Introduction to Reversible and Irreversible Reactions
PCl5 dissocia...
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.[1−αα+1]
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B
P⋅[2α1−α]
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C
P⋅[α1−α]
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D
P⋅[α1+α]
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Solution
The correct option is AP.[1−αα+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=1M,nPCl3=0M and nCl2=0M