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PCl(5) dissociation a closed container a...

`PCl_(5)` dissociation a closed container as :
`PCl_(5(g))hArrPCl_(3(g))+Cl_(2(g))`
If total pressure at equilibrium of the reaction mixture is `P` and degree of dissociation of `PCl_(5)` is `alpha`, the partial pressure of `PCl_(3)` will be:

A

`P[(alpha)/(alpha+1)]`

B

`P[(2alpha)/(1-alpha)]`

C

`P[(alpha)/(alpha-1)]`

D

`P[(alpha)/(1-alpha)]`

Text Solution

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The correct Answer is:
A
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