Question

at temperature t Kelvin PCL5 is 50% dissociated at equilibrium pressure of 4 atmosphere at what pressure it would disassociate to extent of 80% of same temperature

Answer 1

Given data:

At Equilibrium pressure P = 4 atm, dissociation x = 0.5

Chemical reaction behind the given scenario is

PCl5 ---> PCl3 + Cl2

No. of moles initially will be 1  -->  0 , 0

No. of moles at equilibrium will be 1-x ----> x , x

Therefore, total no. of moles will be 1 -x + x + x = 1 + x

Kp = Pp(PCl3) * Pp(PCl2) / Pp(PCl5)

= (x/(1+x))P * (x/(1+x))P / (1-x)/(1+x) P

= x^2 * P / 1 - x^2

= 0.5^2 * 4 / (1 - 0.5^2)

Kp = 1.33 ( For 5.% dissociation)

Therefore for 80% dissociation,

Kp = x^2 * P / (1 - x^2)

1.33 = 0.8^2 * P / ( 1 - 0.8^2)

P = 1.33 * 0.36 / 0.64 = 0.75 atm

Hence, for dissociation of 80% of PCl5 at same temperature, pressure of 0.75 atmosphere is essential.