Q . FOR THE FOLLOWING EQUILIBRIUM :- PCl 5 (g) ---------- ---------PCl 3 (g) + Cl 2 (g) VAPOUR DENSITY IS FOUND TO BE 100 WHEN 1 MOLE OF PCl 5 IS TAKEN IN 10 LITRE FLASK AT 27 0 C . THE PERCENTAGE DEGREE OF DISSOCIATION OF PCl 5 IS- 5.5% 6.25% 2.12% 4.25%

Question

Q . FOR THE FOLLOWING EQUILIBRIUM :- PCl 5 (g) ---------- --------->PCl 3 (g) + Cl 2 (g) VAPOUR DENSITY IS FOUND TO BE 100 WHEN 1 MOLE OF PCl 5 IS TAKEN IN 10 LITRE FLASK AT 27 0 C . THE PERCENTAGE DEGREE OF DISSOCIATION OF PCl 5 IS- 5.5% 6.25% 2.12% 4.25%

Sunil Kumar FP · Answer

ANS-(4)
If the equilibrium reaction is established in a closed vessel, then vapor density will be inversely proportional to the number of moles of the gaseous species as the density of the gaseous mixture (r) is a constant.

inital vapour density/final vapour density =number of moles at equilibrium/initial mole
we have
initial mole=1
initial vapour density= molecular formula of PCL5/2 =208.23/2 =104
final vapour density =100
PCL5----PCL3 +CL2
inital 1 0 0
final 1-d d d
number of moles at equilibrium =1+d,where d is the degree of dissociation
104/100 =1+d/1
1.04 =1+d
d=.04 or 4%

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Q . FOR THE FOLLOWING EQUILIBRIUM :- PCl 5 (g) ---------- --------->PCl 3 (g) + Cl 2 (g) VAPOUR DENSITY IS FOUND TO BE 100 WHEN 1 MOLE OF PCl 5 IS TAKEN IN 10 LITRE FLASK AT 27 0 C . THE PERCENTAGE DEGREE OF DISSOCIATION OF PCl 5 IS- 5.5% 6.25% 2.12% 4.25%

Q . FOR THE FOLLOWING EQUILIBRIUM :-
PCl₅ (g) _---------- --------->PCl₃(g) + Cl₂(g)
VAPOUR DENSITY IS FOUND TO BE 100 WHEN 1 MOLE OF PCl₅ IS TAKEN IN 10 LITRE FLASK AT 27⁰C . THE PERCENTAGE DEGREE OF DISSOCIATION OF PCl₅ IS-

5.5%

6.25%

2.12%

4.25%

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