find out Kc_I and Kc_II for

2) When 0.4 mole of PCl5 is heated to 523 K in a 10 dm3 container, an equilibrium is established in which 0.25 mole Cl2 is present.

Find out K_c for the dissociation of PCl₅ at 523 K.

Initially         0.4 mole         0 mole         0 mole

At equili. (0.4 - 0.25)       0.25 mole      0.25 mole

3) The equilibrium constant KP for the dissociation of PCl₅ at 523 K and 101.3 KP_a pressure is 180.3. Calculate its percent dissociation at 523 K.

Initially 1 mole 0 mole 0 mole

At equilibrium (1 - x) mole x mole x mole

(Total number of moles at equilibrium = 1- x + x + x = 1+ x )

If P is total pressure then partial pressure at equilibrium

180.3 - 180.3x2 = x2 ´ 101.3

101.3x2 + 180.3x2 = 180.3

\ x2 = 0.64 \ x = 0.8

Percent dissociation = 0.8 ´ 100 = 80%

Index

Chapter 12