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Question

# The degree of dissociation of PCl5(α) obeying the equilibrium PCl5⇌PCl3+Cl2 is related to the equilibrium pressure by

A
α1P4
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B
α1P
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C
α1P2
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D
αP
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Solution

## The correct option is C α∝1√PPCl5⇌PCl3+Cl2α= the degree of dissociation of PCl5P= total equilibrium pressure. Suppose initially, 1 mole of PCl5 is present. α moles of PCl5 will dissociate to form α moles of PCl3 and α moles of Cl2.(1−α) moles of PCl5 will remain at equilibrium.Total number of moles =(1−α)+α+α=(1+α)Mole fraction of PCl5=(1−α)(1+α)Mole fraction of PCl3=(α)(1+α)Mole fraction of Cl2=(α)(1+α)Partial pressure of PCl5=(1−α)(1+α)PPartial pressure of PCl3=(α)(1+α)PPartial pressure of Cl2=(α)(1+α)PKp=PPCl3PCl2PPCl5∴Kp=α1+αP×α1+αP1−α1+αP∴Kp=α2P1−α2Assume 1−α2=1 as the degree of dissociation is small.Kp=α2P∴α=√KpPα∝1√P

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