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:)

1 Answer
Oct 2, 2017

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)