Question

Consider the equilibrium between SbCl5, SbCl3 and Cl2. SbCl5(g) -->SbCl3(g) + Cl2(g) K = 2.30×10-2 at 566 K .The reaction is allowed to reach equilibrium in a 7.40-L flask. At equilibrium, [SbCl5] = 0.333 M, [SbCl3] = 8.75×10-2 M and [Cl2] = 8.75×10-2 M.

(a) The equilibrium mixture is transferred to a 14.8-L flask. In which direction will the reaction proceed to reach equilibrium?
(b) Calculate the new equilibrium concentrations that result when the equilibrium mixture is transferred to a 14.8-L flask.

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

Answer 1

Explanation:

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

The equilibrium constant of the reaction in 7.40 L = $K_c$

$K_c=2.30\times 10^{-2}$

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 volume

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

If the volume of the container is increased, the pressure will decrease according to Boyle's Law. Now, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where increase in pressure is taking place. As the number of moles of gas molecules is greater at the product side. So, the equilibrium will shift in the right direction.

b)

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

Concentration of $SbCl_5$ in 7.40 L = 0.333 M

Moles of $SbCl_5$ in 7.40 L:

$=0.333 M\times 7.40 L=2.4642 mol$

Concentration of $SbCl_3$ in 7.40 L = $8.75\times 10^{-2} M$

Moles of $SbCl_3$ in 7.40 L:

$=8.75\times 10^{-2} M\times 7.40 L=0.6453 mol$

Concentration of $Cl_2$ in 7.40 L = $8.75\times 10^{-2} M$

Moles of $Cl_2$ in 7.40 L:

$=8.75\times 10^{-2} M\times 7.40 L=0.6453 mol$

On adding all these compounds in new container of volume of 14.8 L, equilibrium will reestablish and there initial concentration of all the compounds will change;

Concentration of $SbCl_5$ in 14.8 L =$\frac{2.4642 mol}{14.8 L}=0.1665 M$

Concentration of $SbCl_3$ in 14.8 L =$\frac{0.6453 mol}{14.8 L}=0.04360 M$

Concentration of $Cl_2$ in 14.8 L =$\frac{0.6453 mol}{14.8 L}=0.04360 M$

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

Initially

0.1665 M         0.04360 M     0.04360 M

At equilibrium:

(0.1665-x) M         (0.04360+x) M     (0.04360+x) M

The equilibrium constant of the reaction in 14.8 L = $K_c$

$K_c=2.30\times 10^{-2}$

The equilibrium expression is given as:

$K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}$

$2.30\times 10^{-2}=\frac{(0.04360+x)M\times (0.04360+x)M}{(0.1665-x) M}$

On solving for x:

x = 0.01536 M

The new equilibrium concentrations that result when the equilibrium mixture is transferred to a 14.8 L flask:

$[SbCl_5]=(0.1665-x) M=(0.1665-0.01536) M=0.1511 M$

$[SbCl_3]=(0.04360+x) M=(0.04360+0.01536) M=0.05896 M$

$[Cl_2]=(0.04360+x) M=(0.04360+0.01536) M=0.05896 M$