A container whose volume is V contains an equilibrium mixture that consits of 2mol each of PCl_(5) , PCl_(3) and Cl_(2) ( all as gases ). The pressure is 30.3975 kPa and temperature is T. A certain amount of Cl_(2) ( g) is now introduced keeping the pressure and temperature constant until the equilibrium volume is 2V. Calculate the amount of Cl_(2) that was added and the value of K_(p).

A container whose volume is V contains an equilibrium mixture that con

1.	A container whose volume is V contains an equilibrium mixture that consits of 2mol each of PCl_(5) , PCl_(3) and Cl_(2) ( all as gases ). The pressure is 30.3975 kPa and temperature is T. A certain amount of Cl_(2) ( g) is now introduced keeping the pressure and temperature constant until the equilibrium volume is 2V. Calculate the amount of Cl_(2) that was added and the value of K_(p).
Answer» Solution :`PCl_(5(g)) hArr PCl_(3(g)) + Cl_(2(g))` At equilibrium 2mol2 mol2 mol Total pressure = 30.3975 Kpa = 3 atm = P ( say ) `K _(p) = ( P_(PCl_(3)) xx P_(Cl_(2)))/(P_(Cl_(5))) = (( P)/( 3) xx ( P)/( 3))/((P)/( 3))` `K_(p ) = ( P)/( 3) = 1`.....(1) When CHLORINE is added to the system. The system will behave to nullify the effect and hence formation of `PCl_(5)` will be preferred. Since P and T are CONSTANT are `( V_(1))/( V_(2)) = ( n _(1))/( n _(2)) = n _(1) = 6" " V_(1) = V " " V_(2) = 2V ` `:. n _(2) = 12 ` moles Say a moles of `Cl_(2)` were added `PCl_(5) hArr PCl_(3) = Cl_(2)` Initial 222 FInal2`+x`2-x`2+a-x` `n_(T )- 12 = 6+a -x ` `K_(p) = ( (( 2+ax) - x)/( 12) xx ( 2-x)/( 12))/((2+x)/(12)) xxP=((2+a-x)(2-x))/(4( 2+x)) =1 `.....(2) Solving equation ( 1) and ( 2) we get `a = ( 20)/(3)` moles Hence `( 20)/( 3)` moles of `Cl_(2)` was added