Question

At T = 248 °C the reaction

SbCl5(g) SbCl3(g) + Cl2(g)

has an equilibrium constant in terms of pressures Kp = 1.07.

(a) Suppose the initial partial pressure of SbCl5 is 0.566 atm, and PSbCl3 = PCl2 = 0.908 atm. Calculate the reaction quotient Qp and state whether the reaction proceeds to the right or to the left as equilibrium is approached.

Qp = ?

Reaction proceeds to the (right or left).

(b) Calculate the partial pressures at equilibrium.

PSbCl5 = ?atm

PSbCl3 = ?atm

PCl2 = ?atm

(c) If the volume of the system is then decreased, will there be net formation or net dissociation of SbCl5?

There will be net (formation or dissociation).

Answer 1

(a).Sol :-

ICE table is :

.........................SbCl₅(g) --------------------> SbCl₃(g)..................... +...................... Cl₂(g)

Initial (I).............0.566 atm............................0.908 atm.........................................0.908 atm

Change (C)........- y .......................................+ y.........................................................+ y

Equilibrium (E)...(0.566-y) atm ...................(0.908+y) atm.......................................(0.908 + y ) atm

Expression of Reaction quotient (Q_p) at any stage of the reaction is :

Q_p = P_SbCl3.P_Cl2 / P_SbCl5

Q_p = (0.908)² / 0.566

Q_p = 1.46  

Here, Q_p > K_p

Therefore, Reaction proceed towards the left side as equilibrium is approached.

(b). At equilibrium

K_p = P_SbCl3.P_Cl2 / P_SbCl5

K_p = (0.908+y)²/(0.566-y)

1.07 (0.566-y) = 0.824 + y² + 1.816 y

0.606 - 1.07 y = 0.824 + y² + 1.816 y

y² + 2.886 y + 0.218 = 0 On solving, we have

y = - 0.0776

So, Equilibrium pressure is :

P_SbCl5 = 0.566-y = 0.566 + 0.0776 = 0.644 atm

P_SbCl3 = 0.908+y =  0.908 - 0.0776 =  0.830 atm

P_Cl2 = 0.908+y =  0.908 - 0.0776 =  0.830 atm

(c).Volume is decreased i.e. Pressure increased , equilibrium will shifts in that direction where number of moles of gases are less i.e. reactant side. Hence, There will be net formation of SbCl₅.