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Question

PCl5 dissociation a closed container as:
PCl5(g)PCl3(g)+Cl2(g)
If the total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl5 is α. Then partial pressure of PCl5 will be:

A
P.[1αα+1]
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B
P[2α1α]
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C
P[α1α]
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D
P[α1+α]
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Solution

The correct option is A P.[1αα+1]
The equilibrium reaction is PCl5(g)PCl3(g)+Cl2(g).
The value of the equilibrium constant is Kp=[PCl3][Cl2][PCl5].
At the start of the reaction, the number of moles are nPCl5=1 M,nPCl3=0 M and nCl2=0 M
At the equilibrium, the number of moles are
nPCl5=1α,nPCl3=α and nCl2=α.
At the equilibrium, the partial pressures are
PPCl5=1α1+α×P,
PPCl3=α1+α×P and
PCl2=α1+α×P

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