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)