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PCl(5) dissociates into PCl(3) and Cl(2)...

`PCl_(5)` dissociates into `PCl_(3)` and `Cl_(2)`, thus
`PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g)`
If the total pressure of the system in equilibrium is `P` at a density `rho` and temperature `T`, show that the degree of dissociation `alpha=(PM)/(rhoRT)-1`, where `M` is the relative molar mass of `PCl_(5)`. If the vapour density of the gas mixture at equilibrium has the value of `62` when the temperature is `230^(@)C`, what is the value of `P//rho`?

Text Solution

Verified by Experts

We have
`PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g)`
Since `PV=nRT=w/MRT`
`M=(wRT)/(PV)=(rhoRT)/(rho)`
`:. V.D(d)=(rhoRT)/(2P)`
`PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g)`
`{:("Initial",c,,0,0),("At equilibrium",c(1-alpha),,calpha,calpha):}`
Total moles at equilibrium `=c(1+alpha)`
`:. ("Total moles at equilirium")/("Total initial moles")`
`=("Initial vapour density")/("Vapour density at equilibrium")`
`(c(1+alpha))/c=D/d`
`rArr 1+alpha=(M//2)/(rhoRT//2P)=(PM)/(rhoRT)`
If `d=62 (given)`
`M/2=208.5/2=104.25`
`:. alpha=(104.25-62)/62=0.68`
`:. P/(rho)=((1+alpha)RT)/(M)=0.3327 "atm"//(gL^(-1))`
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