Ques.)
2 moles of PCl5 are taken in a closed vessel. At equilibrium the vessel contains 2.5 moles. The degree of dissociation of PCl5 is
1. 0.25
2. 0.5
3. 0.4
4. 0.2
29@ E= mc2
This question is from chemical equilibrium and not ionic..
@manmay
2.5 moles of the mixture of gases including PCl5 , PCl3 , Cl2
and Ans is 1
49PCl5 -----> PCl3+ Cl2
2 moles
2-a a a
no of moles= 2+a
2+a=2.5
a=.5
degree of dissociation=0.5/2=0.25
1@Subhomoy~ : " " degree of dissociation=0.5/2=0.25 " "
This was my dbt ??
Y to half it ??
Isnt a deg of dissoscn
29Well the way subho did it may create some confusion so lemme do it in the actual way..
PCl5 ≡ PCl3 + Cl2
at t=0 C 0 0
at equilibrium C(1-α) Cα Cα
where C is the intial concentration and α is the degree of dissociation
so according to the question C=2 and finally we have 2.5 moles
C - Cα +Cα + Cα = 2.5
2+ 2α = 2.5
2α = 0.5
α = 0.25
i hope the confusion is cleared now
1Ya Govind Thanks a lot
& btw the title was a mistake (In a hurry I was solving miscellaaneous probs so i posted wrong heading )
29Wen did I say that subho did it wrong?
I just told my method to make things clear for E = MC^2 ..since subho was offline by then
and that regarding that concentration and moles thing....take the volume of the container as 1 lt
