Question

Phosphorus pentachloride decomposes according to the chemical equation PCl5(g)↽−−⇀PCl3(g)+Cl2(g)????c=1.80 at 250 ∘C A 0.349 mol sample of PCl5(g) is injected into an empty 4.00 L reaction vessel held at 250 ∘C. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium.

Answer 1

Answer : The concentration of
`PCl_3` and
`PCl_5` at equilibrium are, 0.0834 M and 0.00385 M

Explanation :

First we have to calculate the concentration of
`PCl_5`.

`\text{Concentration of }PCl_5=\frac{\text{Moles of }PCl_5}{\text{Volume of }PCl_5}=(0.349mole)/(4.00L)=0.08725mole/L`

The given equilibrium reaction is,

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

Initial conc. 0.08725 0 0

At equilibrium (0.08725-x) x x

The expression for equilibrium constant will be,

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

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

`1.80=((x)* (x))/((0.08725-x))`

By solving the term 'x', we get:

`x=0.0834M`

The concentration of
`PCl_3` at equilibrium = x = 0.0834 M

The concentration of
`PCl_5` at equilibrium = (0.08725-x) = (0.08725-0.0834) = 0.00385 M

Therefore, the concentration of
`PCl_3` and
`PCl_5` at equilibrium are, 0.0834 M and 0.00385 M

Answer 2

Concentration of
`PCl_(5)` at equilibrium is 0.05385 M and concentration of
`PCl_(3)` at equilibrium is 0.0334 M

Step-by-step explanation:

Construct an ICE table to calculate change in concentration at equilibrium.

Initial concentration of
`PCl_(5)` =
`(0.349)/(4.00)M = 0.08725 M`

`PCl_(5)\rightleftharpoons PCl_(3)+Cl_(2)`

I: 0.08725 0 0

C: -x +x +x

E: 0.08725-x x x

species inside third bracket represent equilibrium concentrations

`([PCl_(3)][Cl_(2)])/([PCl_(5)])=K_(c)`

or,
`(x^(2))/(0.08725-x)=1.80`

or,
`x^(2)+1.8x-0.15705=0`

or,
`x=\frac{-1.8+\sqrt{(1.8)^(2)+(4* 0.15705)}}{2}`

So,
`x=0.0334 M`

So,
`[PCl_(3)]=x=0.0334 M`,
`[PCl_(5)]= 0.08725-x = 0.05385 M`