vaporised to a temperature of 250^(@)C. The pressure observed at this temperature was 1 atm. The possibility exists that some of PCl_(5) may have dissociated according to the equation. PCl_(5)(g) hArr PCl_(3)(g) + Cl_(2)(g). What are partial pressure of PCl_(5), PCl_(3) and Cl_(2) under these experimental conditions ?
Answer» Solution :Let us first calculate the pressure supposing `PCl_(5)` does not underho dissociation. pV = nRT `p xx 1 = (2.69)/(208) xx 0.082 xx 523` [mol. WT. of `PCl_(5)` = 208, R = 0.082 lit. atm/K/mole, T - (273 + 250)K] p = 0.553 atm. But `PCl_(5)` undergoes dissociation in the following way, `{:("a (say)",0,,0),(PCl_(5) hArr,PCl_(3) ,+,Cl_(2)):} (ALPHA -= "deg. of dissociation")` Moles at eqb. : `a(1-alpha)` `therefore` TOTAL no. of moles `= a(1-alpha) + a alpha + a alpha = a(1+alpha)` `because` pressure of a gas is PROPORTIONAL to no. of moles `therefore ("moles before diss.")/("moles after diss.") = (a)/(a(1+alpha)) = (0.553)/(1)` or `alpha = 0.81`. Partial pressure of `PCl_(5) = ("moles of " PCl_(5) " at eqb.")/("total moles")xx "total pressure"` `= (a(1-alpha))/(a(1+alpha)) xx 1 = (1-0.81)/(1+0.81) = 0.104` atm. Partial pressure of `PCl_(5)` = partial pressure of `Cl_(2)` `= ("moles of " PCl_(3) " or " Cl_(2))/("total mole") xx "total pressure"` `= (a alpha)/(a(1+alpha)) xx 1 = (0.81)/(1.81) = 0.447` atm.