Question

An equilibrium mixture of 2 moles each of `PCl_5`, `PCl_3` and `Cl_2` in V ltrs. in T temp. has a total pressure of 3 atm. `Cl_2` is added at the same temp. and pressure till the volume is doubled. What is the number of moles of `Cl_2` added? Thank you:)

Answer 1

`20/3`, or `6.7` moles of `Cl_2` must be added.

Explanation:

The chemical equation is `PCl_5 ⇄PCl_3 + Cl_2`.　

First, we have to find the equilibrium constant `K`:
`K=([PCl_5])/([PCl_3][Cl_2])=(2/V)/((2/V)(2/V))=V/2("L/mol")`

Let `x(mol)` to be the amount of `Cl_2` you need to add.
Adding `Cl_2` will move the equilibrium to the left(`[PCl_5]` increases. `[PCl_3]` and `[Cl_2]` decease.)
Let `y(mol)` to be the amount of `Cl_2` used to reach `color(red)"the new equilibrium"` after adding `Cl_2"`.

At the new equilibrium point, amount of `PCl_5, PCl_3` and `Cl_2` will be `2+y`, `2-y` , and `2+x-y` moles, respectively.

Here you obtain two equations:
(1) Since the volume is doubled at the constant pressure and temparature, the total amount of substance must be doubled too.
`(2+y)+(2-y)+(2+x-y)=6xx2=12`
`x-y=6` ・・・(A)

(2) The equilibrium constant doesn't change since the temparature is the same .Though, the volume is now `2V`.
`K=([PCl_5])/([PCl_3][Cl_2])=((2+y)/(2V))/(((2-y)/(2V))((2+x-y)/(2V)))`
`=((2+y)/(2V))/(((2-y)/(2V))*(8)/(2V))` (using (A))

`=(V(2+y))/(4(2-y)`

`K` must be the same as before:
`(V(2+y))/(4(2-y))=V/2` ・・・(B)
Multiply equation (B) by `(4(2-y))/V` leads to:
`2+y=2(2-y)` and `y=2/3` is obtained.

Substitute `y=2/3` to (A) and you did it!
`x=20/3=6.66…` (mole)