Question

For the equilibrium PCl5(g) PCl3(g) + Cl2(g), Kc = 4.0 at 228°C. If pure PCl5 is placed in a 1.00-L container and allowed to come to equilibrium, and the equilibrium concentration of PCl5(g) is 0.26 M, what is the equilibrium concentration of PCl3?

Answer 1

Answer : The equilibrium concentration of
`PCl_3` is, 1.0 M

Explanation : Given,

Equilibrium concentration of
`PCl_5` = 0.26 M

Volume of solution = 1.00 L

Equilibrium constant
`(K_c)` = 4.0

The balanced equilibrium reaction will be,

`PCl_5\rightleftharpoons PCl_3+Cl_2`

The expression of equilibrium constant for the reaction will be:

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

From the reaction we conclude that the concentration of
`PCl_3` and
`Cl_2` are equal.

Let the concentration of
`PCl_3` be 'X'.

So, concentration of
`Cl_2` = X

Now put all the values in this expression, we get :

`4.0=((X)* (X))/(0.26)`

`4.0=((X)^2)/(0.26)`

`X=1.0M`

Thus,

The concentration of
`PCl_3` at equilibrium = X = 1.0 M

The concentration of
`Cl_2` at equilibrium = X = 1.0 M