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Question

The degree of dissociation of PCl5(α) obeying the equilibrium PCl5PCl3+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 α1P
PCl5PCl3+Cl2
α= the degree of dissociation of PCl5
P= 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.
(1α) moles of PCl5 will remain at equilibrium.
Total number of moles =(1α)+α+α=(1+α)
Mole fraction of PCl5=(1α)(1+α)
Mole fraction of PCl3=(α)(1+α)
Mole fraction of Cl2=(α)(1+α)
Partial pressure of PCl5=(1α)(1+α)P
Partial pressure of PCl3=(α)(1+α)P
Partial pressure of Cl2=(α)(1+α)P
Kp=PPCl3PCl2PPCl5
Kp=α1+αP×α1+αP1α1+αP
Kp=α2P1α2
Assume 1α2=1 as the degree of dissociation is small.
Kp=α2P
α=KpP
α1P

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