Question

Phosphorus pentachloride decomposes according to the chemical equation PCl 5 ( g ) − ⇀ ↽ − PCl 3 ( g ) + Cl 2 ( g ) K c = 1.80 at 250 ∘ C A 0.166 mol sample of PCl 5 ( g ) is injected into an empty 2.15 L reaction vessel held at 250 ∘ C. Calculate the concentrations of PCl 5 ( g ) and PCl 3 ( g ) at equilibrium.

Answer 1

Answer : The concentration of
`PCl_5` and
`PCl_3` at equilibrium is, 0.0031 M and 0.0741 M respectively.

Explanation : Given,

Moles of
`PCl_5` = 0.166 mol

Volume of solution = 2.15 L

First we have to calculate the concentration of
`PCl_5`

`\text{Concentration of }PCl_5=\frac{\text{Moles of }PCl_5}{\text{Volume of solution (in L)}}`

`\text{Concentration of }PCl_5=(0.166mol)/(2.15L)=0.0772M`

Now we have to calculate the concentration of
`PCl_5` and
`PCl_3` at equilibrium.

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

Initial conc. 0.0772 0 0

At eqm. 0.0772-x x x

The expression for equilibrium constant is:

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

`1.80=((x)* (x))/((0.0772-x))`

By solving the term, we get the value of 'x'.

x = 0.0741

Thus,

The concentration of
`PCl_5` at equilibrium = (0.0772-x) = (0.0772-0.0741) = 0.0031 M

The concentration of
`PCl_3` at equilibrium = x = 0.0741 M