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1 mole of N2 and 3 moles of PCl5 are pla...

1 mole of `N_2` and 3 moles of `PCl_5` are placed in a 100 litre vessels heated at ` 227^(@) C ` the equilibrium pressure is 2.05 atm Assuming ideal behaviour,Calculate degree of dissociation of `PCl_5 and K_p ` for the reaction
` PCl_5 (g) hArr PCl_3 (g) + Cl_2(g)`

Text Solution

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`{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),("moles at t=0",3,,0,,1),("moles at equilibrium",(3-x),,x,,x):}`
`:.` Total moles present at equilibrium `=4+x`
Given, total pressure at equilibrium `=2.05`
Now `PV=nRT` at equilibrium
`2.05xx100=(4+x)xx0.0821xx500`
`:. x=0.9939`
Now the degree of dissociation for
`PCl_(5)=("moles dissociated")/("Total moles")`
`=0.9939/3=0.313` or `33.13%`
`:. K_(p)=[(n_(PCl_(3))xxn_(Cl_(2)))/n_(PCl_(5))]xx[P/(Sigman)]^(+1)`
`=x^(2)/((3-x))xx[2.05/(4+x)]`
`=((0.9939)^(2))/((3-0.9939))xx[2.05/(4+0.9939)]=0.20`
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