Question

When this system is at equilibrium at a certain temperature PCl5(g) ⇋ PCl3(g) + Cl2(g), the concentrations are found to be [PCl5] = 0.40 M, [PCl3] = [Cl2] = 0.20. If the volume of the container is suddenly halved at the same temperature, what will be the new equilibrium concentration of PCl5?

Answer 1

The new concentration of
`PCl_5` will be 0.9 M.

Step-by-step explanation:

`PCl_5(g)\rightleftharpoons PCl_3(g) + Cl_2(g)`

The equilibrium concentration of all species:

`[PCl_5]=0.40 M`

`[PCl_3]=[Cl_2]=0.20 M`

The equilibrium constant's expression can be written as:

`K_c=([PCl_3][Cl_2])/([PCl_5])`

`K_c=(0.20 M* 0.20 M)/(0.40 M)=0.1`

On halving the volume of the container, the concentration will get doubled;

`PCl_5(g)\rightleftharpoons PCl_3(g) + Cl_2(g)`

0.80 M 0.40M 0.40 M

`Q_c=(0.40 M* 0.40 M)/(0.80 M)=0.2`

`Q_c>K_c`

Reaction will go backward.

`PCl_5(g)\rightleftharpoons PCl_3(g) + Cl_2(g)`

Initially

0.80 M 0.40M 0.40 M

After reestablishment of an equilibrium

(0.80+x) M (0.40-x)M (0.40-x) M

So, the equilibrium expression for above reaction can be written as :

`K_c=((0.40-x) M* (0.40-x) M)/((0.80+x) M)`

`0.1=((0.40-x) M* (0.40-x) M)/((0.80+x) M)`

x = 0.1 M

The new concentration of
`PCl_5` will be:

`[PCl_5]=(0.80+x) M=(0.80+0.1) M = 0.90`