Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×10-2 M. Additional SbCl5 is added so that [SbCl5]new = 0.370 M and the system is allowed to once again reach equilibrium.

SbCl5(g) <-> SbCl3(g) + Cl2(g) K = 2.50×10-2 at 521 K

(a) In which direction will the reaction proceed to reach equilibrium? to the right to the left
(b) What are the new concentrations of reactants and products after the system reaches equilibrium?

[SbCl5] = M
[SbCl3] = ___M
[Cl2] = ___M

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

Explanation:

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of is  increasing .So, the equilibrium will shift in the right direction.

b)

Concentration of  = 0.195 M

Concentration of  =

Concentration of  =

On adding more to 0.370 M at equilibrium :

Initially

0.370 M            

At equilibrium:

(0.370-x)M    

The equilibrium constant of the reaction  =

The equilibrium expression is given as:

On solving for x:

x = 0.0233 M

The new equilibrium concentrations after reestablishment of the equilibrium :