Question

at same temperature and under a pressure for ATM PCL5 is 10% associated calculate the pressure at PCL5 will be 20% associated temp temperature remaining same

Answer 1

Solution:

We have,                             PCl5        ↔          PCl3        +             Cl2

Mole before dissociation      1                              0                          0

Mole after dissociation        1 – x                       x                             x

Given, x = 0.1 at 4 atm pressure

Therefore,    KP            =             {[nPCl3 × nCl2]/nPCl5} × [P/∑n]Δn

=             [x.x/(1 – x)]×[P/(1+x)]1

=             Px2/(1 – x2)

=             4×(0.1)2/{1 – (0.1)2}

=             0.040 atm

Again when x is desired at 0.2, KP remaining constant and thus

KP            =             Px2/(1 – x2)

Or,             0.040     =             {P×(0.2)2}/{1 – (0.2)2}

Or,       P             =             0.96 atm   (A)      [Ans.]

Hope it helped!!!