Falling Behind in Studies? Join Our Performance Improvement Batch. Enroll For Free
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
At 250 degree c and 1 atm pressure , the vapour density of PCl5 is 57.9 . What will be the dissociation of PCl5 ?
PCl5(g) → PCl3(g) + Cl2(g)We are given the vapour density at equilibrium at 250oC.The initial vapour density will be it would be MPCl5 / 2. ∴ Initial vapour density = (31 + 5 × 35.5) / 2 = 104.25 ∴ Total moles at equilibrium / Total moles initial = 1 + α = Vapour density initial / Vapour density at equilibriumAt 250oC, 1 + α = 104.25 / 57.9 = 1.8 ∴ a = 0.8
PCl5(g) → PCl3(g) + Cl2(g)
We are given the vapour density at equilibrium at 250oC.
The initial vapour density will be
it would be MPCl5 / 2.
∴ Initial vapour density = (31 + 5 × 35.5) / 2 = 104.25
∴ Total moles at equilibrium / Total moles initial = 1 + α = Vapour density initial / Vapour density at equilibrium
At 250oC, 1 + α = 104.25 / 57.9 = 1.8
∴ a = 0.8
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -